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Ta có : \(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)
Suy ra : \(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=0\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-\frac{x^2}{a^2+b^2+c^2}-\frac{y^2}{a^2+b^2+c^2}-\frac{z^2}{a^2+b^2+c^2}=0\)
\(\Leftrightarrow\frac{x^2}{a^2}-\frac{x^2}{a^2+b^2+c^2}+\frac{y^2}{b^2}-\frac{y^2}{a^2+b^2+c^2}+\frac{z^2}{c^2}-\frac{z^2}{a^2+b^2+c^2}=0\)
\(\Leftrightarrow x^2\left(\frac{1}{a^2}-\frac{1}{a^2+b^2+c^2}\right)+y^2\left(\frac{1}{b^2}-\frac{1}{a^2+b^2+c^2}\right)+z^2\left(\frac{1}{c^2}-\frac{1}{a^2+b^2+c^2}\right)=0\)
Vì : \(x^2\left(\frac{1}{a^2}-\frac{1}{a^2+b^2+c^2}\right);y^2\left(\frac{1}{b^2}-\frac{1}{a^2+b^2+c^2}\right);z^2\left(\frac{1}{c^2}-\frac{1}{a^2+b^2+c^2}\right)\ge0\forall x\)
Nên : \(x^2\left(\frac{1}{a^2}-\frac{1}{a^2+b^2+c^2}\right)=0;y^2\left(\frac{1}{b^2}-\frac{1}{a^2+b^2+c^2}\right)=0;z^2\left(\frac{1}{c^2}-\frac{1}{a^2+b^2+c^2}\right)=0\)
=> x = 0 ; y = 0 ; z = 0
Vậy x + y + z = 0 (đpcm)
a ) Đặt A = \(\frac{-a+b+c}{2a}+\frac{a-b+c}{2b}+\frac{a+b-c}{2c}=\frac{1}{2}\left(-1+\frac{b}{a}+\frac{c}{a}+\frac{a}{b}-1+\frac{c}{b}+\frac{a}{c}+\frac{b}{c}-1\right)\)
\(=\frac{1}{2}\left(\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}-3\right)\)
Do a ; b ; c > 0 , áp dụng BĐT Cô - si cho các cặp số dương , ta có :
\(A\ge\frac{1}{2}\left[2\sqrt{\frac{a}{b}.\frac{b}{a}}+2\sqrt{\frac{b}{c}.\frac{c}{b}}+2\sqrt{\frac{a}{c}.\frac{c}{a}}-3\right]=\frac{1}{2}\left(2+2+2-3\right)=\frac{3}{2}\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c\)
b ) \(P=\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=\frac{x^2}{xy+xz}+\frac{y^2}{xy+yz}+\frac{z^2}{xz+yz}\ge\frac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)}\ge\frac{3\left(xy+yz+xz\right)}{2\left(xy+yz+xz\right)}=\frac{3}{2}\)
( áp dụng BĐT Cauchy - Schwarz )
Dấu " = " xảy ra \(\Leftrightarrow x=y=z\)
Từ \(a+b+c=0\)
\(\Rightarrow\hept{\begin{cases}a=-\left(b+c\right)\\b=-\left(a+c\right)\\c=-\left(a+b\right)\end{cases}}\)
Từ \(x+y+z=0\)
\(\Rightarrow\hept{\begin{cases}x=-\left(y+z\right)\\y=-\left(x+z\right)\\z=-\left(x+y\right)\end{cases}}\)
Thay vào \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\)
\(\Leftrightarrow\frac{-\left(b+c\right)}{-\left(y+z\right)}+\frac{b}{y}+\frac{c}{z}=0\)
\(\Rightarrow2byz+2cyz+bz^2+cy^2=0\)
\(\Rightarrow-\left(b+c\right).-\left(y+z\right)^2+by^2+cz^2=0\)
\(\Rightarrow\text{ax}^2+by^2+cz^2=0\)(dpcm)
Suy ngược nha k chắc
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 và 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 2 ( ax2 + by2 + cz2 ) = 0 \(\Rightarrow\)ax2 + by2 + cz2 = 0
chịu khó lắm
Ok
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\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\Rightarrow\frac{ayz+bxz+cxy}{xyz}=0\Rightarrow ayz+bxz+cxy=0\)
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\Rightarrow\frac{xbc+yac+zab}{abc}=1\)
\(\Rightarrow xbc+yac+zab=abc\)
\(\Rightarrow\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2+2.xbc.yac+2.yac.zab+2.xbc.zab=\left(abc\right)^2\)
\(\Rightarrow x^2b^2c^2+y^2a^2c^2+z^2a^2b^2+2abc\left(cxy+ayz+bxz\right)=\left(abc\right)^2\)
\(\Rightarrow x^2b^2c^2+y^2a^2c^2+z^2a^2b^2=a^2b^2c^2\)
\(\Rightarrow\frac{x^2b^2c^2+y^2a^2c^2+z^2a^2b^2}{a^2b^2c^2}=1\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\)
\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\Rightarrow ayz+bxz+cxy=0\)
