(3x-1)^10=(3x-1)
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TA
21 tháng 7 2021
`Q=(3x-1)(9x^2-3x+1)-(1-3x)(1+3x+9x^2)`
`=(3x-1)(9x^2-3x+1)+(3x-1)(9x^2+3x+1)`
`=(3x-1)(9x^2-3x+1+9x^2+3x+1)`
`=(3x-1)(18x^2+2)`
Thay `x=10` vào biểu thức: `Q=(3.10-1)(18 .10^2+2)=52258`
20 tháng 7 2019
\(\frac{3x}{\sqrt{3x+10}}=\sqrt{3x+1-1}\)
<=> 3x\(=\sqrt{3x+1-1}.\sqrt{3x+10}\)
<=> (3x)2 = (\(\sqrt{3x+1-1}.\sqrt{3x+10}\))2
<=> 9x2 = 9x2 + 30
<=> x = 0
=> x = 0
LS
0
25 tháng 11 2021
\(\Rightarrow\left(3x-1\right)^{20}-\left(3x-1\right)^{10}=0\\ \Rightarrow\left(3x-1\right)^{10}\left[\left(3x-1\right)^{10}-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}3x-1=0\\\left(3x-1\right)^{10}=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x-1=0\\3x-1=1\\3x-1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=\dfrac{2}{3}\\x=0\end{matrix}\right.\)
3 tháng 2 2021
a) Ta có: \(x^2-3x+7=1+2x\)
\(\Leftrightarrow x^2-3x+7-1-2x=0\)
\(\Leftrightarrow x^2-3x-2x+6=0\)
\(\Leftrightarrow x\left(x-3\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
Vậy: S={3;2}
b) Ta có: \(x^2-3x-10=0\)
\(\Leftrightarrow x^2-5x+2x-10=0\)
\(\Leftrightarrow x\left(x-5\right)+2\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
Vậy: S={5;-2}
c) Ta có: \(x^2-3x+4=2\left(x-1\right)\)
\(\Leftrightarrow x^2-3x+4=2x-2\)
\(\Leftrightarrow x^2-3x+4-2x+2=0\)
\(\Leftrightarrow x^2-3x-2x+6=0\)
\(\Leftrightarrow x\left(x-3\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
Vậy: S={3;2}
d) Ta có: \(\left(x+1\right)\left(x-2\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\\x=5\end{matrix}\right.\)
Vậy: S={-1;2;5}
e) Ta có: \(2x^2+3x+1=0\)
\(\Leftrightarrow2x^2+2x+x+1=0\)
\(\Leftrightarrow2x\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{-1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{-1;\dfrac{-1}{2}\right\}\)
f) Ta có: \(4x^2-3x=2x-1\)
\(\Leftrightarrow4x^2-3x-2x+1=0\)
\(\Leftrightarrow4x^2-5x+1=0\)
\(\Leftrightarrow4x^2-4x-x+1=0\)
\(\Leftrightarrow4x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\4x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{1;\dfrac{1}{4}\right\}\)
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 7 2021
a.
ĐKXĐL \(x\ge-\dfrac{1}{3}\)
\(\dfrac{3x}{\sqrt{3x+10}}=\dfrac{3x}{\sqrt{3x+1}+1}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\sqrt{3x+10}=\sqrt{3x+1}+1\left(1\right)\end{matrix}\right.\)
Xét (1)
\(\Leftrightarrow3x+10=3x+2+2\sqrt{3x+1}\)
\(\Leftrightarrow\sqrt{3x+1}=4\)
\(\Leftrightarrow x=5\)
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 7 2021
b.
ĐKXĐ: \(-1\le x\le1\)
\(\Leftrightarrow\dfrac{\left(1+x-1\right)}{\sqrt{1+x}+1}\left(\sqrt{1-x}+1\right)=2x\)
\(\Leftrightarrow\dfrac{x\left(\sqrt{1-x}+1\right)}{\sqrt{1+x}+1}=2x\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\dfrac{\sqrt{1-x}+1}{\sqrt{1+x}+1}=2\left(1\right)\end{matrix}\right.\)
Xét (1)
\(\Leftrightarrow\sqrt{1-x}+1=2\sqrt{1+x}+2\)
\(\Leftrightarrow\sqrt{1-x}=2\sqrt{1+x}+1\)
\(\Leftrightarrow1-x=4\left(x+1\right)+1+4\sqrt{x+1}\)
\(\Leftrightarrow4\sqrt{x+1}=-5x-4\) (\(x\le-\dfrac{4}{5}\))
\(\Leftrightarrow16\left(x+1\right)=25x^2+40x+16\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-\dfrac{24}{25}\end{matrix}\right.\)
31 tháng 1 2017
\(\Rightarrow x=0\)
Thử \(\left(3x-1\right)^{10}=\left(3.0-1\right)^{10}=\left(-1\right)^{10}=1\)
\(\left(3x-1\right)^{20}=\left(3.0-1\right)^{20}=\left(-1\right)^{20}=1\)
Suy ra \(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\)
N
2
24 tháng 1 2018
\(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\)
\(\Rightarrow\left(3x-1\right)^{10}=\left(3x-1\right)^{20}=0\)
\(\Rightarrow\left(3x-1\right)^{10}.\left[\left(3x-1\right)^{10}-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(3x-1\right)^{10}=0\\\left(3x-1\right)^{10}-1=0\end{cases}\Rightarrow\orbr{\begin{cases}3x-1=0\\3x-1=\pm0\end{cases}}}\)
\(\left(+\right)3x-1=0\Rightarrow x=\frac{1}{3}\)
\(\left(+\right)\orbr{\begin{cases}3x-1=1\\2x-1=-1\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=0\end{cases}}}\)
\(\Rightarrow x\in\left\{\frac{1}{3};\frac{2}{3};0\right\}\)
BN
0
`(3x-1)^10=(3x-1)`
`=>(3x-1)^10 :(3x-1)=(3x-1):(3x-1)`
`=>(3x-1)^9=1`
`=>3x-1=1`
`=>3x=2`
`=>x=2/3`
(3X - 1)10 = (3X -1)
(3X - 1)10 - (3X - 1) = 0
(3X - 1){(3X -1)9 - 1} = 0
3X - 1 = 0 hoặc (3X -1 )9 = 1
3X - 1 = 0
3X = 1
X = 1/3
(3X - 1 )9 - 1 = 0
(3X - 1)9 = 1
3X - 1 = 1
3X = 2
X = 2/3
vậy X ϵ { 1/3;2/3}