Bạn tham khảo tại đây:

Câu hỏi của trieu dang - Toán lớp 8 - Học toán với OnlineMath

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

\(\Rightarrow\frac{\left(yz+xz+xy\right)}{xyz}=0\)

\(\Rightarrow yz+zx+xy=0\)

Ta có : \(x^2+2yz=x^2+yz+yz\)

                              \(=x^2+yz-zx-xy\)

                              \(=x\left(x-z\right)-y\left(x-z\right)\)

                              \(=\left(x-y\right)\left(x-z\right)\)

Tương tự : \(y^2+2xz=y^2+xz+xz\)

                                    \(=y^2+xz-xy-yz\)

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

                                    \(=\left(x-y\right)\left(z-y\right)\)

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

\(\Rightarrow M=\frac{yz}{\left(x-y\right)\left(x-z\right)}+\frac{xz}{\left(x-y\right)\left(z-y\right)}+\frac{xy}{\left(x-z\right)\left(y-z\right)}\)  \(M=\frac{yz\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}-\frac{xz\left(x-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}+\frac{xy\left(x-y\right)}{\left(x-z\right)\left(y-z\right)\left(x-y\right)}\)

\(M=\frac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\frac{yz\left(y-z\right)-xz\left(x-y+y-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)

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

Tính thế làm gì bạn ê

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Leftrightarrow\frac{xy+yz+zx}{xyz}=0\Leftrightarrow xy+yz+zx=0\)

\(\Leftrightarrow xy=-yz-zx;yz=-xy-zx;zx=-xy-yz\)

Ta có: x²+2yz=x²+yz+yz=x²+yz-xy-zx=x(x-y)-z(x-y)=(x-y)(x-z)

Tương tự: \(y^2+2xz=\left(y-x\right)\left(y-z\right);z^2+2xy=\left(z-x\right)\left(z-y\right)\)

A= \(\frac{yz}{x^2+2yz}+\frac{xz}{y^2+2xz}+\frac{xy}{z^2+2xy}\)=\(\frac{yz}{\left(x-y\right)\left(x-z\right)}+\frac{xz}{\left(y-x\right)\left(y-z\right)}+\frac{xy}{\left(z-x\right)\left(z-y\right)}\)

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

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

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

\(=\frac{x\left(y-z\right)\left(x-y\right)-z\left(x-y\right)\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}=\frac{\left(x-y\right)\left(y-z\right)\left(x-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}=1\)

Câu 29:

a: \(\left(a+b\right)^2\le2\left(a^2+b^2\right)\)

\(\Leftrightarrow-a^2+2ab-b^2\le0\)

\(\Leftrightarrow-\left(a-b\right)^2\le0\)(luôn đúng)

Hả lơp 1 ????????

\(14,P=x^2+xy+y^2-3x-3y+3\\ P=\left(x^2+xy+\dfrac{1}{4}y^2\right)-3\left(x+\dfrac{1}{2}y\right)+\dfrac{3}{4}y^2-\dfrac{3}{2}y+3\\ P=\left(x+\dfrac{1}{2}y\right)^2-3\left(x+\dfrac{1}{2}y\right)+\dfrac{9}{4}+\dfrac{3}{4}\left(y^2-2y+1\right)\\ P=\left(x+\dfrac{1}{2}y-\dfrac{3}{2}\right)^2+\dfrac{3}{4}\left(y-1\right)^2\ge0\)

đây là lớp 4 ư

Đặt x = y = z = 1 . Ta có:

\(\left(x+1\right)\left(y+1\right)\left(z+1\right)\ge8\) 

\(\Leftrightarrow\left(1+1\right)^3\Leftrightarrow2^3\). Mà

\(2^3=8\RightarrowĐPCM\)

bai nay de ma

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)nhân lần lượt với x; y; z, ta có:

\(1+\frac{x}{y}+\frac{x}{z}=0\)(1)

\(1+\frac{y}{z}+\frac{y}{x}=0\)(2)

\(1+\frac{z}{x}+\frac{z}{y}=0\)(3)

Từ: (1); (2) và (3) => \(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+\frac{x}{z}+\frac{y}{x}+\frac{z}{y}=-3\)(*)

Mặt khác: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)quy đồng ta có:

\(\frac{\left(xy+yz+zx\right)}{xyz}=0\)hay xy + yz + zx = 0

Hay: \(\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right).\left(xy+yz+zx\right)=0\)

Khai triển, ta có:

\(\frac{yz}{x^2}+\frac{zx}{y^2}+\frac{xy}{z^2}+\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+\frac{z}{x}+\frac{y}{x}+\frac{z}{y}=0\)

Vậy: \(\frac{yz}{x^2}+\frac{zx}{y^2}+\frac{xy}{z^2}=-\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+\frac{x}{z}+\frac{y}{x}+\frac{z}{y}\right)=3\)

hình như bạn lộn r, đề đâu có biểu tính phép tính đó 

Phan Thị Hà Vy à ừ, nhầm thật