1.Tìm x, biết. a, $|x|$ x = $\dfrac{1}{3}$ b, $|x 2|$ = x c, $\dfrac{2}{\left(x-1\right)\left(x-3\right)}$ $\dfrac{5}{\left(x-3\right)\left(x-8\right)}$ \(\dfrac{12}{\left(x-8\right...

1

NL

Nguyễn Lê Phước Thịnh

3 tháng 10 2022

a: TH1: x>=0

=>x+x=1/3

=>x=1/6(nhận)

TH2: x<0

Pt sẽ là -x+x=1/3

=>0=1/3(loại)

b: $\Leftrightarrow\left\{{}\begin{matrix}x>=0\\x^2-x-2=0\end{matrix}\right.\Leftrightarrow x=2$

c: $\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-8}+\dfrac{1}{x-8}-\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-3}{4}$

$\Leftrightarrow\dfrac{1}{x-1}-\dfrac{2}{x-20}=\dfrac{-3}{4}$

$\Leftrightarrow\dfrac{x-20-2x+2}{\left(x-1\right)\left(x-20\right)}=\dfrac{-3}{4}$

$\Leftrightarrow-3\left(x^2-21x+20\right)=4\left(-x-18\right)$

$\Leftrightarrow3x^2-63x+60=4x+72$

=>3x^2-67x-12=0

hay $x\in\left\{22.51;-0.18\right\}$

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

NH

Nguyễn Hữu Bình

18 tháng 4 2022

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

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

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

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

Đúng(1)

NV

Nhật Văn

2 tháng 2 2023

Giải phương trình:

a) $\dfrac{1}{x-2}+3=\dfrac{x-3}{2-x}$

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

c) $1+\dfrac{1}{x+2}=\dfrac{12}{8+x^3}$

#Toán lớp 8

1

NL

Nguyễn Lê Phước Thịnh

2 tháng 2 2023

a: =>1+3x-6=-x+3

=>3x-5=-x+3

=>4x=8

=>x=2(loại)

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

=>3x-9+2x-4=x-1

=>5x-13=x-1

=>4x=12

=>x=3(loại)

c: =>x^2-2x+4+x^3+8=12

=>x^3+x^2-2x=0

=>x(x^2+x-2)=0

=>x(x+2)(x-1)=0

=>x=0 hoặc x=1

Đúng(2)

NV

Nhật Văn

2 tháng 2 2023

tks

TD

Thầy Đức Anh

Giáo viên VIP

27 tháng 4 2022

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

13

YN

Yen Nhi

21 tháng 6 2022

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

DP

Đông Phương Vô Hàn

7 tháng 8 2017

Tìm x, biết:

$\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}-\dfrac{1}{x-20}=\dfrac{-1}{4}$

#Toán lớp 6

1

NT

Nguyễn Thanh Hằng

7 tháng 8 2017

$\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}-\dfrac{1}{x-20}$

$\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-8}+\dfrac{1}{x-8}-\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-1}{4}$

$\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-1}{4}$

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

$\Leftrightarrow\left(x-1\right)\left(-1\right)=4$

$\Leftrightarrow x-1=-4$

$\Leftrightarrow x=-3\left(tm\right)$

Vậy ..............

N

nguyenviethung

3 tháng 10 2023

$\dfrac{2}{\left(x-1\right)\left(x-3\right)}$+$\dfrac{2}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}-\dfrac{1}{x-20}=\dfrac{-3}{4}$

1

NL

Nguyễn Lê Phước Thịnh

10 tháng 12 2023

Sửa đề:

$\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}-\dfrac{1}{x-20}=-\dfrac{3}{4}$

ĐKXĐ: $x\notin\left\{1;3;8;20\right\}$

PT=>$-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-8}-\dfrac{1}{x-8}+\dfrac{1}{x-20}-\dfrac{1}{x-20}=-\dfrac{3}{4}$

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

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

=>$x=\dfrac{4}{3}+1=\dfrac{7}{3}$(nhận)

NT

Nguyễn Thu Huyền

1 tháng 9 2023

Tìm...

