\(\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}\)

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

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

Áp dụng BĐT AM-GM ta có:

\(\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}\ge2.\sqrt{\frac{x}{z}.\frac{z}{x}}+2.\sqrt{\frac{x}{y}.\frac{y}{x}}+2.\sqrt{\frac{y}{z}.\frac{z}{y}}=2+2+2=6\)

                                                                                                                           đpcm

Svac-xơ 

\(VT=\left(\frac{x+y}{z}+1\right)+\left(\frac{y+z}{x}+1\right)+\left(\frac{z+x}{y}+1\right)-3\)

\(VT=\frac{x+y+z}{x}+\frac{x+y+z}{y}+\frac{x+y+z}{z}-3=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)-3\)

\(\ge\left(x+y+z\right).\frac{\left(1+1+1\right)^2}{x+y+z}-3=9-3=6\)

a) \(A=\frac{97^3+83^3}{180}-97\cdot83\)

\(A=\frac{\left(97+83\right)\left(97^2-97\cdot83+83^2\right)}{180}-97\cdot83\)

\(A=\frac{180\cdot\left(97^2-97\cdot83+83^2\right)}{180}-97\cdot83\)

\(A=97^2-97\cdot83+83^2-97\cdot83\)

\(A=9409-2\cdot8051+6889\)

\(A=196\)

b) \(B=\left(50^2+48^2+...+2^2\right)-\left(49^2+47^2+...+1^2\right)\)

\(B=50^2+48^2+...+2^2-49^2-47^2-...-1^2\)

\(B=\left(50^2-49^2\right)+\left(48^2-47^2\right)+...+\left(2^2-1^2\right)\)

\(B=\left(50+49\right)\left(50-49\right)+\left(48+47\right)\left(48-47\right)+...+\left(2+1\right)\left(2-1\right)\)

\(B=50+49+48+47+...+2+1\)

Số số hạng là : \(\left(50-1\right):1+1=50\)( số )

Tổng B là : \(\left(50+1\right)\cdot50:2=1275\)

Vậy....

a,\(6x-8y=9\)

\(\Rightarrow x=\frac{9+8y}{6}\)

\(y=\frac{6x-9}{8}\)

Vậy....

\(b,11x+18y=120\)

\(\Rightarrow x=\frac{120-18y}{11}\)

\(y=\frac{120-11x}{18}\)

100 nha

______________
_____________
^_^

\(A=a^4-2a^3+3a^2-4a+5\)

    \(=\left(a^4-2a^3+a^2\right)+2\left(a^2-2a+1\right)+3\)

      \(=\left(a^2-a\right)^2+2\left(a-1\right)^2+3\ge3\)

Dấu "='' khi a = 1

Vậy ..........

=7 nhá

học tốt

3+4=7

ngày  18/2 là sinh nhật J-hope