x² + 3² = (9 - x)²

x² + 9 = (9 - x)(9 - x)

x² + 9 = 9(9 - x) - x(9 - x)

x² + 9 = 81 - 9x - 9x + x²

x² + 9 = 81 - 18x + x²

81 - 18x = 9

18x = 81 - 9 

18x = 72

=> x = 4

Vậy x = 4

Ta có :

\(\left(\frac{1}{16}\right)^{200}=\frac{1}{16^{200}}=\frac{1}{\left(2^4\right)^{200}}=\frac{1}{2^{800}}\)

Vì \(\frac{1}{2^{800}}>\frac{1}{2^{1000}}\) nên \(\left(\frac{1}{16}\right)^{200}>\frac{1}{2^{1000}}\)

Bấm máy thì cả 2 đều = 0

=> (1/16)^200 = (1/2)^1000

Tam giác ABC vuông có: \(AB^2+AC^2=BC^2\)

mà trong tam giác vuông ABH có: \(AH^2+BH^2=AB^2\)

tương tự tam giác vuông ACH có: \(HC^2+AH^2=AC^2\)

thay vào biểu thức đầu ta có ĐPCM

mình gợi ý được không

thôi để mình làm luôn:

\(C=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\left(1-\frac{1}{1+2+3+4}\right)...\left(1-\frac{1}{1+2+...+2017}\right)\)

\(=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)...\left(1-\frac{1}{\left(2017.2018\right):2}\right)\)

\(=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}...\frac{\left(2017.2018\right):2-1}{\left(2017.2018\right):2}\)

\(=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}...\frac{\left(2017.2018:2-1\right).2}{2017.2018}\)

\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}...\frac{2016.2019}{2017.2018}\)

\(=\frac{\left(1.2...2016\right)\left(4.5...2019\right)}{\left(2.3...2017\right)\left(3.4...2018\right)}\)

\(=\frac{2019}{2017.3}=\frac{2019}{6051}\)

\(D=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2015.2016.2017}\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2015.2016}-\frac{1}{2016.2017}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2016.2017}\right)=\frac{1}{2}.\left(\frac{2016.2017:2-1}{2016.2017}\right)\)

D=1/1.2.3+1/2.3.4+....+1/2015.2016.2017

D=1/2(1/1.2-1/2.3+1/2.3-1/3.4+.......+1/2015.2016-1/2016.2017)

D=1/2(1/1.2-1/2016.2017)

K

ta có 1/2(3/4x-1/2)^2006=0

=>3/4x-1/2=0

<=>2017/2018|4/5y+6/25|=0

=>34/5y+6/25=0

=>x=2/3

=>y=-3/10

ta có 1/2(3/4x-1/2)^2006=0

=>3/4x-1/2=0

=>2007/2008|4/5y+6/25|=0

=>4/5y+6/25=0

=>x=2/3

=>y=-3/10

Vì \(\left|2x-1\right|\ge0\) Để \(\left|2x-1\right|=x+4\Leftrightarrow x+4\ge0\Rightarrow x\ge-4\)

Ta có : \(\left|2x-1\right|=\hept{\begin{cases}2x-1\Leftrightarrow2x-1\ge0\Rightarrow x\ge\frac{1}{2}\\1-2x\Leftrightarrow2x-1< 0\Rightarrow x< \frac{1}{2}\end{cases}}\)

TH1 : Với x ≥ 1/2 thì 2x - 1 = x + 4 <=> 2x - x = 4 + 1 => x = 5 ( TM )

TH2 : Với x < 1/2 thì 1 - 2x = x + 4 <=> - 2x - x = 4 - 1 <=> - 3x = 3 => x = - 1 ( TM )

vẬY X = { - 1; 5 }

|2x-1|=x+4

<=>2x-1=x+4

=>2x-x=4+1

=>x=5