Ta có: x + y = 9

           x + z = 15

           y + z = 12

=> x + y + x + z + y + z = 9 + 15 + 12

<=> 2( x + y + z ) = 36

<=> x + y + z = 18

=> x = 18 - 12 = 6

     y = 18 - 15 = 3

     z = 18 - 9 = 9

Từ \(x+y=9\), \(x+z=15\), \(y+z=12\)

\(\Rightarrow x+y+x+z+y+z=9+15+12\)\(\Rightarrow2\left(x+y+z\right)=36\)

\(\Rightarrow x+y+z=18\)

Ta có: \(x+y=9\)\(\Rightarrow z=18-9=9\)

         \(x+z=15\)\(\Rightarrow y=18-15=3\)

         \(y+z=12\)\(\Rightarrow x=18-12=6\)

Vậy \(x=6\), \(y=3\), \(z=9\)

Ai giúp mik vs

Vì x² >= 0 với mọi x nên x² - 4 >= -4 với mọi x.

=> \(\frac{-4}{x^2-4}\)=< 1 với mọi x.

Dấu = xra <=> x² = 0 <=> x = 0.

nếu thay x=1; -1 thì biểu thức lớn nhất

Ta có:

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

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

\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\)

\(\Leftrightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right)=0\)

\(\Leftrightarrow\left(x+y\right)\cdot\frac{xy+z\left(x+y+z\right)}{xyz\left(x+y+z\right)}=0\)

\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)

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

Xét \(x=-y\)

Ta có:

\(\frac{1}{x^{2017}}+\frac{1}{y^{2017}}+\frac{1}{z^{2017}}=\frac{1}{x^{2017}}+\frac{1}{-y^{2017}}+\frac{1}{y^{2017}}=\frac{1}{z^{2017}}\)

\(\frac{1}{x^{2017}+y^{2017}+z^{2017}}=\frac{1}{-x^{2017}+y^{2017}+z^{2017}}=\frac{1}{z^{2017}}\)

\(\Rightarrow\frac{1}{x^{2017}}+\frac{1}{y^{2017}}+\frac{1}{z^{2017}}=\frac{1}{x^{2017}+y^{2017}+z^{2017}}\left(dpcm\right)\)

Một cái chặt hơn nè:))

CMR nếu \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\) thì \(\frac{1}{x^n}+\frac{1}{y^n}+\frac{1}{z^n}=\frac{1}{x^n+y^n+z^n}\) với n lẻ.

còn on ko bn

vx on,dag doi bai

\(\left(x+1\right)^2+\left(x+1\right)\left(x-1\right)+\left(x-1\right)^2\)

\(=x^2+2x+1+\left(x+1\right)\left(x-1\right)+\left(x-1\right)^2\)

\(=x^2+2x+1+x^2-1+\left(x-1\right)^2\)

\(=x^2+2x+1+x^2-1+x^2-2x+1\)

\(=3x^2+1\) (1)

Thay x = 100 vào (1), ta có:

\(3x^2+1=\left(3.100\right)^2+1=90001\)

a) Ta có:  A = x² - 4x + 7

A = (x² - 4x + 4) + 3

A = (x - 2)² + 3 \(\ge\)3 \(\forall\)x

Dấu "=" xảy ra <=> x - 2  = 0 <=> x = 2

Vậy MinA = 3 <=> x = 2

b) Xem lại đề

hộ mình với ạ 

\(\Leftrightarrow\frac{\left(x+2\right)\left(x-2\right)}{x\left(x+2\right)}=\frac{A}{x}\)

\(\Leftrightarrow\frac{x-2}{x}=\frac{A}{x}\)

\(\Rightarrow A=x-2\)

^_gggggggg