Question

bài 1: Tìm x biết:

a,\(x+\dfrac{1}{4}=\dfrac{3}{4}\)

b,\(\dfrac{3}{4}-\dfrac{2}{5}x=\dfrac{29}{60}\)

c,\(2x-\dfrac{1}{3}=\dfrac{-5}{6}\)

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

e,\(\dfrac{11}{12}-\dfrac{2}{3}|x|=\dfrac{3}{8}\)

f,\(|2x-1|=5\)

lưu ý: Có câu có 2 trường hợp

Các bn giúp mk với

Answer 1

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

<=>x=\(\frac{3}{4}-\frac{1}{4} \)

<=>\(x=\frac{1}{2} \)

b,\(\frac{3}{4}-\frac{2}{5}x=\frac{29}{60} \)

<=>\(\frac{2}{5}x =\frac{3}{4}-\frac{29}{60} \)

<=>\(\frac{2}{5}x=\frac{4}{15} \)

<=x=\(\frac{2}{3} \)

c,\(2x-\frac{1}{3}=\frac{-5}{6} \)

2x=\(\frac{-5}{6}+\frac{1}{3} \)

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

x=\(\frac{-1}{4} \)

d,2-\(\frac{3}{4x}=\frac{1}{2} \)

<=>\(\frac{3}{4x}=2-\frac{1}{2} \)

<=>\(\frac{3}{4}x=\frac{3}{2} \)

<=>x=2

e,\(\frac{11}{12}- \frac{2}{3}|x| =\frac{3}{8} \)

<=>\(\frac{2}{3}|x|=\frac{13}{24} \)

<=>\(|x|=\frac{13}{16} \)

<=>x=\(\pm\frac{13}{16} \)

f,\(|2x-1|=5\)

<=>2x-1=5 hoặc 2x-1=-5

<=> 2x=6 2x=-4

<=> x=3 x=-2

Answer 2

Giải

a) \(x+\dfrac{1}{4}=\dfrac{3}{4}\)

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

b)\(\dfrac{3}{4}-\dfrac{2}{5}x=\dfrac{29}{60}\)

=>\(\dfrac{2}{5}x=\dfrac{3}{4}-\dfrac{29}{60}\)=>\(\dfrac{2}{5}x=\dfrac{4}{15}\)

=>\(x=\dfrac{4}{15}:\dfrac{2}{5}=\dfrac{2}{3}\)

c)\(2x-\dfrac{1}{3}=\dfrac{-5}{6}\)=>\(2x=\dfrac{-5}{6}+\dfrac{1}{3}\)=\(\dfrac{-1}{2}\)

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

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

=>\(x=\dfrac{3}{2}.\dfrac{3}{4}=\dfrac{9}{8}\)

e)\(\dfrac{11}{12}-\dfrac{2}{3}.\left|x\right|=\dfrac{3}{8}\)=>\(\dfrac{2}{3}.\left|x\right|=\dfrac{11}{12}-\dfrac{3}{8}\)=\(\dfrac{13}{24}\)

=>\(\left|x\right|=\dfrac{13}{24}:\dfrac{2}{3}=\dfrac{13}{16}\)

Vậy: \(x=\dfrac{13}{16}\)hoặc\(x=\dfrac{-13}{16}\)

f)\(\left|2x-1\right|=5\)

*2x-1=5 =>2x=5+1=6 =>x=6:2=3

*2x-1=-5 =>2x=(-5)+1=-4 =>x=-4:2=-2

Vậy: x=3 hoặc x=-2

Tick cho Phong nhé:>

Yêu nhiều>3

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