$\left|x-0,5\right|$=$\dfrac{7}{2}$ $x:\dfrac{3}{8} \dfrac{5}{8}=x$ $\dfrac{5}{6}-\left|x-\dfrac{1}{2}\right|$=$\dfrac{15}{18}$ $\dfrac{-7}{12}$-$\dfrac{2}{3}$:x=$\dfrac{-5}{8}$ \(...

NT

Ngân Thiên

18 tháng 1 2018

$\left|x-0,5\right|$=$\dfrac{7}{2}$ $x:\dfrac{3}{8}+\dfrac{5}{8}=x$ $\dfrac{5}{6}-\left|x-\dfrac{1}{2}\right|$=$\dfrac{15}{18}$ $\dfrac{-7}{12}$-$\dfrac{2}{3}$:x=$\dfrac{-5}{8}$ $\left(5x-3\right)^2-\dfrac{1}{64}=0$ 9.27$\le$$3^x$ $\le$243 $(x-\dfrac{1}{2})^3$...

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#Toán lớp 7

1

NL

Nguyễn Lê Phước Thịnh

15 tháng 6 2022

a: |x-1/2|=7/2

=>x-1/2=7/2 hoặc x-1/2=-7/2

=>x=4 hoặc x=-3

b: $x:\dfrac{3}{8}+\dfrac{5}{8}=x$

=>8/3x-x=-5/8

=>5/3x=-5/8

hay x=-5/8:5/3=-5/8x3/5=-15/40=-3/8

c: $\dfrac{5}{6}-\left|x-\dfrac{1}{2}\right|=\dfrac{15}{18}=\dfrac{5}{6}$

=>|x-1/2|=0

=>x-1/2=0

hay x=1/2

e: $\left(5x-3\right)^2-\dfrac{1}{64}=0$

=>(5x-3)²=1/64

=>5x-3=1/8 hoặc 5x-3=-1/8

=>5x=25/8 hoặc 5x=23/8

=>x=5/8 hoặc x=23/40

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LN

Lê Ngọc Anh

8 tháng 3 2022

Tìm x, biết:a) x+$\dfrac{1}{6}$=$\dfrac{-3}{8}$ b) $2-\left(\dfrac{3}{4}-x\right)=\dfrac{7}{12}$c) $\dfrac{1}{2}x$+$\dfrac{1}{8}x=\dfrac{3}{4}$ d)...

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#Toán lớp 6

2

HP

hưng phúc

8 tháng 3 2022

$a.x+\dfrac{1}{6}=-\dfrac{3}{8}$

$\Leftrightarrow x=-\dfrac{13}{24}$

$b.2-\left(\dfrac{3}{4}-x\right)=\dfrac{7}{12}$

$\Leftrightarrow2-\dfrac{3}{4}+x=\dfrac{7}{12}$

$\Leftrightarrow x=-\dfrac{2}{3}$

$c.\dfrac{1}{2}x+\dfrac{1}{8}x=\dfrac{3}{4}$

$\Leftrightarrow\dfrac{5}{8}x=\dfrac{3}{4}$

$\Leftrightarrow x=\dfrac{6}{5}$

$d.75\%-1\dfrac{1}{2}+0,5:\dfrac{5}{12}-\left(\dfrac{-1}{2}\right)^2$

$=\dfrac{75}{100}-\dfrac{3}{2}+\dfrac{1}{2}:\dfrac{5}{12}-\dfrac{1}{4}$

$=-\dfrac{3}{4}+\dfrac{6}{5}-\dfrac{1}{4}$

$=\dfrac{1}{5}$

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DC

dâu cute

8 tháng 3 2022

a) $x+\dfrac{1}{6}=\dfrac{-3}{8}$

$x=\dfrac{-3}{8}-\dfrac{1}{6}$

$x=\dfrac{-13}{24}$

vậy x =....

b) $2-\left(\dfrac{3}{4}-x\right)=\dfrac{7}{12}$

$\dfrac{3}{4}-x=2-\dfrac{7}{12}$

$\dfrac{3}{4}-x=\dfrac{17}{12}$

$x=\dfrac{3}{4}-\dfrac{17}{12}$

$x=\dfrac{-2}{3}$

vậy x =....

