\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)

= \(\frac{3x\left(x-y\right)}{5.2.\left(x+y\right)\left(x-y\right)}-\frac{x\left(x+y\right)}{10\left(x^2-y^2\right)}\)

= \(\frac{3x^2-3xy-x^2-xy}{10\left(x^2-y^2\right)}\)

= \(\frac{3x\left(x-y\right)}{10\left(x^2-y^2\right)}\)

= \(\frac{3x}{10\left(x+y\right)}\)

Đây là lớp 8 nha các b giúp mk với

Do mk viết nhầm

b) \(\left[\frac{2}{3x}-\frac{2}{x+1}.\left(\frac{x+1}{3x}-x-1\right)\right]:\frac{x-1}{x}\)

\(=\left[\frac{2}{3x}-\frac{2}{x+1}.\left(\frac{x+1}{3x}-\left(x+1\right)\right)\right]:\frac{x-1}{x}\)

\(=\left[\frac{2}{3x}-\frac{2}{x+1}.\left(x+1\right)\left(\frac{1}{3x}-1\right)\right]:\frac{x-1}{x}\)

\(=\left[\frac{2}{3x}-2\left(\frac{1}{3x}-1\right)\right]:\frac{x-1}{x}\)

\(=\left[\frac{2}{3x}-\frac{2}{3x}+2\right]:\frac{x-1}{x}\)

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

a) \(\left(\frac{9}{x^3-9x}+\frac{1}{x+3}\right):\left(\frac{x-3}{x^2+3x}-\frac{x}{3x+9}\right)\)

\(=\left(\frac{9}{x\left(x^2-9\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)

\(=\left(\frac{9}{x\left(x+3\right)\left(x-3\right)}+\frac{x^2-3x}{x\left(x+3\right)\left(x-3\right)}\right)\)

\(:\left(\frac{3x-9}{3x\left(x+3\right)}-\frac{x^2}{3x\left(x+3\right)}\right)\)

\(=\frac{x^2-3x+9}{x\left(x+3\right)\left(x-3\right)}:\frac{-x^2+3x-9}{3x\left(x+3\right)}\)

\(=\frac{x^2-3x+9}{x\left(x+3\right)\left(x-3\right)}.\frac{3x\left(x+3\right)}{-x^2+3x-9}\)

\(=\frac{x^2-3x+9}{x-3}.\frac{3}{x^2+3x-9}\)

\(=\frac{x^2-3x+9}{3-x}.\frac{3}{x^2-3x+9}\)

\(=\frac{3}{3-x}\left(đpcm\right)\)

a) Qui đồng rồi khử mẫu ta được:

   3(3x+2)-(3x+1)=2x.6+5.2

<=> 9x+6-3x-1 = 12x+10

<=> 9x-3x-12x  = 10-6+1

<=> -6x            = 5

<=> x               = -5/6

Vậy ....

b) ĐKXĐ: \(x\ne\pm2\)

Qui đồng rồi khử mẫu ta được:

   (x+1)(x+2)+(x-1)(x-2) = 2(x²+2)

<=> x²+3x+2+x²-3x+2 = 2x²+4

<=> x²+x²-2x²+3x-3x = 4-2-2

<=> 0x             = 0

<=> x vô số nghiệm

Vậy x vô số nghiệm với x khác 2 và x khác -2

c) \(\left(2x+3\right)\left(\frac{3x+7}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\) (ĐKXĐ:x khắc 2/7)

\(\Leftrightarrow\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)-\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)=0\)

\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left[\left(2x+3\right)-\left(x-5\right)\right]=0\)

\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left(x+8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{3x+8}{2-7x}+1=0\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{3x+8}{2-7x}=-1\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}3x+8=-1\left(2-7x\right)\\x=0-8\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x+8=-2+7x\\x=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}-4x=-10\\x=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-8\end{cases}}}\) (nhận)

Vậy ...... 

d) (x+1)²-4(x²-2x+1) = 0

<=> x²+2x+1-4x²+8x-4 = 0

<=> -3x²+10x-3 = 0

giải phương trình

Bài 1:

a) Ta có: \(\frac{4}{5}x-3=\frac{1}{5}x\left(4x-15\right)\)

\(\Leftrightarrow\frac{4x}{5}-3=\frac{4x^2}{5}-3x\)

\(\Leftrightarrow\frac{12x}{15}-\frac{45}{15}-\frac{12x^2}{15}+\frac{45x}{15}=0\)

Suy ra: \(12x-45-12x^2+45x=0\)

\(\Leftrightarrow-12x^2+57x-45=0\)

\(\Leftrightarrow-12x^2+12x+45x-45=0\)

\(\Leftrightarrow-12x\left(x-1\right)+45\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(-12x+45\right)=0\)

\(\Leftrightarrow-3\left(x-1\right)\left(4x-15\right)=0\)

mà \(-3\ne0\)

nên \(\left[{}\begin{matrix}x-1=0\\4x-15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\4x=15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{15}{4}\end{matrix}\right.\)

Vậy: Tập nghiệm \(S=\left\{1;\frac{15}{4}\right\}\)

b) Ta có: \(\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3-x\right)}{4}\)

\(\Leftrightarrow\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}+\frac{\left(x-3\right)^2}{4}=0\)

\(\Leftrightarrow\frac{12\left(x-3\right)}{12}-\frac{2\left(x-3\right)\left(2x-5\right)}{12}+\frac{3\left(x-3\right)^2}{12}=0\)

Suy ra: \(12\left(x-3\right)-2\left(2x^2-11x+15\right)+3\left(x^2-6x+9\right)=0\)

\(\Leftrightarrow12x-36-4x^2+22x-30+3x^2-18x+27=0\)

\(\Leftrightarrow-x^2+16x-39=0\)

\(\Leftrightarrow-\left(x^2-16x+39\right)=0\)

\(\Leftrightarrow x^2-13x-3x+39=0\)

\(\Leftrightarrow x\left(x-13\right)-3\left(x-13\right)=0\)

\(\Leftrightarrow\left(x-13\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-13=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\\x=3\end{matrix}\right.\)

Vậy: Tập nghiệm S={3;13}

c) Ta có: \(\frac{\left(3x+1\right)\left(3x-2\right)}{3}+5\left(3x+1\right)=\frac{2\left(2x+1\right)\left(3x+1\right)}{3}+2x\left(3x+1\right)\)

\(\Leftrightarrow\frac{9x^2-3x-2}{3}+5\left(3x+1\right)-\frac{12x^2+10x+2}{3}-2x\left(3x+1\right)=0\)

\(\Leftrightarrow\frac{9x^2-3x-2-12x^2-10x-2}{3}-6x^2+13x+5=0\)

\(\Leftrightarrow\frac{-3x^2-13x-4}{3}+\frac{3\left(-6x^2+13x+5\right)}{3}=0\)

Suy ra: \(-3x^2-13x-4-18x^2+39x+15=0\)

\(\Leftrightarrow-21x^2+26x+11=0\)

\(\Leftrightarrow-21x^2-7x+33x+11=0\)

\(\Leftrightarrow-7x\left(3x+1\right)+11\left(3x+1\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(-7x+11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\-7x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\-7x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{3}\\x=\frac{11}{7}\end{matrix}\right.\)

Vậy: Tập nghiệm \(S=\left\{-\frac{1}{3};\frac{11}{7}\right\}\)

Mấy câu như vậy bạn tải photomath về dùng nhé :)