\(\frac{4^{1007}.9^{1007}}{3^{2015}.2^{2016}}=\frac{\left(2^2\right)^{1007}.\left(3^2\right)^{1007}}{3^{2015}.2^{2016}}\)

\(=\frac{2^{2014}.3^{2014}}{3^{2015}.2^{2016}}=\frac{2^{2014}.3^{2014}}{3^{2014}.2^{2014}.3.2^2}\)

\(=\frac{1}{3.2^2}=\frac{1}{3.4}=\frac{1}{12}\)

Rút gọn

\(\frac{4^{1007}\cdot9^{1007}}{3^{2015}\cdot2^{2016}}=\frac{\left(2^2\right)^{2007}\cdot\left(3^2\right)^{1007}}{3^{2015}\cdot2^{2016}}\)

\(=\frac{2^{2\cdot1007}\cdot3^{2\cdot1007}}{3^{2015}\cdot2^{2016}}=\frac{2^{2014}\cdot3^{2014}}{3^{2015}\cdot2^{2016}}\)

\(=\frac{1}{3.2^2}=\frac{1}{12}\)

Vậy ...

hok tót .

có thể bn vt sai đề nhưng mk làm theo cái đề nha

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

\(=>\orbr{\begin{cases}x=\sqrt{\frac{1}{6}}-\frac{1}{2}\\x=-\sqrt{\frac{1}{6}}-\frac{1}{2}\end{cases}}\)

Vậy .....

\(\left(\left(x+\frac{1}{2}\right)\cdot2\right)-\frac{1}{6}=0\)

\(\frac{x\cdot2+1}{2}=\frac{2x+1}{2}\)

\(\left(\frac{\left(x\cdot2+1\right)}{2}\cdot2\right)-\frac{1}{6}=0\)

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

\(2x+1=\frac{2x+1}{1}=\frac{\left(2x+1\right)\cdot6}{6}\)

\(\frac{\left(2x+1\right)\cdot6-\left(1\right)}{6}=\frac{12x+5}{6}\)

\(\frac{12x+5}{6}\cdot6=0\cdot6\)

\(12x+5=0\)

\(x=\frac{-5}{12}=0.417\)

Giải:
Đặt \(\frac{x}{2}=\frac{y}{5}=\frac{z}{7}=k\)

\(\Rightarrow x=2k,y=5k,z=7k\)

Ta có: \(A=\frac{x-y+z}{x+2y-z}=\frac{2k-5k+7k}{2k+10k-7k}=\frac{k.\left(2-5+7\right)}{k\left(2+10-7\right)}=\frac{4k}{5k}=\frac{4}{5}\)

Vậy \(A=\frac{4}{5}\)

xem trên mạng

Mình tóm tắt sơ thôi rồi bạn tự làm

Có: \(|2x-1|\ge0;|1-2y|\ge0\)

=> \(|2x-1|+|1-2y|\ge0\)

TH1: \(|2x-1|+|1-2y|=0+4\)

=> \(\hept{\begin{cases}|2x-1|=0\\|1-2y|=4\end{cases}}\)hoặc \(\hept{\begin{cases}|2x-1|=4\\|1-2y|=0\end{cases}}\)

.................................... bạn tìm x;y rồi loại TH không thỏa mãn vì \(x;y\in Z\)

TH2:  ................................................................

TH3:  ................................................................

Tự làm nha. Mình nhátttttttt

ĐẶt A = \(\frac{1}{1+2}+\frac{1}{1+2+3}+....+\frac{1}{1+2+3+....+2014}\)

\(\frac{1}{2}A=\frac{1}{2.3}+\frac{1}{3.4}+........+\frac{1}{2014.2015}\)

\(\frac{1}{2}A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.....-\frac{1}{2014}+\frac{1}{2014}-\frac{1}{2015}\)

\(\frac{1}{2}A=\frac{1}{2}-\frac{1}{2015}=\frac{2013}{4030}\)

A = 2013/4030 : 1/2 = 2013/2015   

a) \(\frac{3}{4}-\frac{2}{5}.x=x\)

\(\Rightarrow\frac{-2}{5}.x-x=\frac{-3}{4}\)

\(x.\left(\frac{-2}{5}-1\right)=\frac{-3}{4}\)

\(x.\frac{-7}{5}=\frac{-3}{4}\)

\(x=\frac{-3}{4}:\left(\frac{-7}{5}\right)\)

\(x=\frac{15}{28}\)

b) (2x-1).(3x-1/5).(4-2x) = 0

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

3x-1/5 = 0 => 3x = 1/5 => x = 1/15

4-2x = 0 => 2x = 4 => x = 2

KL: x = 1/2 hoặc x = 1/15 hoặc x = 2

M = 3⁹⁵ + 3⁹⁶ + 3⁹⁷ + 3⁹⁸

M = (3⁹⁵ + 3⁹⁶) + (3⁹⁷ + 3⁹⁸)

M = 3⁹⁵(1 + 3) + 3⁹⁷(1 + 3)

M = 3⁹⁵.4 + 3⁹⁷.4

M = 4(3⁹⁵ + 3⁹⁷) chia hết cho 4 (đpcm)

M = 3^95 + 3^96 + 3^97 + 3^98 = 3^95(3^3 + 3^2 + 3 + 1) = 3^95.[3^2(3+1) + 3 + 1] = 3^95.(3^2 + 1).4 chia hết cho 4

Xin lỗi nha, bài nhìn hơi hack não, bạn tự dịch nha