( x - 1 )^x+2 = ( x - 1 )^x+6

=> ( x - 1 )^x+6 -(x-1)^x+2 = 0

=> ( x - 1 )^x+2 . [ ( x - 1 )⁴ - 1 ] = 0

\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^{x+2}=0\\\left(x-1\right)^4-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\\left(x-1\right)^4=1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x-1∈\left\{1;-1\right\}\end{cases}}\)

Từ x - 1 ∈ { 1 ; -1 }

\(\Rightarrow\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}\Rightarrow}\orbr{\begin{cases}x=2\\x=0\end{cases}}\)

đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)

ta có : 

\(\frac{9a^2+4b^2}{9a^2-4b^2}=\frac{9b^2k^2+4b^2}{9b^2k^2-4b^2}=\frac{b^2\left(9k^2+4\right)}{b^2\left(9k^2-4\right)}=\frac{9k^2+4}{9k^2-4}\)

\(\frac{9c^2+4d^2}{9c^2-4d^2}=\frac{9d^2k^2+4d^2}{9d^2k^2-4d^2}=\frac{d^2\left(9k^2+4\right)}{d^2\left(9k^2-4\right)}=\frac{9k^2+4}{9k^2-4}\)

=> đpcm

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow\hept{\begin{cases}\frac{a}{b}=k\Rightarrow a=bk\Rightarrow a^2=b^2k^2\\\frac{c}{d}=k\Rightarrow c=dk\Rightarrow c^2=d^2k^2\end{cases}}\)

đặt VT = \(\frac{9a^2+4b^2}{9a^2-4b^2}=\frac{9b^2k^2+4b^2}{9b^2k^2-4b^2}\) \(=\frac{b^2\left(9k^2+4\right)}{b^2\left(9k^2-4\right)}=\frac{9k^2+4}{9k^2-4}\)

đặt VP = \(\frac{9c^2+4d^2}{9c^2-4d^2}=\frac{9d^2k^2+4d^2}{9d^2k^2-4d^2}\) \(=\frac{d^2\left(9k^2+4\right)}{d^2\left(9k^2-4\right)}=\frac{9k^2+4}{9k^2-4}\)

=> VT = VP

vậy 2 bt trên = nhau

ta có 5^-3/2.-2^2/5 =-2/625

=> x=1

Bài làm

Ta có: ˆxOy=ˆxOm+ˆyOn+ˆmOz+ˆzOn

Mà ˆxOm=ˆyOn=2ˆxOm

Oz là tia phân giác của ˆmOn

=> ˆmOz=ˆzOn=2ˆmOz

=> ˆxOy=2ˆxOm+2ˆmOz

Hay 1800=2ˆxOm+2ˆmOz

=> 1800=2(ˆxOm+ˆmOz)

=> ˆxOm+ˆmOz=1800:2

=> ˆxOm+ˆmOz=900xOm^+mOz^=90

Hay ˆxOz=900

=> Oz⊥xy

Vậy Oz⊥xy( đpcm )

HT NHÉ BẠN

\(M=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}.\)

\(3M=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+....+\frac{100}{3^{99}}\)

\(3M-M=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}-\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+....+\frac{100}{3^{100}}\)

\(2M=1+\left(\frac{2}{3}-\frac{1}{3}\right)+\left(\frac{3}{3^2}-\frac{2}{3^2}\right)+...+\left(\frac{100}{3^{99}}-\frac{100}{3^{99}}\right)-\frac{100}{3^{100}}\)

\(2M=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

Đặt :

 \(N=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{99}}\)

\(3N=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{98}}\)

\(3N-N=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)

\(2N=1-\frac{1}{3^{99}}\)

\(N=\frac{1-\frac{1}{3^{99}}}{2}\)

Thay vào ta có :

\(2M=1+\frac{1-\frac{1}{3^{99}}}{2}-\frac{100}{3^{100}}\)

\(2M=1+\frac{1}{2}-\frac{1}{2\times3^{100}}-\frac{100}{3^{100}}< 1+\frac{1}{2}=\frac{3}{2}\)

Ta có : \(\frac{3}{4}< \frac{3}{2}\)

\(\Rightarrow\)\(2M=1+\frac{1}{2}-\frac{1}{2\times3^{100}}-\frac{100}{3^{100}}< \frac{3}{4}\)

\(\Rightarrow\)\(M< \frac{3}{4}\)

* Sai thì xin lỗi ạ ! *