\(5x\left(2x-\frac{1}{2}\right)+2\left(2x-\frac{1}{2}\rig...">
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11 tháng 7 2019

\(5x\left(2x-\frac{1}{2}\right)+2\left(2x-\frac{1}{2}\right)=0\)

\(\Rightarrow\left(2x-\frac{1}{2}\right)\left(5x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{2}=0\\5x+2=0\end{cases}\Rightarrow}\orbr{\begin{cases}2x=\frac{1}{2}\\5x=-2\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{4}\\x=\frac{-2}{5}\end{cases}}\)

11 tháng 7 2019

5x.(2x - 1/2) + 2.(2x - 1/2) = 0

<=> 5x.2x + 5x.(-1/2) + 2.2x + 2.(-1/2) = 0

<=> 10x2 - 5/2x + 4x - 1 = 0

<=> 10x2 - 13/2x - 1 = 0

=> x = 1/4 hoặc x = -2/5

9 tháng 9 2019

\(2^{x+2}+2^{x+1}-2^x=40\)

\(\Rightarrow2^x\left(2^2+2-1\right)=40\)

\(\Rightarrow2^x=8\)

\(\Rightarrow x=3\)

9 tháng 9 2019

2x+2 + 2x+1 - 2x = 40

2x.22+2x.2-2x=40

2x.(4+2-1)=40

2x.5=40

2x=8

2x=23

x=3

vậy x=3

20 tháng 7 2018

1.

a)\(\left(x-\frac{1}{2}\right)^2=0\)

\(\Leftrightarrow x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)

b)\(\left(x-2\right)^2=1\Leftrightarrow\orbr{\begin{cases}x-2=\sqrt{1}\\x-2=-\sqrt{1}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{1}+2\\x=-\sqrt{1}+2\end{cases}}\)

Mấy câu kia tương tự,bạn tự làm nha :)) 

6 tháng 7 2018

2 bạn kia làm hơi tắt, mình sẽ làm lại cho đầy đủ nha Nguyễn Phạm Thy Vân:

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

\(\Rightarrow2x-1=5x-4\)

\(\Rightarrow2x-5x=1-4\)

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

\(\Rightarrow x=1\)

Vậy x = 1

6 tháng 7 2018

vì (3/4)^2x-1=(3/4)^5x-4=> 2x-1=5x-4=> 2x=5x-3

3=5x-2x=3x=> x=1

27 tháng 9 2019

a)\(\left(\frac{4}{5}\right)^{2x+7}=\left(\frac{4}{5}\right)^4\)

=> 2x + 7 = 4 

     2x        = 4 - 7 

     2x        = -3

       x        = -3 : 2

       x         = -1,5

   Vậy x = -1,5

Bài 1

\(a,\left(\frac{3}{5}\right)^2-\left[\frac{1}{3}:3-\sqrt{16}.\left(\frac{1}{2}\right)^2\right]-\left(10.12-2014\right)^0\)

\(=\frac{9}{25}-\left[\frac{1}{9}-4.\frac{1}{4}\right]-1\)

\(=\frac{9}{25}-\left(-\frac{8}{9}\right)-1\)

\(=\frac{9}{25}+\frac{8}{9}-1\)

\(=\frac{56}{225}\)

\(b,|-\frac{100}{123}|:\left(\frac{3}{4}+\frac{7}{12}\right)+\frac{23}{123}:\left(\frac{9}{5}-\frac{7}{15}\right)\)

\(=\frac{100}{123}:\left(\frac{4}{3}\right)+\frac{23}{123}:\frac{4}{3}\)

\(=\left(\frac{100}{123}+\frac{23}{123}\right):\frac{4}{3}\)

\(=1:\frac{4}{3}=\frac{3}{4}\)

Phần c đăng riêng vì mk chưa tìm đc cách giải bt mỗi đáp án :v 

\(c,\frac{\left(-5\right)^{32}.20^{43}}{\left(-8\right)^{29}.125^{25}}\)

\(=\frac{\left(-5\right)^{32}.\left(4.5\right)^{43}}{\left[4.\left(-2\right)\right]^{29}.\left(-5^3\right)^{25}}\)

\(=\frac{-5^{32}.4^{43}.5^{43}}{4^{29}.\left(-2\right)^{29}.\left(5\right)^{75}}\)

\(=\frac{\left(-5^4\right)^8.4^{43}.5^{43}}{4^{29}.\left(-2\right)^{29}.\left(5^3\right)^{25}}\)

\(=-\frac{1}{2}\)

A=\(\left(\frac{1}{4}-1\right).\left(\frac{1}{9}-1\right).\left(\frac{1}{16}-1\right).............\left(\frac{1}{9801}-1\right).\left(\frac{1}{10000}-1\right)\)

A=\(\left(\frac{1-4}{4}\right).\left(\frac{1-9}{9}\right).\left(\frac{1-16}{16}\right).............\left(\frac{1-9801}{9801}\right).\left(\frac{1-10000}{10000}\right)\)

A=\(\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}.....................\frac{-9800}{9801}.\frac{-9999}{10000}\)

A=\(\frac{-1.3}{2^2}.\frac{-2.4}{3^2}.\frac{-3.5}{4^2}.....................\frac{-98.100}{99^2}.\frac{-99.101}{100^2}\)

A=\(\frac{\left[\left(-1\right).\left(-2\right).\left(-3\right)....................\left(-98\right).\left(-99\right)\right].\left(3.4.5............100.101\right)}{\left(2.3.4.........99.100\right).\left(2.3.4...............99.100\right)}\)

A=\(\frac{1.101}{100.2}\)=\(\frac{101}{200}\)

2

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.................+\frac{2}{x.\left(x+1\right)}=\frac{2015}{2017}\)

\(\frac{1}{3.2}+\frac{1}{6.2}+\frac{1}{10.2}+.................+\frac{2}{2.x.\left(x+1\right)}=\frac{1}{2}.\frac{2015}{2017}\)

\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.................+\frac{1}{x.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+..................+\frac{1}{x.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+..............+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2017}.\frac{1}{2}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{2017}.\frac{1}{2}\)

\(\frac{x+1}{2.\left(x+1\right)}-\frac{2}{2.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)

\(\frac{\left(x+1\right)-2}{2.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)

\(\frac{x-1}{2.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)

=>\(\frac{x-1}{x+1}=\frac{2015}{2017}.\frac{1}{2}:\frac{1}{2}\)

\(\frac{x-1}{x+1}=\frac{2015}{2017}\)

=>x+1=2017

=>x=2018-1

=>x=2016

Vậy x=2016

Còn bài 3 em ko biết làm em ms lớp 6

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