1.a, cho a,b,c và x,y,z là các số khác 0, thỏa mãn đk a+b+c=0, x+y+z=0,\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=0\). chứng minh rằng:
\(a^2x+b^2y+c^2z=0\)
b, cho a,b,c là các hằng số và a,b,c≠-1. chứng minh rằng nếu x=by+cz, y=ax+cz, z=ax+by, x+y+z≠0 thì\(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=2\)
2. giả sử \(a_1,b_1,c_1,a_2,b_2,c_2\) là các số khác 0 thỏa mãn các đk: \(\frac{a_1}{a_2}+\frac{b_1}{b_2}+\frac{c_1}{c_2}=0\) và \(\frac{a_2}{a_1}+\frac{b_2}{b_1}+\frac{c_2}{c_1}=1\)
cmr \(\frac{a\frac{2}{2}}{a\frac{2}{1}}+\frac{b\frac{2}{2}}{b\frac{2}{1}}+\frac{c\frac{2}{2}}{c\frac{2}{1}}=1\)
3. a, biết x,y,z khác 0 và \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\). tính gt bt
M=\(\frac{yz}{x^2}+\frac{xz}{y^2}+\frac{xy}{z^2}\)
b, biết x,y,z khác 0 và x+y+z=\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\). cmr
y(\(x^2-yz\))\(\left(1-xz\right)=x\left(1-yz\right)\left(y^2-xz\right)\)
4. cho x,y,z khác 0 và \(\frac{y^2+z^2-x^2}{2yz}+\frac{z^2+x^2-y^2}{2xz}+\frac{x^2+y^2-z^2}{2xy}=1\)
chứng minh rằng trong 3 phân thức đã cho có 1 phân thức bằng -1 và hai phân thức còn lại đều bằng 1
Bài 1:
a) Từ đkđb:
$x+y+z=0\Rightarrow x+y=-z; y+z=-x; z+x=-y$
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=0\Rightarrow xbc+yac+zab=0$
$a+b+c=0\Rightarrow a=-(b+c)\Rightarrow a^2=(b+c)^2$
$\Rightarrow a^2x=(b+c)^2x$.
Tương tự: $b^2y=(a+c)^2y; c^2z=(a+b)^2z$
Do đó:
$a^2x+b^2y+c^2z=(b+c)^2x+(a+c)^2y+(a+b)^2z=a^2(y+z)+b^2(z+x)+c^2(x+y)+2(xbc+yac+zab)$
$=a^2(-x)+b^2(-y)+c^2(-z)+2.0=-(a^2x+b^2y+c^2z)$
$\Rightarrow 2(a^2x+b^2y+c^2z=0$
$\Rightarrow a^2x+b^2y+c^2z=0$ (đpcm)
b)
\(\left\{\begin{matrix} x=by+cz\\ y=ax+cz\\ z=ax+by\end{matrix}\right.\Rightarrow \frac{x+y+z}{2}=ax+by+cz\)
\(\Rightarrow \left\{\begin{matrix} ax=\frac{x+y+z}{2}-x=\frac{y+z-x}{2}\\ by=\frac{x+y+z}{2}-y=\frac{x+z-y}{2}\\ cz=\frac{x+y+z}{2}-z=\frac{x+y-z}{2}\end{matrix}\right.\) \(\Rightarrow \left\{\begin{matrix} a=\frac{y+z-x}{2x}\\ b=\frac{x+z-y}{2y}\\ c=\frac{x+y-z}{2z}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} a+1=\frac{y+z+x}{2x}\\ b+1=\frac{x+z+y}{2y}\\ c+1=\frac{x+y+z}{2z}\end{matrix}\right.\)
\(\Rightarrow \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=\frac{2x}{x+y+z}+\frac{2y}{x+y+z}+\frac{2z}{x+y+z}=2\) (đpcm)
Bài 2:
Đặt $\frac{a_2}{a_1}=x; \frac{b_2}{b_1}=y; \frac{c_2}{c_1}=z$
Khi đó bài toán trở thành: Cho $x,y,z\neq 0$ thỏa mãn \(\left\{\begin{matrix} \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\\ x+y+z=1\end{matrix}\right.\)
CMR: $x^2+y^2+z^2=1$
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Thật vậy:
Ta có: \(\left\{\begin{matrix} \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\\ x+y+z=1\end{matrix}\right.\Rightarrow \left\{\begin{matrix} xy+yz+xz=0\\ x+y+z=1\end{matrix}\right.\)
Khi đó: $x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)=1^2-2.0=1$ (đpcm)
Vậy........
