Tìm x : $\dfrac{x 1}{2x 1}=\dfrac{0,5 2}{x 3}$

1

NL

Nguyễn Lê Phước Thịnh

24 tháng 9 2022

=>x^2+4x+3=x^2+4x+0,5x+0,5

=>4x+3=4,5x+0,5

=>-0,5x=-2,5

=>x=5

N

nanako

21 tháng 10 2021

Tìm TXĐ:

a) y=$\left(1-x\right)^{\dfrac{-1}{3}}$

b) $y=\sqrt{\log_{0,5}\dfrac{2x+1}{x+5}-2}$

c) $y=\log_{10}\sqrt{x^2-x-12}$

d) $y=\sqrt{\log_{10}x-1+\log_{10}x+1}$

#Toán lớp 12

0

LN

Lê Ngọc Anh

17 tháng 3 2022

Tìm x, biết:

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

b) $\dfrac{1}{3}.x-0,5.x=0,75$

#Toán lớp 6

1

NL

Nguyễn Lê Phước Thịnh

27 tháng 1

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

=>$\left(\dfrac{3}{4}x+\dfrac{5}{2}\right)=\dfrac{1}{8}:\dfrac{-2}{3}=\dfrac{-3}{16}$

=>$\dfrac{3}{4}x=-\dfrac{3}{16}-\dfrac{5}{2}=-\dfrac{3}{16}-\dfrac{40}{16}=-\dfrac{43}{16}$

=>$x=-\dfrac{43}{16}:\dfrac{3}{4}=\dfrac{-43}{16}\cdot\dfrac{4}{3}=\dfrac{-43}{12}$

b: $\dfrac{1}{3}\cdot x-0,5x=0,75$

=>$x\left(\dfrac{1}{3}-\dfrac{1}{2}\right)=0,75$

=>$x\cdot\dfrac{-1}{6}=0,75$

=>$x=-0,75\cdot6=-4,5$

Tìm x, biết:a) $\left(5x+1\right)^2=\dfrac{36}{49}$b) $\left[\left(-0,5\right)^3\right]^x=\dfrac{1}{64}$c) $2020^{\left(x-2\right).\left(2x+3\right)}=1$d) $\left(x+1\right)^{x+10}=\left(x+1\right)^{x+4}$ với \(x\in...

NM

Nguyễn Minh An

23 tháng 11 2021

1

NH

Nguyễn Hoàng Minh

23 tháng 11 2021

$a,\Rightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}5x=\dfrac{1}{7}\\5x=-\dfrac{13}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{35}\\x=-\dfrac{13}{35}\end{matrix}\right.\\ b,\Rightarrow\left(-\dfrac{1}{8}\right)^x=\dfrac{1}{64}=\left(-\dfrac{1}{8}\right)^2\Rightarrow x=2\\ c,\Rightarrow\left(x-2\right)\left(2x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\\ d,\Rightarrow\left(x+1\right)^{x+10}-\left(x+1\right)^{x+4}=0\\ \Rightarrow\left(x+1\right)^{x+4}\left[\left(x+1\right)^6-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\\left(x+1\right)^6=1\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x+1=0\\x+1=1\\x+1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=-2\end{matrix}\right.\\ e,\Rightarrow\dfrac{3}{4}\sqrt{x}=\dfrac{5}{6}\left(x\ge0\right)\\ \Rightarrow\sqrt{x}=\dfrac{10}{9}\Rightarrow x=\dfrac{100}{81}$

Đúng(2)

O

Ɲɠọç⁀²ᵏ⁹

21 tháng 6 2021

Tìm $x$$\in$$Q$:

$a$$)$ $|0,5.x-2|-|x+\dfrac{2}{3}|=0$

$b)$ $2x-|x+1|=\dfrac{-1}{2}$

1

D

๖ۣۜDũ๖ۣۜN๖ۣۜG

21 tháng 6 2021

a)

TH1: $x< \dfrac{-2}{3}$

<=> $\left\{{}\begin{matrix}\left|0,5x-2\right|=2-0,5x\\\left|x+\dfrac{2}{3}\right|=-x-\dfrac{2}{3}\end{matrix}\right.$

PT <=> $2-0,5x+x+\dfrac{2}{3}=0< =>x=\dfrac{-16}{3}\left(c\right)$

TH2: $\dfrac{-2}{3}\le x< 4$

<=> $\left\{{}\begin{matrix}\left|0,5x-2\right|=2-0,5x\\\left|x+\dfrac{2}{3}\right|=x+\dfrac{2}{3}\end{matrix}\right.$

