Question

a)(x+1/3)²=1/4

b)(2.x+3)²=9/121

c)(3.x-1)³= -8/27

d) 4^x+4^x+2= 112

Answer 1

a,

\(\left(x+\frac{1}{3}\right)^2=\frac{1}{4}\)

⇒ \(\left[{}\begin{matrix}x+\frac{1}{3}=\frac{1}{2}\\x+\frac{1}{3}=-\frac{1}{2}\end{matrix}\right.\)

⇒ \(\left[{}\begin{matrix}x=\frac{3}{6}-\frac{2}{6}=\frac{1}{6}\\x=-\frac{3}{6}-\frac{2}{6}=-\frac{5}{6}\end{matrix}\right.\)

Vậy.....

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

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

⇒ \(\left[{}\begin{matrix}2x=\frac{3}{11}-\frac{33}{11}=-\frac{30}{11}\\2x=-\frac{3}{11}-\frac{33}{11}=-\frac{36}{11}\end{matrix}\right.\)

⇒ \(\left[{}\begin{matrix}x=-\frac{15}{11}\\x=-\frac{12}{11}\end{matrix}\right.\)

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

⇒ \(3x-1=-\frac{2}{3}\)

⇒ \(3x=-\frac{2}{3}+1=\frac{1}{3}\)

⇒ \(x=\frac{1}{9}\)

d, \(4^x+4^{x+2}=112\)

⇒ \(4^x.\left(1+16\right)=112\)

=> \(4^x.17=112\)

Chỗ này kì nè :33 bạn xem lại đềcoi

Answer 2

giúp mình với các bạn ơi

Answer 3

a,(x + \(\frac{1}{3})^2\) = \(\frac{1}{4}\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x+\frac{1}{3}=\frac{1}{2}\\x+\frac{1}{3}=\frac{-1}{2}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{6}\\x=\frac{-5}{6}\end{matrix}\right.\)

Vậy x\(\in\left\{\frac{1}{6},\frac{-5}{6}\right\}\)