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TD

Thầy Đức Anh

Giáo viên VIP

27 tháng 4 2022 - olm

Tìm $x$, biết :
a) $\left(\dfrac{1}{2}+1,5\right) \cdot x=\dfrac{1}{5}$
b) $\left(-1 \dfrac{3}{5}+x\right): \dfrac{12}{13}=2 \dfrac{1}{6}$
c) $\left(x: 2 \dfrac{1}{3}\right) \cdot \dfrac{1}{7}=\dfrac{-3}{8}$
d) $\dfrac{-4}{7} \cdot x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1 \dfrac{2}{3}\right)$

#Toán lớp 7

13

YN

Yen Nhi

21 tháng 6 2022

$a)\left(\dfrac{1}{2}+1,5\right)x=\dfrac{1}{5}$

$\Rightarrow2x=\dfrac{1}{5}$

$\Rightarrow x=\dfrac{1}{10}$

$b)\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}$

$\Leftrightarrow-\dfrac{8}{5}+x=\dfrac{13}{6}.\dfrac{12}{13}$

$\Leftrightarrow-\dfrac{8}{5}+x=2$

$\Leftrightarrow x=\dfrac{18}{5}$

$c)\left(x:2\dfrac{1}{3}\right).\dfrac{1}{7}=-\dfrac{3}{8}$

$\Leftrightarrow x:\dfrac{7}{3}=-\dfrac{3}{8}:\dfrac{1}{7}$

$\Leftrightarrow x=-\dfrac{21}{8}.\dfrac{7}{3}$

$\Leftrightarrow x=-\dfrac{49}{8}$

$d)-\dfrac{4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)$

$\Leftrightarrow-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}$

$\Leftrightarrow-\dfrac{4}{7}x=-\dfrac{59}{40}$

$\Leftrightarrow x=\dfrac{413}{160}$

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T

TINH

13 tháng 7 2022

a) $\left(\dfrac{1}{2}+1,5\right) \cdot x=\dfrac{1}{5}$

$\cdot x=\dfrac{1}{5}$

$x=\dfrac{1}{5}: 2$

$x=\dfrac{1}{10}$
b) $\left(-1 \dfrac{3}{5}+x\right): \dfrac{12}{13}=2 \dfrac{1}{6}$

$\dfrac{3}{5}+x=\dfrac{13}{6} \cdot \dfrac{12}{13}$
$\dfrac{3}{5}$

$\dfrac{3}{5}$
c) $\left(x: 2 \dfrac{1}{3}\right) \cdot \dfrac{1}{7}=\dfrac{-3}{8}$

$\cdot \dfrac{3}{7} \cdot \dfrac{1}{7}=\dfrac{-3}{8}$

$x=\dfrac{-3}{8}: \dfrac{3}{49}$
$x=\dfrac{-49}{8}=-6 \dfrac{1}{8}$
d) $\dfrac{-4}{7} \cdot x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1 \dfrac{2}{3}\right)$

$\dfrac{-4}{7} x+\dfrac{7}{5}=\dfrac{1}{8} \cdot \dfrac{-3}{5}$
$-\dfrac{4}{7} x=\dfrac{-3}{40}-\dfrac{7}{5} \\ x=\dfrac{-59}{40}: \dfrac{-4}{7}=\dfrac{413}{160}=2 \dfrac{93}{160}$

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2S

2012 SANG

30 tháng 8 2023

$\left(6\right)\dfrac{3\sqrt{x}}{5\sqrt{x}-1}\le-3$

$\left(7\right)\dfrac{8\sqrt{x}+8}{6\sqrt{x}+9}>\dfrac{8}{3}$

$\left(8\right)\dfrac{\sqrt{x}-2}{2\sqrt{x}-3}< -4$

$\left(9\right)\dfrac{4\sqrt{x}+6}{5\sqrt{x}+7}\le-\dfrac{2}{3}$

$\left(10\right)\dfrac{6\sqrt{x}-2}{7\sqrt{x}-1}>-6$

#Toán lớp 9

1

NL

Nguyễn Lê Phước Thịnh

30 tháng 8 2023

6:ĐKXĐ: x>=0; x<>1/25

BPT=>$\dfrac{3\sqrt{x}}{5\sqrt{x}-1}+3< =0$

=>$\dfrac{3\sqrt{x}+15\sqrt{x}-5}{5\sqrt{x}-1}< =0$

=>$\dfrac{18\sqrt{x}-5}{5\sqrt{x}-1}< =0$

=>$\dfrac{1}{5}< \sqrt{x}< =\dfrac{5}{18}$

=>$\dfrac{1}{25}< x< =\dfrac{25}{324}$

7:

