Ta có: 2^x+2^y=128

=>2^x(1+2^x-y)=128

=>1+2^x-y E Ư(128)=(1;2;2²;2³;2⁴;2⁵;2⁶;2⁷}

Với x#y

=>1+2^x-y lẻ

=>x,y E rỗng

Với x=y

=>1+2^x-y=2  (TM)

=>2^x.2=128

=>2^x=64=2⁶

=>x=6

=>y=6

=>x+y=12

=>z=0

         Giai

Vi 2^x+2^y=128 => 2^(x+y)=128

2 ^7 =128

=> y=12-7=5

d/s y=5

ta có x+y+z=12 \(\Rightarrow x+y=12-z\)

ta có \(2^x.2^y=128\)\(\Leftrightarrow2^{x+y}=2^7\)\(\Leftrightarrow2^{12-z}=2^7\)\(\Leftrightarrow12-z=7\)\(\Rightarrow z=5\)

Ta có: \(x^2+y^2-z^2\)

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

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

\(=-2xy\)

Ta có: \(x^2+z^2-y^2\)

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

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

\(=-2xz\)

Ta có: \(y^2+z^2-x^2\)

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

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

\(=-2yz\)

Ta có: \(\dfrac{xy}{x^2+y^2-z^2}+\dfrac{xz}{x^2+z^2-y^2}+\dfrac{yz}{y^2+z^2-x^2}\)

\(=\dfrac{xy}{-2xy}+\dfrac{xz}{-2xz}+\dfrac{yz}{-2yz}\)

\(=\dfrac{1}{-2}+\dfrac{1}{-2}+\dfrac{1}{-2}\)

\(=\dfrac{-3}{2}\)

Đáp án A

abcdacscas

Bài 2: 

Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3k\\y=4k\end{matrix}\right.\)

Ta có: xy=12

\(\Leftrightarrow12k^2=12\)

\(\Leftrightarrow k^2=1\)

Trường hợp 1: k=1

\(\Leftrightarrow\left\{{}\begin{matrix}x=3k=3\\y=4k=4\end{matrix}\right.\)

Trường hợp 2: k=-1

\(\Leftrightarrow\left\{{}\begin{matrix}x=3k=-3\\y=4k=-4\end{matrix}\right.\)

\(x+y+z=6\Rightarrow\left(x+y+z\right)^2=36\Rightarrow x^2+y^2+z^2+2\left(xy+yz+xz\right)=36\)

\(\Rightarrow xy+yz+zx=\frac{36-\left(x^2+y^2+z^2\right)}{2}=\frac{36-12}{2}=12=x^2+y^2+z^2\)(1)

Mặt khác ta luôn có: \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)

hay: \(2x^2+2y^2+2z^2-2\left(xy+yz+zx\right)\ge0\)

hay: \(x^2+y^2+z^2\ge xy+yz+zx\)

Vậy để đẳng thức (1) xảy ra thì x = y = z = 2.