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\Leftrightarrow\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1-2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)\)
\(=1-2.\frac{cxy+bxz+ayz}{abc}=1-2.0=1\)
đặt a = 2x+y+z ; b = 2y+z+x ; c = 2z+x+y => a+b+c = 4x+4y+4z
=> a - (a+b+c)/4 = x => x = (3a-b-c)/4 ; tương tự y = (3b-c-a)/4 ; z = (3c-a-b)/4
thay vào vế trái ta có
P = (3a-b-c)/4a + (3b-c-a)/4b + (3c-a-b)/4c =
= 9/4 - (b/4a + c/4a + c/4b + a/4b + a/4c + b/4c)
= 9/4 - (1/4)(b/a+a/b + c/a+a/c + c/b+b/c)
Côsi cho từng cặp ta có: b/a+a/b ≥ 2 ; c/a+a/c ≥ 2 ; c/b+b/c ≥ 2
=> b/a+a/b + c/a+a/c + c/b+b/c ≥ 6
=> -(1/4)(b/a+a/b +c/a+a/c + c/b+b/c) ≤ -6/4 thay vào P ta có:
P ≤ 9/4 - 6/4 = 3/4 (đpcm) ; dấu "=" khi a = b = c hay x = y = z
cách này tuy biến đổi dài nhưng dễ hiểu)
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Cách khác:
P = x/(2x+y+z) -1 + y/(2y+z+x) -1 + z/(2z+x+y) - 1 + 3
= -(x+y+z)/(2x+y+z) -(x+y+z)/(2y+z+x) -(x+y+z)/(2z+x+y) + 3
= -(x+y+z).[1/(2x+y+z) + 1/(2y+z+x) + 1/(2z+x+y)] + 3
- - -
Côsi cho 3 số:
2x+y+z + 2y+z+x + 2z+x+y ≥ 3.³√(2x+y+z)(2y+z+x)(2z+x+y)
=> 4(x+y+z) ≥ 3.³√(2x+y+z)(2y+z+x)(2z+x+y) (1*)
Côsi cho 3 số:
1/(2x+y+z)+1/(2y+z+x)+1/(2z+x+y) ≥ 3³√1/(2x+y+z)(2y+z+x)(2z+x+y) (2*)
Lấy (1*) *(2*) ta có:
4(x+y+z)[1/(2x+y+z) + 1/(2y+z+x) + 1/(2z+x+y)] ≥ 9
=> -(x+y+z).[1/(2x+y+z) + 1/(2y+z+x) + 1/(2z+x+y)] ≤ -9/4
thay vào P ta có:
P ≤ -9/4 + 3 = 3/4 (đpcm) ; dấu "=" khi x = y = z
Bạn ơi vì sao lại nhân với 9/4 mình tưởng chỉ nhân với 3/4 thôi chứ nhỉ
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\Rightarrow\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1^2\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)=1\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{cxy+ayz+bxz}{abc}\right)=1\)
Mà \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\Rightarrow\frac{ayz+bxz+cxy}{xyz}=0\Rightarrow ayz+bxz+cxy=0\)
\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2.\frac{0}{abc}=1\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2.0=1\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\left(dpcm\right)\)
Chúc bạn học tốt
1 cái T I C K nha cảm ơn
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........
\(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)
\(\Leftrightarrow\)\(\frac{x^2}{a^2+b^2+c^2}+\frac{y^2}{a^2+b^2+c^2}+\frac{z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)
\(\Leftrightarrow\)\(\left(\frac{x^2}{a^2+b^2+c^2}-\frac{x^2}{a^2}\right)+\left(\frac{y^2}{a^2+b^2+c^2}-\frac{y^2}{b^2}\right)+\left(\frac{z^2}{a^2+b^2+c^2}-\frac{z^2}{c^2}\right)=0\)
\(\Leftrightarrow\)\(x^2\left(\frac{1}{a^2+b^2+c^2}-\frac{1}{a^2}\right)+y^2\left(\frac{1}{a^2+b^2+c^2}-\frac{1}{b^2}\right)+z^2\left(\frac{1}{a^2+b^2+c^2}-\frac{1}{c^2}\right)=0\)
Ta có :
\(\frac{1}{a^2+b^2+c^2}< \frac{1}{a^2};\frac{1}{b^2};\frac{1}{c^2}\)
\(\Rightarrow\)\(\frac{1}{a^2+b^2+c^2}-\frac{1}{a^2}< 0;\frac{1}{a^2+b^2+c^2}-\frac{1}{b^2}< 0;\frac{1}{a^2+b^2+c^2}-\frac{1}{c^2}< 0\)
\(\Rightarrow\)\(x^2=y^2=z^2=0\)\(\Rightarrow\)\(x=y=z=0\) ( đpcm )
Chúc bạn học tốt ~
Ta có x+y +z =0 =>x^2 =(y+z)^2 ;y^2=(x+z)^2;z^2=(y+x)^2
=>ax^2+by^2+cz^2=a(y+z)^2+b(x+z)^2+c(y+x)^2
=>(b+c)x^2+(a+c)y^2+(a+b)z^2+2(ayz+bxz+cyz) (1)
Tu a+b+c=0=>-a=b+c;-b=a+c;-c=a+b (2)
Tu a/x+b/y+c/x =0=>ayz+bxz+cxy/xyz=0=>ayz+bxz+cxy = 0 (3)
Thay (2) va (3 ) va (1) ta dc :ax^2+by^2+cz^2=-(ax^2+by^2+cz^2)=>ax^2+by^2+cz^2=0
(Hai số đối nhau mà bằng nhau chỉ có số 0)
hình như bạn làm sai bạn ạ