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1/
\(\dfrac{\left(x-y\right)^3-3xy\left(x+y\right)+y^3}{x-6y}\)
\(=\dfrac{x^3-3x^2y+3xy^2-y^3-3x^2y-3xy^2+y^3}{x-6y}\)
\(=\dfrac{x^3-6x^2y}{x-6y}\)
\(=\dfrac{x^2\left(x-6y\right)}{x-6y}\)
\(=x^2\)
\(2\)/
\(\dfrac{x^2+y^2+z^2-2xy+2xz-2yz}{x^2-2xy+y^2-z^2}\)
\(=\dfrac{\left(x-y+z^{ }\right)^2}{\left(x-y\right)^2-z^2}\)
\(=\dfrac{\left(x-y+z\right)^2}{\left(x-y-z\right)\left(x-y+z\right)}\)
\(=\dfrac{x-y+z}{x-y-z}\)
3/
\(\dfrac{\left(n+1\right)!}{n!\left(n+2\right)}\)
\(=\dfrac{n!\left(n+1\right)}{n!\left(n+2\right)}\)
\(=\dfrac{n+1}{n+2}\)
4/
\(\dfrac{n!}{\left(n+1\right)!-n!}\)
\(=\dfrac{n!}{n!\left(n+1\right)-n!}\)
\(=\dfrac{n!}{n!\left[\left(n+1\right)-1\right]}\)
\(=\dfrac{n!}{n!.n}\)
\(=\dfrac{1}{n}\)
5/
\(\dfrac{\left(n+1\right)!-\left(n+2\right)!}{\left(n+1\right)!+\left(n+2\right)!}\)
\(=\dfrac{\left(n+1\right)!-\left(n+1\right)!\left(n+2\right)}{\left(n+1\right)!+\left(n+1\right)!\left(n+2\right)}\)
\(=\dfrac{\left(n+1\right)!\left(-n-1\right)}{\left(n+1\right)!\left(n+3\right)}\)
\(=\dfrac{-n-1}{n+3}\)
1: \(=\dfrac{\left(x^2+2xy+y^2\right)-1}{\left(x^2+2x+1\right)-y^2}\)
\(=\dfrac{\left(x+y+1\right)\left(x+y-1\right)}{\left(x+1-y\right)\left(x+1+y\right)}=\dfrac{x+y-1}{x-y+1}\)
2: \(=\dfrac{\left(x^2-y^2\right)\left(x^2+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}\)
\(=\dfrac{\left(x-y\right)\left(x^2+y^2\right)}{x^2-xy+y^2}\)
3: \(=\dfrac{\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz}{2x^2+2y^2+2z^2-2xy-2yz-2xz}\)
\(=\dfrac{\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)}{2\left(x^2+y^2+z^2-xy-yz-xz\right)}\)
\(=\dfrac{x+y+z}{2}\)
\(1.\)
\(a.\)
\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^3+x^2+3x+3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^2\left(x+1\right)+3\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{\left(x^2+3\right)\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=x-1\)
\(b.\)
\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)
\(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2\left(x^2-y^2\right)}-\dfrac{\left(x-y\right)^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2}{2\left(x^2-y^2\right)}-\dfrac{x^2-2xy+y^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4xy+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4y\left(x+y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2y}{\left(x-y\right)}\)
Tương tự các câu còn lại
a: \(A=2x^2-2xy-y^2+2xy=2x^2-y^2\)
\(=2\cdot\dfrac{4}{9}-\dfrac{1}{9}=\dfrac{7}{9}\)
b: \(B=5x^2-20xy-4y^2+20xy=5x^2-4y^2\)
\(=5\cdot\dfrac{1}{25}-4\cdot\dfrac{1}{4}\)
=1/5-1=-4/5
c \(C=x^3+6x^2+12x+8=\left(x+2\right)^3=\left(-9\right)^3=-729\)
d: \(D=20x^3-10x^2+5x-20x^2+10x+4\)
\(=20x^3-30x^2+15x+4\)
\(=20\cdot5^3-30\cdot5^2+15\cdot2+4=1784\)
a) \(\dfrac{x^2-4}{2x^2-4x}\)
\(=\dfrac{\left(x-2\right)\left(x+2\right)}{2x\left(x-2\right)}\)
\(=\dfrac{x+2}{2x}\)
b) \(\dfrac{2x-x^2}{x^2-4x+4}\)
\(=\dfrac{x\left(2-x\right)}{\left(x-2\right)^2}\)
\(=\dfrac{x\left(2-x\right)}{\left(2-x\right)^2}\)
\(=\dfrac{x}{2-x}\)
c) \(\dfrac{x^2-xy-x+y}{x^2+xy-x-y}\)
\(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}\)
\(=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}\)
\(=\dfrac{x-y}{x+y}\)
d) \(\dfrac{5x^2+10x+5}{x+x^2}\)
\(\dfrac{5\left(x^2+2x+1\right)}{x\left(1+x\right)}\)
\(=\dfrac{5\left(x+1\right)^2}{x\left(x+1\right)}\)
\(=\dfrac{5\left(x+1\right)}{x}\)
e) \(\dfrac{3x^2+3x}{\left(x+1\right)\left(2x+6\right)}\)
\(=\dfrac{3x\left(x+1\right)}{\left(x+1\right).2\left(x+3\right)}\)
