a) 99= 22+33+44

b) 285=57+95+133

c) 2A5 là  cái gì ?

d) 465= 60+105+120+180

ta có: \(\frac{10}{5^8}=\frac{2\cdot5}{5^8}=\frac{2}{5^7}\)=\(\frac{2}{78125}\)

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

<=> \(x^2+y^2+z^2-2xy-2yz+2zx=x^2-y^2+z^2\)

<=> \(2y^2-2xy-2yz+2zx=0\)

<=> \(\left(2y^2-2yz\right)-\left(2xy-2xz\right)=0\)

<=>\(2y\left(y-z\right)-2x\left(y-z\right)=0\)

<=>\(2\left(y-x\right)\left(y-z\right)=0\)

<=> \(\left[\begin{array}{nghiempt}y-x=0\\y-z=0\end{array}\right.\)

<=> \(\left[\begin{array}{nghiempt}y=x\\y=z\end{array}\right.\)

Với y=x thì mọi giá trị của z đều thỏa mãn.

Với y=z  ta có: \(\left(x-2y\right)^2=x^2\)

<=> \(\left[\begin{array}{nghiempt}x-2y=-x\\x-2y=x\end{array}\right.\)

<=> \(\left[\begin{array}{nghiempt}x=y\\x=-y\end{array}\right.\)

=> x=y=z hoặc  -x=y=z.

a) ĐKXĐ: x\(\ne\) 0;4

Ta có: Q= \(\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{2}{\sqrt{x}}\right)\)

                   = \(\frac{4\sqrt{x}\cdot\left(2-\sqrt{x}\right)+8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}:\frac{\sqrt{x}-1-2\cdot\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

                   =\(\frac{8\sqrt{x}+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)= \(\frac{4\sqrt{x}\cdot\left(2+\sqrt{x}\right)}{2+\sqrt{x}}\cdot\frac{-\sqrt{x}}{3-\sqrt{x}}\)=\(\frac{-4}{3-\sqrt{x}}\)=\(\frac{4}{\sqrt{x}-3}\)

b) Q=-1 => \(\frac{4}{\sqrt{x}-3}=-1\)

<=> \(4=3-\sqrt{x}\)

<=> \(\sqrt{x}=-1\) (vô lí)

Vậy ko tìm được x.

Ta có: \(a^2=bc\)

=> \(bc-a^2=a^2-bc\)

<=> \(bc-a^2+ac-ab=a^2-bc+ac-ab\)

<=> \(\left(ac-a^2\right)+\left(bc-ab\right)=\left(a^2-ab\right)+\left(ac-bc\right)\)

<=> \(a\left(c-a\right)+b\left(c-a\right)=a\left(a-b\right)+c\left(a-b\right)\)

<=> \(\left(a+b\right)\left(c-a\right)=\left(a+c\right)\left(a-b\right)\)

<=> \(\frac{a+b}{a-b}=\frac{a+c}{c-a}\)(đpcm)