( 1/2 +1 ) x ( 1/3 + 1) x ( 1/4 + 1 ) x ( 1/5 + 1) x ( 1/6 + 1) . Gíup t vs . Mơn rất nhìu :)
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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 ) \(\frac{3-x}{5-x}=\left(\frac{-3}{5}\right)^2\)
\(\frac{3-x}{5-x}=\frac{9}{25}\Leftrightarrow\frac{3-x}{5-x}-\frac{9}{25}\Leftrightarrow\frac{75-25x}{125-25x}-\frac{45-9x}{125-5x}=0\)
\(\Rightarrow\frac{75-25-45+9x}{125-25x}=0\Leftrightarrow5+9x=0\Leftrightarrow x=\frac{-5}{9}\)
PN
3
1 tháng 7 2017
Ta có : (2x + 1)4 = (2x + 1)6
=> (2x + 1)4 - (2x + 1)6 = 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 \(-\frac{1}{2};0;-1\)
TM
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
TM
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
PT
1
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) \(\left(3x-1\right)^6=\left(3x-1\right)^4\)
\(\Leftrightarrow\left(3x-1\right)^6-\left(3x-1\right)^4=0\)
\(\Leftrightarrow\left(3x-1\right)^4\cdot\left[\left(3x-1\right)^2-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(3x-1\right)^4=0\\\left(3x-1\right)^2=1\end{cases}}\Leftrightarrow x\in\left\{0;\frac{1}{3};\frac{2}{3}\right\}\)
\(\left(\dfrac{1}{2}+1\right)\times\left(\dfrac{1}{3}+1\right)\times\left(\dfrac{1}{4}+1\right)\times\left(\dfrac{1}{5}+1\right)\times\left(\dfrac{1}{6}+1\right)\)
\(=\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\times\dfrac{6}{5}\times\dfrac{7}{6}\)
\(=\dfrac{3\times4\times5\times6\times7}{2\times3\times4\times5\times6}\)
\(=\dfrac{7}{2}\)
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