Question

[Ngô Ngọc Tâm Anh]

giúp em với anh/chị ơi

Answer 1

[Nguyễn Ngọc Huy Toàn]

a.\(\dfrac{x}{3}+\dfrac{2x-1}{5}=2-\dfrac{x}{3}\)

\(\Leftrightarrow\dfrac{5x+3\left(2x-1\right)}{15}=\dfrac{30-5x}{15}\)

\(\Leftrightarrow5x+3\left(2x-1\right)=30-5x\)

\(\Leftrightarrow5x+6x-3-30+5x=0\)

\(\Leftrightarrow16x-33=0\)

\(\Leftrightarrow16x=33\)

\(\Leftrightarrow x=\dfrac{33}{16}\)

b.\(\left(2x-1\right)\left(3x+2\right)=\left(5x-8\right)\left(2x-1\right)\)

\(\Leftrightarrow\left(2x-1\right)\left(3x+2\right)-\left(5x-8\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(3x+2-5x+8\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(10-2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=5\end{matrix}\right.\)

c.\(ĐK:x\ne-1;3\)

\(\Rightarrow\dfrac{x-1}{x+1}+\dfrac{x+5}{x-3}=\dfrac{8}{\left(x+1\right)\left(x-3\right)}\)

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

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)+\left(x+5\right)\left(x+1\right)=8\)

\(\Leftrightarrow x^2-3x-x+3+x^2+x+5x+5-8=0\)

\(\Leftrightarrow2x^2+2x=0\)

\(\Leftrightarrow2x\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)

Answer 2

[Nguyễn Lê Phước Thịnh]

a: \(\Leftrightarrow5x+3\left(2x-1\right)=30-5x\)

=>5x+6x-3=30-5x

=>11x-3=30-5x

=>16x=33

hay x=33/16

b: \(\Leftrightarrow\left(2x-1\right)\left(5x-8\right)-\left(2x-1\right)\left(3x+2\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(5x-8-3x-2\right)=0\)

=>(2x-1)(2x-10)=0

=>x=1/2 hoặc x=5

c: \(\Leftrightarrow\dfrac{x-1}{x+1}+\dfrac{x+5}{x-3}=\dfrac{8}{\left(x+1\right)\left(x-3\right)}\)

\(\Leftrightarrow x^2-4x+3+x^2+6x+5=8\)

\(\Leftrightarrow2x^2+2x=0\)

=>2x(x+1)=0

=>x=0(nhận) hoặc x=-1(loại)