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5 tháng 7 2023

\(\left(\dfrac{1}{2}+1\right)\times\left(\dfrac{1}{3}+1\right)\times\left(\dfrac{1}{4}+1\right)\times\left(\dfrac{1}{5}+1\right)\times\left(\dfrac{1}{6}+1\right)\)

\(=\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\times\dfrac{6}{5}\times\dfrac{7}{6}\)

\(=\dfrac{3\times4\times5\times6\times7}{2\times3\times4\times5\times6}\)

\(=\dfrac{7}{2}\)

5 tháng 7 2023

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14 tháng 2 2022

chào bn mik đến từ năm 2022

25 tháng 7 2020

1) \(x^2+\frac{8}{9}=\frac{41}{36}\)\(\Leftrightarrow x^2=\frac{1}{4}\Leftrightarrow x=\pm\frac{1}{2}\)

2) \(\left(x-3\right)^2+-\frac{9}{25}=\frac{2}{5}.\frac{8}{5}\)

\(\Leftrightarrow\left(x-3\right)^2=1\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=1\\x-3=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=2\end{cases}}\)

3) \(\frac{3}{11}.\frac{22}{6}-\left(x-1\right)^2=\frac{7}{16}\)

\(\Leftrightarrow\left(x-1\right)^2=\frac{9}{16}\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=\frac{3}{4}\\x-1=-\frac{3}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{4}\\x=\frac{1}{4}\end{cases}}\)

4) \(1+\left(x+1\right)^3=\frac{37}{64}\)

\(\Leftrightarrow\left(x+1\right)^3=-\frac{27}{64}\)

\(\Rightarrow x+1=-\frac{3}{4}\)

\(\Leftrightarrow x=-\frac{7}{4}\)

5) \(\left(x-\frac{1}{2}\right)^2-\frac{9}{16}=1\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=\frac{25}{16}\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{2}=\frac{5}{4}\\x-\frac{1}{2}=-\frac{5}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{4}\\x=-\frac{3}{4}\end{cases}}\)

6) Bn ghi rõ đề nha mk ko hiểu 

25 tháng 7 2020

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

\(\frac{3-x}{5-x}=\frac{9}{25}\Leftrightarrow\frac{3-x}{5-x}-\frac{9}{25}\Leftrightarrow\frac{75-25x}{125-25x}-\frac{45-9x}{125-5x}=0\)

\(\Rightarrow\frac{75-25-45+9x}{125-25x}=0\Leftrightarrow5+9x=0\Leftrightarrow x=\frac{-5}{9}\)

1 tháng 7 2017

Ta có : (2x + 1)4 = (2x + 1)6

=> (2x + 1)- (2x + 1)= 0

<=> (2x + 1)4[1 - (2x + 1)2] = 0

\(\Leftrightarrow\orbr{\begin{cases}\left(2x+1\right)^4=0\\1-\left(2x+1\right)^2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\\left(2x+1\right)^2=1\end{cases}}\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\2x=0;-2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=0;-1\end{cases}}\)

Vậy x thuộc \(-\frac{1}{2};0;-1\)

1 tháng 7 2017

hk hỉu j hết bn ạ

3 tháng 7 2017

3) \(\left(x-3\right)\left(x+3\right)\left(x^2+9\right)-\left(x^2-2\right)\left(x^2+2\right)\)

\(=\left(x^2-9\right)\left(x^2+9\right)-\left(x^4-4\right)\)

\(=\left(x^4-81\right)-\left(x^4-4\right)\)

\(=x^4-81-x^4+4\)

=-77 =>đpcm

4)\(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)

\(=\left[\left(3x+1\right)-\left(3x+5\right)\right]^2\)

\(=\left(3x+1-3x-5\right)^2\)

=(-4)2

=16 => đpcm

3 tháng 7 2017

1)\(\left(x-2\right)^2-\left(x-3\right)\left(x-1\right)=\left(x^2-4x+4\right)-\left(x^2-4x+3\right)=1\)

=>đpcm

2)\(\left(x-1\right)^3-\left(x+1\right)^3+6\left(x+1\right)\left(x-1\right)\)

\(=\left(x-1-x-1\right)\left[\left(x-1\right)^2+\left(x-1\right)\left(x+1\right)+\left(x+1\right)^2\right]+6\left(x^2-1\right)\)

\(=\left(-2\right)\left(x^2-2x+1+x^2-1+x^2+2x+1\right)+6x^2-6\)

\(=\left(-2\right)\left(3x^2+1\right)+6x^2-6=-6x^2-2+6x^2-6=-8\) => đpcm

27 tháng 10 2020

a) \(9\left(x-1\right)^2-\frac{4}{9}\div\frac{2}{9}=\frac{1}{4}\)

\(\Leftrightarrow9\left(x-1\right)^2-2=\frac{1}{4}\)

\(\Leftrightarrow9\left(x-1\right)^2=\frac{9}{4}\)

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

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

b) \(\left(3x-1\right)^6=\left(3x-1\right)^4\)

\(\Leftrightarrow\left(3x-1\right)^6-\left(3x-1\right)^4=0\)

\(\Leftrightarrow\left(3x-1\right)^4\cdot\left[\left(3x-1\right)^2-1\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(3x-1\right)^4=0\\\left(3x-1\right)^2=1\end{cases}}\Leftrightarrow x\in\left\{0;\frac{1}{3};\frac{2}{3}\right\}\)