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\(\left(x+1\right)\left(x-2\right)^2+x^2\left(4-x\right)=13\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-4x+4\right)+4x^2-x^3=13\)
\(\Leftrightarrow x^3-4x^2+4x+x^2-4x+4+4x^2-x^3=13\)
\(\Leftrightarrow x^2=9\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=3\\x=-3\end{array}\right.\)
Vậy x={-3;3}
\(\frac{x+1}{2015}+\frac{x+2}{2014}=\frac{x+3}{2013}+\frac{x+4}{2012}\)
\(\Leftrightarrow\frac{x+1}{2015}+1+\frac{x+2}{2014}+1=\frac{x+3}{2013}+1+\frac{x+4}{2012}+1\)
\(\Leftrightarrow\frac{x+2016}{2015}+\frac{x+2016}{2014}=\frac{x+2016}{2013}+\frac{x+2016}{2012}\)
\(\Leftrightarrow\frac{x+2016}{2015}+\frac{x+2016}{2014}-\frac{x+2016}{2013}-\frac{x+2016}{2012}=0\)
\(\Leftrightarrow\left(x+2016\right)\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)
\(\Leftrightarrow x+2016=0\).Do \(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\ne0\)
\(\Leftrightarrow x=-2016\)
\(\)\(\dfrac{x+1}{2015}+\dfrac{x+2}{2014}=\dfrac{x+3}{2013}+\dfrac{x+4}{2012}\)
\(\Rightarrow\dfrac{x+1}{2015}+1+\dfrac{x+2}{2014}+1=\dfrac{x+3}{2013}+1+\dfrac{x+4}{2012}+1\)
\(\Rightarrow\dfrac{x+2016}{2015}+\dfrac{x+2016}{2014}=\dfrac{x+2016}{2013}+\dfrac{x+2016}{2012}\)
\(\Rightarrow\dfrac{x+2016}{2015}+\dfrac{x+2016}{2014}-\dfrac{x+2016}{2013}-\dfrac{x+2016}{2012}=0\)
\(\Rightarrow\left(x+2016\right)\left(\dfrac{1}{2015}+\dfrac{1}{2014}-\dfrac{1}{2013}-\dfrac{1}{2012}\right)=0\)
Vì \(\dfrac{1}{2015}+\dfrac{1}{2014}-\dfrac{1}{2013}-\dfrac{1}{2012}\ne0\)
Nên:
\(x+2016=0\Rightarrow x=-2016\)
\(3f\left(x\right)+2f\left(1-x\right)=2x+9\)
\(\left\{\begin{matrix}3f\left(2\right)+2f\left(-1\right)=2.2+9=13\left(1\right)\\3f\left(-1\right)+2f\left(2\right)=2.\left(-1\right)+9=7\left(2\right)\end{matrix}\right.\)
Lấy (1) nhân 3 trừ đi (2) nhân 2:
\(\left(3.3-2.2\right)f\left(2\right)+\left(6-6\right)f\left(-1\right)=13.3-7.2\)
\(f\left(2\right)=\frac{39-14}{9-4}=\frac{25}{5}=5\)
Câu hỏi của Phạm Mai Chi - Toán lớp 8 - Học toán với OnlineMath
Theo đề bài ta có:
\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=24\)
\(\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]=24\)
\(\left(x^2+5x+4\right)\left(x^2+5x+6\right)=24\)
Đặt \(y=x^2+5x+5\) ,ta được:
\(\left(y-1\right)\left(y+1\right)=24\)
\(y^2-1-24=0\)
\(y^2-25=0\)
\(\left(y+5\right)\left(y-5\right)=0\)
\(\left\{{}\begin{matrix}y+5=0\\y-5=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}y=-5\\y=5\end{matrix}\right.\)
Với y = 5 ta được : \(x^2+5x+5=5\)
\(\Rightarrow x=0\)
Với y = -5 ta được : \(x^2+5x+5=-5\)
\(\Rightarrow x=-5\)
Vậy \(x\in\left\{-5;0\right\}\)
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