ĐKXĐ: x>=0

BPT $\Leftrightarrow\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}>\dfrac{8}{3}:\dfrac{8}{3}=1$

=>$\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}-1>=0$

=>$\dfrac{\sqrt{x}+1-2\sqrt{x}-3}{2\sqrt{x}+3}>=0$

=>$-\sqrt{x}-2>=0$(vô lý)

8:

ĐKXĐ: x>=0; x<>9/4

BPT $\Leftrightarrow\dfrac{\sqrt{x}-2}{2\sqrt{x}-3}+4< 0$

=>$\dfrac{\sqrt{x}-2+8\sqrt{x}-12}{2\sqrt{x}-3}< 0$

=>$\dfrac{9\sqrt{x}-14}{2\sqrt{x}-3}< 0$

TH1: 9căn x-14>0 và 2căn x-3<0

=>căn x>14/9 và căn x<3/2

=>14/9<căn x<3/2

=>196/81<x<9/4

TH2: 9căn x-14<0 và 2căn x-3>0

=>căn x>3/2 hoặc căn x<14/9

mà 3/2<14/9

nên trường hợp này Loại

9:

ĐKXĐ: x>=0

$BPT\Leftrightarrow\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}< =-\dfrac{1}{3}$

=>$\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}+\dfrac{1}{3}< =0$

=>$\dfrac{6\sqrt{x}+9+5\sqrt{x}+7}{3\left(5\sqrt{x}+7\right)}< =0$

=>$\dfrac{11\sqrt{x}+16}{3\left(5\sqrt{x}+7\right)}< =0$(vô lý)

10:

ĐKXĐ: x>=0; x<>1/49

$BPT\Leftrightarrow\dfrac{6\sqrt{x}-2}{7\sqrt{x}-1}+6>0$

=>$\dfrac{6\sqrt{x}-2+42\sqrt{x}-6}{7\sqrt{x}-1}>0$

=>$\dfrac{48\sqrt{x}-8}{7\sqrt{x}-1}>0$

=>$\dfrac{6\sqrt{x}-1}{7\sqrt{x}-1}>0$

TH1: 6căn x-1>0 và 7căn x-1>0

=>căn x>1/6 và căn x>1/7

=>căn x>1/6

=>x>1/36

TH2: 6căn x-1<0 và 7căn x-1<0

=>căn x<1/6 và căn x<1/7

=>căn x<1/7

=>0<=x<1/49

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2S

2012 SANG

30 tháng 8 2023

câu 9 nhầm đề bài r bạn

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QN

quan nguyen hoang

13 tháng 2 2022

$a,\dfrac{-8}{5}:\left(1+\dfrac{2}{3}\right)$ $b,\dfrac{7}{5}x\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):\dfrac{11}{5}$$c,\dfrac{1}{3}:\left(\dfrac{2}{9}-\dfrac{7}{8}\right)$ $d,\left(\dfrac{1}{6}-\dfrac{4}{5}\right):\dfrac{7}{5}$Giúp mik...

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#Toán lớp 6

1

DD

Đinh Đức Anh

13 tháng 2 2022

A -$\dfrac{24}{25}$

B -$\dfrac{5}{21}$

C -$\dfrac{24}{47}$

D -$\dfrac{19}{42}$

tick cho mk

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I

ILoveMath

13 tháng 2 2022

trả lời hẳn ra sao bạn cứ chỉ ghi kết quả thế

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HM

Hồng Minh

25 tháng 11 2017

Giúp mk nha Giải các phương trình 1/ $\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}=\dfrac{1}{6}$ 2/ $\dfrac{1}{x^2-6x+8}+\dfrac{1}{x^2-10x+24}+\dfrac{1}{x^2-14x+48}=\dfrac{1}{9}$ 3/ $\dfrac{1}{x^2-3x+3}+\dfrac{2}{x^2-3x+4}=\dfrac{6}{x^2-3x+5}$ 4/ $\dfrac{6}{\left(x+1\right)\left(x+2\right)}+\dfrac{8}{\left(x-1\right)\left(x+4\right)}=1$ 5/ $4\left(x^3+\dfrac{1}{x^3}\right)=13\left(x+\dfrac{1}{x}\right)$ 6/...

