\(\left(2x+1\right)^3=9\times81\)

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15 tháng 2 2017

\(\left(2x+1\right)^3=9.81\)

10 tháng 2 2017
trả lời nhanh
12 tháng 2 2017

\(\left(2x+1\right)^3=9\times81\)

\(\left(2x+1\right)^3=9^3\)

\(\Rightarrow2x+1=9\)

 \(\Rightarrow2x=9-1\)

\(\Rightarrow2x=8\)

\(\Rightarrow x=4\)

Vậy \(x=4\)

12 tháng 2 2017

\(\left(2x+1\right)^3=9.9.9=9^3\Leftrightarrow\left(2x+1\right)=9\Leftrightarrow2x=8\Rightarrow x=4\)

3 tháng 10 2018

\(\frac{64^2.81^3.34}{2^{13}.3^9.17}=\frac{2^{12}.3^{12}.2.17}{2^{13}.3^9.17}=\frac{2^{13}.3^{12}.17}{2^{13}.3^9.17}=3^3=27\)

Chúc bạn học tốt!

30 tháng 1 2017

bài tập tết nâng cao phải ko

mk cũng có nhưng chưa làm dc

27 tháng 1 2020

tìm 2 số nguyên a và b biết :a+b=-1 và a.b=-12.Giup mình nha

23 tháng 5 2019

\(\left(3x-1\right)\left(\frac{-1}{2}x+5\right)=0\)

\(\orbr{\begin{cases}3x-1=0\\\frac{-1}{2}x+5=0\end{cases}}\)

\(\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}\)

24 tháng 5 2019

\(\frac{1}{4}+\frac{1}{3}:(2x-1)=-5\)

\(\Rightarrow\frac{1}{3}:(2x-1)=-5-\frac{1}{4}\)

\(\Rightarrow\frac{1}{3}:(2x-1)=\frac{-21}{4}\)

\(\Rightarrow2x-1=\frac{1}{3}:-\frac{21}{4}\)

\(\Rightarrow2x-1=\frac{1}{3}\cdot-\frac{4}{21}\)

\(\Rightarrow2x-1=\frac{-4}{63}\)

\(\Rightarrow2x=-\frac{4}{63}+1\)

\(\Rightarrow2x=\frac{59}{63}\Leftrightarrow x=\frac{59}{126}\)

20 tháng 10 2017

a) \(A=\frac{6^3.9^7.4^3-16^2.81^4}{\left(24.3\right)^2}\)

\(A=\frac{2^3.3^3.3^{14}.2^6-2^8.3^{16}}{24^2.3^2}\)

\(A=\frac{2^9.3^{17}-2^8.3^{16}}{3^2.8^2.3^2}\)

\(A=\frac{2^8.3^{16}.\left(2.3-1\right)}{3^4.2^6}\)

\(A=\frac{2^8.3^{16}.5}{3^4.2^6}=2.3^{12}.5=10.3^{12}\)

b) \(B=2198+123-198-1123+42\)

\(B=\left(2198-198\right)+\left(123-1123\right)+42\)

\(B=2000-1000+42=1042\)

12 tháng 6 2017

K chép lại đề, lm luôn nhé:

*\(\Rightarrow\) \(\left(\dfrac{7}{2}+2x\right)\cdot\dfrac{8}{3}=\dfrac{16}{3}\)

\(\Rightarrow\dfrac{7}{2}+2x=\dfrac{16}{3}:\dfrac{8}{3}=2\)

\(\Rightarrow2x=2-\dfrac{7}{2}=-\dfrac{3}{2}\)

\(\Rightarrow x=-\dfrac{3}{4}\)

* \(\Rightarrow\left|2x-\dfrac{2}{3}\right|=\dfrac{\dfrac{3}{4}-2}{2}=-\dfrac{5}{8}\)

=> K có gt x nào t/m đề

* Đề sai

* \(\Rightarrow\left[{}\begin{matrix}3x-1=0\\-\dfrac{1}{2}x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=10\end{matrix}\right.\)

*\(\Rightarrow\dfrac{1}{3}:\left(2x-1\right)=-5-\dfrac{1}{4}=-\dfrac{21}{4}\)

\(\Rightarrow2x-1=\dfrac{1}{3}:\left(-\dfrac{21}{4}\right)=-\dfrac{4}{63}\)

\(\Rightarrow2x=-\dfrac{4}{63}+1=\dfrac{59}{63}\)

\(\Rightarrow x=\dfrac{59}{63}:2=\dfrac{59}{126}\)

* \(\Rightarrow\left(2x+\dfrac{3}{5}\right)^2=\dfrac{9}{25}\)

\(\Rightarrow\left[{}\begin{matrix}2x+\dfrac{3}{5}=\dfrac{3}{5}\\2x+\dfrac{3}{5}=-\dfrac{3}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=0\Rightarrow x=0\\2x=-\dfrac{6}{5}\Rightarrow x=-\dfrac{3}{5}\end{matrix}\right.\)

