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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\(x,y,z,a,b,c\ne0\)\(\left\{{}\begin{matrix}\dfrac{a}{x}=A\\\dfrac{b}{y}=B\\\dfrac{c}{z}=C\end{matrix}\right.\) \(\Rightarrow A,B,C\ne0\)
\(\left\{{}\begin{matrix}A+B+C=2\\\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}A^2+B^2+C^2+2\left(AB+BC+AC\right)=4\\\dfrac{ABC}{A}+\dfrac{ABC}{B}+\dfrac{ABC}{C}=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}AB+BC+AC=0\\A^2+B^2+C^2=4\end{matrix}\right.\)
\(\left(\dfrac{a}{x}\right)^2+\left(\dfrac{b}{y}\right)^2+\left(\dfrac{c}{z}\right)^2=4\)
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
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
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...
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:
\(A=\left(\dfrac{2}{3}\right)^3+\left(\dfrac{1}{3}\right)^3=\dfrac{8}{27}+\dfrac{1}{27}=\dfrac{9}{27}=\dfrac{1}{3}\)
Lời giải:
Ta có:
\(x^3+y^3+z^3=3xyz\Leftrightarrow x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow (x+y+z)(x^2+y^2+z^2-xy-yz-xz)=0\)
Vì \(x+y+z\neq 0\Rightarrow x^2+y^2+z^2-xy-yz-xz=0\)
\(\Leftrightarrow 2(x^2+y^2+z^2-xy-yz-xz)=0\)
\(\Leftrightarrow (x-y)^2+(y-z)^2+(z-x)^2=0\)
Ta thấy \((x-y)^2; (y-z)^2;(z-x)^2\geq 0\)
\(\Rightarrow (x-y)^2+(y-z)^2+(z-x)^2\geq 0\). Dấu bằng xảy ra khi
\((x-y)^2=(y-z)^2=(z-x)^2=0\Leftrightarrow x=y=z\)
Khi đó:
\(P=(1+\frac{x}{y})(1+\frac{y}{z})(1+\frac{z}{x})=(1+1)(1+1)(1+1)=8\)
vi a/x + b/y + c/z =0 suy ra ayz/xyz + bxz/xyz + cxy/xyz =0 suy ra ayz+bxz+cxy /xyz =0 suy ra ayz + bxz + cxy =0
vi x/a + y/b =z/c =0 suy ra (x/a + y/b + z/c )^2 =0 suy ra x^2/a^2 +y^2/b^2 + z^2/c^2 + 2(xy/ab + xz/ac + yz/bc) =0
suy ra x^2/a^2 + y^2/b^2 + z^2/c^2 + 2(cxy+ bxz +ayz /abc) =0
suy ra x^2/a^2 + y^2/b^2 + z^2/c^2 =0
suy ra x^2/a^2 + y^2/b^2 + z^2/c^2 +2011 = 2011
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: \(A=\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}= \frac{x+y}{z}+1+\frac{x+z}{y}+1+ \frac{y+z}{x}+1 -3 \)
\(=\frac{x+y+z}{z}+\frac{x+y+z}{y}+\frac{x+y+z}{x}-3=(x+y+z)( \frac{1}{z}+\frac{1}{x}+\frac{1}{y})-3=0-3=-3 \)