\(=\dfrac{3x}{2\left(x+3\right)}\)
f) \(\dfrac{\left(2-3x\right)\left(x+1\right)}{x^2+2x+1}\)
\(=\dfrac{\left(2-3x\right)\left(x+1\right)}{\left(x+1\right)^2}\)
\(=\dfrac{2-3x}{x+1}\)
a) \(5x-10x^2\) = \(5x\left(1-2x\right)\)
b) Mạn phép sửa đề:
\(\dfrac{1}{2}x\left(x^2-4\right)+4\left(x+2\right)\) = \(\left(x+2\right)\left[\dfrac{1}{2}x\left(x-2\right)+4\right]\)
= \(\left(x+2\right)\left(\dfrac{1}{2}x^2-x+4\right)\)
c) \(x^4-y^6=\left(x^2-y^3\right)\left(x^2+y^3\right)\)
e) \(x^3-4x^2+4x-1=x^3-x^2-3x^2+3x+x-1\)
= \(x^2\left(x-1\right)-3x\left(x-1\right)+\left(x-1\right)\)
= \(\left(x-1\right)\left(x^2-3x+1\right)\)
g) \(x^4+6x^3-12x^2-8x\)
= \(x\left(x^3-2x^2+8x^2-16x+4x-8\right)\)
= \(x\left[x^2\left(x-2\right)+8x\left(x-2\right)+4\left(x-2\right)\right]\)
= \(x\left(x-2\right)\left(x^2+8x+4\right)\)
h) \(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\) (*)
Đặt \(x^2+4x+8=a\) => (*) trở thành:
\(a^2+3ax+2x^2\) = \(a^2+ãx+2ax+x^2\)
= \(a\left(a+x\right)+2x\left(a+x\right)\)
= \(\left(a+x\right)\left(a+2x\right)\) (1)
Thay \(a=x^2+4x+8\) vào (1) ta được:
\(\left(x^2+5x+8\right)\left(x^2+6x+8\right)\)
=\(\left(x^2+5x+8\right)\left(x^2+2x+4x+8\right)\)
= \(\left(x^2+5x+8\right)\left[x\left(x+2\right)+4\left(x+2\right)\right]\)
= \(\left(x+2\right)\left(x+4\right)\left(x^2+5x+8\right)\)
P/s: Còn câu f đang suy nghĩ!
\(1.\text{ }\text{ }\text{ }\dfrac{\left(x^2+2\right)^2-4x^2}{y\left(x^2+2\right)-2xy-\left(x-1\right)^2-1}\\ =\dfrac{\left(x^2+2-2x\right)\left(x^2+2+2x\right)}{x^2y+2y-2xy-x^2+2x-1-1}\\ =\dfrac{\left(x^2+2-2x\right)\left(x^2+2+2x\right)}{\left(x^2y-x^2\right)-\left(2xy-2x\right)+\left(2y-2\right)}\\ =\dfrac{\left(x^2+2-2x\right)\left(x^2+2+2x\right)}{x^2\left(y-1\right)-2x\left(y-1\right)+2\left(y-1\right)}\\ =\dfrac{\left(x^2+2-2x\right)\left(x^2+2+2x\right)}{\left(x^2-2x+2\right)\left(y-1\right)}\\ =\dfrac{x^2+2x+2}{y-1}\)
\(2.\text{ }\text{ }\text{ }\text{ }\dfrac{x^2+5x+6}{x^2+3x+2}\\ =\dfrac{x^2+3x+2x+6}{x^2+2x+x+2}\\ =\dfrac{\left(x^2+3x\right)+\left(2x+6\right)}{\left(x^2+2x\right)+\left(x+2\right)}\\ =\dfrac{x\left(x+3\right)+2\left(x+3\right)}{x\left(x+2\right)+\left(x+2\right)}\\ =\dfrac{\left(x+2\right)\left(x+3\right)}{\left(x+2\right)\left(x+1\right)}\\ =\dfrac{x+3}{x+1}\)
\(3.\text{ }\text{ }\text{ }\dfrac{x^2+y^2-z^2-2zt+2xy-t^2}{x^2-y^2+z^2-2yt+2xz-t^2}\text{ ( Chữa đề ) }\\ =\dfrac{\left(x^2+2xy+y^2\right)-\left(z^2+2zt+t^2\right)}{\left(x^2+2xz+z^2\right)-\left(y^2+2yt+t^2\right)}\\ =\dfrac{\left(x+y\right)^2-\left(z+t\right)^2}{\left(x+z\right)^2-\left(y+t\right)^2}\\ =\dfrac{\left(x+y+z+t\right)\left(x+y-z-t\right)}{\left(x+z+y+t\right)\left(x+z-y-t\right)}\\ =\dfrac{x+y-z-t}{x+z-y-t}\)
\(4.\text{ }\text{ }\text{ }\dfrac{\left(n+1\right)!}{\left(n+1\right)!+\left(n+2\right)!}=\dfrac{\left(n+1\right)!}{\left(n+1\right)!\left(1+n+2\right)}=\dfrac{1}{n+3}\)
\(5.\text{ }\text{ }\text{ }\dfrac{x^2+5x+4}{x^2-1}\\ =\dfrac{x^2+x+4x+4}{\left(x+1\right)\left(x-1\right)}\\ =\dfrac{\left(x^2+x\right)+\left(4x+4\right)}{\left(x+1\right)\left(x-1\right)}\\ =\dfrac{x\left(x+1\right)+4\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\\ =\dfrac{\left(x+1\right)\left(x+4\right)}{\left(x+1\right)\left(x-1\right)}\\ =\dfrac{x+4}{x-1}\)
\(6.\text{ }\text{ }\text{ }\dfrac{x^2-3x}{2x^2-7x+3}\\ =\dfrac{x\left(x-3\right)}{2x^2-6x-x+3}\\ =\dfrac{x\left(x-3\right)}{\left(2x^2-6x\right)-\left(x-3\right)}\\ =\dfrac{x\left(x-3\right)}{2x\left(x-3\right)-\left(x-3\right)}\\ =\dfrac{x\left(x-3\right)}{\left(2x-1\right)\left(x-3\right)}\\ =\dfrac{x}{2x-1}\)