Tìm x, biết: \(\left(3x-7\right)^{2015}\)=\(\left(3x-7\right)^{2017}\)
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Lời giải:
Ta thấy: \(f(x)=\frac{x^3}{1-3x+3x^2}\Rightarrow f(1-x)=\frac{(1-x)^3}{1-3(1-x)+3(1-x)^2}=\frac{(1-x)^3}{3x^2-3x+1}\)
\(\Rightarrow f(x)+f(1-x)=\frac{x^3}{1-3x+3x^2}+\frac{(1-x)^3}{3x^2-3x+1}=\frac{x^3+(1-x)^3}{3x^2-3x+1}=1\)
Do đó:
\(f\left(\frac{1}{2017}\right)+f\left(\frac{2016}{2017}\right)=1\)
\(f\left(\frac{2}{2017}\right)+f\left(\frac{2015}{2017}\right)=1\)
............
\(f\left(\frac{1008}{2017}\right)+f\left(\frac{1009}{2017}\right)=1\)
Cộng theo vế:
\(\Rightarrow A=f\left(\frac{1}{2017}\right)+f\left(\frac{2}{2017}\right)+f\left(\frac{3}{2017}\right)+...f\left(\frac{2015}{2017}\right)+f\left(\frac{2016}{2017}\right)\)
\(=\underbrace{1+1+1...+1}_{1008}=1008\)
Bạn xem lại đề nhé.
a) \(A=x^2+5y^2+2xy-4x-8y+2015\)
\(A=x^2-4x+4-2y\left(x-2\right)+y^2+2011+4y^2\)
\(A=\left(x-2\right)^2-2y\left(x-2\right)+y^2+2011+4y^2\)
\(A=\left(x-2-y\right)^2+4y^2+2011\)
Vì \(\left(x-y-2\right)^2\ge0;4y^2\ge0\)
\(\Rightarrow A_{min}=2011\)
Dấu bằng xảy ra : \(\Leftrightarrow\left\{{}\begin{matrix}x-y-2=0\\4y^2=0\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)
đầu bài trên tớ làm luôn nhá !!!
a, / 3x+1/= 5-3
/ 3x+1/= 2
3x+1=2
x+1 = 2:3
x+1 = 2 phần 3
x= 2/3 -1
x= -1/3
\(y'=7\left(-x^2+3x+7\right)^6.\left(-x^2+3x+7\right)'\)
\(=7\left(-2x+3\right)\left(-x^2+3x+7\right)^6\)
( 3x - 24 ) . 75 = 2.76 .1/20170
( 3x - 24 ) . 75 =235298
( 3x - 24 ) = 235298 : 75
( 3x - 24 ) =14
3x = 14 + 24
3x = 30
x = 0
dung 100%
\(\left(3x-2^4\right).7^5=2.7^6.\frac{1}{2017^0}\)
\(\Leftrightarrow\left(3x-16\right).7^5=2.7^6.1\)
\(\Leftrightarrow3x-16=\frac{2.7^6}{7^5}\)
\(\Leftrightarrow3x-16=2.7\)
\(\Leftrightarrow3x-16=14\)
\(\Leftrightarrow3x=30\)
\(\Leftrightarrow x=10\)
pt <=> \(\left(3x-1\right)^{2013}-\left(3x-1\right)^{2015}=0\)
\(\Leftrightarrow\left(3x-1\right)^{2013}\left(\left(3x-1\right)^2-1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)^{2013}\left(3x-2\right)3x=0\)
\(\orbr{\begin{cases}x=0\\3x-1=0,3x-2=0\end{cases}}\)
Vậy x=0, x=1/3,hoặc x=2/3
b)\(\left|21x-5\right|=\left|3x-7\right|\)
\(\Leftrightarrow\begin{cases}21x-5=3x-7\\21x-5=7-3x\end{cases}\)
\(\Leftrightarrow\begin{cases}9x=-1\\24x=12\end{cases}\)
\(\Leftrightarrow\begin{cases}x=-\frac{1}{9}\\x=\frac{1}{2}\end{cases}\)
a)\(\left|2x-7\right|=3\)
\(\Rightarrow2x-7=\pm3\)
Nếu \(2x-7=3\)
\(\Rightarrow2x=10\)
\(\Rightarrow x=5\)
Nếu \(2x-7=-3\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
(3x - 7)2015 = (3x - 7)2017
(3x - 7)2017 - (3x - 7)2015 = 0
(3x - 7)2017[(3x - 7)2 - 1] = 0
=> (3x - 7)2017 = 0 hoặc (3x - 7)2 = 1
=> 3x - 7 = 0 hoặc 3x - 7 = ± 1
=> x = 7/3 hoặc x = { 8/3 ; 2 }
Vậy x = { 2; 7/3; 8/3 }
\(y\left(y^2-1\right)=0\Leftrightarrow\orbr{\begin{cases}y=0\\y^2-1=0\end{cases}}\)