đề yêu cầu tìm x hả

a) ( x - 1/5 )² = 0

<=> x - 1/5 = 0

<=> x = 1/5

b) ( x - 2 )² = 1

<=> ( x - 2 )² = ( ±1 )²

<=> x - 2 = 1 hoặc x - 2 = -1

<=> x = 3 hoặc x = 1

c) ( 2x - 1 )³ = -8

<=> ( 2x - 1 )³ = (-2)³

<=> 2x - 1 = -2

<=> 2x = -1

<=> x = -1/2

d) ( x⁴ )² = x¹²/x⁵

<=> x⁸ = x⁷

<=> x⁸ - x⁷ = 0

<=> x⁷( x - 1 ) = 0

<=> x⁷ = 0 hoặc x - 1 = 0

<=> x = 0 hoặc x = 1

e) x¹⁰ = 25x⁸

<=> x¹⁰ - 25x⁸ = 0

<=> x⁸( x² - 25 ) = 0

<=> x⁸ = 0 hoặc x² - 25 = 0

<=> x = 0 hoặc x = ±5

f) ( 2x + 3 )² = 9/121

<=> ( 2x + 3 )² = ( ±3/11 )²

<=> 2x + 3 = 3/11 hoặc 2x + 3 = -3/11

<=> x = -15/11 hoặc x = -18/11

a) \(\left(x-\frac{1}{5}\right)^2=0\Leftrightarrow x-\frac{1}{5}=0\Leftrightarrow x=\frac{1}{5}\)

b) \(\left(x-2\right)^2=1\)

\(\Leftrightarrow\left(x-2\right)^2-1=0\)

\(\Leftrightarrow\left(x-2-1\right)\left(x-2+1\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

c) \(\left(2x-1\right)^3=-8\)

\(\Leftrightarrow\left(2x-1\right)^3+8=0\)

\(\Leftrightarrow\left(2x-1+8\right)\left[\left(2x-1\right)^2-8\left(2x-1\right)+64\right]=0\)

\(\Leftrightarrow2x+7=0\)

\(\Leftrightarrow x=\frac{-7}{2}\)

d) ĐKXĐ : \(x\ne0\)

 \(\left(x^4\right)^2=\frac{x^{12}}{x^5}\)

\(\Leftrightarrow x^8=x^7\)

\(\Leftrightarrow x^8-x^7=0\)

\(\Leftrightarrow x^7\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\left(ktm\right)\\x=1\left(tm\right)\end{cases}\Leftrightarrow x=1}\)

e) ĐKXĐ : x khác 0 

 \(x^{10}=25x^8\)

\(\Leftrightarrow x^2=25\Leftrightarrow x=5\)

f) \(\left(2x+3\right)^2=\frac{9}{121}\)

\(\Leftrightarrow\left(2x+3+\frac{3}{11}\right)\left(2x+3-\frac{3}{11}\right)=0\)

\(\Leftrightarrow\left(2x+\frac{36}{11}\right)\left(2x+\frac{30}{11}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-18}{11}\\x=-\frac{15}{11}\end{cases}}\)

a,(x-\(\frac{1}{5}\))²=0

(x - \(\frac{1}{5}\))²=0²

=>x- \(\frac{1}{5}\)=0

              x=0+\(\frac{1}{5}\)

              x=\(\frac{1}{5}\)

b,(x-2)² =1

=>(x-2)² = 1²

=>x-2=1 

=>x=1+2=3

a: \(\Leftrightarrow\left|2x-3\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

a, \(\sqrt{\left(2x-3\right)^2}=7\\ \Rightarrow\left|2x-3\right|=7\\ \Rightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

c, \(\sqrt{x^2-9}-3\sqrt{x-3}=0\\ \Rightarrow\sqrt{x-3}\sqrt{x+3}-3\sqrt{x-3}=0\\ \Rightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x+3}-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x-3=0\\x+3=9\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)

Bài 1 : \(M=\frac{8^{20}+4^{20}}{4^{25}+64^5}=\frac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\frac{2^{60}+2^{40}}{2^{50}+2^{30}}=\frac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}=2^{10}=1024\)

Bài 2 : a) \(\left(x^4\right)^2=\frac{x^{12}}{x^5}\)=> \(x^8=x^7\)

=> \(x^8-x^7=0\)

=> \(x^7\left(x-1\right)=0\)