Bài 3:
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow xy+yz+xz=0\)
\(\Rightarrow yz+xz=-xy\)
Khi đó:
\(M=\frac{(yz)^3+(xz)^3+(xy)^3}{x^2y^2z^2}=\frac{(yz+xz)^3-3yz.xz(yz+xz)+(xy)^3}{x^2y^2z^2}\)
\(=\frac{(-xy)^3-3yz.xz(-xy)+(xy)^3}{x^2y^2z^2}=\frac{3x^2y^2z^2}{x^2y^2z^2}=3\)
Bài 4:
ĐKĐB \(\Leftrightarrow (\frac{y^2+z^2-x^2}{2yz}+1)+(\frac{z^2+x^2-y^2}{2xz}-1)+(\frac{x^2+y^2-z^2}{2xy}-1)=0\)
\(\Leftrightarrow \frac{(y+z)^2-x^2}{2yz}+\frac{(z-x)^2-y^2}{2xz}+\frac{(x-y)^2-z^2}{2xy}=0\)
\(\Leftrightarrow \frac{(y+z-x)(y+z+x)}{2yz}+\frac{(z-x-y)(z-x+y)}{2xz}+\frac{(x-y-z)(x-y+z)}{2xy}=0(*)\)
Đặt \(\left\{\begin{matrix} y+z-x=a\\ x+z-y=b\\ x+y-z=c\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x+y+z=a+b+c\\ z=\frac{a+b}{2}\\ x=\frac{b+c}{2}\\ y=\frac{a+c}{2}\end{matrix}\right.\)
Khi đó:
\((*)\Leftrightarrow \frac{2a(a+b+c)}{(a+b)(a+c)}+\frac{-2ac}{(a+b)(b+c)}+\frac{-2ab}{(b+c)(a+c)}=0\)
\(\Rightarrow 2a(a+b+c)(b+c)-2ac(a+c)-ab(a+b)=0\)
\(\Rightarrow abc=0\)
Nếu $a=0$ thì: \(\left\{\begin{matrix} \frac{y^2+z^2-x^2}{2yz}+1=\frac{2a(a+b+c)}{(a+b)(a+c)}=0\\ \frac{z^2+x^2-y^2}{2xz}-1=\frac{-2ac}{(a+b)(b+c)}=0\\ \frac{x^2+y^2-z^2}{2xy}-1=\frac{-2ab}{(b+c)(a+c)}=0\end{matrix}\right.\Rightarrow \left\{\begin{matrix} \frac{y^2+z^2-x^2}{2yz}=-1\\ \frac{z^2+x^2-y^2}{2xz}=1\\ \frac{x^2+y^2-z^2}{2xy}=1\end{matrix}\right.\) (đpcm)
Tương tự với $b=0; c=0$
Bài 1:
a) Từ đkđb:
$x+y+z=0\Rightarrow x+y=-z; y+z=-x; z+x=-y$
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=0\Rightarrow xbc+yac+zab=0$
$a+b+c=0\Rightarrow a=-(b+c)\Rightarrow a^2=(b+c)^2$
$\Rightarrow a^2x=(b+c)^2x$.
Tương tự: $b^2y=(a+c)^2y; c^2z=(a+b)^2z$
Do đó:
$a^2x+b^2y+c^2z=(b+c)^2x+(a+c)^2y+(a+b)^2z=a^2(y+z)+b^2(z+x)+c^2(x+y)+2(xbc+yac+zab)$
$=a^2(-x)+b^2(-y)+c^2(-z)+2.0=-(a^2x+b^2y+c^2z)$
$\Rightarrow 2(a^2x+b^2y+c^2z=0$
$\Rightarrow a^2x+b^2y+c^2z=0$ (đpcm)
b)
\(\left\{\begin{matrix} x=by+cz\\ y=ax+cz\\ z=ax+by\end{matrix}\right.\Rightarrow \frac{x+y+z}{2}=ax+by+cz\)
\(\Rightarrow \left\{\begin{matrix} ax=\frac{x+y+z}{2}-x=\frac{y+z-x}{2}\\ by=\frac{x+y+z}{2}-y=\frac{x+z-y}{2}\\ cz=\frac{x+y+z}{2}-z=\frac{x+y-z}{2}\end{matrix}\right.\) \(\Rightarrow \left\{\begin{matrix} a=\frac{y+z-x}{2x}\\ b=\frac{x+z-y}{2y}\\ c=\frac{x+y-z}{2z}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} a+1=\frac{y+z+x}{2x}\\ b+1=\frac{x+z+y}{2y}\\ c+1=\frac{x+y+z}{2z}\end{matrix}\right.\)
\(\Rightarrow \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=\frac{2x}{x+y+z}+\frac{2y}{x+y+z}+\frac{2z}{x+y+z}=2\) (đpcm)