1 + ( x - 1 ) ^2 = 26
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\(\left(1+\dfrac{2}{3}\right).\left(1+\dfrac{2}{4}\right).\left(1+\dfrac{2}{5}\right)....\left(1+\dfrac{2}{2020}\right).\left(1+\dfrac{2}{2021}\right)\)
= \(\dfrac{5}{3}.\dfrac{6}{4}.\dfrac{7}{5}.\dfrac{8}{6}.\dfrac{9}{7}....\dfrac{2022}{2020}.\dfrac{2023}{2021}\)
= \(\dfrac{1}{3}.\dfrac{1}{4}.2022.2023\)
= \(\dfrac{337.2023}{2}\)
= \(\dfrac{\text{681751}}{2}\)
\(\dfrac{1}{2}\left(x-2\right)+\dfrac{1}{3}\left(2-x\right)=x\\ \Leftrightarrow\dfrac{1}{2}\left(x-2\right)-\dfrac{1}{3}\left(x-2\right)=x\\ \Leftrightarrow\left(x-2\right).\left(\dfrac{1}{2}-\dfrac{1}{3}\right)=x\\ \Leftrightarrow\left(x-2\right).\left(\dfrac{3-2}{6}\right)=x\\ \Leftrightarrow\left(x-2\right).\dfrac{1}{6}=x\\ \Leftrightarrow\dfrac{1}{6}x-\dfrac{1}{3}-x=0\\ \Leftrightarrow\left(\dfrac{1}{6}-1\right)x=\dfrac{1}{3}\\ \Leftrightarrow\left(\dfrac{1-6}{6}\right)x=\dfrac{1}{3}\\ \Leftrightarrow\dfrac{-5}{6}x=\dfrac{1}{3}\\ \Leftrightarrow x=\dfrac{1}{3}:\left(-\dfrac{5}{6}\right)\\ \Leftrightarrow x=-\dfrac{2}{5}\)
Vậy \(x=-\dfrac{2}{5}\)
a: x/3-1/6=1/5
=>x/3=11/30
hay x=11/90
b: =>1/2x=2
hay x=4
c: =>2/3:x=-7-1/3=-22/3
=>x=-1/11
\(x^3+x^2+\dfrac{1}{3}x+\dfrac{1}{27}\)
\(=x^3+3\cdot\dfrac{1}{3}\cdot x^2+3\cdot\left(\dfrac{1}{3}\right)^2\cdot x+\left(\dfrac{1}{3}\right)^3\)
\(=\left(x+\dfrac{1}{3}\right)^3\)
\(\frac{2}{5}\times\frac{1}{2}-\frac{2}{5}\times\frac{1}{3}-\frac{2}{5}\times\)\(\frac{1}{6}\)
\(=\frac{2}{5}\times\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
\(=\frac{2}{5}\times\frac{0}{6}\)
\(=\frac{2}{5}\times0\)
\(=0\)
\(\frac{2}{5}\times\frac{1}{2}-\frac{2}{5}\times\frac{1}{3}-\frac{2}{5}\times\frac{1}{6}\)
\(=\frac{2}{5}\times\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
\(=\frac{2}{5}\times0\)
\(=0\)
\(a,\Rightarrow2^{2x-5}=2^5\Rightarrow2x-5=5\Rightarrow x=5\\ b,\Rightarrow4^{2x-1}=4^3\Rightarrow2x-1=3\Rightarrow x=2\\ c,\Rightarrow\left[{}\begin{matrix}2x+2=11\\2x+2=-11\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{13}{2}\end{matrix}\right.\\ d,\Rightarrow2^{x^3}=256=2^8\Rightarrow x^3=8\Rightarrow x=2\)
a)
\(2^{2x-5}=2^5\)
2x-5=5
2x=10
x=5
b)
\(4^{2x-1}=4^3\)
2x-1=3
2x=4
x=2
c)
\(\left(2x+2\right)^2=11^2\)
2x+2=11
2x=9
x=9/2
5(x-1)=125-25
5(x-1)=100
x-1=100:5
x-1=20
x=20+1
x=21
12(x-1):3=64+8
12(x-1):3=72
12(x-1)=72.3
12(x-1)=216
x-1=216:12
x-1=18
x=18+1
x=19
(x-1)^3=5^3
=>x-1=5
x=5+1
x=6
\(1+\left(x-1\right)^2=26\Rightarrow\left(x-1\right)^2=26-1=25\)
\(\Rightarrow\left(x-1\right)^2=5^2\Leftrightarrow x-1=5\Rightarrow x=5+1\)
\(x=6\) Vậy \(x=6\)
1+(x+1)2=26
(x+1)2=26-1
(x+1)2=25
Ta có: 52=25
=> x+1=5 ; x+1=-5
x=5-1 x=-5-1
x=4 x=-6
Vậy: \(x\in\){4;-6}