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21 tháng 7 2019

\(\frac{2^{4-x}}{16^5}=32^6\)

=> \(\frac{2^{4-x}}{\left(2^4\right)^5}=\left(2^5\right)^6\)

=> \(\frac{2^{4-x}}{2^{20}}=2^{30}\)

=> \(2^{4-x}=2^{30}.2^{20}\)

=> \(2^{4-x}=2^{50}\)

=> 4  - x = 50

=> x = 4 - 50 = -46

\(\frac{3^{2x+3}}{9^3}=9^{14}\)

=> \(\frac{3^{2x+3}}{\left(3^2\right)^3}=\left(3^2\right)^{14}\)

=> \(\frac{3^{2x+3}}{3^6}=3^{28}\)

=> \(3^{2x+3}=3^{28}.3^6\)

=> \(3^{2x+3}=3^{34}\)

=> 2x + 3 = 34

=> 2x = 34 - 3

=> 2x = 31

=> x = 31/2

29 tháng 8 2015

x = 1                                  

Bài 1:...
Đọc tiếp

Bài 1: Tính

a. \(\left(1+\frac{1}{1\cdot3}\right)\cdot\left(1+\frac{1}{2\cdot4}\right)\cdot\left(1+\frac{1}{3\cdot5}\right)+\left(1+\frac{1}{4\cdot6}\right).....\left(1+\frac{1}{99\cdot101}\right)\)

b. \(\left[\sqrt{0,64}+\sqrt{0,0001}-\sqrt{\left(-0,5\right)^2}\right]\div\left[3\cdot\sqrt{\left(0,04\right)^2}-\sqrt{\left(-2\right)^4}\right]\)

c. \(\frac{5.4^{15}\cdot9^9-4.3^{20}\cdot8^9}{5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6}-\frac{2^{19}\cdot6^{15}-7\cdot6^{10}\cdot2^{20}\cdot3^6}{9\cdot6^{19}\cdot2^9-4\cdot3^{17}\cdot2^{26}}+0,\left(6\right)\)

Bài 2: Tìm x, y, z biết :
a. \(\left(x-10\right)^{1+x}=\left(x-10\right)^{x+2009}\left(x\in Z\right)\)

b. \(\left|x-2007\right|+\left|x-2008\right|+\left|y-2009\right|+\left|x-2010\right|=3\left(x,y\in N\right)\) 

c. \(25-y^2=8\left(x-2009\right)^2\left(x,y\in Z\right)\)

d. \(2008\left(x-4\right)^2+2009\left|x^2-16\right|+\left(y+1\right)^2\le0\)

e. \(2x=3y\) ; \(4z=5x\) và \(3y^2-z^2=-33\)

Bài 3: Chứng minh rằng

a. \(1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{2009^2}>\frac{1}{2009}\)

b. \(\left[75\cdot\left(4^{2008}+4^{2007}+4^{2006}+...+4+1\right)+25\right]⋮100\)

Bài 4: 

a. Tìm giá trị nhỏ nhất của biểu thức : \(M=\left(x^2+2\right)+\left|x+y-2009\right|+2005\)

b. So sánh: \(31^{11}\) và \(\left(-17\right)^{14}\)

c. So sánh: \(\left(\frac{9}{11}-0,81\right)^{2012}\) và \(\frac{1}{10^{4024}}\)

1

Bài 1 :\(a,=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{100^2}{99.101}\)

           \(=\frac{2.3.4...100}{1.2.3...99}.\frac{2.3.4...100}{3.4...101}\)

          \(=100.\frac{2}{101}=\frac{200}{101}\)

10 tháng 10 2015

Phần c khó để tớ giải cho

22 tháng 9 2016

oho nhiều quá trời, lm chắc mỏi tay luôn

23 tháng 9 2016

\(\left(\frac{1}{2}\right)^5\times x=\left(\frac{1}{2}\right)^7\) 

              \(x=\left(\frac{1}{2}\right)^7\div\left(\frac{1}{2}\right)^5\)

             \(x=\left(\frac{1}{2}\right)^{7-5}=\left(\frac{1}{2}\right)^2=\frac{1}{4}\) .

