Question

[Ha Thù]

Mg giải giúp mình ạ

Answer 1

[Nguyễn Lê Phước Thịnh]

Bài 2:

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

=>\(\dfrac{3}{5}x=\dfrac{4}{5}-\dfrac{2}{3}=\dfrac{12-10}{15}=\dfrac{2}{15}\)

=>\(x=\dfrac{2}{15}:\dfrac{3}{5}=\dfrac{2}{15}\cdot\dfrac{5}{3}=\dfrac{10}{45}=\dfrac{2}{9}\)

b: \(\dfrac{1}{3}+\dfrac{2}{3}:x=-2\)

=>\(\dfrac{2}{3}:x=-2-\dfrac{1}{3}=-\dfrac{7}{3}\)

=>\(x=-\dfrac{2}{3}:\dfrac{7}{3}=\dfrac{-2}{3}\cdot\dfrac{3}{7}=-\dfrac{2}{7}\)

c: \(\left|x+\dfrac{1}{2}\right|=1,5\)

=>|x+0,5|=1,5

=>\(\left[{}\begin{matrix}x+0,5=1,5\\x+0,5=-1,5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

d: \(\left(2x-1\right)^2=-\dfrac{16}{25}\)

mà \(\left(2x-1\right)^2>=0\forall x\)

nên \(x\in\varnothing\)

e: \(\left(2x-\dfrac{4}{3}\right)\left(x+\dfrac{1}{2}\right)=0\)

=>\(\left[{}\begin{matrix}2x-\dfrac{4}{3}=0\\x+\dfrac{1}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{2}{3}=0\\x+\dfrac{1}{2}=0\end{matrix}\right.\)

=>x=2/3 hoặc x=-1/2

Answer 2

[Ha Thù]

Mình cần giải luôn cả bài 1,3 nx á