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Lời giải:
a)
\(A=\frac{x^2y(y-x)-xy^2(x-y)}{3y^2-2x^2}=\frac{x^2y(y-x)+xy^2(y-x)}{3y^2-2x^2}=\frac{(xy^2+x^2y)(y-x)}{3y^2-2x^2}\)
\(=\frac{xy(x+y)(y-x)}{3y^2-2x^2}=\frac{xy(y^2-x^2)}{3y^2-2x^2}\)
Với $x=-3; y=\frac{1}{2}$ thì:
$xy=\frac{-3}{2}; x^2=9; y^2=\frac{1}{4}$
Do đó $A=\frac{-35}{46}$
b)
\(B=\frac{(8x^3-y^3)(4x^2-y^2)}{(2x+y)(4x^2-4xy+y^2)}=\frac{(2x-y)(4x^2+2xy+y^2)(2x-y)(2x+y)}{(2x+y)(2x-y)^2}\)
\(=4x^2+2xy+y^2=4.2^2+2.2.\frac{-1}{2}+(\frac{-1}{2})^2=\frac{57}{4}\)
4x2−4xy+y2y3−6xy2+12x2y−8x34x2-4xy+y2y3-6xy2+12x2y-8x3
=4x2−4xy+y2y3+3.(−2x).y2−3.(−2x)2.y−(−2x)3=4x2-4xy+y2y3+3.(-2x).y2-3.(-2x)2.y-(-2x)3
=(2x−y)2(−2x+y)3=(2x-y)2(-2x+y)3
=−(2x−y)2(2x−y)3=-(2x-y)2(2x-y)3
=−12x−y
a) \(x^2+\frac{1}{3}+\frac{1}{36}=\left(x+\frac{1}{6}\right)^2\)
Thay \(x=\frac{-7}{6}\)vào biểu thức ta được: \(\left(\frac{-7}{6}+\frac{1}{6}\right)^2=\left(-1\right)^2=1\)
b) \(x^3-9x^2+27x-27=\left(x-3\right)^3\)
Thay \(x=103\)vào biểu thức ta được: \(\left(103-3\right)^2=100^2=10000\)
c) \(4x^2-y^2-2y-1=4x^2-\left(y^2+2y+1\right)\)
\(=4x^2-\left(y+1\right)^2=\left(2x-y-1\right)\left(2x+y+1\right)\)
Thay \(x=234\)và \(y=465\)vào biểu thức ta được:
\(\left(2.234-465-1\right)\left(2.234+465+1\right)=2.934=1868\)
a) Ta có: \(x^2+\frac{1}{3}x+\frac{1}{36}=x^2+2\cdot\frac{1}{6}\cdot x+\left(\frac{1}{6}\right)^2\)
\(=\left(x+\frac{1}{6}\right)^2\) , tại \(x=-\frac{7}{6}\) thì giá trị của BT là:
\(\left(-\frac{7}{6}+\frac{1}{6}\right)^2=1^2=1\)
b) Ta có: \(x^3-9x^2+27x-27=\left(x-3\right)^3\)
Tại x = 103 thì giá trị của BT là:
\(\left(103-3\right)^3=100^3=1000000\)
c) Ta có: \(4x^2-y^2-2y-1\)
\(=\left(2x\right)^2-\left(y+1\right)^2\)
\(=\left(2x-y-1\right)\left(2x+y+1\right)\)
Tại x = 234, y = 465 thì giá trị của BT là:
\(\left(2\cdot234-465-1\right)\left(2\cdot234+465+1\right)\)
\(=2\cdot934=1868\)
Bài 1:
a) Ta có: \(\left(\frac{1}{\left(2x-y\right)^2}+\frac{2}{4x^2-y^2}+\frac{1}{\left(2x+y\right)^2}\right):\frac{16x}{4x^2+4xy+y^2}\)
\(=\left(\frac{\left(2x+y\right)^2}{\left(2x-y\right)^2\cdot\left(2x+y\right)^2}+\frac{2\cdot\left(2x+y\right)\left(2x-y\right)}{\left(2x+y\right)^2\cdot\left(2x-y\right)^2}+\frac{\left(2x-y\right)^2}{\left(2x+y\right)^2\cdot\left(2x-y\right)^2}\right)\cdot\frac{\left(2x+y\right)^2}{16x}\)
\(=\frac{\left(2x+y+2x-y\right)^2}{\left(2x-y\right)^2\cdot\left(2x+y\right)^2}\cdot\frac{\left(2x+y\right)^2}{16x}\)
\(=\frac{\left(4x\right)^2}{\left(2x-y\right)^2}\cdot\frac{1}{16x}\)
\(=\frac{16x^2}{16x\cdot\left(2x-y\right)^2}\)
\(=\frac{x}{\left(2x-y\right)^2}\)
