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a)\(x\left(x+4\right)+3x=x^2+3x-4\)
\(\Leftrightarrow x^2+4x+3x=x^2+3x-4\)
\(\Leftrightarrow x^2+7x=x^2+3x-4\)
\(\Leftrightarrow4x=-4\Rightarrow x=-1\)
b)\(3x\left(4-2x\right)+2x\left(3x+1\right)=5+2x\)
\(\Leftrightarrow12x-6x^2+6x^2+2x=5+2x\)
\(\Leftrightarrow14x=5+2x\)
\(\Leftrightarrow12x=5\Rightarrow x=\dfrac{5}{12}\)
a, \(x\left(x+4\right)+3x=x^2+3x-4\)
\(\Leftrightarrow x^2+4x+3x=x^2+3x-4\)
\(\Leftrightarrow4x=-4\Leftrightarrow x=-1\)
Vậy x = -1
b, \(3x\left(4-2x\right)+2x\left(3x+1\right)=5+2x\)
\(\Leftrightarrow12x-6x^2+6x^2+2x=5+2x\)
\(\Leftrightarrow12x=5\Leftrightarrow x=\dfrac{5}{12}\)
Bài 1.
\( a)\dfrac{{4x - 8}}{{2{x^2} + 1}} = 0 (x \in \mathbb{R})\\ \Leftrightarrow 4x - 8 = 0\\ \Leftrightarrow 4x = 8\\ \Leftrightarrow x = 2\left( {tm} \right)\\ b)\dfrac{{{x^2} - x - 6}}{{x - 3}} = 0\left( {x \ne 3} \right)\\ \Leftrightarrow \dfrac{{{x^2} + 2x - 3x - 6}}{{x - 3}} = 0\\ \Leftrightarrow \dfrac{{x\left( {x + 2} \right) - 3\left( {x + 2} \right)}}{{x - 3}} = 0\\ \Leftrightarrow \dfrac{{\left( {x + 2} \right)\left( {x - 3} \right)}}{{x - 3}} = 0\\ \Leftrightarrow x - 2 = 0\\ \Leftrightarrow x = 2\left( {tm} \right) \)
Bài 2.
\(c)\dfrac{{x + 5}}{{3x - 6}} - \dfrac{1}{2} = \dfrac{{2x - 3}}{{2x - 4}}\)
ĐK: \(x\ne2\)
\( Pt \Leftrightarrow \dfrac{{x + 5}}{{3x - 6}} - \dfrac{{2x - 3}}{{2x - 4}} = \dfrac{1}{2}\\ \Leftrightarrow \dfrac{{x + 5}}{{3\left( {x - 2} \right)}} - \dfrac{{2x - 3}}{{2\left( {x - 2} \right)}} = \dfrac{1}{2}\\ \Leftrightarrow \dfrac{{2\left( {x + 5} \right) - 3\left( {2x - 3} \right)}}{{6\left( {x - 2} \right)}} = \dfrac{1}{2}\\ \Leftrightarrow \dfrac{{ - 4x + 19}}{{6\left( {x - 2} \right)}} = \dfrac{1}{2}\\ \Leftrightarrow 2\left( { - 4x + 19} \right) = 6\left( {x - 2} \right)\\ \Leftrightarrow - 8x + 38 = 6x - 12\\ \Leftrightarrow - 14x = - 50\\ \Leftrightarrow x = \dfrac{{27}}{5}\left( {tm} \right)\\ d)\dfrac{{12}}{{1 - 9{x^2}}} = \dfrac{{1 - 3x}}{{1 + 3x}} - \dfrac{{1 + 3x}}{{1 - 3x}} \)
ĐK: \(x \ne -\dfrac{1}{3};x \ne \dfrac{1}{3}\)
\( Pt \Leftrightarrow \dfrac{{12}}{{1 - 9{x^2}}} - \dfrac{{1 - 3x}}{{1 + 3x}} - \dfrac{{1 + 3x}}{{1 - 3x}} = 0\\ \Leftrightarrow \dfrac{{12}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} - \dfrac{{1 - 3x}}{{1 + 3x}} - \dfrac{{1 + 3x}}{{1 - 3x}} = 0\\ \Leftrightarrow \dfrac{{12 - {{\left( {1 - 3x} \right)}^2} - {{\left( {1 + 3x} \right)}^2}}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} = 0\\ \Leftrightarrow \dfrac{{12 + 12x}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} = 0\\ \Leftrightarrow 12 + 12x = 0\\ \Leftrightarrow 12x = - 12\\ \Leftrightarrow x = - 1\left( {tm} \right) \)
a) \(\left(3x-4\right)^2+2\left(3x-4\right)\left(2x+4\right)+\left(2x+4\right)^2\)\(=\left(3x-4+2x+4\right)^2=\left(5x\right)^2=25x^2\)
b)\(\left(3x+4\right)^2+\left(7+3x\right)^2-\left(6x+8\right)\left(3x+7\right)\)
\(=\left(3x+4\right)^2-2\left(3x+4\right)\left(7+3x\right)+\left(7+3x\right)^2\)
\(=\left[3x+4-\left(7+3x\right)\right]^2=\left(3x+4-7-3x\right)^2=\left(-3\right)^2=9\)
c)\(\left(2x+1\right)^2+2\left(4x^2-1\right)+\left(2x-1\right)^2\)
\(=\left(2x+1\right)^2+2\left(\left(2x\right)^2-1^2\right)+\left(2x-1\right)^2\)
\(=\left(2x+1\right)^2+2\left(2x+1\right)\left(2x-1\right)+\left(2x-1\right)^2\)
\(=\left(2x+1+2x-1\right)^2=\left(4x\right)^2=14x^2\)
xong rồi đấy,bạn k cho mình nhé