mk chi lam duoc cau c thoi

suy ra 2x-1=5 hoac 2x-1=-5

          2x=6            2x=-4

          x=3                 x=-2

vay x=3 hoac x=-2

a. 3/4.x+1/3=-1/2

=>3/4.x=-1/2-1/3=-5/6

=>x=-5/6 chia 3/4=-10/9

b. -x/4=-9/x =>-x*x=4*-9

=>-2x=-36 =>x=18

c./2x-1/=5

=> 2x-1=5 =>2x=5+1=6 =>x=3

  hoặc 2x-1=-5 =>2x=-5+1=-4 =>x=-2

d,e: Sai đề rồi

1. 

$(3^2-2^3)x+3^2.2^2=4^2.3$

$\Leftrightarrow x+36=48$

$\Leftrightarrow x=48-36=12$

2.

$x^5-x^3=0$

$\Leftrightarrow x^3(x^2-1)=0$

$\Leftrightarrow x^3(x-1)(x+1)=0$

$\Leftrightarrow x^3=0$ hoặc $x-1=0$ hoặc $x+1=0$

$\Leftrightarrow x=0$ hoặc $x=\pm 1$
3.

$(x-1)^2+(-3)^2=5^2(-1)^{100}$

$\Leftrightarrow (x-1)^2+9=25$

$\Leftrightarrow (x-1)^2=25-9=16=4^2=(-4)^2$

$\Rightarrow x-1=4$ hoặc $x-1=-4$

$\Leftrightarrow x=5$ hoặc $x=-3$

4.

$(2x-1)^2-(2x-1)=0$

$\Leftrightarrow (2x-1)(2x-1-1)=0$

$\Leftrightarrow (2x-1)(2x-2)=0$

$\Leftrightarrow 2x-1=0$ hoặc $2x-2=0$

$\Leftrightarrow x=\frac{1}{2}$ hoặc $x=1$

$\Lef

`@` `\text {Ans}`

`\downarrow`

\((3^2-2^3)x+3^2.2^2=4^2.3\)

`=> x + (3*2)^2 = 48`

`=> x+6^2 = 48`

`=> x + 36 = 48`

`=> x = 48 - 36`

`=> x=12`

Vậy, `x=12`

\(x^5-x^3=0\)

`=> x^3(x^2 - 1)=0`

`=>`\(\left[{}\begin{matrix}x^3=0\\x^2-1=0\end{matrix}\right.\)

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

`=>`\(\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)

Vậy, `x \in {0; +- 1 }`

\(\left(x-1\right)^2+\left(-3\right)^2=5^2\cdot\left(-1\right)^{100}\)

`=> (x-1)^2 + 9 = 25*1`

`=> (x-1)^2 + 9 = 25`

`=> (x-1)^2 = 25 - 9`

`=> (x-1)^2 = 16`

`=> (x-1)^2 = (+-4)^2`

`=>`\(\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=4+1\\x=-4+1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

Vậy, `x \in {5; -3}`

\((2x-1)^2-(2x-1)=0\)

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

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

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

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

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

Vậy, `x \in {1; 1/2}`

b) x + 2/5 = 2-3x/3

x + 2/5 = 2 -x

x +x = 2 - 2/5

2x= 8/5

x = 8/5 x 1/2

x = 4/5

c) 2. 3^x . 3^2 = 18

3^x+2 = 18:2

3^ x+2 = 9

3^ x+2 = 3^2

=> x+2 = 2

=> x=0

a) TH1: 2x-1 = x + 2

2x-x = 2+1

x =3

TH2 : 2x-1 = -(x+2)

2x-1 = -x-2

2x +x = -2 +1

3x = -1

x = -1/3

`#040911`

`a,`

`15 + 25 \div (2x - 1) = 20`

`\Rightarrow 25 \div (2x - 1) = 20 - 15`

`\Rightarrow 25 \div (2x - 1) = 5`

`\Rightarrow 2x - 1 = 25 \div 5`

`\Rightarrow 2x - 1 = 5`

`\Rightarrow 2x = 6`

`\Rightarrow x = 3`

Vây, `x = 3.`

`b,`

\(3^{x-1}+2\cdot3^x=21\)