PT <=> $2-0,5x-x-\dfrac{2}{3}=0< =>x=\dfrac{8}{9}\left(c\right)$

TH3: $x\ge4$

<=> $\left\{{}\begin{matrix}\left|0,5x-2\right|=0,5x-2\\\left|x+\dfrac{2}{3}\right|=x+\dfrac{2}{3}\end{matrix}\right.$

PT <=> $0,5x-2-x-\dfrac{2}{3}=0< =>x=\dfrac{-16}{3}\left(l\right)$

KL: x $\left\{\dfrac{-16}{3};\dfrac{8}{9}\right\}$

b) TH1: $x\ge-1< =>\left|x+1\right|=x+1$

PT <=> 2x - x -1 = $\dfrac{-1}{2}$

<=> x = $\dfrac{1}{2}$ (c)

TH2: x < -1 <=> $\left|x+1\right|=-x-1$

PT <=> 2x + x + 1 = $\dfrac{-1}{2}$

<=> x = $\dfrac{-1}{2}$ (l)

KL: x $\in\left\{\dfrac{1}{2}\right\}$

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)...

LN

Lê Ngọc Anh

8 tháng 3 2022

#Toán lớp 6

2

HP

hưng phúc

8 tháng 3 2022

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

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

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

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

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

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

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

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

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

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

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

$=\dfrac{1}{5}$

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 =....

HT

Hà Trí Kiên

15 tháng 6 2023

$\dfrac{x-3}{3}=\dfrac{2x+1}{5}$

$\dfrac{x+1}{22}=\dfrac{6}{x}$

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

$\dfrac{2x-1}{21}=\dfrac{3}{2x+1}$

$\dfrac{2x+1}{9}=\dfrac{5}{x+1}$

Tìm x

2

KT

『Kuroba ム Tsuki Ryoo ︵²ᵏ⁷』

15 tháng 6 2023

`@` `\text {Ans}`

`\downarrow`

$\dfrac{x-3}{3}=\dfrac{2x+1}{5}$

`=> (x-3)5 = (2x+1)3`

`=> 5x-15 = 6x+3`

`=> 5x-6x = 15+3`

`=> -x=18`

`=> x=-18`

$\dfrac{x+1}{22}=\dfrac{6}{x}$

`=> (x+1)x = 22*6`

`=> (x+1)x = 132`

`=> x^2 + x = 132`

`=> x^2+x-132=0`

`=> (x-11)(x+12)=0`

`=>`$\left[{}\begin{matrix}x-11=0\\x+12=0\end{matrix}\right.$

`=>`$\left[{}\begin{matrix}x=11\\x=-12\end{matrix}\right.$

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

`=> (2x-1)x = 2*5`

`=> 2x^2 - x =10`

`=> 2x^2 - x - 10 =0`

`=> 2x^2 + 4x - 5x - 10 =0`

`=> (2x^2 + 4x) - (5x+10)=0`

`=> 2x(x+2) - 5(x+2)=0`

`=> (2x-5)(x+2)=0`

`=>`$\left[{}\begin{matrix}2x-5=0\\x+2=0\end{matrix}\right.$

`=>`$\left[{}\begin{matrix}2x=5\\x=-2\end{matrix}\right.$

`=>`$\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-2\end{matrix}\right.$

