CMR
a)(3x-5)(3x+5)=9x^2-25
b)x^3-y^3=(x-y)(x^2+xy+y^2)
c)x^2+y^2=(x+y)^2-2xy
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a: Ta có: \(\left(3x-5\right)\left(3x+5\right)\)
\(=9x^2+15x-15x-25\)
\(=9x^2-25\)
b: Ta có: \(\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=x^3+x^2y+xy^2-x^2y-xy^2-y^3\)
\(=x^3-y^3\)
c: Ta có: \(\left(x+y\right)^2-2xy\)
\(=x^2+2xy+y^2-2xy\)
\(=x^2+y^2\)
1)2xy+3z+6y+xz
= x(2y + z) + 3(z + 2y)
= (x + 3)(2y + z)
2)x^4-9x^3+x^2-9x
= x^2(x^2 + 1) - 9x(x^2 + 1)
= (x^2 + 1)(x^2 - 9x)
= x(x - 9)(x^2 + 1)
3)x^2-xy+x-y
= x(x - y) + (x - y)
= (x + 1)(x - y)
4)xz+yz-5(x+y)
= z(x + y) - 5(x + y)
= (z - 5)(x + y)
5)3x^2-3xy-5x+5y
= 3x(x - y) - 5(x - y)
= (3x - 5)(x - y)
6)x^2+4x-y^2+4y
= (x - y)(x + y) + 4(x + y)
= (x - y + 4)(x + y)
a: \(F=-\left(2x-y\right)^3-x\left(2x-y\right)^2-y^3\)
\(=-\left(2x-y\right)^2\cdot\left[2x-y+x\right]-y^3\)
\(=-\left(2x-y\right)^2\cdot\left(3x-y\right)-y^3\)
\(=\left(-4x^2+4xy-y^2\right)\left(3x-y\right)-y^3\)
\(=-12x^3+4x^2y+12x^2y-4xy^2-3xy^2+y^3-y^3\)
\(=-12x^3+16x^2y-7xy^2\)
\(\left(x-2\right)^2+y^2=0\)
mà \(\left(x-2\right)^2+y^2>=0\forall x,y\)
nên dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-2=0\\y=0\end{matrix}\right.\)
=>x=2 và y=0
Thay x=2 và y=0 vào F, ta được:
\(F=-12\cdot2^3+16\cdot2^2\cdot0-7\cdot2\cdot0^2\)
\(=-12\cdot2^3\)
\(=-12\cdot8=-96\)
b: \(G=\left(x+y\right)\left(x^2-xy+y^2\right)+3\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=x^3+y^3+3\left(2x-y\right)\left[\left(2x\right)^2+2x\cdot y+y^2\right]\)
\(=x^3+y^3+3\left(8x^3-y^3\right)\)
\(=x^3+y^3+24x^3-3y^3\)
\(=25x^3-2y^3\)
Ta có: \(\left\{{}\begin{matrix}x+y=2\\y=-3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-3\\x=2-y=2-\left(-3\right)=2+3=5\end{matrix}\right.\)
Thay x=5 và y=-3 vào G, ta được:
\(G=25\cdot5^3-2\cdot\left(-3\right)^3\)
\(=25\cdot125-2\cdot\left(-27\right)\)
\(=3125+54=3179\)
c: \(H=\left(x+3y\right)\left(x^2-3xy+9y^2\right)+\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)
\(=\left(x+3y\right)\left[x^2-x\cdot3y+\left(3y\right)^2\right]+\left(3x-y\right)\left[\left(3x\right)^2+3x\cdot y+y^2\right]\)
\(=x^3+27y^3+27x^3-y^3\)
\(=28x^3-26y^3\)
Ta có: \(\left\{{}\begin{matrix}3x-y=5\\x=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2\\y=3x-5=3\cdot2-5=1\end{matrix}\right.\)
Thay x=2 và y=1 vào H, ta được:
\(H=28\cdot2^3-26\cdot1^3\)
\(=28\cdot8-26\)
=198
a: \(x^2+x-2x-2\)
\(=x\left(x+1\right)-2\left(x+1\right)\)
\(=\left(x+1\right)\left(x-2\right)=\left(-1+1\right)\left(-1-2\right)=0\)
b: \(3x^2-2x+9x-6\)
\(=x\left(3x-2\right)+3\left(3x-2\right)\)
\(=\left(3x-2\right)\left(x+3\right)=\left(3\cdot7-2\right)\left(7+3\right)\)
\(=19\cdot10=190\)
c: \(2x^2-3xy-xy^2\)
\(=x\left(2x-3y-y^2\right)\)
\(=2\left(2\cdot2-3\cdot3-9\right)\)
\(=2\cdot\left(4-18\right)=-28\)
ĐKXĐ: \(\left\{{}\begin{matrix}3x\ne-y\\3x\ne y\end{matrix}\right.\)
a. \(\dfrac{x}{3x+y}+\dfrac{x}{3x-y}-\dfrac{2xy}{y^2-9x^2}\)
\(=\dfrac{x.\left(3x-y\right)}{\left(3x+y\right).\left(3x-y\right)}+\dfrac{x.\left(3x+y\right)}{\left(3x+y\right).\left(3x-y\right)}+\dfrac{2xy}{9x^2-y^2}\)
\(=\dfrac{x.\left(3x+y+3x-y\right)+2xy}{\left(3x-y\right).\left(3x+y\right)}\)
\(=\dfrac{6x^2+2xy}{\left(3x-y\right).\left(3x+y\right)}\)
\(=\dfrac{2x\left(3x+y\right)}{\left(3x+y\right).\left(3x-y\right)}\)
\(=\dfrac{2x}{3x-y}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x\ne0\\x\ne-5\end{matrix}\right.\)
b. \(\dfrac{4x+5}{x^2+5x}-\dfrac{3}{x+5}\)
\(=\dfrac{4x+5}{x.\left(x+5\right)}-\dfrac{3x}{x.\left(x+5\right)}\)
\(=\dfrac{x+5}{x.\left(x+5\right)}\)
\(=\dfrac{1}{x}\)
a) \(\left(3x-5\right)\left(3x+5\right)=9x^2-25\Leftrightarrow9x^2+15x-15x-25=9x^2-25\Leftrightarrow9x^2-25=9x^2-25\)(đúng)
b) \(x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)\Leftrightarrow x^3-y^3=x^3+x^2y+xy^2-x^2y-xy^2-y^3\Leftrightarrow x^3-y^3=x^3-y^3\)(đúng)
c) \(x^2+y^2=\left(x+y\right)^2-2xy\Leftrightarrow x^2+y^2=x^2+y^2+2xy-2xy\Leftrightarrow x^2+y^2=x^2+y^2\)(đúng)
a: \(\left(3x-5\right)\left(3x+5\right)\)
\(=9x^2+15x-15x-25\)
\(=9x^2-25\)
b: \(\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=x^3+x^2y+xy^2-x^2y-xy^2-y^3\)
\(=x^3-y^3\)
c: \(\left(x+y\right)^2-2xy\)
\(=x^2+2xy+y^2-2xy\)
\(=x^2+y^2\)