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a) \(\frac{3x^2-6xy+3y^2}{5x^2-5xy+5y^2}:\frac{10x-10y}{x^3+y^3}\)
\(=\frac{3x^2-6xy+3y^2}{5x^2-5xy+5y^2}.\frac{x^3+y^3}{10x-10y}\)
\(=\frac{3\left(x^2-2xy+y^2\right)}{5\left(x^2-xy+y^2\right)}.\frac{\left(x+y\right)\left(x^2-xy+y^2\right)}{10\left(x-y\right)}\)
\(=\frac{3\left(x^2-2xy+y^2\right)}{5}.\frac{x+y}{10\left(x-y\right)}\)
\(=\frac{3\left(x-y\right)^2}{5}.\frac{x+y}{10\left(x-y\right)}\)
\(=\frac{3\left(x-y\right)}{5}.\frac{x+y}{10}\)
\(=\frac{3x^2-3y^2}{50}\)
c) \(\frac{2}{xy}:\left(\frac{1}{x}-\frac{1}{y}\right)-\frac{x^2-y^2}{\left(x-y\right)^2}\)
\(=\frac{2}{xy}:\frac{y-x}{xy}-\frac{\left(x+y\right)\left(x-y\right)}{\left(x-y\right)^2}\)
\(=\frac{2}{y-x}-\frac{x+y}{x-y}\)
\(=\frac{2}{y-x}+\frac{x+y}{y-x}\)
\(=\frac{x+y+2}{y-x}\)
a, mình nghĩ đề là cm đẳng thức nhé
\(VT=\left(5x^4-3x^3+x^2\right):3x^2=\frac{5x^4}{3x^2}-\frac{3x^3}{3x^2}+\frac{x^2}{3x^2}=\frac{5}{3}x^2-x+\frac{1}{3}=VP\)
Vậy ta có đpcm
b, \(VT=\left(5xy^2+9xy-x^2y^2\right):\left(-xy\right)=\frac{5xy^2}{-xy}+\frac{9xy}{-xy}-\frac{x^2y^2}{-xy}\)
\(=-5y-9+xy=VP\)
Vậy ta có đpcm
c, \(VT=\left(x^3y^3-x^2y^3-x^3y^2\right):x^2y^2=\frac{x^3y^3}{x^2y^2}-\frac{x^2y^3}{x^2y^2}-\frac{x^3y^2}{x^2y^2}=xy-y-x=VP\)
Vậy ta có đpcm
Answer:
Câu đầu bạn xem lại.
\(\left(3x+4\right)^2+\left(4x-3\right)^2+\left(2+5x\right).\left(2-5x\right)\)
\(=\left(3x\right)^2+2.2x.4+4^2+\left(4x\right)^2-2.4x.3+3^2+2^2-\left(5x\right)^2\)
\(=9x^2+24x+16+16x^2-24x+9+4-25x^2\)
\(=\left(9x^2+16x^2-25x^2\right)+\left(24x-24x\right)+\left(16+9+4\right)\)
\(=29\)
\(\left(5x+y\right).\left(25x^2-5xy+y^2\right)-\left(5x-y\right).\left(25x^2+5xy+y^2\right)\)
\(=\left(5x+y\right).[\left(5x\right)^2-5x.y+y^2]-\left(5x-y\right).[\left(5x\right)^2+5x.y+y^2]\)
\(=\left(5x\right)^3+y^3-[\left(5x\right)^3-y^3]\)
\(=\left(5x\right)^3+y^3-\left(5x\right)^3+y^3\)
\(=2y^3\)
\(\left(\frac{5x+y}{x^2-5xy}+\frac{5x-y}{x^2+5xy}\right).\frac{x^2-25y^2}{x^2+y^2}\)
\(=\left(\frac{5x+y}{x\left(x-5y\right)}+\frac{5x-y}{x\left(x+5y\right)}\right).\frac{\left(x-5y\right)\left(x+5y\right)}{x^2+y^2}\)
\(=\frac{\left(5x+y\right)\left(x+5y\right)+\left(5x-y\right)\left(x-5y\right)}{x\left(x-5y\right)\left(x+5y\right)}.\frac{\left(x-5y\right)\left(x+5y\right)}{x^2+y^2}\)
\(=\frac{10\left(x^2+y^2\right)}{x\left(x^2+y^2\right)}=\frac{10}{x}\)
\(\left(\frac{5x+y}{x^2-5xy}+\frac{5x-y}{x^2+5xy}\right).\frac{x^2-25y^2}{x^2+y^2}\)
\(=\left(\frac{5x+y}{x\left(x-5y\right)}+\frac{5x-y}{x\left(x+5y\right)}\right)\frac{\left(x-5y\right)\left(x+5y\right)}{x^2+y^2}\)
\(=\frac{\left(5x+y\right)\left(x+5y\right)+\left(5x-y\right)\left(x-5y\right)}{x\left(x-5y\left(x+4y\right)\right)}.\frac{\left(x-5y\right)\left(x+5y\right)}{x^2+y^2}\)
\(=\frac{10\left(x^2+y^2\right)}{x\left(x^2+y^2\right)}=\frac{10}{x}\)