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\(a,\Leftrightarrow2x^2+10x-2x^2=12\Leftrightarrow x=\dfrac{12}{10}=\dfrac{6}{5}\\ b,\Leftrightarrow\left(5-2x-4\right)\left(5-2x+4\right)=0\\ \Leftrightarrow\left(1-2x\right)\left(9-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\\ d,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ e,\Leftrightarrow4x^2-4x+1-4x^2+196=0\\ \Leftrightarrow-4x=-197\Leftrightarrow x=\dfrac{197}{4}\)
\(f,\Leftrightarrow x^2+8x+16-x^2+1=16\Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\\ g,Sửa:\left(3x+1\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(3x+1-x-1\right)\left(3x+1+x+1\right)=0\\ \Leftrightarrow2x\left(4x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\\ h,\Leftrightarrow x^2+8x-x-8=0\\ \Leftrightarrow\left(x+8\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-8\end{matrix}\right.\\ i,\Leftrightarrow2x^2-13x+15=0\\ \Leftrightarrow2x^2+2x-15x-15=0\\ \Leftrightarrow\left(x+1\right)\left(2x-15\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{15}{2}\end{matrix}\right.\)
\(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\\ \Leftrightarrow2x\cdot\left(4x+3\right)-15\cdot\left(6x-2\right)=35\cdot\left(5x+4\right)+315\\ \Leftrightarrow80x+63-90x+30=175x+140+315\\ \\\Leftrightarrow-6x+93=175x+455\\ \Leftrightarrow93=175x+455+6x\\ \Leftrightarrow93=181x+45\\ \Leftrightarrow-362=181x\\ \Rightarrow x=-\frac{362}{181}=-2\)
cái này trong toán violympic tiếng anh cấp tỉnh vong 9 do
\(\frac{x^3-x^2-x-2}{x^5-3x^4+4x^3-5x^2+3x-2}\)
\(=\frac{x^3-2x^2+x^2-2x+x-2}{x^5-2x^4-x^4+2x^3+2x^3-4x^2-x^2+2x+x-2}\)
\(=\frac{\left(x^3-2x^2\right)+\left(x^2-2x\right)+\left(x-2\right)}{\left(x^5-2x^4\right)-\left(x^4-2x^3\right)+\left(2x^3-4x^2\right)-\left(x^2-2x\right)+\left(x-2\right)}\)
\(=\frac{x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)}{x^4\left(x-2\right)-x^3\left(x-2\right)+2x^2\left(x-2\right)-x\left(x-2\right)+\left(x-2\right)}\)
\(=\frac{\left(x-2\right)\left(x^2+x+1\right)}{\left(x-2\right)\left(x^4-x^3+2x^2-x+1\right)}=\frac{x^2+x+1}{x^4-x^3+2x^2-x+1}\)
Ta có : \(\left(3x-1\right)^2-\left(x+7\right)^2=0\)
\(\left(3x-1+x+7\right)\left(3x-1-x-7\right)=0\)
\(\left(4x+6\right)\left(2x-8\right)=0\)
Nên : 4x + 6 = 0 hoặc 2x - 8 = 0
4x = -6 hoặc 2x = 8
x = \(\frac{-3}{2}\) hoặc x = 4
Vậy x = \(\frac{-3}{2}\) hoặc x = 4
\(Chúc bạn học tốt!!!\)
g/ x(x-2)-x2=5x-7
\(\Leftrightarrow x^2-2x-x^2=5x-7\\ \Leftrightarrow7x=7\Leftrightarrow x=1\)
h/\(3x\left(x-7\right)+2\left(x-7\right)=0\\ \Leftrightarrow3x^2-19x-14=0\\ \)
\(\Leftrightarrow\left(x-7\right)\left(3x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\frac{2}{3}\end{matrix}\right.\)