152+35x-3x3-11x=
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4 tháng 10 2021
1: \(\dfrac{2x^3+11x^2+18x-3}{2x+3}\)
\(=\dfrac{2x^3+3x^2+8x^2+12x+6x+9-12}{2x+3}\)
\(=x^2+4x+3-\dfrac{12}{2x+3}\)
24 tháng 12 2021
\(A=BQ+R\\ \Leftrightarrow A:B=Q\left(\text{dư }R\right)\)
Ta có \(A:B=\left(2x^4+3x^3-5x^2-11x+8\right):\left(x^3-3x+1\right)\)
\(\Leftrightarrow A:B=\left(2x^4-6x^2+2x+3x^3-9x^2+3x+10x^2-16x+8\right):\left(x^3-3x+1\right)\\ \Leftrightarrow A:B=\left[\left(x^3-3x+1\right)\left(2x+3\right)+10x^2-16x+8\right]:\left(x^3-2x+1\right)\\ =2x+3\left(\text{dư }10x^2-16x+8\right)\\ \Leftrightarrow\left\{{}\begin{matrix}Q=2x+3\\R=10x^2-16x+8\end{matrix}\right.\)
9 tháng 12 2019
Ta có:
\(\frac{x}{x^2+x+1}=-\frac{1}{4}\Rightarrow x^2+x+1=-4x\)
\(\Rightarrow x^2+5x+1=0\Rightarrow x^2=5x+1\)
Với x2=5x+1 ta được:
\(P=\frac{2x\left(5x+1\right)^2+10\left(5x+1\right)^2+2x\left(5x+1\right)-7\left(5x+1\right)-35x+2009}{2029+60x+11\left(5x+1\right)-5x\left(5x+1\right)-\left(5x+1\right)^2}\)
\(P=\frac{2x\left(25x^2+10x+1\right)+10\left(25x^2+10x+1\right)+10x^2+2x-35x-7-35x+2009}{2029+60x+55x+11-25x^2-5x-\left(25x^2+10x+1\right)}\)
\(P=\frac{50x^3+20x^2+2x+250x^2+100x+10+10x^2+2x-35x-7-35x+2009}{2029+60x+55x+11-25x^2-5x-25x^2-10x-1}\)
\(P=\frac{50x^3+280x^2+34x+2012}{2039+100x-50x^2}\)
\(P=\frac{50x\left(5x+1\right)+280\left(5x+1\right)+34x+2012}{2039+100x-50\left(5x+1\right)}\)
\(P=\frac{250x^2+50x+1400x+280+34x+2012}{2039+100x-250x-50}\)
\(P=\frac{250\left(5x+1\right)+50x+1400x+280+34x+2012}{1989-150x}\)
\(P=\frac{1250x+250+50x+1400x+280+34x+2012}{1989-150x}\)
5 tháng 9 2021
\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)
\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
Pt vô nghiệm do ko có 2 bình phương số nguyên có tổng là 11
5 tháng 9 2021
e: Ta có: \(x^2-6x+y^2+4y+2=0\)
\(\Leftrightarrow x^2-6x+9+y^2+4y+4-11=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
Dấu '=' xảy ra khi x=3 và y=-2