Ta có:

1/2+1/4+1/8+1/16+1/32=31/32

Vì có 5 tổng=> ta gọi phần còn lại là 5X

=>5X+31/32=2

=>5X+31/32=64/32

=>5X=64-32-31/32

=>5X=33/32

=>X=33/32:5

=>X=33/160

  64/32 chứ không phải là 64-32 đâu nha

<=>4x+15/16=1

<=>4x=1-15/16

<=>4x=1/16

<=>x=(1/16)/4

<=>x=1/64

\(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)=11\)

\(\Rightarrow\left(4x+4\right)^2+\left(2x-1\right)^2-\left(8x-8\right)\left(x+1\right)=11\)

\(\Rightarrow\left(16x^2+32x+16\right)+\left(4x^2-4x+1\right)-\left(8x^2+8x-8x-8\right)=11\)\(\Rightarrow16x^2+32x+16+4x^2-4x+1-8z^2-8x+8x-8=11\)\(\Rightarrow\left(16x^2+4x^2\right)+\left(32x-4x-8x+8x\right)+\left(16+1-8\right)=11\)\(\Rightarrow20x^2+28x+9=11\)

\(\Rightarrow20x^2+28x=2\)

Xét ước

a)=5/4 

b)=38/15

c)=56/33

d)=16/9

Bài 1:

a)\(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}=\frac{\left(0,2\cdot4\right)^5}{\left(0,2\cdot2\right)^6}=\frac{\left(0,2\right)^5\cdot\left(2^2\right)^5}{\left(0,2\right)^6\cdot2^6}=\frac{\left(0,2\right)^5\cdot2^{10}}{\left(0,2\right)^6\cdot2^6}=\frac{2^4}{0,2}=\frac{16}{\frac{2}{10}}=80\)

b)\(\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}=\frac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(1+2^{10}\right)}=\frac{2^{20}}{2^{12}}=256\)

Bài 2:

a)\(2^{x-1}=16\)

\(\Rightarrow2^{x-1}=2^4\)

\(\Rightarrow x-1=4\Rightarrow x=5\)

b)\(\left(x-1\right)^2=25\)

\(\Rightarrow\left(x-1\right)^2=5^2=\left(-5\right)^2\)

\(\Rightarrow x-1=5\) hoặc \(x-1=-5\)

\(\Rightarrow x=6\) hoặc \(x=-4\)

Vậy \(x=6\) hoặc \(x=-4\)

c)\(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+6}\)

\(\Rightarrow\left(x-1\right)^{x+2}-\left(x-1\right)^{x+6}=0\)

\(\Leftrightarrow\left(x-1\right)^{x+2}\left[1-\left(x-1\right)^4\right]\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}\left(x-1\right)^{x+2}=0\\1-\left(x-1\right)^4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\1=\left(x-1\right)^4\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\\left(x-1\right)^4=\left(-1\right)^4=1^4\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x-1=1\\x-1=-1\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=2\\x=0\end{array}\right.\)

d)\(\left(x+20\right)^{100}+\left|y+4\right|=0\left(1\right)\)

Ta thấy: \(\begin{cases}\left(x+20\right)^{100}\ge0\\\left|y+4\right|\ge0\end{cases}\)

\(\Rightarrow\left(x+20\right)^{100}+\left|y+4\right|\ge0\left(2\right)\)

Từ (1) và (2) suy ra \(\begin{cases}\left(x+20\right)^{100}=0\\\left|y+4\right|=0\end{cases}\)

\(\Rightarrow\begin{cases}x+20=0\\y+4=0\end{cases}\)\(\Rightarrow\begin{cases}x=-20\\y=-4\end{cases}\)

`x xx 2/3 xx 3/4 xx 4/5 xx ... xx 2010/2011 = 2/2012`

`<=> x/2011 = 1/1006`

`=> x = 2011/1006`