Question

a) \(^{\left(x-\dfrac{2}{3}\right)^2}\)=\(\dfrac{25}{6}\)

b)\(\dfrac{x-2}{4}=\dfrac{5+x}{3}\)

Answer 1

Xem lại đề câu 1 em nhé xem có phải là `25/9` không ?

Answer 2

b) \(\dfrac{x-2}{4}=\dfrac{5+x}{3}\)

=> \(3\left(x-2\right)=4\left(5+x\right)\)

\(3x-6=20+4x\)

\(3x-4x=20+6\)

\(-x=26\)

=> \(x=26\)

Vậy \(x=26\)

Answer 3

Giải

b) \(\dfrac{x-2}{4}=\dfrac{5+x}{3}\)

⇒ \(\left(x-2\right).3=\left(5+x\right).4\)

⇒ \(3x-6=20+4x\)

⇒ \(-6-20=4x-3x\)

⇒ \(x=-26\)

 

Answer 4

Giải

a) \(\left(x-\dfrac{2}{3}\right)^2=\dfrac{25}{6}\)

       \(x-\dfrac{4}{9}=\dfrac{25}{6}\)

       \(x\)         \(=\)\(\dfrac{25}{6}+\dfrac{4}{9}\)

       \(x\)         \(=\dfrac{83}{18}\)

Answer 5

a) Ta có: \(\left(x-\dfrac{2}{3}\right)^2=\dfrac{25}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{2}{3}=\dfrac{5\sqrt{6}}{6}\\x-\dfrac{2}{3}=-\dfrac{5\sqrt{6}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4+5\sqrt{6}}{6}\\x=\dfrac{4-5\sqrt{6}}{6}\end{matrix}\right.\)

b) Ta có: \(\dfrac{x-2}{4}=\dfrac{x+5}{3}\)

nên 4x+20=3x-6

\(\Leftrightarrow x=-26\)