=> \(x-1=0\Rightarrow x=1\)(vì x⁷ = 0 => x = 0 mà x \(\ne\)0 nên loại)

b) \(x^{10}-25x^8=0\)

=> \(x^8\left(x^2-25\right)=0\)

=> x⁸ = 0 hoặc x² - 25 = 0

=> x = 0 hoặc x² = 25

=> x = 0 hoặc x = \(\pm\)5

Bài 3 : a) \(\left(2x+3\right)^2=\frac{9}{121}=\left(\pm\frac{3}{11}\right)^2\)

=> \(\orbr{\begin{cases}2x+3=\frac{3}{11}\\2x+3=-\frac{3}{11}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{15}{11}\\x=-\frac{18}{11}\end{cases}}\)

b) \(\left(3x-1\right)^3=-\frac{8}{27}=\left(-\frac{2}{3}\right)^3\)

=> 3x - 1 = -2/3

=> 3x = 1/3

=> x = 1/3 : 3 = 1/9

1) Ta có \(M=\frac{8^{20}+4^{20}}{4^{25}+64^5}=\frac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\frac{2^{60}+2^{40}}{2^{50}+2^{30}}=\frac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{30}+1\right)}=2^{10}=1024\)

2) a) \(\left(x^4\right)^2=\frac{x^{12}}{x^5}\)

=> x⁸ = x⁷

=> x⁸ - x⁷ = 0

=> x⁷(x - 1) = 0

=> \(\orbr{\begin{cases}x^7=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

Vậy x \(\in\left\{0;1\right\}\)

b) x¹⁰ = 25x⁸

=> x¹⁰ - 25x⁸ = 0

=> x⁸(x² - 25) = 0

=> \(\orbr{\begin{cases}x^8=0\\x^2-25=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm5\end{cases}}\)

Vậy \(x\in\left\{0;5;-5\right\}\)

3) \(\left(2x+3\right)^2=\frac{9}{121}\)

=> \(\left(2x+3\right)^2=\left(\frac{3}{11}\right)^2\)

=> \(\orbr{\begin{cases}2x+3=\frac{3}{11}\\2x+3=-\frac{3}{11}\end{cases}}\Rightarrow\orbr{\begin{cases}2x=\frac{-30}{11}\\2x=-\frac{36}{11}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{15}{11}\\x=-\frac{18}{11}\end{cases}}\)

Vậy \(x\in\left\{-\frac{15}{11};-\frac{18}{11}\right\}\)

b) \(\left(3x-1\right)^3=-\frac{8}{27}\)

=> \(\left(3x-1\right)^3=\left(-\frac{2}{3}\right)^3\)

=> \(3x-1=-\frac{2}{3}\)

=> \(3x=\frac{1}{3}\)

=> \(x=\frac{1}{9}\)

Vậy \(x=\frac{1}{9}\)

cảm ơn các bạn nhiều lắm nuôn.I love u

a: ĐKXĐ: \(x\in R\)

\(\sqrt{\left(x+3\right)^2}=12\)

=>\(\left|x+3\right|=12\)

=>\(\left[{}\begin{matrix}x+3=12\\x+3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-15\end{matrix}\right.\)

b: ĐKXĐ: x>=1

\(\sqrt{25x-25}-\sqrt{9x-9}=10\)

=>\(5\sqrt{x-1}-3\sqrt{x-1}=10\)

=>\(2\sqrt{x-1}=10\)

=>x-1=25

=>x=26(nhận)

5: =>4x^2-1/9=0

=>(2x-1/3)(2x+1/3)=0

=>x=1/6 hoặc x=-1/6

6: =>x-1=2

=>x=3

7:=>(2x-1)^3=-27

=>2x-1=-3

=>2x=-2

=>x=-1

8: =>1/8(x-1)^3=-125

=>(x-1)^3=-1000

=>x-1=-10

=>x=-9

3: =>(5x-5)^2-4=0

=>(5x-7)(5x-3)=0

=>x=3/5 hoặc x=7/5

4: =>(5x-1)^2=0

=>5x-1=0

=>x=1/5

1: =>(3x-1)(2x-1)=0

=>x=1/3 hoặc x=1/2

2: =>x^2(2x-3)-4(2x-3)=0

=>(2x-3)(x^2-4)=0

=>(2x-3)(x-2)(x+2)=0

=>x=3/2;x=2;x=-2

`@` `\text {Answer}`

`\downarrow`

`1,`

\(2x\left(3x-1\right)+1-3x=0\)