\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{9}{21}\right)^2\) 

 \(\left(\frac{3}{7}\right)^2\times x=\left(\frac{3}{7}\right)^4\)            

              \(x=\left(\frac{3}{7}\right)^4\div\left(\frac{3}{7}\right)^2\)

              \(x=\left(\frac{3}{7}\right)^{4-2}=\left(\frac{3}{7}\right)^2=\frac{9}{49}\)

\(2^x=2\Rightarrow x=1\)

\(3^x=3^4\Rightarrow x=4\)

\(7^x=7^7\Rightarrow x=7\)

\(\left(-3\right)^x=\left(-3\right)^5\Rightarrow x=5\)

\(\left(-5\right)^x=\left(-5\right)^4\Rightarrow x=4\)

\(2^x=4\Leftrightarrow2^x=2^2\Rightarrow x=2\)

\(2^x=8\Leftrightarrow2^x=2^3\Rightarrow x=3\)

\(2^x=16\Leftrightarrow2^x=2^4\Rightarrow x=4\)

\(3^{x+1}=3^2\Leftrightarrow x+1=2\Leftrightarrow x=2-1\Rightarrow x=1\)

\(5^{x-1}=5\Leftrightarrow x-1=1\Leftrightarrow x=1+1\Rightarrow x=2\)

\(6^{x+4}=6^{10}\Leftrightarrow x+4=10\Leftrightarrow x=10-4\Rightarrow x=6\)

\(5^{2x-7}=5^{11}\Leftrightarrow2x-7=11\Leftrightarrow2x=11+7\Leftrightarrow2x=18\Leftrightarrow x=18\div2\Rightarrow x=9\)

\(\left(-2\right)^{4x+2}=64\)

\(2^{-4x+2}=2^6\Leftrightarrow-4x+2=6\Leftrightarrow-4x=6-2\Leftrightarrow-4x=4\Leftrightarrow x=4\div\left(-4\right)\Rightarrow x=-1\)

\(\left(\frac{1}{2}\right)^x=\left(\frac{1}{2}\right)^5\Rightarrow x=5\)

\(\left(\frac{5}{6}\right)^{2x}=\left(\frac{5}{6}\right)^5\Rightarrow2x=5\Rightarrow x=\frac{5}{2}\)

\(\left(\frac{3}{4}\right)^{2x-1}=\left(\frac{3}{4}\right)^{5x-4}\Rightarrow2x-1=5x-4\)

                                      \(2x-5x=-4+1\) 

                                           \(-3x=-3\Rightarrow x=1\)

\(\left(\frac{-1}{10}\right)^x=\frac{1}{100}\)

 \(\left(\frac{1}{10}\right)^{-x}=\left(\frac{1}{10}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)

\(\left(\frac{-3}{2}\right)^x=\frac{9}{4}\)

\(\left(\frac{3}{2}\right)^{-x}=\left(\frac{3}{2}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)

\(\left(\frac{-3}{5}\right)^{2x}=\frac{9}{25}\)

 \(\left(\frac{3}{5}\right)^{-2x}=\left(\frac{3}{5}\right)^2\Rightarrow-2x=2\Rightarrow x=-1\)

\(\left(\frac{-2}{3}\right)^x=\frac{-8}{27}\)

\(\left(\frac{-2}{3}\right)^x=\left(\frac{-2}{3}\right)^3\Rightarrow x=3\).

hehe.heheoho đánh tới què tay, hoa mắt lun r nekkk!!hum

8 tháng 8 2019

c) \(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)

\(\Leftrightarrow\left(\frac{x-1}{2009}-1\right)+\left(\frac{x-2}{2008}-1\right)=\left(\frac{x-3}{2007}-1\right)+\left(\frac{x-4}{2006}-1\right)\)

\(\Leftrightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)

\(\Leftrightarrow\left(x-2010\right).\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)=0\)

\(\Leftrightarrow x-2010=0\)

\(\Leftrightarrow x=0+2010\)

\(\Rightarrow x=2010\)

Vậy \(x=2010.\)

Mình chỉ làm câu c) thôi nhé.

Chúc bạn học tốt!

27 tháng 3 2018

Ta có: \(\frac{3}{\left(x+2\right)\left(x+5\right)}=\frac{1}{x+2}-\frac{1}{x+5}\);  \(\frac{5}{\left(x+5\right)\left(x+10\right)}=\frac{1}{x+5}-\frac{1}{x+10}\)

\(\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{1}{x+10}-\frac{1}{x+17}\)

=> Phương trình tương đương:

\(\frac{1}{x+2}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\frac{1}{x+2}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)<=> \(\frac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

<=> \(\frac{15}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

=> x=15

Đáp số: x=15