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1)
\(\Leftrightarrow\left(x^2-2+\dfrac{1}{x^2}\right)+\left(y^2-2+\dfrac{1}{y^2}\right)+z^2=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2+z^2=0\)
\(\left\{{}\begin{matrix}x-\dfrac{1}{x}=0\Rightarrow\left|x\right|=1\\y-\dfrac{1}{y}=0\Rightarrow\left|y\right|=1\\z=0\end{matrix}\right.\)
dk\(x,y,z,a,b,c\ne0\)\(\left\{{}\begin{matrix}\dfrac{a}{x}=A\\\dfrac{b}{y}=B\\\dfrac{c}{z}=C\end{matrix}\right.\) \(\Rightarrow A,B,C\ne0\)
\(\left\{{}\begin{matrix}A+B+C=2\\\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}A^2+B^2+C^2+2\left(AB+BC+AC\right)=4\\\dfrac{ABC}{A}+\dfrac{ABC}{B}+\dfrac{ABC}{C}=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}AB+BC+AC=0\\A^2+B^2+C^2=4\end{matrix}\right.\)
\(\left(\dfrac{a}{x}\right)^2+\left(\dfrac{b}{y}\right)^2+\left(\dfrac{c}{z}\right)^2=4\)
\(a^2-2a+6b+b^2=-10\\ \Leftrightarrow a^2-2a+1+b^2+6b+9=0\\ \Leftrightarrow\left(a-1\right)^2+\left(b+3\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}a=1\\b=-3\end{matrix}\right.\)
Vậy \(\left(a;b\right)=\left(1;-3\right)\)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\\ \Leftrightarrow xy+yz+zx=0\\ \Rightarrow\left\{{}\begin{matrix}xy+yz=-zx\\xy+zx=-yz\\yz+zx=-xy\end{matrix}\right.\)
Ta có:
\(A=\dfrac{xz+yz}{z^2}+\dfrac{xy+yz}{y^2}+\dfrac{xy+xz}{x^2}\\ =\dfrac{-xy}{z^2}+\dfrac{-xz}{y^2}+\dfrac{-yz}{x^2}\\ =-xyz\cdot\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)\\ =-xyz\cdot\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}-\dfrac{2}{xy}-\dfrac{2}{yz}-\dfrac{2}{xz}\right)\\ =0\)
b)\(N=\dfrac{yz}{x^2}+\dfrac{zx}{y^2}+\dfrac{xy}{z^2}\)
\(N=\dfrac{xyz}{x^3}+\dfrac{xyz}{y^3}+\dfrac{xyz}{z^3}\)
\(N=xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)\)
Ta cm đẳng thức sau:\(x^3+y^3+z^3=3xyz\Leftrightarrow x+y+z=0\)
ĐT\(\Leftrightarrow x^3+y^3-3xyz=-z^3\)
\(\Leftrightarrow\left(x+y\right)\left(x^2-xy+y^2\right)-3xy=-z^3\)
\(\Leftrightarrow-zx^2+xyz-zy^2-3xyz=-z^3\)
\(\Leftrightarrow x^2+2xy+y^2=z^2\)
\(\Leftrightarrow\left(x+y\right)^2=z^2\)
\(\Leftrightarrow\left(-z\right)^2=z^2\)(luôn đúng)
Áp dụng\(\Rightarrow N=xyz.\dfrac{3}{xyz}=3\)
a, (M-1)/70-71=m
m=(71^9+71^8....71+1)
71m=71^10+...71^2+71
70m=71^10-1
(M-1)/70=71^10+70
M-1=70(71^10+70)
M=70(71^10+70)-1
Bài 1:
\(A=\left(x-y\right)\left(x^2+xy+y^2\right)+2y^3\)
\(A=x^3-y^3+2y^3\)
\(A=x^3+y^3\)
Thay \(x=\dfrac{2}{3},y=\dfrac{1}{3}\) vào A, ta có:
\(A=\left(\dfrac{2}{3}\right)^3+\left(\dfrac{1}{3}\right)^3=\dfrac{8}{27}+\dfrac{1}{27}=\dfrac{9}{27}=\dfrac{1}{3}\)
Sửa đề :
\(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=0\)
Bài làm
đề có sai chỗ nào ko bn,mk thấy chỗ giả thiết sai sai thì phải,bn kt lại giúp mk
1, Ta có: \(x+y=9\Rightarrow\left(x+y\right)^2=81\)
\(\Rightarrow x^2+2xy+y^2=81\)
\(\Rightarrow x^2+y^2=45\)
\(\Rightarrow x^2+y^2-2xy=9\)
\(\Rightarrow\left(x-y\right)^2=9\Rightarrow\left[{}\begin{matrix}x-y=3\\x-y=-3\end{matrix}\right.\)
\(A=x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(\Rightarrow\left[{}\begin{matrix}A=3.63=189\\A=-3.63=-189\end{matrix}\right.\)
Vậy...
1) Đặt \(B=x^2+y^2+z^2\)
\(C=\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=2\left(x^2+y^2+z^2\right)-2\left(xy+yz+xz\right)\)
Ta có: \(x+y+z=0\Rightarrow\left(x+y+z\right)^2=0\)
\(\Leftrightarrow-2\left(xy+yz+xz\right)=x^2+y^2+z^2\)
Suy ra: \(C=2\left(x^2+y^2+z^2\right)-2\left(xy+yz+xz\right)=2\left(x^2+y^2+z^2\right)+x^2+y^2+z^2=3\left(x^2+y^2+z^2\right)\)
\(\Rightarrow A=\dfrac{B}{C}=\dfrac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)}=\dfrac{1}{3}\)
2) \(x^2-2y^2=xy\Leftrightarrow x^2-xy-2y^2=0\)
\(\Leftrightarrow x^2+xy-2xy-2y^2=0\)
\(\Leftrightarrow x\left(x+y\right)-2y\left(x+y\right)=0\)
\(\Leftrightarrow\left(x-2y\right)\left(x+y\right)=0\)
Do \(x+y\ne0\) nên \(x-2y=0\Leftrightarrow x=2y\)
Do đó: \(A=\dfrac{2y-y}{2y+y}=\dfrac{y}{3y}=\dfrac{1}{3}\)
Bài 1
\(a^2-2a+6b+b^2=-10\)
<=>\(a^2-2a+1+b^2+6b+9=0\)
<=>\((a-1)^2+(b+3)^2=0\)
Ta lại có: \((a-1)^2\ge0 \)
\((b+3)^2\ge0\)
=> \((a-1)^2+(b+3)^2\ge0\)
Mà\((a-1)^2+(b+3)^2=0\)
=>(a-1)2=0=>a=1
(b+3)2=0=>b=-3
Vậy a=1,b=-3
Bài 2
Ta có: \(A=\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}= \frac{x+y}{z}+1+\frac{x+z}{y}+1+ \frac{y+z}{x}+1 -3 \)
\(=\frac{x+y+z}{z}+\frac{x+y+z}{y}+\frac{x+y+z}{x}-3=(x+y+z)( \frac{1}{z}+\frac{1}{x}+\frac{1}{y})-3=0-3=-3 \)