P/s: Em mới lớp 7 thôi nên có gì sai mong anh/chị thông cảm ạ.

Khai triển ra ta được: \(Q=x^2+y^2+z^2+3\left(xy+xz+yz\right)\)

\(P=2\left(x^2+y^2+z^2\right)+2\left(xy+yz+zx\right)\)

Do P = Q nên P - Q = 0.Hay:\(x^2+y^2+z^2-xy-yz-zx=0\)

Nhân 2 vào hai vế suy ra \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

Ta có: \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(y-z\right)^2\ge0\\\left(z-x\right)^2\ge0\end{cases}}\) .Suy ra \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)

Dấu "=' xảy ra khi x = y = z (đpcm)

chứng minh ngược lại bạn ơi

chứng minh x=y=z thì p=q 

Đặt \(x+y=a;y+z=b;z+x=c\)thì P=Q có nghĩa là:

\(a^2+b^2+c^2-ab-bc-ac=0\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\)

\(\Leftrightarrow a=b=c\Leftrightarrow x+y=y+z=z+x\Leftrightarrow x=y=z\)

Ta có:

\(\left(x+y+z\right)\left(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\right)=\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}+x+y+z\)

\(\Leftrightarrow x+y+z=\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}+x+y+z\)

\(\Leftrightarrow\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}=0\)

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1 câu đăng tận 3 lần thế b

Câu hỏi của Vũ Anh Quân - Toán lớp 8 | Học trực tuyến nè nhé b .

Câu hỏi của Vũ Anh Quân - Toán lớp 8 | Học trực tuyến

Bài này mình làm 2 cách cho bạn dễ hiểu nha

C1:\(P=\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}=1\Leftrightarrow x\left(z+x\right)\left(x+y\right)+y\left(y+z\right)\left(x+y\right)+z\left(z+x\right)\left(y+z\right)=\left(y+z\right)\left(x+y\right)\left(z+x\right) \)\(\Leftrightarrow x^2\left(y+z\right)+y^2\left(x+z\right)+z^2\left(x+y\right)+x^3+y^3+z^3+3xyz=x^2\left(y+z\right)+y^2\left(x+z\right)+z^2\left(x+y\right)+2xyz\)

\(\Leftrightarrow x^3+y^3+z^3+xyz=0\)

\(\Rightarrow\left(x^3+y^3+z^3+xyz\right)\left(x+y+z\right)=0 \)

Ta cũng thấy Q=\(Q=\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}=\dfrac{x^2\left(z+x\right)\left(x+y\right)+y^2\left(y+z\right)\left(x+y\right)+z^2\left(y+z\right)\left(z+x\right)}{\left(y+z\right)\left(x+z\right)\left(x+y\right)}=\dfrac{\left(x^3+y^3+z^3+xyz\right)\left(x+y+z\right)}{\left(y+z\right)\left(x+z\right)\left(x+y\right)}=0\)

C2 nè :
\(P=\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}=1\)

\(P=\left(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\right)\left(x+y+z\right)=x+y+z .\)

\(\Leftrightarrow\dfrac{x^2+x\left(y+z\right)}{y+z}+\dfrac{y^2+y\left(x+z\right)}{z+x}+\dfrac{z^2+z\left(x+y\right)}{x+y}=x+y+z.\)

\(\Leftrightarrow\dfrac{x^2}{y+z}+x+\dfrac{y^2}{z+x}+y+\dfrac{z^2}{x+y}+z=x+y+z \left(ĐPCM\right)\)

Q = \(\dfrac{x^2}{y+z}\) + \(\dfrac{y^2}{x+z}\) + \(\dfrac{z^2}{x+y}\)

= \(\dfrac{x\left[\left(x+y+z\right)-\left(y+z\right)\right]}{y+z}\) + \(\dfrac{y\left[\left(x+y+z\right)-\left(x+z\right)\right]}{x+z}\) + \(\dfrac{z\left[\left(x+y+z\right)-\left(x+y\right)\right]}{x+y}\)

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

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

= (x + y + z)\(\left[\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}\right]\) - (x + y + z)

= (x + y +z) . P - (x + y + z)

= ( x + y +z) .1 - (x + y +z)

= 0 (đpcm)

Ta có:

\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=\left(y+z-2x\right)^2+\left(z+x-2y\right)^2+\left(x+y-2z\right)^2\)

\(\Rightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=6x^2+6y^2+6z^2-6xy-6yz-6zx\)

\(\Rightarrow4x^2+4y^2+4z^2-4xy-4yz-4zx=0\)

\(\Rightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=0\)

\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}\left(x-y\right)^2=0\\\left(y-z\right)^2=0\\\left(z-x\right)^2=0\end{cases}}\Rightarrow x=y=z\)