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\(1,\Leftrightarrow\left\{{}\begin{matrix}\Delta=\left(-3\right)^2-4\left(-2\right)\left(-m+1\right)>0\\x_1+x_2=\dfrac{3}{-2}< 0\\x_1x_2=\dfrac{-m+1}{-2}>0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}17-8m>0\\-m+1< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m< \dfrac{17}{8}\\m>1\end{matrix}\right.\Leftrightarrow1< m< \dfrac{17}{8}\)
\(2,\Leftrightarrow\left\{{}\begin{matrix}\Delta=\left(-4\right)^2-4\left(-3\right)\left(-2m+1\right)\ge0\\x_1+x_2=\dfrac{4}{-3}< 0\\x_1x_2=\dfrac{-2m+1}{-3}>0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}28-24m\ge0\\-2m+1< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\le\dfrac{7}{6}\\m>\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\dfrac{1}{2}< m\le\dfrac{7}{6}\)
Bài 2:
a: =>2x^2-4x+1=x^2+x+5
=>x^2-5x-4=0
=>\(x=\dfrac{5\pm\sqrt{41}}{2}\)
b: =>11x^2-14x-12=3x^2+4x-7
=>8x^2-18x-5=0
=>x=5/2 hoặc x=-1/4
\(a,ĐK:...\\ PT\Leftrightarrow x^2-6x=x^2-7x+10\\ \Leftrightarrow x=10\left(tm\right)\\ b,ĐK:...\\ PT\Leftrightarrow2x\left(4-x\right)-\left(2-2x\right)\left(8-x\right)=\left(8-x\right)\left(4-x\right)\\ \Leftrightarrow8x-2x^2+16+18x-2x^2=32-12x+x^2\\ \Leftrightarrow3x^2-38x+16=0\left(casio\right)\\ c,ĐK:...\\ PT\Leftrightarrow2x\left(x-4\right)-4x=0\\ \Leftrightarrow2x^2-12x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)
\(\left|2x-5\right|+\left|2x^2-7x+5\right|=0\)
\(\left\{{}\begin{matrix}2x-5=0\\2x^2-7x+5=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x-5=0\\\left(2x-5\right)\left(x-1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{5}{2}\)
\(\Leftrightarrow\)\(x^2\left(y-2\right)+x\left(y-2\right)-x+4=0\)
\(\Leftrightarrow x\left(x+1\right)\left(y-2\right)-\left(x+1\right)=-5\)
\(\Leftrightarrow\left(x+1\right)\left(xy-2x-1\right)=-5\)
\(x;y\in Z\Rightarrow\left\{{}\begin{matrix}x+1\in Z\\xy-2x-1\in Z\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+1\inƯ\left(-5\right)\\xy-2x-1\inƯ\left(-5\right)\end{matrix}\right.\)
Bạn kẻ bảng sẽ tìm được (x;y) tương ứng
b) \(3\left(x^2+2x+1\right)=10\)
\(\Leftrightarrow\left(x+1\right)^2=\frac{10}{3}\)
Chia 2 TH tiếp .