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Ta có (a+b+c)*(x^2+y^2+z^2)=0
vì a+b+c=0 suy ra (a+b+c)*(x^2+y^2+z^2)=0
suy ra ax^2+by^2+cz^2=0
Ta có (a+b+c)*(x^2+y^2+z^2)=0
vì a+b+c=0 suy ra (a+b+c)*(x^2+y^2+z^2)=0
suy ra ax^2+by^2+cz^2=0
từ x + y + z = 0 suy ra x2 = ( y + z )2 , y2 = ( x + z )2 , z2 = ( x + y )2
Do đó : ax2 + by2 + cz2 = a ( y + z )2 + b ( x + z )2 + c ( x + y )2
= a ( y2 + 2yz + z2 ) + b ( x2 + 2xz + z2 ) + c ( x2 + 2xy + y2 )
= x2 ( b + c ) + y2 ( a + c ) + z2 ( a + b ) + 2 ( ayz + bxz + cxy ) ( 1 )
Thay b + c = -a, a + c = -b , a + b = -c do a + b + c = 0
Thay ayz + bxz + cxy = 0 do \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\) vào ( 1 ), ta được :
ax2 + by2 + cz2 = -ax2 - by2 - cz2
nên 2ax2 + 2by2 + 2cz2 = 0 \(\Rightarrow\)ax2 + by2 + cz2 = 0
Áp dụng : A = - A => A = 0
Từ \(a+b+c=0\Rightarrow\hept{\begin{cases}a=-\left(b+c\right)\\b=-\left(c+a\right)\\c=-\left(a+b\right)\end{cases}}\)
\(x+y+z=0\Rightarrow\hept{\begin{cases}x=-\left(y+z\right)\\y=-\left(x+z\right)\\z=-\left(x+y\right)\end{cases}\Rightarrow\hept{\begin{cases}x^2=\left(y+z\right)^2\\y^2=\left(x+z\right)^2\\z^2=\left(x+y\right)^2\end{cases}}}\)
Và \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=\frac{ayz+bxz+cxy}{xyz}=0\)\(\Rightarrow ayz+bxz+cxy=0\)
Ta có : \(x^2a+y^2b+x^2c=\)\(\left(y+z\right)^2a+\left(x+z\right)^2b+\left(x+y\right)^2c\)
= \(x^2\left(b+c\right)+y^2\left(c+a\right)+z^2\left(a+b\right)\)\(+2\left(ayz+bxz+cxy\right)\)
= \(-\left(x^2a+y^2b+z^2c\right)\) => \(x^2a+y^2b+x^2c=\) 0
Ta có: \(\hept{\begin{cases}a=-b-c\\x=-y-z\end{cases}}\)
\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\)
\(\Leftrightarrow\frac{\left(-b-c\right)}{\left(-y-z\right)}+\frac{b}{y}+\frac{c}{z}=0\)
\(\Leftrightarrow2byz+2cyz+bz^2+cy^2=0\)
Ta lại có:
\(ax^2+by^2+cz^2=\left(-b-c\right)\left(-y-z\right)^2+by^2+cz^2\)
\(=-2byz-2cyz-bz^2-cy^2=0\)
Ta có : \(x+y+z=0\)
\(\Rightarrow\hept{\begin{cases}x=-\left(y+z\right)\\y=-\left(z+x\right)\\z=-\left(x+y\right)\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x^2=\left(y+z\right)^2\\y^2=\left(z+x\right)^2\\z=\left(x+y\right)^2\end{cases}}\)
\(\Rightarrow ax^2+by^2+cz^2=a\left(y+z\right)^2+b\left(z+x\right)^2+c\left(x+y\right)^2\)
\(=ay^2+az^2+bz^2+bx^2+cx^2+cy^2+2\left(ayz+bzx+cxy\right)\)
\(=x^2\left(b+c\right)+y^2\left(c+a\right)+z^2\left(a+b\right)+2\left(ayz+bzx+cxy\right)\left(1\right)\)
Từ \(a+b+c=0\) \(\Rightarrow\hept{\begin{cases}b+c=-a\\c+a=-b\\a+b=-c\end{cases}}\)
Thay vào \(\left(1\right)\), ta được :
\(ax^2+by^2+cz^2=-ax^2-by^2-cz^2+2\left(ayz+bzx+cxy\right)\)
Ta có : \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\)\(\Rightarrow ayz+bzx+cxy=0\)
\(\Rightarrow ax^2+by^2+cz^2=-ax^2-by^2-cz^2\)
\(\Rightarrow2\left(ax^2+by^2+cz^2\right)=0\)
\(\Rightarrow ax^2+by^2+cz^2=0\left(đpcm\right)\)
Câu hỏi của Momozono Nanami - Toán lớp 8 - Học toán với OnlineMath
ta có x+y+z=0 =>x^2=(y+z)^2
y^2=(x+z)^2
z^2=(x+y)^2
do đó ax^2+by^2+cz^2
=a(y+z)^2+b(x+z)^2+c(x+y)^2
=a(y^2+2yz+z^2)+b(x^2+2xz+z^2)
+c(x^2+2xy+y^2)
=x^2(b+c)+y^2(a+c)+z^2(a+b)
+2(ayz+bxz+cxy) (1)
thay b+c=-a ,a+c=-b , a+b=-c do a+b+c=0
và ayz+bxz+cxy=0 do a/x+b/y+c/z=0 vào (1) ta được
ax^2+by^2+cz^2 = -(ax^2+by^2+cz^2)
=> ax^2+by^2+cz^2=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)
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........
\(\frac{\left(ax+by+cz\right)^2}{x^2+y^2+z^2}=a^2+b^2+c^2\)
\(\Leftrightarrow\left(ax+by+cz\right)^2=\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\)
\(\Leftrightarrow\left(ax+by+cz\right)^2=a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2\)
Mà \(\left(ax+by+cz\right)^2=a^2x^2+b^2y^2+c^2x^2+2abxy+2acxz+2bcyz\)
Nên \(a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2=2abxy+2acxz+2bcyz\)
\(\Leftrightarrow a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2-2abxy-2acxz-2bcyz=0\)
\(\Leftrightarrow\left(a^2y^2-2abxy+b^2x^2\right)+\left(a^2z^2-2acxz+c^2x^2\right)+\left(b^2z^2-2bcyz+c^2y^2\right)=0\)
\(\Leftrightarrow\left(ay-bx\right)^2+\left(az-cx\right)^2+\left(bz-cy\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}ay-bx=0\\az-cx=0\\bz-cy=0\end{cases}\Rightarrow\hept{\begin{cases}ay=bx\\az=cx\\bz=cy\end{cases}\Rightarrow}\hept{\begin{cases}\frac{a}{x}=\frac{b}{y}\\\frac{a}{x}=\frac{c}{z}\\\frac{b}{y}=\frac{c}{z}\end{cases}}}\)
\(\Rightarrow\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\) (đpcm)
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