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#Toán lớp 8

1

MP

Mysterious Person

6 tháng 2 2018

1) điều kiện xác định : $x\notin\left\{-1;-2;-3;-4\right\}$

ta có : $\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}=\dfrac{1}{6}$

$\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}$ $\Leftrightarrow\dfrac{\left(x+3\right)\left(x+4\right)+\left(x+1\right)\left(x+4\right)+\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}$

$\Leftrightarrow\dfrac{x^2+7x+12+x^2+5x+4+x^2+3x+2}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}$

$\Leftrightarrow\dfrac{3x^2+15x+18}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}$

$\Leftrightarrow6\left(3x^2+15x+18\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)$

$\Leftrightarrow18\left(x^2+5x+6\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)$

$\Leftrightarrow18\left(x+2\right)\left(x+3\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)$

$\Leftrightarrow18=\left(x+1\right)\left(x+4\right)$ ( vì điều kiện xác định )

$\Leftrightarrow18=x^2+5x+4\Leftrightarrow x^2+5x-14=0$

$\Leftrightarrow\left(x-2\right)\left(x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+7=0\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-7\end{matrix}\right.\left(tmđk\right)$

vậy $x=2$ hoặc $x=-7$ mấy câu kia lm tương tự nha bn

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LT

Lê Thị Ngọc Duyên

20 tháng 8 2017

ĐỀ 1: 1. Tính: a. $\dfrac{7}{23}\left[\left(\dfrac{-8}{6}\right)-\dfrac{45}{18}\right]$ b. $\dfrac{1}{5}:\dfrac{1}{10}-\dfrac{1}{3}\left(\dfrac{6}{5}-\dfrac{9}{4}\right)$ c. $\dfrac{3}{5}.\left(\dfrac{-8}{3}\right)-\dfrac{3}{5}:\left(-6\right)$ d. $\dfrac{1}{2}\left(\dfrac{4}{3}+\dfrac{2}{5}\right)-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)$ e. $\dfrac{6}{7}:\left(\dfrac{3}{26}-\dfrac{3}{13}\right)+\dfrac{6}{7}\left(\dfrac{1}{10}-\dfrac{8}{5}\right)$ 2. Tìm x,...

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#Toán lớp 7

1

TT

Thanh Trà

20 tháng 8 2017

1.Tính

a.$\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}$

b.$\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}$

c.$\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}$

d.$\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}$

e.$\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}$

Bài 2

a.$1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}$

$x=\dfrac{13}{49}$

b.$\left|x-1,5\right|=2$

Xảy ra 2 trường hợp

TH1

$x-1,5=2$

$x=3,5$

TH2

$x-1,5=-2$

$x=-0,5$

Vậy $x=3,5$ hoặc $x=-0,5$ .

Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.

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LT

Lê Thị Ngọc Duyên

20 tháng 8 2017

Ths bn nhé

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TL

Thùyy Lynhh

24 tháng 8 2021

a) (X-2)(x+3)-3(4x-2)=(x-4)$^{^{ }2}$b) $\dfrac{2x^2+1}{8}-\dfrac{7x-2}{12}=\dfrac{x^2-1}{4}-\dfrac{x-3}{6}$c) $x-\dfrac{2x-2}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}$d) $\left(2x+5\right)^2=\left(x+2\right)^2$e) $x^2-5+6=0$g) $2x^3+6x^2=x^2+3x$h) $\left(x+\dfrac{1}{2}\right)^2+2\left(x+\dfrac{1}{x}\right)-8=0$mọi người giúp e với...

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#Toán lớp 8

4

NH

Nguyễn Hoàng Minh

24 tháng 8 2021

$a,\left(x-2\right)\left(x-3\right)-3\left(4x-2\right)=\left(x-4\right)^2\\ \Leftrightarrow x^2-5x+6-12x+6=x^2-8x+16\\ \Leftrightarrow-9x-4=0\\ \Leftrightarrow x=-\dfrac{4}{9}$

$b,\dfrac{2x^2+1}{8}-\dfrac{7x-2}{12}=\dfrac{x^2-1}{4}-\dfrac{x-3}{6}\\ \Leftrightarrow6x^2+3-14x+4=6x^2-6-4x+12\\ \Leftrightarrow10x=1\\ \Leftrightarrow x=\dfrac{1}{10}$

$c,x-\dfrac{2x-2}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\\ \Leftrightarrow30x-12x+12+5x+40=210+10x-10\\ \Leftrightarrow13x=148\\ \Leftrightarrow x=\dfrac{148}{13}$