$\dfrac{2x-1}{21}=\dfrac{3}{2x+1}$

`=> (2x-1)(2x+1)=21*3`

`=> 4x^2 + 2x - 2x - 1 = 63`

`=> 4x^2 - 1=63`

`=> 4x^2 - 1 - 63=0`

`=> 4x^2 - 64 = 0`

`=> 4(x^2 - 16)=0`

`=> 4(x^2 + 4x - 4x - 16)=0`

`=> 4[(x^2+4x)-(4x+16)]=0`

`=> 4[x(x+4)-4(x+4)]=0`

`=> 4(x-4)(x+4)=0`

`=>`$\left[{}\begin{matrix}x-4=0\\x+4=0\end{matrix}\right.$

`=>`$\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.$

$\dfrac{2x+1}{9}=\dfrac{5}{x+1}$

`=> (2x+1)(x+1) = 9*5`

`=> (2x+1)(x+1)=45`

`=> 2x^2 + 2x + x + 1 = 45`

`=> 2x^2 + 3x + 1 =45`

`=> 2x^2 + 3x + 1 - 45 =0`

`=> 2x^2+3x-44=0`

`=> 2x^2 + 11x - 8x - 44=0`

`=> (2x^2 +11x) - (8x+44)=0`

`=> x(2x+11) - 4(2x+11)=0`

`=> (x-4)(2x+11)=0`

`=>`$\left[{}\begin{matrix}x-4=0\\2x+11=0\end{matrix}\right.$

`=>`$\left[{}\begin{matrix}x=4\\2x=-11\end{matrix}\right.$

`=>`$\left[{}\begin{matrix}x=4\\x=-\dfrac{11}{2}\end{matrix}\right.$

BD

boi đz

15 tháng 6 2023

$\dfrac{x-3}{3}=\dfrac{2x+1}{5}\\ \left(x-3\right)\cdot5=\left(2x+1\right)\cdot3\\ x5-15=6x+3\\ x5-6x=3+15\\ -x=18\\ \Rightarrow x=-18$

$\dfrac{x+1}{22}=\dfrac{6}{x}\\ \left(x+1\right)\cdot x=6\cdot22\\ \left(x+1\right)\cdot x=2\cdot3\cdot2\cdot11\\ \left(x+1\right)\cdot x=12\cdot11\\ \Rightarrow x=11$

$\dfrac{2x-1}{21}=\dfrac{3}{2x+1}\\ \left(2x-1\right)\cdot\left(2x+1\right)=21\cdot3\\ \left(2x-1\right)\cdot\left(2x+1\right)=7\cdot3\cdot3\\ \left(2x-1\right)\cdot\left(2x+1\right)=7\cdot9\\ \Rightarrow2x+1=9\\ 2x=8\\ x=4$

tìm x$\dfrac{-4}{x-1}$ $\dfrac{3}{x-1}$ $\dfrac{2x+1}{x-3}$ $\dfrac{x+3}{x-2}$ $\dfrac{4x-1}{3-x}$ $\dfrac{3x+3}{x-1}$ $\dfrac{x-2}{x+3}$ ...

TB

Thảo Bùi

3 tháng 5 2021

#Toán lớp 6

1

NT

Nguyễn Trần Thành Đạt

3 tháng 5 2021

Không có dấu "=" hay như nào đâu giải tìm x được

TB

Thảo Bùi

3 tháng 5 2021

ko có dấu bằng

72x+72x+3=344 (5-x)(9x2-4)=0 $\left|2-2x\right|$-3,75=(-0,5)2 $\sqrt{x-1}$+$\dfrac{2}{3}$=1 ($\dfrac{1}{3}$-$\dfrac{3}{2}$x)2=2$\dfrac{1}{4}$ giúp mình với!!!! cảm ơn...

TD

Trần Đình Hoàng Quân

24 tháng 5 2023

7

NT

Nguyễn Thị Thương Hoài

Giáo viên VIP

24 tháng 5 2023

(5 - $x$)(9$x^2$- 4) =0

$\left[{}\begin{matrix}5-x=0\\9x^2-4=0\end{matrix}\right.$

$\left[{}\begin{matrix}x=5\\9x^2=4\end{matrix}\right.$

$\left[{}\begin{matrix}x=5\\x^2=\dfrac{4}{9}\end{matrix}\right.$

$\left[{}\begin{matrix}x=5\\x=-\dfrac{2}{3}\\x=\dfrac{2}{3}\end{matrix}\right.$

Vậy $x$ $\in$ { - $\dfrac{2}{3}$; $\dfrac{2}{3}$; $5$}

Đúng(2)

NT

Nguyễn Thị Thương Hoài

Giáo viên VIP

24 tháng 5 2023

7^2$x$ + 7^2$x$^{+ 3} = 344

7²^$x$$\times$( 1 + 7³) = 344

7^2$x$ $\times$ (1 + 343) = 344

7^2$x$ $\times$ 344 = 344

7²^$x$ = 344 : 344

7^2$x$ = 1

7^2$x$ = 7⁰

$2x$ = 0

$x$ = 0

Kết luận: $x$ = 0

1. tìm xa, 2x - $\dfrac{1}{2}$ : $\dfrac{3}{4}$ = $\dfrac{12}{9}$b, x$^2$ - 2 = 02. Cho A = $\dfrac{1,11+0,19-13.2}{2,06+0,54}$ - $\left(\dfrac{1}{2}+\dfrac{1}{4}\right)$ : 2B = $\left(5\dfrac{7}{8}-2\dfrac{1}{4}-0,5\right)$ : $2\dfrac{23}{26}$a, Rút gọn A, Bb, Tìm x ϵ Z để A < x <...

