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b: Ta có: \(\left(4x-y\right)\left(4x+y\right)-2\left(3x-2y\right)^2+\left(x-3y\right)^2\)
\(=16x^2-y^2-2\left(9x^2-12xy+4y^2\right)+x^2-6xy+9y^2\)
\(=17x^2-6xy+8y^2-18x^2+24xy-8y^2\)
\(=-x^2+18xy\)
c: Ta có: \(\left(2a-3b+4c\right)\left(2a-3b-4c\right)\)
\(=\left(2a-3b\right)^2-16c^2\)
\(=4a^2-12ab+9b^2-16c^2\)
\(a.\dfrac{2\left(x-y\right)}{3y-3x}=\dfrac{-2\left(y-x\right)}{3\left(y-x\right)}=\dfrac{-2}{3}\)
\(b.\dfrac{x-2}{-x}=\dfrac{2-x}{x}=\dfrac{\left(2-x\right)\left(x^2+2x+4\right)}{x\left(x^2+2x+4\right)}=\dfrac{8-x^3}{x\left(x^2+2x+4\right)}\)
\(\dfrac{3x}{x+y}=\dfrac{3x\left(x-y\right)}{\left(x+y\right)\left(x-y\right)}=\dfrac{-3x\left(x-y\right)}{\left(x+y\right)\left(y-x\right)}=\dfrac{-3x\left(x-y\right)}{y^2-x^2}\)
c: \(\dfrac{-3x\left(x-y\right)}{y^2-x^2}=\dfrac{3x\left(x-y\right)}{\left(x+y\right)\left(x-y\right)}=\dfrac{3x}{x+y}\)
a: \(\dfrac{2\left(x-y\right)}{3y-3x}=\dfrac{2\left(x-y\right)}{-3\left(x-y\right)}=\dfrac{-2}{3}\)
b: \(\dfrac{8-x^3}{x\left(x^2+2x+4\right)}=\dfrac{\left(2-x\right)\left(x^2+2x+4\right)}{x\left(x^2+2x+4\right)}=\dfrac{2-x}{x}\)
a)2(x-y)/(-3)(x-y)=-2/3
b)8-x^3=(2-x)(x^2+2x+4) => Vế phải =(2-x)/x=(x-2)/-x
c)y^2-x^2=(y+x)(y-x) bạn đổi dấu rồi rút gọn là được,cũng tương tự như trên ý
\(=\dfrac{\left(x+y\right)^2}{x}.\dfrac{x}{\left(x+y\right)^2}-\dfrac{\left(x+y\right)^2}{x}.\dfrac{x}{\left(x+y\right)\left(x-y\right)}-\dfrac{5x-3y}{y-x}\)
\(=1-\dfrac{x+y}{x-y}+\dfrac{5x-3y}{x-y}\)
\(=\dfrac{x-y-x-y+5x-3y}{x-y}=\dfrac{5x-5y}{x-y}=5\)
1A,B,D
2 M=2
3 \(=\dfrac{3}{4x}\)
4 \(=\dfrac{4\left(x+y\right)}{x-y}=\dfrac{4x+4y}{x-y}\)
5 K rút gọn đc
6 \(=\dfrac{4\left(x-1\right)+2\left(x-1\right)}{6\left(x-1\right)}=\dfrac{6\left(x-1\right)}{6\left(x-1\right)}=1\)
\(=\left[\dfrac{\left(3x+y\right)\left(x+3y\right)+\left(3x-y\right)\left(x-3y\right)}{x\left(x-3y\right)\left(x+3y\right)}\right].\dfrac{\left(x-3y\right)\left(x+3y\right)}{x^2+y^2}\)
\(=\dfrac{\left(3x+y\right)\left(x+3y\right)+\left(3x-y\right)\left(x-3y\right)}{x.\left(x^2+y^2\right)}\)
\(=\dfrac{3x^2+3xy+xy+3y^2+3x^2-3xy-xy+3y^2}{x\left(x^2+y^2\right)}\)
\(=\dfrac{6x^2+6y^2}{x\left(x^2+y^2\right)}=\dfrac{6\left(x^2+y^2\right)}{x\left(x^2+y^2\right)}=\dfrac{6}{x}\)
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