Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Sửa lại đề nha: x+y+z=0
a)
Xét x+y+z=0
(x+y+z)2=02
x2+y2+z2+2xy+2yz+2zx=0
=> x2+y2+z2=-2xy-2yz-2zx
Xét \(\dfrac{x^2+y^2+z^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
= \(\dfrac{x^2+y^2+z^2}{\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)}\)
=\(\dfrac{x^2+y^2+z^2}{x^2-2xy+y^2+y^2-2yz+z^2+z^2-2zx+x^2}\)
=\(\dfrac{x^2+y^2+z^2}{2x^2+2y^2+2z^2-2xy-2yz-2zx}\)(1)
Thay x2+y2+z2=-2xy-2yz-2zx vào (1)
=>\(\dfrac{x^2+y^2+z^2}{2x^2+2y^2+2z^2+x^2+y^2+z^2}\\=\dfrac{x^2+y^2+z^2}{3x^2+3y^2+3z^2}\\ =\dfrac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)}\\ =\dfrac{1}{3}\)
b)
Xét x+y+z=0 ba lần:
- Lần 1:x+y+z=0
<=> x+y=0-z
<=>(x+y)2=(0-z)2
<=>x2+2xy+y2=z2
<=>x2+y2-z2=-2xy(1)
-Lần 2: x+y+z=0
<=> y+z=0-x
<=>(y+z)2=(0-x)2
<=>y2+2yz+z2=x2
<=>y2+z2-x2=-2yz(2)
-Lần 3: x+y+z=0
<=>z+x=0-y
<=>(z+x)2=(0-y)2
<=>z2+2zx+x2=y2
<=> z2+x2-y2=-2zx(3)
Thay (1),(2),(3) vào Q, ta có:
=>\(\dfrac{\left(x^2+y^2-z^2\right)\left(y^2+z^2-x^2\right)\left(z^2+x^2-y^2\right)}{16xyz}=\dfrac{\left(-2xy\right)\left(-2yz\right)\left(-2zx\right)}{16xyz}\\=\dfrac{\left(-2yz\right)\left(-2zx\right)}{-8z}\\ =\dfrac{y\left(-2zx\right)}{4}\\ =\dfrac{-2xyz}{4}\\ =-\dfrac{xyz}{2}\)
\(=\dfrac{y^2+xy+x^2}{y^2}:\dfrac{x^2-y^2}{x^2}\cdot\dfrac{y^2}{x^2-y^2}\)
\(=\dfrac{x^2+xy+y^2}{y^2}\cdot\dfrac{x^2}{x^2-y^2}\cdot\dfrac{y^2}{x^2-y^2}\)
\(=\dfrac{x^2+xy+y^2}{\left(x^2-y^2\right)^2}\cdot x^2\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\\\sqrt{y}=b\end{matrix}\right.\), ta có:
\(A=\left[\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\times\dfrac{2}{a+b}+\dfrac{1}{a^2}+\dfrac{1}{b^2}\right]\)\(\times\dfrac{a^3+ab^2+a^2b+b^3}{ab^3+a^3b}\)
\(=\left(\dfrac{b+a}{ab}\times\dfrac{2}{a+b}+\dfrac{b^2+a^2}{a^2b^2}\right)\)\(\times\dfrac{a^2\left(a+b\right)+b^2\left(a+b\right)}{ab\left(a^2+b^2\right)}\)
\(=\dfrac{2ab+b^2+a^2}{a^2b^2}\times\dfrac{\left(a+b\right)\left(a^2+b^2\right)}{ab\left(b^2+a^2\right)}\)
\(=\dfrac{\left(a+b\right)^3}{a^3b^3}\)
\(=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^3}{\sqrt{\left(xy\right)^3}}\)
\(\left[\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{2}{x+y}.\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\right]:\dfrac{x^3+y^3}{x^2y^2}-\dfrac{x+y}{x^2-xy+y^2}\)
