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

Trường hợp 1: \(2x+3=\frac{3}{11}\)

\(2x=\frac{3}{11}-3=-\frac{30}{11}\)

\(x=-\frac{30}{11}:2=-\frac{15}{11}\)

Trường hợp 2: \(2x+3=-\frac{3}{11}\)

\(2x=-\frac{3}{11}-3=-\frac{36}{11}\)

\(x=-\frac{36}{11}:2=-\frac{18}{11}\)

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

b,(3x-1)3= -8/27= (-2/3)^3

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

<=>x=1/9 Mjk thấy phần a có bạn lm rồi nên bổ sung phần b

Chúc các bạn học tốt nhé^^

a) (x² - 121) . (2x + 3) = 0

=>x²-121=0 hoặc 2x+3=0

+)Nếu x²-121=0

=>x²=0+121=121

=>x²=(-11)² hoặc x²=11²

=>x=-11 hoặc x=11

+)Nếu 2x+3=0

=>2x=0-3=-3

=>x=(-3):2=\(\frac{-3}{2}\)

Vậy x=-11 hoặc x=11 hoặc x=\(\frac{-3}{2}\)
b) 2x² - 8x = 0

=>2x(x-4)=0

=>x=0 hoặc x-4=0

Nếu x-4=0

=>x=0+4=4

Vậy x=0 hoặc x=4
c) (3x + 1)⁵ = (3x + 1)⁴

=>(3x+1)⁵:(3x+1)⁴=(3x+1)⁴:(3x+1)⁴

=>3x+1=1

=>3x=1-1=0

=>x=0:3=0

Vậy x=0

a)(x² - 121) . (2x + 3) = 0

=>x²-121=0 hoặc 2x+3=0

<=>x²=121 <=>x=±11

<=>2x=-3 <=>x=-3/2

b) 2x² - 8x = 0

=>2x(x-4)=0

=>2x=0 hoặc x-4=0

=>x=0 hoặc x=4

a) \(\left(3x-1\right)^4=\frac{1}{16}\)

\(\Rightarrow\orbr{\begin{cases}3x-1=\frac{1}{2}\\3x-1=-\frac{1}{2}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}3x=\frac{1}{2}+1=\frac{3}{2}\\3x=-\frac{1}{2}+1=\frac{1}{2}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{1}{6}\end{cases}}\)

b) \(\left(5x-2\right)^2=\frac{25}{121}\)

\(\Rightarrow\orbr{\begin{cases}5x-2=\frac{5}{11}\\5x-2=-\frac{5}{11}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}5x=\frac{5}{11}+2=\frac{27}{11}\\5x=-\frac{5}{11}+2=\frac{17}{11}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{27}{55}\\x=\frac{17}{55}\end{cases}}\)

c) \(\left(5x+6\right)^3=\frac{-8}{27}=\left(\frac{-2}{3}\right)^3\)

\(\Rightarrow5x+6=-\frac{2}{3}\)

\(\Rightarrow5x=-\frac{2}{3}-6=-\frac{20}{3}\)

\(\Rightarrow x=-\frac{20}{3}:5=-\frac{4}{3}\)

a) (3x — 1) ^4 = 1/16

=>3x-1=-1/2 hoặc 1/2

=>3x=1/2

=>x=1/6

=>3x=3/2

=>x=1/2

b)(5x — 2 )^2 = 25/121

=>5x-2=-5/11 hoặc 5/11

=>5x=17/11

=>x=17/55

=>5x=27/11

=>x=27/55

c)(5x + 6)^3= -8/27

=>5x+6=-2/3

=>5x=-20/3

=>x=-4/3

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

-3*(x - 7) - 2(3x + 1) = 5x - 121

=> -3x + 21 - 6x - 2 - 5x + 121 = 0

=> -14x + 140 = 0

=> -14x = -140

=> x = 10

Vậy x = 10

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!

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

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

\(\Rightarrow2x+3=\frac{3}{11}\text{ hoặc }2x+3=\frac{-3}{11}\)

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

\(2x=\frac{3}{11}-\frac{33}{11}\text{ hoặc }2x=\frac{-3}{11}-\frac{33}{11}\)

\(2x=\frac{-30}{11}\text{ hoặc }2x=\frac{-36}{11}\)

\(x=\frac{-30}{11}:2\text{ hoặc }x=\frac{-36}{11}:2\)

\(x=\frac{-30}{11}.\frac{1}{2}\text{ hoặc }x=\frac{-36}{11}.\frac{1}{2}\)

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