Question

[Cuồng Sơn Tùng M-tp]

giải giúp mk vs , mk đg cần gấp ...

Answer 1

[Rain Tờ Rym Te]

a) \(-1\dfrac{4}{5}x-1\dfrac{1}{10}=25\%\)

\(\Leftrightarrow\dfrac{-9}{5}x-\dfrac{11}{10}=\dfrac{1}{4}\)

\(\Leftrightarrow x=\dfrac{-3}{4}\)

b) \(\left(2x+\dfrac{3}{5}\right)^2-\dfrac{9}{25}=0\)

\(\Leftrightarrow\left(2x+\dfrac{3}{5}\right)^2-\left(\dfrac{3}{5}\right)^2=0\)

\(\Leftrightarrow\left(2x+\dfrac{3}{5}+\dfrac{3}{5}\right)\left(2x+\dfrac{3}{5}-\dfrac{3}{5}\right)=0\)

\(\Leftrightarrow2x\left(2x+\dfrac{6}{5}\right)=0\)

\(\Leftrightarrow4x^2+\dfrac{12}{5}=0\)

\(\Leftrightarrow x^2=\dfrac{-3}{5}\)

\(\Rightarrow x\in\varnothing\)

c) \(1,25-\left|\dfrac{x}{3}+2\right|=\left(-\dfrac{1}{2}\right)^3\)

\(\Leftrightarrow\dfrac{5}{4}-\left|\dfrac{x}{3}+2\right|=\dfrac{-1}{8}\)

\(\Leftrightarrow\dfrac{11}{8}-\left|\dfrac{x}{3}+2\right|=0\)

\(\left|\dfrac{x}{3}+2\right|=\dfrac{x}{3}+2\) khi \(\dfrac{x}{3}+2\ge0\Rightarrow x\ge-6\)

Với \(x\ge-6\) ta có: \(\dfrac{11}{8}-\dfrac{x}{3}-2=0\)

\(\Rightarrow x=\dfrac{-15}{8}\left(TM\right)\)

\(\left|\dfrac{x}{3}+2\right|=-\dfrac{x}{3}-2\) khi \(\dfrac{x}{3}+2< 0\Rightarrow x< -6\)

Với \(x< -6\) ta có: \(\dfrac{11}{8}+\dfrac{x}{3}+2=0\)

\(\Rightarrow x=-\dfrac{81}{8}\) (TM)

Vậy..

d) \(\dfrac{5}{6}\left(4x-\dfrac{2}{5}\right)-3\left(\dfrac{1}{12}x-\dfrac{1}{2}\right)=2x-\dfrac{5}{9}\)

\(\Leftrightarrow\dfrac{10}{3}x-\dfrac{1}{3}-\dfrac{1}{4}x+\dfrac{3}{2}=\dfrac{18x-5}{9}\)

\(\Leftrightarrow\dfrac{120x-12-9x+54-72x+20}{36}=0\)

\(\Rightarrow39x+62=0\)

\(\Rightarrow x=\dfrac{-62}{39}\)

e) \(\left|7x-4\right|-5x=28\)

\(\left|7x-4\right|=7x-4\) khi \(7x-4\ge0\Rightarrow x\ge\dfrac{4}{7}\)

Với \(x\ge\dfrac{4}{7}\) ta có: \(7x-4-5x=28\)

\(\Rightarrow x=16\left(TM\right)\)

\(\left|7x-4\right|=4-7x\) khi \(7x-4< 0\Rightarrow x< \dfrac{4}{7}\)

Với \(x< \dfrac{4}{7}\) ta có: \(4-7x-5x=28\)

\(\Rightarrow x=-2\left(TM\right)\)

Vậy...

Answer 2

[Hoàng Thị Ngọc Anh]

d) \(\dfrac{5}{6}\left(4x-\dfrac{2}{5}\right)-3\left(\dfrac{1}{12}x-\dfrac{1}{2}\right)=2x-\dfrac{5}{9}\)

\(\Rightarrow\dfrac{10}{3}x-\dfrac{1}{3}-\dfrac{1}{4}x+\dfrac{3}{2}=2x-\dfrac{5}{9}\)

\(\Rightarrow\dfrac{10}{3}x-\dfrac{1}{4}x-2x=\dfrac{1}{3}-\dfrac{3}{2}-\dfrac{5}{9}\)

\(\Rightarrow\left(\dfrac{10}{3}-\dfrac{1}{4}-2\right)x=\dfrac{-31}{18}\)

\(\Rightarrow\dfrac{13}{12}x=\dfrac{-31}{18}\)

\(\Rightarrow x=\dfrac{-62}{39}.\)

Vậy \(x=\dfrac{-62}{39}.\)

e) \(\left|7x-4\right|-5x=28\)

\(\Rightarrow\left|7x-4\right|=5x+28\)

+) \(TH1:7x-4\ge0\Rightarrow7x\ge4\Rightarrow x\ge\dfrac{4}{7}\)

\(7x-4=5x+28\)

\(\Rightarrow7x-5x=4+28\)

\(\Rightarrow2x=32\)

\(\Rightarrow x=16\) (nhận)

+) \(TH2:7x-4< 0\Rightarrow7x< 4\Rightarrow x< \dfrac{4}{7}\)

\(-7x+4=5x+28\)

\(\Rightarrow-7x-5x=-4+28\)

\(\Rightarrow-12x=24\)

\(\Rightarrow x=-2\) (nhận)

Vậy \(x\in\left\{16;-2\right\}\).

Mấy câu kia dễ tự làm.