HG

Hoàng Giang

25 tháng 12 2023

2

TD

Trần Đức Hải Phong

25 tháng 12 2023

a)
=$2x-\dfrac{1}{2}=\dfrac{12}{9}\cdot\dfrac{3}{4}=1$
=$2x=1+\dfrac{1}{2}=1.5$
=$x=1.5:2=0.75$
b)
=$x^2=0+2=2$
TH1:$x=2$
TH2:$x=-2$

NL

Nguyễn Lê Phước Thịnh

31 tháng 12 2023

Bài 1:

a: $2x-\dfrac{1}{2}:\dfrac{3}{4}=\dfrac{12}{9}$

=>$2x-\dfrac{1}{2}\cdot\dfrac{4}{3}=\dfrac{4}{3}$

=>$2x=\dfrac{4}{3}+\dfrac{2}{3}=\dfrac{6}{3}=2$

=>x=2/2=1

b: $x^2-2=0$

=>$x^2=2$

=>$\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.$

Bài 2:

a: $A=\dfrac{1,11+0,19-13\cdot2}{2,06+0,54}-\left(\dfrac{1}{2}+\dfrac{1}{4}\right):2$

$=\dfrac{1,3-26}{2,6}-\dfrac{3}{4}:2$

$=-9,5-\dfrac{3}{8}=-\dfrac{79}{8}$

$B=\left(5\dfrac{7}{8}-2\dfrac{1}{4}-0,5\right):\left(2\dfrac{23}{26}\right)$

$=\left(5+\dfrac{7}{8}-2-\dfrac{1}{4}-\dfrac{1}{2}\right):\dfrac{75}{26}$

$=\left(3+\dfrac{1}{8}\right)\cdot\dfrac{26}{75}=\dfrac{25}{8}\cdot\dfrac{26}{75}=\dfrac{13}{12}$

b: A<x<B

=>$-\dfrac{79}{8}< x< \dfrac{13}{12}$

mà $x\in Z$

nên $x\in\left\{-9;-8;...;0;1\right\}$

NT

ngoc tranbao

3 tháng 8 2021

Giải...

#Toán lớp 9

3

AH

Akai Haruma

Giáo viên

3 tháng 8 2021

a. ĐKXĐ: $x\geq 1$

PT $\Leftrightarrow \frac{1}{2}\sqrt{x-1}-\frac{3}{2}.\sqrt{9}.\sqrt{x-1}+24.\sqrt{\frac{1}{64}}.\sqrt{x-1}=-17$

$\Leftrightarrow \frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17$

$\Leftrightarrow -\sqrt{x-1}=-17$

$\Leftrightarrow \sqrt{x-1}=17$

$\Leftrightarrow x-1=289$

$\Leftrightarrow x=290$

b. ĐKXĐ: $x\geq \frac{1}{2}$

PT $\Leftrightarrow \sqrt{9}.\sqrt{2x-1}-0,5\sqrt{2x-1}+\frac{1}{2}.\sqrt{25}.\sqrt{2x-1}+\sqrt{49}.\sqrt{2x-1}=24$

$\Leftrightarrow 3\sqrt{2x-1}-0,5\sqrt{2x-1}+2,5\sqrt{2x-1}+7\sqrt{2x-1}=24$
$\Leftrightarrow 12\sqrt{2x-1}=24$

$\Leftrihgtarrow \sqrt{2x-1}=2$

$\Leftrightarrow x=2,5$ (tm)

AH

Akai Haruma

Giáo viên

3 tháng 8 2021

c. ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow \sqrt{36}.\sqrt{x-2}-15\sqrt{\frac{1}{25}}\sqrt{x-2}=4(5+\sqrt{x-2})$

$\Leftrightarrow 6\sqrt{x-2}-3\sqrt{x-2}=20+4\sqrt{x-2}$

$\Leftrightarrow \sqrt{x-2}=-20< 0$ (vô lý)

Vậy pt vô nghiệm