* \(\Rightarrow-5x-1-\dfrac{1}{2}x+\dfrac{1}{3}=\dfrac{3}{2}x-\dfrac{5}{6}\)

\(\Rightarrow-5x-\dfrac{1}{2}x-\dfrac{3}{2}x=-\dfrac{5}{6}+1-\dfrac{1}{3}\)

\(\Rightarrow-7x=-\dfrac{1}{6}\)

\(\Rightarrow x=-\dfrac{1}{6}:7=-\dfrac{1}{42}\)

12 tháng 6 2017

a)\(\left(3\dfrac{1}{2}+2x\right).2\dfrac{2}{3}=5\dfrac{1}{3}\)

\(\left(\dfrac{7}{2}+2x\right).\dfrac{8}{3}=\dfrac{16}{3}\)

\(\dfrac{7}{2}+2x=\dfrac{16}{3}:\dfrac{8}{3}=2\)

\(2x=2-\dfrac{7}{2}=\dfrac{-3}{2}\Rightarrow x=\dfrac{-3}{4}\)

b)\(\dfrac{3}{4}-2.\left|2x-\dfrac{2}{3}\right|=2\)

\(2.\left|2x-\dfrac{2}{3}\right|=\dfrac{3}{4}-2=\dfrac{-1}{4}\)

\(\Rightarrow\left|2x-3\right|=\dfrac{-1}{8}\)

\(\Rightarrow x\in\varnothing\)

c) Đề sai,bạn có viết chữ x đâu,đó là phép tính mà.

d)\(\left(3x-1\right)\left(\dfrac{-1}{2}x+5\right)=0\)

\(\Leftrightarrow3x-1=0\Rightarrow x=\dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{-1}{2}x+5=0\Rightarrow x=10\)

e)\(\dfrac{1}{4}+\dfrac{1}{3}:\left(2x-1\right)=-5\)

\(\dfrac{1}{3}:\left(2x-1\right)=-5-\dfrac{1}{4}=\dfrac{-21}{4}\)

\(2x-1=\dfrac{1}{3}:\dfrac{-21}{4}=\dfrac{-4}{63}\)

\(\Rightarrow2x=\dfrac{59}{63}\Rightarrow x=\dfrac{59}{126}\)

g)\(\left(2x+\dfrac{3}{5}\right)^2-\dfrac{9}{25}=0\)

\(\left(2x+\dfrac{3}{5}\right)^2=0+\dfrac{9}{25}=\dfrac{9}{25}\)

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

\(th1:x=0\)

\(th2:x=\dfrac{-3}{5}\)

h)\(-5\left(x+\dfrac{1}{5}\right)-\dfrac{1}{2}\left(x-\dfrac{2}{3}\right)=\dfrac{3}{2}x-\dfrac{5}{6}\)

\(-5x+-1-\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{3}{2}x-\dfrac{5}{6}\)

\(\Leftrightarrow-5x+-1+\dfrac{5}{6}-\dfrac{1}{3}=2x\)

\(-5x+\dfrac{-1}{2}=2x\)

\(\dfrac{-1}{2}=2x+5x\)

\(\dfrac{-1}{2}=7x\Rightarrow x=\dfrac{-1}{14}\)

22 tháng 11 2016

a) \(\left(2x+1\right)^3=125\)

\(\Rightarrow2x+1=5\)

\(\Rightarrow2x=4\)

\(\Rightarrow x=2\)

Vậy \(x=2\)

b) \(1999^{2x-6}=1\)

\(\Rightarrow1999^{2x-1}=1999^0\)

\(\Rightarrow2x-1=0\)

\(\Rightarrow2x=1\)

\(\Rightarrow x=\frac{1}{2}\)

Vậy \(x=\frac{1}{2}\)

c) \(x^{2002}=x\)

\(\Rightarrow x^{2002}-x=0\)

\(\Rightarrow x.\left(x^{2001}-1\right)=0\)

\(\Rightarrow x=0\) hoặc \(x^{2001}-1=0\)

+) \(x=0\)

+) \(x^{2001}-1=0\Rightarrow x^{2001}=1\Rightarrow x=1\)

Vậy \(x\in\left\{0;1\right\}\)

d) \(\left(x-1\right)^2=9\)

\(\Rightarrow x-1=\pm3\)

+) \(x-1=3\Rightarrow x=4\)

+) \(x-1=-3\Rightarrow x=-2\)

Vậy \(x\in\left\{4;-2\right\}\)

e) \(\left(2x-3\right)^2=81\)

\(\Rightarrow2x-3=\pm9\)

+) \(2x-3=9\Rightarrow2x=12\Rightarrow x=6\)

+) \(2x-3=-9\Rightarrow2x=-6\Rightarrow x=-3\)

Vậy \(x\in\left\{6;-3\right\}\)

Các phần khác làm tương tự

22 tháng 11 2016

Dễ nhưng bận r