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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}
\(\)\(\dfrac{x+1}{2015}+\dfrac{x+2}{2014}=\dfrac{x+3}{2013}+\dfrac{x+4}{2012}\)
\(\Rightarrow\dfrac{x+1}{2015}+1+\dfrac{x+2}{2014}+1=\dfrac{x+3}{2013}+1+\dfrac{x+4}{2012}+1\)
\(\Rightarrow\dfrac{x+2016}{2015}+\dfrac{x+2016}{2014}=\dfrac{x+2016}{2013}+\dfrac{x+2016}{2012}\)
\(\Rightarrow\dfrac{x+2016}{2015}+\dfrac{x+2016}{2014}-\dfrac{x+2016}{2013}-\dfrac{x+2016}{2012}=0\)
\(\Rightarrow\left(x+2016\right)\left(\dfrac{1}{2015}+\dfrac{1}{2014}-\dfrac{1}{2013}-\dfrac{1}{2012}\right)=0\)
Vì \(\dfrac{1}{2015}+\dfrac{1}{2014}-\dfrac{1}{2013}-\dfrac{1}{2012}\ne0\)
Nên:
\(x+2016=0\Rightarrow x=-2016\)
Câu 5 :
Ta có :\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+....+\frac{1}{\left(x+99\right)\left(x+100\right)}=\frac{k}{x\left(x+100\right)}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+....+\frac{1}{x+99}-\frac{1}{x+100}=\frac{k}{x}-\frac{k}{x+100}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+100}=\frac{k}{x\left(x+100\right)}\)
\(\Leftrightarrow\frac{x+100}{x\left(x+100\right)}-\frac{x}{x\left(x+100\right)}=\frac{k}{x\left(x+100\right)}\)
\(\Leftrightarrow\frac{100}{x\left(x+100\right)}=\frac{k}{x\left(x+100\right)}\)
Vậy k = 100
Câu 6 :
Ta có : \(\left(x+5\right)^2-\left(x+2\right)\left(x-3\right)=-2\)
\(\Leftrightarrow x^2+10x+25-x^2+3x-2x+6+2=0\)
\(\Leftrightarrow11x+33=0\)
\(\Leftrightarrow11x=-33\)
\(\Leftrightarrow x=\frac{-33}{11}\)
\(\Leftrightarrow x=-3\)
Vậy x = -3
\(x^2 + y^2 +2x +1 = 0\)
\(<=> (x+1)^2 + y^2 = 0\)
\(mà (x+1)^2 \)\(\ge\) 0
\(y^2 \) \(\ge\) \(0\)
\(<=>\) \(\begin{cases} (x + 1)^2 = 0\\ y^2 = 0 \end{cases}\)
\(<=>\) \(\begin{cases} x = -1 \\ y= 0 \end{cases}\)
\(Vậy (x;y) = ( -1;0)\)
Ta có: \(x^2+y^2+2x+1=0\)
\(\Rightarrow\) \(\left(x+1\right)^2+y^2=0\)
\(\Rightarrow\) \(\left[\begin{matrix}\left(x+1\right)^2=0\\y^2=0\end{matrix}\right.\)
\(\Rightarrow\) \(\left[\begin{matrix}x=-1\\y=0\end{matrix}\right.\)
Vậy: Phương trình \(x^2+y^2+2x+1=0\) có nghiệm\(\left(x;y\right)=\left(-1;0\right)\)
Câu 8:
Ta có: \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{49.51}=\frac{6x-5}{10x+1}\)
\(\Rightarrow\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{49.51}\right)=\frac{6x-5}{10x+1}\)
\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{49}-\frac{1}{51}=\frac{6x-5}{10x+1}.2\)
\(\Rightarrow1-\frac{1}{51}=\frac{12x-10}{10x+1}\)
\(\Rightarrow\frac{50}{51}=\frac{12x-10}{10x+1}\)
\(\Rightarrow612x-510=500x+50\)
\(\Rightarrow112x=660\)
\(\Rightarrow x=5\)
Vậy x = 5
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 \(x\in\left\{-5;0\right\}\)
-5;0