b) Ta có: \(\frac{3}{3x+3}+\frac{10}{5-5x}+\frac{5x-1}{x^2-1}\)
\(=\frac{1}{x+1}-\frac{2}{x-1}+\frac{5x-1}{x^2-1}\)
\(=\frac{x-1}{\left(x+1\right)\left(x-1\right)}-\frac{2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{5x-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x-1-2\left(x+1\right)+5x-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x-1-2x-2+5x-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{4x-4}{\left(x-1\right)\left(x+1\right)}=\frac{4\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{4}{x+1}\)
c) Ta có: \(A=\left(x^4-x^2+2x-1\right):\left(x^2+x-1\right)-\left(x^2-x\right)\)
\(=\frac{\left(x^2\right)^2-\left(x^2-2x+1\right)}{x^2+x-1}-x^2+x\)
\(=\frac{\left(x^2\right)^2-\left(x-1\right)^2}{x^2+x-1}-x^2+x\)
\(=\frac{\left(x^2-x+1\right)\left(x^2+x-1\right)}{x^2+x-1}-x^2+x\)
\(=x^2-x+1-x^2+x\)
=1
\(\left[\frac{1}{\left(2x-y\right)^2}+\frac{2}{4x^2-y^2}+\frac{1}{\left(2x+y\right)^2}\right].\frac{4x^2+4xy+y^2}{16x}\)
\(=\frac{\left(2x+y\right)^22\left(4x^2-y^2\right)+\left(2x-y\right)^2}{\left(2x-y\right)^2\left(2x+y\right)^2}.\frac{\left(2x+y\right)^2}{16x}\)
\(=\frac{16x^2}{16x\left(2x-y\right)^2}=\frac{x}{\left(2x-y\right)^2}\)
\(\left[\frac{1}{\left(2x-y\right)^2}+\frac{2}{4x^2-4^2}+\frac{1}{\left(2x+y\right)^2}\right].\frac{4x^2+4xy+y^2}{16x}\)
\(=\frac{\left(2x+y\right)^22\left(4x^2-y^2\right)+\left(2x-y\right)^2}{\left(2x-y\right)^2\left(2x+y\right)^2}.\frac{\left(2x+y\right)^2}{16x}\)
\(=\frac{16x^2}{16x\left(2x-y\right)^2}=\frac{x}{\left(2x-y\right)^2}\)
\(1,\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}=\frac{3}{2x+6}-\frac{x-6}{x\left(2x-6\right)}=\frac{3x-x+6}{x\left(2x-6\right)}=\frac{2x+6}{x\left(2x-6\right)}\)
\(2,\frac{1}{1-x}+\frac{2x}{x^2-1}=\frac{-1\left(x+1\right)+2x}{x^2-1}=\frac{x-1}{x^2-1}=\frac{1}{x+1}\)
\(3,\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{1}{x\left(y-x\right)}-\frac{1}{y\left(y-x\right)}=\frac{y-x}{xy\left(y-x\right)}=\frac{1}{xy}\)
\(4,\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}=\frac{5\left(x+2\right)}{4\left(x-2\right)}.\frac{2\left(2-x\right)}{x+2}=\frac{-5}{2}\)
\(5,\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}=\frac{\left(1-2x\right)\left(1+2x\right)}{x\left(x+4\right)}.\frac{3x}{2\left(1-2x\right)}=\frac{3\left(1+2x\right)}{2x\left(x+4\right)}\)
\(6,\frac{12x}{5y^3}.\frac{15y^4}{8x^3}=\frac{9y}{2x^2}\)
Ta có: \(VT=\frac{4x^2-4xy+y^2}{y^3-6y^2x+12ỹ^2-8x^3}\)
\(=\frac{\left(2x-y\right)^2}{\left(y-2x\right)^3}=-\frac{\left(2x-y\right)^2}{\left(2x-y\right)^3}=\frac{-1}{2x-y}=VP\)(đpcm)