Question

Tìm x thuộc Q, biết :

A) 11phân12 - (2phan5+x) =2phan3

B) 2x ×( x - 1phan7) = 0

C) 3phan4 + 1phan4 ÷ x = 2phan5

Answer 1

\(a,\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)

\(\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)

\(\dfrac{2}{5}+x=\dfrac{1}{4}\)

\(x=\dfrac{1}{4}-\dfrac{2}{5}\)

\(x=\dfrac{-3}{20}\)

Vậy ...

b, \(2x\left(x-\dfrac{1}{7}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)

Vậy ...

c, \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)

\(\dfrac{1}{4}:x=\dfrac{-7}{20}\)

\(x=\dfrac{1}{4}:\dfrac{-7}{20}\)

\(x=\dfrac{-5}{7}\)

Vậy ...

Answer 2

a)\(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)

<=>\(\dfrac{2}{5}+x=\dfrac{1}{4}\)

<=>\(x=\dfrac{1}{4}-\dfrac{2}{5}\)

<=>\(x=-\dfrac{3}{20}\)

b)\(2x\left(x-\dfrac{1}{7}\right)=0\)

=>\(\text{2x=0 hoặc }x-\dfrac{1}{7}=0\)

=>\(x=0hoặcx=\dfrac{1}{7}\)

Vậy....

c)\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

=>\(\dfrac{1}{4}:x=-\dfrac{7}{20}\)

=>x=\(-\dfrac{5}{7}\)

Vậy