3

T

Toru

1 tháng 9 2023

$a,3-x=x+1,8$

$\Rightarrow-x-x=1,8-3$

$\Rightarrow-2x=-1,2$

$\Rightarrow x=0,6$

$b,2x-5=7x+35$

$\Rightarrow2x-7x=35+5$

$\Rightarrow-5x=40$

$\Rightarrow x=-8$

$c,2\left(x+10\right)=3\left(x-6\right)$

$\Rightarrow2x+20=3x-18$

$\Rightarrow2x-3x=-18-20$

$\Rightarrow-x=-38$

$\Rightarrow x=38$

$d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x$

$\Rightarrow8x-3+1=1+6x+x$

$\Rightarrow8x-3=7x$

$\Rightarrow8x-7x=3$

$\Rightarrow x=3$

$e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x$

$\Rightarrow-3x+x=\dfrac{4}{3}-\dfrac{2}{9}$

$\Rightarrow-2x=\dfrac{10}{9}$

$\Rightarrow x=-\dfrac{5}{9}$

Đúng(3)

T

Toru

1 tháng 9 2023

$g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}$

$\Rightarrow\dfrac{1}{2}x-\dfrac{3}{4}x=-\dfrac{1}{2}-\dfrac{5}{6}$

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

$\Rightarrow x=\dfrac{16}{3}$

$h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)$

$\Rightarrow x-4=5-x$

$\Rightarrow x+x=5+4$

$\Rightarrow2x=9$

$\Rightarrow x=\dfrac{9}{2}$

$k,7x^2-11=6x^2-2$

$\Rightarrow7x^2-6x^2=-2+11$

$\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.$

$m,5\left(x+3\cdot2^3\right)=10^2$

$\Rightarrow5\left(x+24\right)=100$

$\Rightarrow x+24=20$

$\Rightarrow x=-4$

$n,\dfrac{4}{9}-\left(\dfrac{1}{6^2}\right)=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}$

$\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{4}{9}-\dfrac{1}{36}$

$\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{5}{12}$

$\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2=0$

$\Rightarrow x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}$

#$Urushi\text{☕}$

Đúng(3)

KJ

❖ Khang/GD❄ 『ʈєɑɱ❖Hoàng๖ۣۜGia☆』

22 tháng 2 2022

Tìm x, biết:

$a,\dfrac{1}{3}:\left(2x-1\right)=\dfrac{-1}{6}$

$b,\left(3x+2\right)\left(\dfrac{-2}{5}x-7\right)=0$

$c,\dfrac{x}{8}=\dfrac{9}{4}$

$d,\dfrac{x-3}{2}=\dfrac{18}{x-3}$

$e,4,5x-6,2x=6,12$

$h,11,4-\left(x-3,4\right)=-16,2$

#Toán lớp 6

2

NL

Nguyễn Lê Phước Thịnh

22 tháng 2 2022

a: =>2x-1=-2

=>2x=-1

hay x=-1/2

b: $\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\-\dfrac{2}{5}x-7=0\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{2}{3};-\dfrac{35}{2}\right\}$

c: x/8=9/4

nên x/8=18/8

hay x=18

d: $\Leftrightarrow\left(x-3\right)^2=36$

=>x-3=6 hoặc x-3=-6

=>x=9 hoặc x=-3

e: =>-1,7x=6,12

hay x=-3,6

h: =>x-3,4=27,6

hay x=31

Đúng(1)

R

Rhider

22 tháng 2 2022

a) $\dfrac{1}{3}\div\left(2x-1\right)=\dfrac{-1}{6}$

$\left(2x-1\right).\dfrac{1}{3}\div\left(2x-1\right)=\left(2x-1\right)\left(-\dfrac{1}{6}\right)$

