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a: ĐKXĐ: x<>y; x<>1/2; x<>-2
b:
\(\left(\frac{x+y}{y}-\frac{2y}{y-x}\right):\frac{x^2+y^2}{y-x}\)
\(=\frac{\left(y+x\right)\left(y-x\right)-2y^2}{y\left(y-x\right)}\cdot\frac{y-x}{x^2+y^2}\)
\(=\frac{-x^2-y^2}{y\left(x^2+y^2\right)}=\frac{-1}{y}\)
\(\left(\frac{x^2+1}{2x-1}-\frac{x}{2}\right)\cdot\frac{1-2x}{x+2}\)
\(=\frac{2\left(x^2+1\right)-x\left(2x-1\right)}{2\left(2x-1\right)}\cdot\frac{-\left(2x-1\right)}{x+2}\)
\(=\frac{2x^2+2-2x^2+x}{2}\cdot\frac{-1}{x+2}=\frac{x+2}{-2\left(x+2\right)}=\frac{-1}{2}\)
Ta có: \(A=\left(\frac{x+y}{y}-\frac{2y}{y-x}\right):\frac{x^2+y^2}{y-x}+\left(\frac{x^2+1}{2x-1}-\frac{x}{2}\right)\cdot\frac{1-2x}{x+2}\)
\(=\frac{-1}{y}+\frac{-1}{2}=\frac{-y-2}{2y}\)
Bài 1:
a: ĐKXĐ: x<>0; x<>3; x<>1
\(A=\left(\frac{x-3}{x}-\frac{x}{x-3}+\frac{9}{x^2-3x}\right):\frac{2x-2}{x}\)
\(=\frac{\left(x-3\right)^2-x^2+9}{x\left(x-3\right)}:\frac{2\left(x-1\right)}{x}\)
\(=\frac{x^2-6x+9-x^2+9}{x\cdot\left(x-3\right)}:\frac{2\left(x-1\right)}{x}=\frac{-6\left(x-3\right)}{x\left(x-3\right)}:\frac{2\left(x-1\right)}{x}=\frac{-6}{x}\cdot\frac{x}{2\left(x-1\right)}=\frac{-3}{x-1}\)
b: Để A là số nguyên thì -3⋮x-1
=>x-1∈{1;-1;3;-3}
=>x∈{2;0;4;-2}
Kết hợp ĐKXĐ, ta được: x∈{2;4;-2}
Bài 2:
a: \(x^3-2x^2=x^2\cdot x-x^2\cdot2=x^2\left(x-2\right)\)
b: \(y^2-2y-x^2+1\)
\(=\left(y^2-2y+1\right)-x^2\)
\(=\left(y-1\right)^2-x^2=\left(y-1-x\right)\left(y-1+x\right)\)
c: \(\left(x+1\right)^2-25\)
=(x+1+5)(x+1-5)
=(x-4)(x+6)
a) ĐKXĐ là \(x^2-y^2\)khác 0
b) A=\(\frac{x^2+2x-y^2-2y}{x^2-y^2}=\frac{\left(x^2-y^2\right)+2\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}=\frac{\left(x+y\right)\left(x-y\right)+2\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}=\frac{\left(x-2\right)\left(x+y+2\right)}{\left(x-y\right)\left(x+y\right)}=\frac{x+y+2}{x+y}=1+\frac{2}{x+y}\)
c) Thay x=5,y=6 vào biểu thức A ta được
A=\(2+\frac{2}{5-6}=2+\frac{2}{-1}=2-2=0\)
Vậy A=0
DKXD x và Y khác 0
B) rút gọn x^2 và y^2 ta dc \(\frac{2x-2y}{1}\)
C. \(2.5-2.6=10-12=-2\)
\(\left(x+4\right)^2-81=0\Leftrightarrow\left(x+4\right)^2-9^2=0\)
\(\Leftrightarrow\left(x+4+9\right)\times\left(x+4-9\right)=0\)
\(\Leftrightarrow\left(x+13\right)\times\left(x-5\right)=0\)
\(\left[{}\begin{matrix}x+13=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-13\\x=5\end{matrix}\right.\)
\(A=\dfrac{x^2-y^2+2y^2}{y\left(x-y\right)}\cdot\dfrac{-\left(x-y\right)}{x^2+y^2}+\dfrac{2x^2+2-2x^2+x}{2\left(2x-1\right)}\cdot\dfrac{-\left(2x-1\right)}{x+2}\)
\(=\dfrac{-1}{y}+\dfrac{-1}{2}=\dfrac{-2-y}{2y}\)
\(1)A=2x\left(x-y\right)-y\left(y-2x\right)\)
\(=2x^2-2xy-y^2+2xy\)
\(=2x^2-y^2=2.\left(-\dfrac{2}{3}\right)^2-\left(-\dfrac{1}{3}\right)^2\)
\(=\dfrac{8}{9}-\dfrac{1}{9}=\dfrac{7}{9}\)
\(2)B=5x\left(x-4y\right)-4y\left(y-5x\right)\)
\(=5x^2-20xy-4y^2+20xy\)
\(=5x^2-4y^2=5.\left(-\dfrac{1}{5}\right)^2-4.\left(-\dfrac{1}{2}\right)^2=\dfrac{1}{5}-1=-\dfrac{4}{5}\)
\(3)C=\text{x.(x^2-y^2)-x^2(x+y)+y(x^2-x)}\)
\(=x^3-xy^2-x^3-x^2y+x^2y-xy\)
\(=-xy\left(x+1\right)\)
a: ĐKXĐ: x<>y; x<>1/2; x<>-2
b:
\(\left(\frac{x+y}{y}-\frac{2y}{y-x}\right):\frac{x^2+y^2}{y-x}\)
\(=\frac{\left(y+x\right)\left(y-x\right)-2y^2}{y\left(y-x\right)}\cdot\frac{y-x}{x^2+y^2}\)
\(=\frac{-x^2-y^2}{y\left(x^2+y^2\right)}=\frac{-1}{y}\)
\(\left(\frac{x^2+1}{2x-1}-\frac{x}{2}\right)\cdot\frac{1-2x}{x+2}\)
\(=\frac{2\left(x^2+1\right)-x\left(2x-1\right)}{2\left(2x-1\right)}\cdot\frac{-\left(2x-1\right)}{x+2}\)
\(=\frac{2x^2+2-2x^2+x}{2}\cdot\frac{-1}{x+2}=\frac{x+2}{-2\left(x+2\right)}=\frac{-1}{2}\)
Ta có: \(A=\left(\frac{x+y}{y}-\frac{2y}{y-x}\right):\frac{x^2+y^2}{y-x}+\left(\frac{x^2+1}{2x-1}-\frac{x}{2}\right)\cdot\frac{1-2x}{x+2}\)
\(=\frac{-1}{y}+\frac{-1}{2}=\frac{-y-2}{2y}\)