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5 tháng 8 2020

7.2^x = 2^8(2+5)=7.2^8

=>x=8

học tốt

5 tháng 8 2020

\(7\times2^x=2^9+5\times2^8\)

\(\Rightarrow7\times2^x=2^8\left(2+5\right)\)

\(\Rightarrow7\times2^x=7\times2^8\)

\(\Rightarrow2^x=2^8\)

\(\Rightarrow x=8\)

10 tháng 8 2020

Ta có :\(\frac{6^8.2^4-4^5.18^4}{27^3.8^4-3^9.2^{13}}=\frac{\left(2.3\right)^8.2^4-\left(2^2\right)^5.\left(3^2.2\right)^4}{\left(3^3\right)^3.\left(2^3\right)^4-3^9.2^{13}}=\frac{2^{12}.3^8-2^{14}.3^8}{3^9.2^{12}-3^9.2^{13}}=\frac{3^8.2^{12}.\left(2^2-1\right)}{3^9.2^{12}.\left(1-2\right)}\)

\(=\frac{3^9.2^{12}}{-3^9.2^{12}}=-1\)

10 tháng 8 2020

\(\frac{6^8\cdot2^2-4^5\cdot18^4}{27^3\cdot8^4-3^9\cdot2^{13}}\)

\(=\frac{\left(2.3\right)^8.2^4-\left(2^2\right)^5.\left(3^2.2\right)^4}{\left(3^3\right)^3.\left(2^3\right)^4-3^9.2^{13}}\)

\(=\frac{2^{12}.3^8-2^{14}.3^8}{3^9.2^{12}-3^9.2^{14}}\)

\(=\frac{3^8.2^{12}.\left(2^2-1\right)}{3^9.2^{12}.\left(1-2\right)}\)

\(=\frac{3^9.2^{12}}{-3^9.2^{12}}=-1\)

25 tháng 12 2018

\(-\frac{5}{9}\left(\frac{3}{10}-\frac{2}{5}\right)=-\frac{5}{9}\left(\frac{3}{10}-\frac{4}{10}\right)=-\frac{5}{9}.\frac{-1}{10}=\frac{1}{18}\)

\(\frac{1}{2}\sqrt{64}-\sqrt{\frac{9}{25}}+1^{2016}=\frac{1}{2}.8-\frac{3}{5}+1=4+\frac{2}{5}=\frac{22}{5}\)

\(2^8:2^5+3^2.2-12=2^3+9.2-12=8+18-12=8+6=14\)

25 tháng 12 2018

\(3^x+\sqrt{\frac{16}{81}}-\sqrt{9}+\frac{\sqrt{81}}{3}=9\frac{4}{9}\)

\(3^x+\frac{4}{9}-3+\frac{9}{3}=9\frac{4}{9}\)

\(3^x+\frac{4}{9}-3+3=9\frac{4}{9}\)

\(3^x+\frac{4}{9}=9+\frac{4}{9}\)

\(\Rightarrow3^x=9+\frac{4}{9}-\frac{4}{9}\)

\(3^x=9\)

\(3^x=3^2\)

\(\Rightarrow x=2\)

Vậy \(x=2\)

29 tháng 1 2016

\(\Leftrightarrow\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+..+\frac{2}{8.9.10}\right).x=\frac{44}{45}\)

\(\Leftrightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+..+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{44}{45}\)

\(\Leftrightarrow\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{44}{45}\Leftrightarrow\frac{22}{45}.x=\frac{44}{5}\Leftrightarrow x=2\)

vậy x=2

29 tháng 1 2016

x=2 nhé

olm đang duyệt

5 tháng 8 2020

\(\left(\frac{1}{3}+\frac{1}{6}\right).2^{x+4}-2^x=2^{13}-2^{10}\)

\(\frac{1}{2}.2^x.2^4-2^x=8192-1024\)

\(2^x.8-2^x=7168\)

\(2^x\left(8-1\right)=7168\)

\(2^x.7=7168\)

\(2^x=7168\div7\)

\(2^x=1024\)

\(2^x=2^{10}\)

\(\Rightarrow x=10\)

Vậy \(x=10\).

(1/3+1/6).2^x.2^4-2^x=8192-1024

(1/3+1/6).2^x.2^4-2^x=7168

1/2.2^x.2^4-2^x=7168

1/2.2^x.(2^4-1)=7168

1/2.2^x.(8-1)=7168

1/2.2^x.7=7168

1/2.2^x=7168:7

1/2.2^x=1024

      2^x=1024:1/2

     2^x=2048

2^x=2^11

x=11

vậy x=11

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

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

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