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NH

Nguyễn Hoàng Minh

24 tháng 8 2021

$d,\left(2x+5\right)^2=\left(x+2\right)^2\\ \Leftrightarrow\left(2x+5\right)^2-\left(x+2\right)^2=0\\ \Leftrightarrow\left(2x+5-x-2\right)\left(2x+5+x+2\right)=0\\ \Leftrightarrow\left(x+3\right)\left(3x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{7}{3}\end{matrix}\right.$

$e,x^2-5x+6=0\\ \Leftrightarrow\left(x-2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.$

$g,2x^3+6x^2=x^2+3x\\ \Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow x\left(2x-1\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-3\end{matrix}\right.$

$h,\left(x+\dfrac{1}{x}\right)^2+2\left(x+\dfrac{1}{x}\right)-8=0\left(x\ne0\right)$

Đặt $x+\dfrac{1}{x}=t$, pt trở thành:

$t^2+2t-8=0\\ \Leftrightarrow\left(t-2\right)\left(t+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t=2\\t=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=2\\x+\dfrac{1}{x}=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1-2x=0\\x^2+1+4x=0\left(1\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\Delta\left(1\right)=16-4=12>0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\\left[{}\begin{matrix}x=-2+\sqrt{3}\\x=-2-\sqrt{3}\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2+\sqrt{3}\\x=-2-\sqrt{3}\end{matrix}\right.$

Tick plzz

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K

Kfkfj

15 tháng 4 2017

tìm x:

a)$\left(x+\dfrac{1}{2}\right).\left(\dfrac{2}{3}-2x\right)=0$

b) $\left(x.6\dfrac{2}{7}+\dfrac{3}{7}\right).2\dfrac{1}{5}-\dfrac{3}{7}=-2$

c) $x.3\dfrac{1}{4}+\left(\dfrac{-7}{6}\right).x-1\dfrac{2}{3}=\dfrac{5}{12}$

d) $5\dfrac{8}{17}:x+\left(-\dfrac{4}{17}\right):x+3\dfrac{1}{7}:17\dfrac{1}{3}=\dfrac{4}{11}$

#Toán lớp 6

1

NL

Nguyễn Lê Phước Thịnh

15 tháng 5 2022

a: (x+1/2)(2/3-2x)=0

=>x+1/2=0 hoặc 2/3-2x=0

=>x=-1/2 hoặc x=1/3

b: undefined

c: $\Leftrightarrow x\cdot\left(\dfrac{13}{4}-\dfrac{7}{6}\right)=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{5}{12}+\dfrac{20}{12}=\dfrac{25}{12}$

$\Leftrightarrow x=\dfrac{25}{12}:\dfrac{39-14}{12}=\dfrac{25}{25}=1$

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NH

Nguyễn Hữu Bình

18 tháng 4 2022

Tìm x biết: a) $\left(x-\dfrac{1}{2}\right)\left(-3-\dfrac{x}{2}\right)=0$ b) $x-\dfrac{1}{8}=\dfrac{5}{8}$c) $-\dfrac{1}{2}-\left(\dfrac{3}{2}+x\right)=-2$ ...

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#Toán lớp 6

1

O

★彡✿ทợท彡★

18 tháng 4 2022

a) $\left(x-\dfrac{1}{2}\right)\left(-3-\dfrac{x}{2}\right)=0$

Th1 : $x-\dfrac{1}{2}=0$

$x=0+\dfrac{1}{2}$

$x=\dfrac{1}{2}$

Th2 : $-3-\dfrac{x}{2}=0$

$\dfrac{x}{2}=-3$

$x=\left(-3\right)\cdot2$

$x=-6$

Vậy $x$ = $\left(\dfrac{1}{2};-6\right)$

b) $x-\dfrac{1}{8}=\dfrac{5}{8}$

$x=\dfrac{5}{8}+\dfrac{1}{8}$

$x=\dfrac{3}{4}$

c) $-\dfrac{1}{2}-\left(\dfrac{3}{2}+x\right)=-2$

$\dfrac{3}{2}+x=-\dfrac{1}{2}-\left(-2\right)$

$\dfrac{3}{2}+x=\dfrac{3}{2}$

$x=\dfrac{3}{2}-\dfrac{3}{2}$

$x=0$

d) $x+\dfrac{1}{3}=\dfrac{-12}{5}\cdot\dfrac{10}{6}$

$x+\dfrac{1}{3}=-4$

$x=-4-\dfrac{1}{3}$

$x=-\dfrac{13}{3}$

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