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Ta có: \(x^2< 3x\)
\(\Leftrightarrow x^2-3x< 0\)
\(\Leftrightarrow x\left(x-3\right)< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x-3< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x-3>0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x< 3\end{matrix}\right.\\\left[{}\begin{matrix}x< 0\\x>3\end{matrix}\right.\left(loại\right)\end{matrix}\right.\Leftrightarrow0< x< 3\)
Vậy: S={x|0<x<3}
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(\frac{1}{2}x+y\right)\left(\frac{1}{2}x+y\right)\)
\(=\frac{1}{2}x.\left(\frac{1}{2}x+y\right)+y.\left(\frac{1}{2}x+y\right)\)
\(=\frac{1}{4}x^2+\frac{1}{2}xy+\frac{1}{2}xy+y^2\)
\(=\frac{1}{4}x^2+xy+y^2\)
b) \(\left(x^3-2x^2+x\right).\left(5-x\right)\)
\(=x^3.\left(5-x\right)-2x^2.\left(5-x\right)+x.\left(5-x\right)\)
\(=5x^3-x^4-10x^2+2x^3+5x-x^2\)
\(=7x^3-x^4-11x^2+5x\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(y=-4\\ \Rightarrow-4=3x^2-7\\ \Rightarrow3x^2=3\\ \Rightarrow x^2=1\\ \Rightarrow x=\pm1\)
\(y=5\\ \Rightarrow5=3x^2-7\\ \Rightarrow3x^2=12\\ \Rightarrow x^2=4\\ \Rightarrow x=\pm2\)
\(y=-6\dfrac{2}{3}\\ \Rightarrow-6\dfrac{2}{3}=3x^2-7\\ \Rightarrow3x^2=\dfrac{1}{3}\\ \Rightarrow x^2=\dfrac{1}{9}\\ \Rightarrow x=\pm\dfrac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(11x^2-15x+4=0\)
\(\Leftrightarrow11x^2-11x-4x+4=0\)
\(\Leftrightarrow11x\left(x-1\right)-4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(11x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\11x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{4}{11}\end{matrix}\right.\)
\(S=\left\{1,\dfrac{4}{11}\right\}\)
Đặt C(x)=0
\(\Leftrightarrow11x^2-15x+4=0\)
\(\Leftrightarrow11x^2-11x-4x+4=0\)
\(\Leftrightarrow11x\left(x-1\right)-4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(11x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\11x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\11x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{4}{11}\end{matrix}\right.\)
Vậy: Nghiệm của đa thức \(C\left(x\right)=11x^2-15x+4\) là 1 và \(\dfrac{4}{11}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: x+y+1=0
nên x+y=-1
Ta có: \(N=x^2\left(x+y\right)-y^2\left(x+y\right)+x^2-y^2+2\left(x+y\right)+3\)
\(=\left(x+y\right)\left(x^2-y^2\right)+\left(x^2-y^2\right)+2\left(x+y\right)+3\)
\(=\left(x^2-y^2\right)\left(x+y+1\right)+2\left(x+y\right)+3\)
\(=\left(x^2-y^2\right)\cdot0+2\cdot\left(-1\right)+3\)
=-2+3=1
đăng từ từ, ok nhìu thế này ai rỗi mà giải