a,1/4

b,8

a, 1/4
b,8

a) x=1/4

b) -12

c) 50

d)Vậy x e thuộc {-3, -2, -1}

\(a,\Leftrightarrow-\dfrac{1}{2}x=\dfrac{1}{4}\Leftrightarrow x=-\dfrac{1}{2}\\ b,\Leftrightarrow\dfrac{1}{6}:x=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\Leftrightarrow x=\dfrac{1}{6}:\dfrac{5}{6}=\dfrac{1}{5}\\ c,\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=3\\x+\dfrac{1}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}\\x=-\dfrac{16}{5}\end{matrix}\right.\)

\(d,\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{22}{9}-\dfrac{7}{3}=\dfrac{1}{9}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{3}\\x+\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{6}\\x=-\dfrac{5}{6}\end{matrix}\right.\\ e,\Leftrightarrow2\left|x\right|=2-\dfrac{1}{2}=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{3}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)

\(f,\Leftrightarrow\left|x+\dfrac{1}{2}\right|=1+\dfrac{1}{6}=\dfrac{7}{6}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{7}{6}\\x+\dfrac{1}{2}=-\dfrac{7}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)

e: ta có: \(2\left|x\right|+\dfrac{1}{2}=2\)

\(\Leftrightarrow2\left|x\right|=\dfrac{3}{2}\)

\(\Leftrightarrow\left|x\right|=\dfrac{3}{4}\)

hay \(x\in\left\{\dfrac{3}{4};-\dfrac{3}{4}\right\}\)

a: \(5^{\left(x-2\right)\left(x+3\right)}=1\)

=>\(\left(x-2\right)\left(x+3\right)=0\)

=>\(\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

c: \(\left|x^2+2x\right|+\left|y^2-9\right|=0\)

mà \(\left\{{}\begin{matrix}\left|x^2+2x\right|>=0\forall x\\\left|y^2-9\right|>=0\forall y\end{matrix}\right.\)

nên \(\left\{{}\begin{matrix}x^2+2x=0\\y^2-9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\left(x+2\right)=0\\\left(y-3\right)\left(y+3\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\in\left\{0;-2\right\}\\y\in\left\{3;-3\right\}\end{matrix}\right.\)

d: \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=120\)

=>\(2^x\left(1+2+2^2+2^3\right)=120\)

=>\(2^x\cdot15=120\)

=>\(2^x=8\)

=>x=3

e: \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)

=>\(\left(x-7\right)^{x+11}-\left(x-7\right)^{x+1}=0\)

=>\(\left(x-7\right)^{x+1}\left[\left(x-7\right)^{10}-1\right]=0\)

=>\(\left[{}\begin{matrix}x-7=0\\x-7=1\\x-7=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\\x=6\end{matrix}\right.\)

1A

2C

a, A

b, C

A,C

1. A

2. D

3. A

4. A

4A

3A

2D

1D

b: =>2x-6=2

hay x=4

\(a,\left(x-1\right)^2=1\\ \Leftrightarrow\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\\ b,7^2x-6=49\\ \Leftrightarrow49x=55\\ \Leftrightarrow x=\dfrac{55}{49}\)

\(a,x-\dfrac{3}{4}=\dfrac{1}{7}\\ x=\dfrac{1}{7}+\dfrac{3}{4}\\ x=\dfrac{25}{28}\\ b,x+\dfrac{7}{5}=\dfrac{9}{8}.\dfrac{4}{27}\\ x+\dfrac{7}{5}=\dfrac{1}{6}\\ x=\dfrac{1}{6}-\dfrac{7}{5}\\ x=\dfrac{-37}{30}\\ c,\dfrac{2}{5}-\dfrac{3}{7}=\dfrac{x}{70}\\ \dfrac{-1}{35}=\dfrac{x}{70}\\ \dfrac{-2}{70}=\dfrac{x}{70}\\ x=-2\\ d,\dfrac{2}{9}-\dfrac{7}{8}.x=1\\ \dfrac{7}{8}.x=\dfrac{2}{9}-1\\ \dfrac{7}{8}.x=\dfrac{-7}{9}\\ x=\dfrac{-7}{9}:\dfrac{7}{8}\\ x=\dfrac{-8}{9}\)

\(a,x-\dfrac{3}{4}=\dfrac{1}{7}\)

\(\Rightarrow x=\dfrac{1}{7}+\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{25}{28}\)

\(b,x+\dfrac{7}{5}=\dfrac{9}{8}.\dfrac{4}{27}\)

\(\Rightarrow x+\dfrac{7}{5}=\dfrac{1}{6}\)

\(\Rightarrow x=\dfrac{1}{6}-\dfrac{7}{5}\)

\(\Rightarrow x=-\dfrac{37}{30}\)

\(c,\dfrac{2}{5}-\dfrac{3}{7}=\dfrac{x}{70}\)

\(\Rightarrow\dfrac{-1}{35}=\dfrac{x}{70}\)

\(\Rightarrow35x=-70\)

\(\Rightarrow x=-2\)

\(d,\dfrac{2}{9}-\dfrac{7}{8}.x=1\)

\(\Rightarrow\dfrac{7}{8}x=\dfrac{2}{9}-1\)

\(\Rightarrow\dfrac{7}{8}x=-\dfrac{7}{9}\)

\(\Rightarrow x=-\dfrac{8}{9}\)

`a)x-3/4=1/7`

`=>x=1/7+3/4`

`=>x=4/28+21/28`

`=>x=25/28`

`b)x+7/5=9/8 . 4/27`

`=>x+7/5=(9.4)/(8.27)`

`=>x+7/5=1/6`

`=>x=1/6+7/5`

`=>x=5/30+42/30`

`=>x=47/30`

`c)2/5-3/7=x/70`

`=>14/35-15/35=x/70`

`=>-1/35=x/70`

`=>-2/70=x/70`

`=>x=-2`

`d)2/9-7/8x=1`

`=>7/8x=2/9-1`

`=>7/8x=-7/9`

`=>x=-7/9:7/8`

`=>x=-7/9 . 8/7`

`=>x=-8/9`

`#lv`

Bài 1: 

a: \(\Leftrightarrow x-1\in\left\{1;-1;3;-3\right\}\)

hay \(x\in\left\{2;0;4;-2\right\}\)

Bài 2: 

a: =>x-1=1 hoặc x-1=-1

=>x=2 hoặc x=0

b: =>x+1=-1

hay x=-2

c: =>(135-7x):9=8

=>135-7x=72

=>7x=63

hay x=9

d: =>(x+7)(x-3)<0

=>-7<x<3

e: \(\Leftrightarrow3^{x-3}=18+9=27\)

=>x-3=3

hay x=6

f: =>4-2x=0

hay x=2

a. 5 - 3(x + 4) = -1

⇔ 5 - 3x - 12 = -1

⇔ 3x = -1 - 5 + 12

⇔ 3x = 6

⇔ x = 2

\(d,2x^2-3=5\)

\(\Leftrightarrow2x^2=8\)

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow x=\pm2\)

\(e,x\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=0\end{matrix}\right.\)

a)\(=>3\left(x+4\right)=6=>x+4=2=>x=-2\)

b)\(=>x-1-x-2=0\)

\(=>-3=0\left(vl\right)\) => x ko tồn tại