a: x-y-z=0

=>x=y+z; y=x-z; z=x-y

\(K=\dfrac{x-z}{x}\cdot\dfrac{y-x}{y}\cdot\dfrac{z+y}{z}=\dfrac{y\cdot\left(-z\right)\cdot x}{xyz}=-1\)

b: Tham khảo:

1)\(A=\frac{b\left(2a\left(a+5b\right)+\left(a+5b\right)\right)}{a-3b}.\frac{a\left(a-3b\right)}{ab\left(a+5b\right)}=\frac{b\left(a+5b\right)\left(2a+1\right).a\left(a-3b\right)}{\left(a-3b\right).ab\left(a+5b\right)}\)

\(A=2a+1\)=>lẻ với mọi a thuộc z=> dpcm 

2) từ: x+y+z=1=> xy+z=xy+1-x-y=x(y-1)-(y-1)=(y-1)(x-1)

tường tự: ta có tử của Q=(x-1)^2.(y-1)^2.(z-1)^2=[(x-1)(y-1)(z-1)]^2=[-(z+y).-(x+y).-(x+y)]^2=Mẫu=> Q=1

3) kiểm tra lại xem đề đã chuẩn chưa

Câu 3: 

a: \(G=\dfrac{a^2}{b\left(a+b\right)}-\dfrac{b^2}{a\left(a-b\right)}+\dfrac{-\left(a^2+b^2\right)}{ab}\)

\(=\dfrac{a^3\left(a-b\right)-b^3\left(a+b\right)-\left(a^2+b^2\right)\left(a^2-b^2\right)}{ab\left(a-b\right)\left(a+b\right)}\)

\(=\dfrac{a^4-a^3b-ab^3-b^4-a^4+b^4}{ab\left(a-b\right)\left(a+b\right)}\)

\(=\dfrac{-ab\left(a^2+b^2\right)}{ab\left(a-b\right)\left(a+b\right)}=\dfrac{-a^2-b^2}{a^2-b^2}\)

b: \(\dfrac{a}{b}=\dfrac{a+1}{b+5}\)

nên ab+5a=ab+b

=>5a=b

\(G=\dfrac{-a^2-\left(5a\right)^2}{a^2-\left(5a\right)^2}=\dfrac{-a^2-25a^2}{a^2-25a^2}=\dfrac{-26}{-24}=\dfrac{13}{12}\)

Bài làm

a) \(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x-4}\)

\(\Leftrightarrow\frac{3x+2}{3x-2}-\frac{6}{3x+2}=\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

\(\Leftrightarrow\frac{(3x+2)\left(3x+2\right)}{(3x-2)\left(3x+2\right)}-\frac{6\left(3x-2\right)}{(3x+2)\left(3x-2\right)}=\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

\(\Rightarrow\left(3x+2\right)^2-\left(18x-12\right)=9x^2\)

\(\Leftrightarrow9x^2+12x+4-18x+12x-9x^2=0\)

\(\Leftrightarrow6x+4=0\)

\(\Leftrightarrow x=-\frac{4}{6}\)

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

Vậy x = -2/3 là nghiệm.

@Tao Ngu :))@ 9x-4 không tách thành (3x+4)(3x-4) được đâu bạn. Chỗ đó phải là: 9x²-4

Bài thiếu đkxđ của x \(\hept{\begin{cases}3x-2\ne0\\2+3x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}3x\ne2\\3x\ne-2\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne\frac{2}{3}\\x\ne\frac{-2}{3}\end{cases}\Leftrightarrow}x\ne\pm\frac{2}{3}}\)

khó thể xem trên mạng

bài 1 câu a bỏ x= nhé !

a) ĐKXĐ: \(x\notin\left\{\frac{1}{3};\frac{-11}{3}\right\}\)

Ta có: \(\frac{2}{\left(1-3x\right)\left(3x+11\right)}=\frac{1}{9x^2-6x+1}-\frac{3}{\left(3x+11\right)^2}\)

\(\Leftrightarrow\frac{2\left(1-3x\right)\left(3x+11\right)}{\left(1-3x\right)^2\cdot\left(3x+11\right)^2}=\frac{\left(3x+11\right)^2}{\left(1-3x\right)^2\cdot\left(3x+11\right)^2}-\frac{3\left(1-3x\right)^2}{\left(1-3x\right)^2\cdot\left(3x+11\right)^2}\)

\(\Leftrightarrow-18x^2-60x+22=9x^2+66x+121-3\left(1-6x+9x^2\right)\)

\(\Leftrightarrow-18x^2-60x+22-9x^2-66x-121+3\left(1-6x+9x^2\right)=0\)

\(\Leftrightarrow-27x^2-126x-99+3-18x+27x^2=0\)

\(\Leftrightarrow-144x-96=0\)

\(\Leftrightarrow-144x=96\)

hay \(x=\frac{-2}{3}\)(tm)

Vậy: \(x=\frac{-2}{3}\)

Câu 1 : 

a, \(\frac{3}{x+3}-\frac{x-6}{x^2+3x}=\frac{3x-x+6}{x\left(x+3\right)}=\frac{2x+6}{x\left(x+3\right)}=\frac{2}{x}\)

b, \(\frac{2x^2-x}{x-1}+\frac{x+1}{1-x}+\frac{2-x^2}{x-1}=\frac{2x^2-x-x-1+2-x^2}{x-1}\)

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

Bài 2 : 

a, Với \(x\ne\pm2\)

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

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

\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{2}=\frac{-3}{x-2}\)

b, Thay x = -4 vào biểu thức trên ta được : 

\(-\frac{3}{-4-2}=-\frac{3}{-6}=\frac{1}{2}\)

c, Để A \(\inℤ\Rightarrow x-2\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)