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

từ đó suy ra 

\(2x+3=\frac{3}{11}\)

2x=3/11-3

2x=-2/8/11

x=-2/8/11:2

x=-1/4/11

b,

(3x-1)^3=-8/27

(3x-1)^3=(-2/3)^3

Vậy suy ra 

3x-1=-2/3

3x=-2/3+1

3x=1/3

x=1/3:3

x=1/9

Ko thấy x             

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

Ta có: \(\frac{9}{121}=\left(\pm\frac{3}{11}\right)^2\)

\(\Rightarrow2x+3\in\left\{\frac{3}{11};\frac{-3}{11}\right\}\)

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

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

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

Ta có: \(\frac{-8}{27}=\left(\frac{-2}{3}\right)^3\)

\(\Rightarrow3x-1=\frac{-2}{3}\)

\(\Rightarrow x=\frac{1}{9}\)

Vậy \(x=\frac{1}{9}\)

a.

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

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

\(2x+3=\pm\frac{3}{11}\)

TH1:

\(2x+3=\frac{3}{11}\)

\(2x=\frac{3}{11}-3\)

\(2x=-\frac{30}{11}\)

\(x=-\frac{30}{11}\div2\)

\(x=-\frac{15}{11}\)

TH2:

\(2x+3=-\frac{3}{11}\)

\(2x=-\frac{3}{11}-3\)

\(2x=-\frac{36}{11}\)

\(x=-\frac{36}{11}\div2\)

\(x=-\frac{18}{11}\)

Vậy \(x=-\frac{15}{11}\) hoặc \(x=-\frac{18}{11}\)

b.

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

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

\(3x-1=-\frac{2}{3}\)

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

\(3x=\frac{1}{3}\)

\(x=\frac{1}{3}\div3\)

\(x=\frac{1}{9}\)

Chúc bạn học tốt ^^

a)\(\left(2x+3\right)^2=\frac{9}{121}\\ \Leftrightarrow\left(2x+3\right)^2=\left(\pm\frac{3}{11}\right)^2\\ \Rightarrow\left\{{}\begin{matrix}2x+3=\frac{3}{11}\\2x+3=\frac{-3}{11}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{-15}{11}\\x=\frac{-18}{11}\end{matrix}\right.\)

Vậy...

b)\(\left(3x-1\right)^3=\frac{-8}{27}\\ \Leftrightarrow\left(3x-1\right)^3=\left(\frac{-2}{3}\right)^3\\ 3x-1=\frac{-2}{3}\\ \Rightarrow x=\frac{1}{9}\)

Vậy...

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

\(\Rightarrow2x+3=\pm\frac{3}{11}\)

\(\Rightarrow\left[{}\begin{matrix}2x+3=\frac{3}{11}\\2x+3=-\frac{3}{11}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=\frac{3}{11}-3=-\frac{30}{11}\\2x=\left(-\frac{3}{11}\right)-3=-\frac{36}{11}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\left(-\frac{30}{11}\right):2\\x=\left(-\frac{36}{11}\right):2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\frac{15}{11}\\x=-\frac{18}{11}\end{matrix}\right.\)

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

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

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

\(\Rightarrow3x-1=-\frac{2}{3}\)

\(\Rightarrow3x=\left(-\frac{2}{3}\right)+1\)

\(\Rightarrow3x=\frac{1}{3}\)

\(\Rightarrow x=\frac{1}{3}:3\)

\(\Rightarrow x=\frac{1}{9}\)

Vậy \(x=\frac{1}{9}.\)

Chúc bạn học tốt!

sao giống bài thi quá vậy

biết giải bài 2

x/12=y/14=x.y/12.24=98/288=49/144

=> x/12=49/144=> 49/12

=> y/14=49/144=> 343/72

mới lớp 2 thôi

bài 1 a

suy ra xx=x^2=-15*(-60)=900=30^2

suy ra x=30

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) x = 4                

b) x = 7     

c) x = 2                

d) x = 5

e) x = 2                

f) x= 1. 

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a) x = 4

b) x = 7

c) x = 2

d) x = 5

e) x = 2

 f) x= 1

a)  \(\left(2x+3\right)^2=\frac{9}{21}\)

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

<=>  \(\orbr{\begin{cases}x=-1\frac{4}{11}\\x=-1\frac{7}{11}\end{cases}}\)

Vậy...

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....