x+y+z=0

nên x+y=-z; y+z=-x; x+z=-y

\(\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)\)

\(=\dfrac{x+y}{y}\cdot\dfrac{y+z}{z}\cdot\dfrac{x+z}{x}=-1\)

ố la la

=x+y+z/2x+2y+2z

=1/2=x+y+z

x/y+z+5=1/2

=>2x=y+z+5

làm thế này thôi

tham khảo nhé bạn!!!!!

Bài này ez thôi, làm mãi rồi.

Theo đề bài, ta có: \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)

=>\(\dfrac{xy+yz+xz}{xyz}=0\)

=> xy+yz+zx=0

=> \(\left\{{}\begin{matrix}xy=-yz-zx\\yz=-xy-zx\\zx=-xy-yz\end{matrix}\right.\)

Ta có: x²+2yz=x²+yz-xy-zx=(x-y)(x-z)

           y²+2xz=y²+xz-xy-yz=(x-y)(z-y)

           z²+2xy=z²+xy-yz-xz=(x-z)(y-z)

=> \(\dfrac{yz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xz}{\left(x-y\right)\left(z-y\right)}+\dfrac{xy}{\left(x-z\right)\left(y-z\right)}=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\dfrac{\left(x-y\right)\left(x-z\right)\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=1\)

a, ĐKXĐ:\(\left\{{}\begin{matrix}x\ne0\\y\ne0\end{matrix}\right.\)

Đặt \(\dfrac{1}{x}=a,\dfrac{1}{y}=b\)

Hệ \(\Leftrightarrow\left\{{}\begin{matrix}a+b=\dfrac{1}{6}\\8a+5y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{18}\\b=\dfrac{1}{9}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{18}\\\dfrac{1}{y}=\dfrac{1}{9}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=18\\y=9\left(tm\right)\end{matrix}\right.\)

\(b,\left\{{}\begin{matrix}\dfrac{x-1}{2}-y=1\\2x+y=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}-\dfrac{2y}{2}=\dfrac{2}{2}\\2x+y=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-1-2y=2\\2x+y=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-2y=3\\2x+y=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

a.\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{6}\\\dfrac{8}{x}+\dfrac{5}{y}=1\end{matrix}\right.\)

\(ĐK:x;y\ne0\)

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{1}{y}=b\end{matrix}\right.\) 

hpt trở thành:

\(\left\{{}\begin{matrix}a+b=\dfrac{1}{6}\\8a+5b=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{18}\\b=\dfrac{1}{9}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{18}\\\dfrac{1}{y}=\dfrac{1}{9}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=18\\y=9\end{matrix}\right.\) ( tm )

Vậy nghiệm hpt: \(\left\{{}\begin{matrix}x=18\\y=9\end{matrix}\right.\)

b.\(\left\{{}\begin{matrix}\dfrac{x-1}{2}-y=1\\2x+y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}+2x=2\\2x+y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x-1+4x}{2}=\dfrac{4}{2}\\2x+y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x=5\\2x+y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\2.1+y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

a.\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{6}\\\dfrac{8}{x}+\dfrac{5}{y}=1\end{matrix}\right.\)

\(ĐK:x;y\ne0\)

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{1}{y}=b\end{matrix}\right.\) 

hpt trở thành:

\(\left\{{}\begin{matrix}a+b=\dfrac{1}{6}\\8a+5b=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{18}\\b=\dfrac{1}{9}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{18}\\\dfrac{1}{y}=\dfrac{1}{9}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=18\\y=9\end{matrix}\right.\) ( tm )

Vậy nghiệm hpt: \(\left\{{}\begin{matrix}x=18\\y=9\end{matrix}\right.\)

b.\(\left\{{}\begin{matrix}\dfrac{x-1}{2}-y=1\\2x+y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}+2x=2\\2x+y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x-1+4x}{2}=\dfrac{4}{2}\\2x+y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x=5\\2x+y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\2.1+y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

Từ x=\(\dfrac{1}{2}\)a+\(\dfrac{1}{2}\)b+\(\dfrac{1}{2}\)c=\(\dfrac{1}{2}\).(a+b+c)\(\Rightarrow\)2x=(a+b+c)

M=(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)+x\(^2\)

= x\(^2\)-xb-ax+ab+x\(^2\)-xc-bx+bc+x\(^2\)-ax-cx+ac+x\(^2\)

= 4x\(^2\)-2ac-2bx-2cx+ab+bc+ac

= 4x\(^2\)-2x(a+b+c)+ab+bc+ca

Thay 2x=a+b+c,ta được:

M= 4x\(^2\)-2x.2c+ab+bc+ca

M= 4x\(^2\)-4x\(^2\)+ab+bc+ca

M= ab+bc+ca

ahihi

Cái này dễ lắm. Mình giải luôn nhé!

a) \(\left[{}\begin{matrix}\dfrac{1}{7}x-\dfrac{2}{7}=0\Leftrightarrow x=\dfrac{2}{7}:\dfrac{1}{7}\Leftrightarrow x=2\\-\dfrac{1}{5}x+\dfrac{3}{5}=0\Leftrightarrow x=-\dfrac{3}{5}:\left(-\dfrac{1}{5}\right)\Leftrightarrow x=3\\\dfrac{1}{3}x+\dfrac{4}{3}=0\Leftrightarrow x=-\dfrac{4}{3}:\dfrac{1}{3}\Leftrightarrow x=-4\end{matrix}\right.\)

Vậy x=2 hoặc x=3 hoặc x=-4

b)\(x\left(\dfrac{1}{6}+\dfrac{1}{10}-\dfrac{4}{15}\right)+1=0\)

\(x.0+1=0\)

\(1=0\) ( vô lí)

Vậy không có giá trị của x nào thỏa mãn

Theo bài ra :

\(\dfrac{x^2-yz}{x\left(1-yz\right)}=\dfrac{y^2-xz}{y\left(1-xz\right)}\)

\(\Leftrightarrow\left(x^2-yz\right)\left(y-xyz\right)=\left(y^2-xz\right)\left(x-xyz\right)\)

\(\Leftrightarrow x^2y-x^3yz-y^2z+xx^2z^2=xy^2-xy^3z-x^2z+x^2yz^2=0\)

\(\Leftrightarrow xy\left(x-y\right)-xyz\left(x^2-y^2\right)+z\left(x^2-y^2\right)+xyz^2\left(x-x\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left[\left(xy-xyz\left(x+y\right)+z\left(x+y\right)-xyz^2\right)\right]=0\)

\(\Leftrightarrow\left(x-y\right)\left(xy+xz+yz-xyz\left(x+y+z\right)\right)=0\)

Mà theo đề bài :

\(x\ne y\Rightarrow xy+xz+yz-xyz\left(x+y+z\right)=0\)

\(\Leftrightarrow xy+xz+yz=xyz\left(x+y+z\right)\)

\(\Leftrightarrow\dfrac{xy}{xyz}+\dfrac{xz}{xyz}+\dfrac{yz}{xyz}=\dfrac{xyz\left(z+y+x\right)}{xyz}\)

\(\Leftrightarrow\dfrac{1}{z}+\dfrac{1}{y}+\dfrac{1}{x}=x+y+z\left(đpcm\right)\)