`\Rightarrow 3^x \div 3 + 2. 3^x = 21`

`\Rightarrow 3^x . \frac{1}{3} + 2. 3^x = 21`

`\Rightarrow 3^x . (\frac{1}{3} + 2) = 21`

`\Rightarrow 3^x . \frac{7}{3} = 21`

`\Rightarrow 3^x = 21 \div \frac{7}{3}`

`\Rightarrow 3^x = 9`

`\Rightarrow 3^x = 3^2`

`\Rightarrow x = 2`

Vậy, `x = 2.`

`c,`

\(2^{x-3}+2^{x+1}=17\)

`\Rightarrow 2^x \div 2^3 + 2^x . 2 = 17`

`\Rightarrow 2^x . \frac{1}{8} + 2^x . 2 = 17`

`\Rightarrow 2^x . (\frac{1}{8} + 2) = 17`

`\Rightarrow 2^x . \frac{17}{8} = 17`

`\Rightarrow 2^x = 17 \div \frac{17}{8}`

`\Rightarrow 2^x = 8`

`\Rightarrow 2^x = 2^3`

`\Rightarrow x = 3`

Vậy, `x = 3`

`d,`

\(5^x-5^{x-1}=20\)

`\Rightarrow 5^x - 5^x \div 5 = 20`

`\Rightarrow 5^x - 5^x . \frac{1}{5} = 20`

`\Rightarrow 5^x . (1 - \frac{1}{5} = 20`

`\Rightarrow 5^x . \frac{4}{5} = 20`

`\Rightarrow 5^x = 20 \div \frac{4}{5}`

`\Rightarrow 5^x = 25`

`\Rightarrow 5^x = 5^2`

`\Rightarrow x = 2`

Vậy, `x = 2.`

\(a.25:\left(2x-1\right)=5\)

\(2x-1=5\Leftrightarrow2x=6\Leftrightarrow x=3\)

\(b.3^x:3+2.3^x=21\)\(\Leftrightarrow3^x.\dfrac{1}{3}+2.3^x=21\)

\(\Leftrightarrow3^x\left(\dfrac{1}{3}+2\right)=21\)

\(\Leftrightarrow3^x.\dfrac{7}{3}=21\)

\(\Leftrightarrow3^x=9\Leftrightarrow x=2\)

\(c.2^x:2^3+2^x.2=17\Leftrightarrow2^x.\dfrac{1}{8}+2^x.2=17\)

\(\Leftrightarrow2^x.\dfrac{17}{8}=17\Leftrightarrow2^x=8\Leftrightarrow x=3\)

\(d.5^x-5^x:5=20\Leftrightarrow5^x-5^x.\dfrac{1}{5}=20\)

\(\Leftrightarrow5^x\left(1-\dfrac{1}{5}\right)=20\Leftrightarrow5^x=20:\dfrac{4}{5}\Leftrightarrow5^x=25\Leftrightarrow x=2\)

Bài 1: 

a) Ta có: \(-5x+32=\left(-2\right)^3\)

\(\Leftrightarrow-5x+32=-8\)

\(\Leftrightarrow-5x=-40\)

hay x=8

Vậy: x=8

b) Ta có: \(-2x+36=6\)

\(\Leftrightarrow-2x=6-36=-30\)

hay x=15

Vậy: x=15

e) Ta có: \(\left(x+9\right)\left(x^2-25\right)=0\)

\(\Leftrightarrow\left(x+9\right)\left(x-5\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+9=0\\x-5=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=5\\x=-5\end{matrix}\right.\)

Vậy: \(x\in\left\{-9;5;-5\right\}\)

b,-2x+36=6

tương đương -2x=-30

tương đương x=15

a, -5x+32=(-2)^3

tương đương -5x+32=8

tương đương -5x=-24

tương đương x=24/5

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

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

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

\(\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)

\(x.\left(x-3\right)-3x+9=0\)

\(x.\left(x-3\right)-3.\left(x-3\right)=0\)

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

a,\(x^2-3x+2\left(x-3\right)=0.\)

\(\Leftrightarrow x^2-3x+2x-6=0\)

\(\Leftrightarrow x^2+x-6=0\)

\(\Leftrightarrow\left(x^2-2x\right)+\left(3x-6\right)=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)

a) Rút gọn VT = 45x + 8. Từ đó tìm được x = 2 15 .  

b) Rút gọn VT = -25x – 8. Từ đó tìm được x = − 11 25 .