\(=\left[\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{2}{x+y}.\dfrac{x+y}{xy}\right].\dfrac{x^2y^2}{x^3+y^3}-\dfrac{x+y}{x^2-xy+y^2}\)
\(=\left[\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{2}{xy}\right].\dfrac{x^2y^2}{\left(x+y\right)\left(x^2-xy+y^2\right)}-\dfrac{x+y}{x^2-xy+y^2}\)
\(=\dfrac{y^2+x^2+2xy}{x^2y^2}.\dfrac{x^2y^2}{\left(x+y\right)\left(x^2-xy+y^2\right)}-\dfrac{x+y}{x^2-xy+y^2}\)
\(=\dfrac{\left(x+y\right)^2}{\left(x+y\right)\left(x^2-xy+y^2\right)}-\dfrac{x+y}{x^2-xy+y^2}\)
=\(=\dfrac{x+y}{x^2-xy+y^2}-\dfrac{x+y}{x^2-xy+y^2}=0\)
a: \(\left(\dfrac{1}{\left(2x-y\right)^2}+\dfrac{2}{\left(2x-y\right)\left(2x+y\right)}+\dfrac{1}{\left(2x+y\right)^2}\right)\cdot\dfrac{\left(2x+y\right)^2}{16x}\)
\(=\dfrac{4x^2+4xy+y^2+2\left(4x^2-y^2\right)+4x^2-4xy+y^2}{\left(2x-y\right)^2\cdot\left(2x+y\right)^2}\cdot\dfrac{\left(2x+y\right)^2}{16x}\)
\(=\dfrac{8x^2+2y^2+8x^2-2y^2}{\left(2x-y\right)^2}\cdot\dfrac{1}{16x}\)
\(=\dfrac{16x^2}{16x}\cdot\dfrac{1}{\left(2x-y\right)^2}=\dfrac{x}{\left(2x-y\right)^2}\)
b: \(\left(\dfrac{2}{x+2}-\dfrac{4}{x^2+4x+4}\right):\left(\dfrac{2}{x^2-4}+\dfrac{1}{2-x}\right)\)
\(=\dfrac{2x+4-4}{\left(x+2\right)^2}:\left(\dfrac{2}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}\right)\)
\(=\dfrac{2x}{\left(x+2\right)^2}:\dfrac{2-x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{2x}{\left(x+2\right)^2}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{-x}=\dfrac{-2\left(x-2\right)}{x+2}\)
câu này post hồi học lớp 8 = )) giờ tốt nghiệp c3 thì có người trả lời :'))
khbiet nên cười hay khóc đây
Lời giải:
a) Ta có:
\(Q=\left[\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{x+y}\left(\frac{1}{x}+\frac{1}{y}\right)\right].\frac{x^2y^2}{x^3+y^3}\)
\(=\left[\frac{x^2+y^2}{x^2y^2}+\frac{2}{x+y}.\frac{x+y}{xy}\right].\frac{x^2y^2}{x^3+y^3}\)
\(=\left[\frac{x^2+y^2}{x^2y^2}+\frac{2}{xy}\right].\frac{x^2y^2}{x^3+y^3}\)
\(=\frac{x^2+y^2}{x^2y^2}.\frac{x^2y^2}{x^3+y^3}+\frac{2x^2y^2}{xy(x^3+y^3)}\)
\(=\frac{x^2+y^2}{x^3+y^3}+\frac{2xy}{x^3+y^3}=\frac{x^2+y^2+2xy}{x^3+y^3}\)
\(=\frac{(x+y)^2}{x^3+y^3}=\frac{(x+y)^3}{(x+y)(x^2-xy+y^2)}=\frac{x+y}{x^2-xy+y^2}\)
b)
Khi \(x=1,y=2\Rightarrow Q=\frac{1+2}{1^2-1.2+2^2}=1\)
\(M=\left(\dfrac{x}{y}+1-5\right)^3=\left(\dfrac{x}{y}-4\right)^3\)
\(=\left(\dfrac{12}{2}-4\right)^3=\left(6-4\right)^3=2^3=8\)