![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1.\)
Theo đề ra, ta có:
\(ax+by=c\)
\(bx+cy=a\Leftrightarrow ax+by+bx+cy+cx+ay=c+a+b\)
\(cx+by=b\)
\(\Leftrightarrow x\left(a+b+c\right)+y\left(a+b+c\right)=a+b+c\)
\(\Leftrightarrow\left(x+y-1\right)\left(a+b+c\right)=0\)
Ta có: \(x,y\)thỏa mãn \(\Rightarrow a+b+c=0\Rightarrow a+b=\left(-c\right)\)
Khi đó ta có:
\(a^3+b^3+c^3=a^3+3ab\left(a+b\right)+b^3-3ab\left(a+b\right)+c^3\)
\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3=\left(-c\right)^3-3ab\left(-c\right)+c^3=3abc\)\(\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\Leftrightarrow ayz+bzx+cxy=0\).
Do đó: \(ax^2+by^2+cz^2=\left(ax+by+cz\right)\left(x+y+z\right)-axy-axz-byz-byx-czx-czy=0-xy\left(a+b\right)-yz\left(b+c\right)-zx\left(c+a\right)=0+xyc+yza+zxb=0\).
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)=\left(ax+by+cz\right)^2\)\(\Rightarrow a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2-\left(a^2x^2+b^2y^2+c^2z^2+2axby+2bycz+2czax\right)=0\)\(\Rightarrow a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2z^2-a^2x^2-b^2y^2-c^2z^2-2axby-2bycz-2czax=0\)\(\Rightarrow a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2-2axby-2bycz-2czax=0\)\(\Rightarrow\left(a^2y^2-2axby+b^2x^2\right)+\left(a^2z^2-2axcz+c^2x^2\right)+\left(b^2z^2-2bycz+c^2y^2\right)=0\)\(\Rightarrow\left(ax-by\right)^2+\left(az-cx\right)^2+\left(bz-cy\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(ay-bx\right)^2=0\\\left(az-cx\right)^2=0\\\left(bz-cy\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}ay=bx\\az=cx\\bz=cy\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{a}{x}=\dfrac{b}{y}\\\dfrac{a}{x}=\dfrac{c}{z}\\\dfrac{b}{y}=\dfrac{c}{z}\end{matrix}\right.\)\(\Rightarrow\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\Rightarrowđpcm\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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:
\(ax^2+by^2+cz^2=\left(-b-c\right)\left(-y-z\right)^2+by^2+cz^2\)
\(=-2byz-2cyz-bz^2-cy^2=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Cho a+b+c = 0 ; x+y+z = 0 và \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)
CMR : \(ax^2+by^2+cz^2=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Có:
\(x+y+z=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}-x=y+z\\-y=x+z\\-z=x+y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2=\left(y+z\right)^2\\y^2=\left(x+z\right)^2\\z^2=\left(x+y\right)^2\end{matrix}\right.\)
\(\Rightarrow ax^2+by^2+cz^2\)
\(=a\left(y+z\right)^2+b\left(x+z\right)^2+c\left(x+y\right)^2\)
\(=x^2\left(b+c\right)+y^2\left(a+c\right)+z^2\left(a+b\right)+2\left(ayz+bxz+cxy\right)\)
Mà \(a+b+c=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}b+c=-a\\a+c=-b\\a+b=-c\end{matrix}\right.\)
Đồng thời có: \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)
\(\Leftrightarrow\dfrac{ayz+bxz+cxy}{xyz}=0\)
\(\Leftrightarrow ayz+bxz+cxy=0\)
Từ đây ta có:)
\(ax^2+by^2+cz^2=-ax^2-by^2-cz^2\)
\(\Rightarrow2\left(ax^2+by^2+cz^2\right)=0\)
\(\Rightarrow ax^2+by^2+cz^2=0\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
Bạn tham khảo tại đây:
Câu hỏi của Son go Ku - Toán lớp 8 - Học toán với OnlineMath
Bấm vô dòng màu xanh:v