$\dfrac{1}{3}=\left(2x-1\right)\left(-\dfrac{1}{6}\right)$

$\dfrac{1}{3}=-1\left(2x-1\right)\div6$

$\dfrac{1}{3}=-2x+1\div6$

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

b) $\left(3x+2\right)\left(\dfrac{-2}{5}x-7\right)=0$

$TH1:3x+2=0$

$3x=0-2$

$3x=-2$

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

$TH2:\left(-\dfrac{2}{5}x-7\right)=0$

$\left(\dfrac{-2}{5}x-7\right)=0$

$\left(\dfrac{-2x}{5}+\dfrac{5\left(-7\right)}{5}\right)=0$

$\left(\dfrac{-2x-35}{5}\right)=0$

$-2x-35=0$

$-2x=0+35$

$x=-\dfrac{35}{2}$

c) $\dfrac{x}{8}=\dfrac{9}{4}$

$\Leftrightarrow x=\dfrac{9.8}{4}=\dfrac{72}{4}=18$

$x=18$

d) $\dfrac{x-3}{2}=\dfrac{18}{x-3}$

$x-3=18+2$

$x=20-3$

$x=17$

e) $4,5x-6,2x=6,12$

$\dfrac{9x}{2}-6,2.x=6,12$

$\dfrac{9x}{2}+\dfrac{-31x}{5}=6,12$

$\dfrac{5.9x}{10}+\dfrac{2\left(-31\right)x}{10}=6.12$

$\dfrac{45x-62x}{10}=6.12$

$=-17x\div10=6.12$

$-17x=10.6.12$

$x=-3,6$

h) $11,4-\left(x-3,4\right)=-16,2$

$x-3,4=-16,2+11,4$

$x-3,4=-4,8$

$x=-1,4$

Đúng(2)

DP

Đỗ Phương Anh

12 tháng 9 2021

Có mấy giá trị của x thỏa mãn bt dưới đây:

$\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}-\dfrac{1}{x-20}=\dfrac{-3}{4}$

0

NW

Nguyên Walker (Walker Of Alan)

9 tháng 5 2023

P=$\left(\dfrac{3\left(x+2\right)}{2x^2+8}-\dfrac{2x^2-x-10}{\left(x+1\right)\left[\left(x+1\right)^2-2x\right]}\right):\left(\dfrac{5}{x^2+1}+\dfrac{3}{2\left(x+1\right)}-\dfrac{3}{x-1}\right)\cdot\dfrac{2}{x-1}$

a) rút gọn P

b)tìm tất cả các giá trị nguyên của x để P có giá trị là bội của 4

#Toán lớp 8

1

NL

Nguyễn Lê Phước Thịnh

9 tháng 5 2023

a: $P=\left(\dfrac{3x+6}{2\left(x^2+4\right)}-\dfrac{2x^2-x-10}{\left(x+1\right)\left(x^2+1\right)}\right):\left(\dfrac{10\left(x^2-1\right)+3\left(x^2+1\right)\left(x-1\right)-6\left(x+1\right)\left(x^2+1\right)}{\left(x^2+1\right)\left(x+1\right)\left(x-1\right)\cdot2}\right)\cdot\dfrac{2}{x-1}$

$=\left(\dfrac{\left(3x+6\right)\left(x^3+x^2+x+1\right)-\left(2x^2+8\right)\left(2x^2-x-10\right)}{2\left(x^2+4\right)\left(x+1\right)\left(x^2+1\right)}\right)\cdot\dfrac{\left(x^2+1\right)\left(x-1\right)\left(x+1\right)\cdot2}{-3x^3+x^2-3x-13}\cdot\dfrac{2}{x-1}$

$=\dfrac{-x^4+11x^3+13x^2+17x+16}{\left(x^2+4\right)}\cdot\dfrac{2}{-3x^3+x^2-3x-13}$

Giải các pt sau: a)$x^2+\dfrac{4x^2}{\left(x+2\right)^2}=12$ b) $\dfrac{x^2}{3}+\dfrac{48}{x^2}=5\left(\dfrac{x}{3}+\dfrac{4}{x}\right)$ c) $\left(\dfrac{x}{x-1}\right)^2+\left(\dfrac{x}{x+1}\right)^2=\dfrac{10}{9}$ d) $\left(\dfrac{x-1}{x}\right)^2+\left(\dfrac{x-1}{x-2}\right)^2=\dfrac{40}{9}$ e) $x^2+\left(\dfrac{x}{x-1}\right)^2=8$ g) $x^3+\dfrac{1}{x^3}=6\left(x+\dfrac{1}{x}\right)$ f)...

TD

Trần Đức Mạnh

17 tháng 1 2018