`<=> 2x(3x - 1) - 3x + 1 = 0`

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

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

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

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

`<=>`\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy,  `S = {1/2; 1/3}`

`2,`

\(x^2\left(2x-3\right)+12-8x=0\)

`<=> x^2(2x - 3) - 8x + 12 =0`

`<=> x^2(2x - 3) - (8x - 12) = 0`

`<=> x^2(2x - 3) - 4(2x - 3) = 0`

`<=> (x^2 - 4)(2x - 3) = 0`

`<=>`\(\left[{}\begin{matrix}x^2-4=0\\2x-3=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x^2=4\\2x=3\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x^2=\left(\pm2\right)^2\\x=\dfrac{3}{2}\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=\pm2\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy, `S = {+-2; 3/2}`

`3,`

\(25\left(x-1\right)^2-4=0\)

`<=> 25(x-1)(x-1) - 4 = 0`

`<=> 25(x^2 - 2x + 1) - 4 = 0`

`<=> 25x^2 - 50x + 25 - 4 = 0`

`<=> 25x^2 - 15x - 35x + 21 = 0`

`<=> (25x^2 - 15x) - (35x - 21) = 0`

`<=> 5x(5x - 3) - 7(5x - 3) = 0`

`<=> (5x - 7)(5x - 3) = 0`

`<=>`\(\left[{}\begin{matrix}5x-7=0\\5x-3=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}5x=7\\5x=3\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=\dfrac{7}{5}\\x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy, `S = {7/5; 3/5}`

`4,`

\(25x^2-10x+1=0\)

`<=> 25x^2 - 5x - 5x + 1 = 0`

`<=> (25x^2 - 5x) - (5x - 1) = 0`

`<=> 5x(5x - 1) - (5x - 1) = 0`

`<=> (5x - 1)(5x-1)=0`

`<=> (5x-1)^2 = 0`

`<=> 5x - 1 = 0`

`<=> 5x = 1`

`<=> x = 1/5`

Vậy,` S = {1/5}.`

`@` `\text {Ans}`

`\downarrow`

`5,`

`-4x^2 + 1/9 = 0`

`<=> -4x^2 = 0 - 1/9`

`<=> -4x^2 = -1/9`

`<=> 4x^2 = 1/9`

`<=> x^2 = 1/9 \div 4`

`<=> x^2 = 1/36`

`<=> x^2 = (+-1/6)^2`

`<=> x = +-1/36`

Vậy, `S = {1/36; -1/36}`

`6,`

`(x-1)^3 = 8`

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

`<=> x-1=2`

`<=> x = 2 + 1`

`<=> x = 3`

Vậy, `S = {3}`

`7,`

`(2x-1)^3 + 27 = 0`

`<=> (2x - 1)^3 = -27`

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

`<=> 2x - 1 = -3`

`<=> 2x = -3 + 1`

`<=> 2x = -2`

`<=> x = -1`

Vậy,` S = {-1}`

`8,`

`125 + 1/8(x-1)^3 = 0`

`<=> 1/8(x-1)^3 = - 125`

`<=> (x-1)^3 = -125 \div 1/8`

`<=> (x-1)^3 = -1000`

`<=> (x-1)^3 = (-10)^3`

`<=> x - 1 = - 10`

`<=> x = -10+1`

`<=> x = -9`

Vậy, `S = {-9}.`

a: =18/70+4/5=18/70+56/70=74/70=37/35

b: =4/5-3/4=16/20-15/20=1/20

a) \(\dfrac{2}{7}\times\dfrac{9}{10}+\dfrac{4}{5}=\dfrac{18}{70}+\dfrac{4}{5}=\dfrac{9}{35}+\dfrac{4}{5}=\dfrac{9}{35}+\dfrac{28}{35}=\dfrac{37}{35}\)

b) \(\dfrac{12}{15}-\dfrac{1}{2}:\dfrac{2}{3}=\dfrac{12}{15}-\dfrac{1}{2}\times\dfrac{3}{2}=\dfrac{12}{15}-\dfrac{3}{4}=\dfrac{4}{5}-\dfrac{3}{4}=\dfrac{16}{20}-\dfrac{15}{20}=\dfrac{1}{20}\)