1) \(x=\frac{99}{196}\)

2) \(x=-2\)

3) \(x\approx-0,59\)

giup mk giải rõ dc ko

a VT=.\(\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\left(\frac{1}{x+1}-\frac{x}{1-x}+\frac{2}{x^2-1}\right)\)

=\(\frac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}:\frac{x-1+x\left(x-1\right)+2}{\left(x+1\right)\left(x-1\right)}\)

\(=\frac{x^2+2x+1-x^2+2x-1}{\left(x+1\right)\left(x-1\right)}.\frac{\left(x+1\right)\left(x-1\right)}{x^2+2x+1}\)

\(=\frac{4x}{\left(x+1\right)^2}\)=VP

b.VT\(=\frac{2+x}{2-x}.\frac{\left(2-x\right)^2}{4x^2}.\left(\frac{2}{2-x}-\frac{4}{\left(x+2\right)\left(x^2-2x+4\right)}.\frac{4-2x+x^2}{2-x}\right)\)

=\(\frac{4-x^2}{4x^2}.\left(\frac{2}{2-x}-\frac{4}{4-x^2}\right)=\frac{4-x^2}{4x^2}.\frac{2\left(2+x\right)-4}{4-x^2}\)

=\(\frac{2x}{4x^2}=\frac{1}{2x}\)=VP

c VT=.\(\left[\left(\frac{3}{x-y}+\frac{3x}{x^2-y^2}\right).\frac{\left(x+y\right)^2}{2x+y}\right].\frac{x-y}{3}\)

\(=\left[\frac{3\left(x+y\right)+3x}{\left(x+y\right)\left(x-y\right)}.\frac{\left(x+y\right)^2}{2x+y}\right].\frac{x-y}{3}\)

\(=\frac{3\left(2x+y\right)\left(x+y\right)^2}{\left(x+y\right)\left(x-y\right)\left(2x+y\right)}.\frac{x-y}{3}\)

\(=x+y=\)VP

Vậy các đẳng thức được chứng minh

=

C là xy mà ko phải x+y

1.

\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)

\(MC:12\)

Quy đồng :

\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)

\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)

\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)

\(\Leftrightarrow6x+9-3x=-4-9+16\)

\(\Leftrightarrow-7x=3\)

\(\Leftrightarrow x=\frac{-3}{7}\)

2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)

\(MC:20\)

Quy đồng :

\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)

\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)

\(\Leftrightarrow30x+15-20=15x-2\)

\(\Leftrightarrow15x=3\)

\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)

Đây là lớp 8 nha các b giúp mk với

Do mk viết nhầm

\(\frac{\left(2-3x\right)^2}{3}-\frac{\left(1+2x\right)^2}{2}=\frac{3}{4}-2\left(x-1\right)\left(x+2\right)+x\left(1+x\right)\)

\(\frac{2^2-12x-3x^2}{3}-\frac{1^2+4x+2x^2}{2}=\frac{3}{4}-\left(x^2+x-2\right)+3x\)

\(\frac{2.\left(4-12x-3x^2\right)}{6}-\frac{3.\left(1+4x+2x^2\right)}{6}=\frac{11}{4}-x^2+2x\)

\(\frac{8-24x-6x^2}{6}-\frac{3+12x+2x^2}{6}=\frac{11}{4}-x^2+2x\)

\(\frac{8-24x-6x^2-3-12x-2x^2}{6}=\frac{11}{4}-x^2+2x\)

\(\frac{5-36x-8x^2}{6}=\frac{11}{4}-x^2+2x\)

Chỗ đây thì mk chịu

\(\Leftrightarrow20\left(x^2-4x+3\right)-24\left(4x^2-4x+1\right)=15\left(9x^2+6x+1\right)+90x\left(x-1\right)\)

\(\Leftrightarrow20x^2-80x+60-96x^2+96x-24=135x^2+90x+15+90x^2-90x\)

\(\Leftrightarrow-301x^2+16x+21=0\)

\(\text{Δ}=16^2-4\cdot\left(-301\right)\cdot21=25540\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là 

\(\left\{{}\begin{matrix}x_1=\dfrac{-16-\sqrt{25540}}{-602}=\dfrac{16+\sqrt{25540}}{602}\\x_2=\dfrac{16-\sqrt{25540}}{602}\end{matrix}\right.\)

máy tính tính dc vô nghiệm

Giải ra mk vs