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1) \(\left(5x-4\right)\left(4x-5\right)+\left(5x-1\right)\left(x+4\right)+3\left(3x-2\right)\)
\(=20x^2-41x+20+\left(5x-1\right)\left(x+4\right)+3\left(3x-2\right)\)
\(=20x^2-41+20+5x^2+19x-4+3\left(3x-2\right)\)
\(=20x^2-41x+20+5x^2+19x-4+9x-4\)
\(=25x^2-13x+10\)
2) \(\left(5x-4\right)^2+\left(16-25x^2\right)+\left(5x+4\right)\left(3x+2\right)\)
\(=\left(5x-4\right)^2+16-25x^2+\left(5x-4\right)\left(3x+2\right)\)
\(=25x^2-40x+16^2-25x^2+\left(5x-4\right)\left(3x+2\right)\)
\(=25x^2-40x+16^2-25x^2+15x^2-2x-8\)
\(=15x^2-42x+24\)
1: \(=x^3-6x^2+12x-8-8x^3-36x^2-54x-27+7\left(x-1\right)^3\)
\(=-7x^3-42x^2-42x-35+7x^3-21x^2+21x-7\)
\(=-63x^2-21x-42\)
2: \(=x^3+125-\left(x^3-8\right)=125+8=133\)
3: \(=8x^3-27-8x^3-12x^2-6x-1=-12x^2-6x-28\)
1) \(\left(x-y-z\right)^2-\left(y+z\right)^2=\left(x\right).\left(x-2y-2z\right)=x^2-2yx-2zx\) 2) \(\left(2x+y\right)^2-4x\left(2x+y\right)+4x^2\Leftrightarrow\left(2x+y\right)\left(2x+y-4x\right)+4x^2\)
\(=\left(2x+y\right)\left(y-2x\right)+4x^2=\left(y^2-4x^2\right)+4x^2=y^2-4x^2+4x^2=y^2\)
3) \(\left(x+y\right)^2-2\left(x^2-y^2\right)+\left(x-y\right)^2\)
\(=x^2+2xy+y^2-2x^2+2y^2+x^2-2xy+y^2\)
\(=4y^2=\left(2y\right)^2\)
1. \(\left(x+5\right)\left(x^2-5x+25\right)-\left(x-2\right)\left(x^2+2x+4\right)\)
\(=x^3+125-\left(x^3-8\right)=x^3+125-x^3+8=133\)
1,
\(\left(x+5\right)\left(x^2-5x+25\right)-\left(x-2\right)\left(x^2+2x+4\right)\\ =\left(x^3+5^3\right)-\left(x^3-2^3\right)\\ =x^3+125-x^3+8\\ =\left(x^3-x^3\right)+\left(125+8\right)\\ =133\)
b,
\(\left(2x-3\right)\left(4x^2+6x+9\right)-\left(2x+1\right)^3\\ =\left[\left(2x\right)^3-3^3\right]-\left[\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot1+3\cdot2x+1+1\right]\\ =\left(8x^3-27\right)-\left(8x^3+12x^2+6x+1\right)\\ =8x^3-27-8x^3-12x^2-6x-1\\ =\left(8x^3-8x^3\right)-\left(12x^2+6x\right)-\left(27+1\right)\\ =-6x\left(2x+1\right)-28\\ =\left(-2\right)\left[3x\left(2x+1\right)+14\right]\)
\(a,\left(2x+1\right)^2-3x^2+4=\left(1-x\right)\left(1+x\right)\)
\(\Leftrightarrow4x^2+4x+1-3x^2+4=1-x^2\)
\(\Leftrightarrow4x^2+4x+1-3x^2+4-1+x^2=0\)
\(\Leftrightarrow2x^2+4x+4=0\)
\(\Leftrightarrow2\left(x^2+2x+1\right)+2=0\)
\(\Leftrightarrow2\left(x+1\right)^2=-2\)
\(\Leftrightarrow\left(x+1\right)^2=-1\Rightarrow\) pt vô nghiệm
\(b,\left(4x-3\right)\left(4x+3\right)-2\left(x+2\right)^2=14x^2\)
\(\Leftrightarrow16x^2-9-2\left(x^2+4x+4\right)-14x^2=0\)
\(\Leftrightarrow16x^2-9-2x^2-8x-8-14x^2=0\)
\(\Leftrightarrow-8x-17=0\)
\(\Leftrightarrow-8x=17\)
\(\Leftrightarrow x=\dfrac{-17}{8}\)
\(c,\left(2x-1\right)\left(x+1\right)-x^2+1=\dfrac{1}{2}\left(x-1\right)^2\)
\(\Leftrightarrow2x^2+2x-x-1-x^2+1=\dfrac{1}{2}\left(x^2-2x+1\right)\)
\(\Leftrightarrow2x^2+2x-x-1-x^2+1-\dfrac{1}{2}x^2+x-\dfrac{1}{2}=0\)\(\Leftrightarrow\dfrac{1}{2}x^2+2x-\dfrac{1}{2}=0\)
\(\Leftrightarrow\dfrac{1}{2}\left(x^2+4x+4\right)-\dfrac{5}{2}=0\)
\(\Leftrightarrow\dfrac{1}{2}\left(x+2\right)^2=\dfrac{5}{2}\)
\(\Rightarrow\left(x+2\right)^2=5\)
\(\Rightarrow\left[{}\begin{matrix}x+2=-\sqrt{5}\\x+2=\sqrt{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\sqrt{5}-2\\x=\sqrt{5}-2\end{matrix}\right.\)
a) \(\left(2x+1\right)^2-3x^2+4=\left(1-x\right)\left(1+x\right)\)
\(\Leftrightarrow4x^2+4x+1-3x^2+4=1-x^2\)
\(\Leftrightarrow4x^2+4x+1-3x^2+4-1+x^2=0\)
\(\Leftrightarrow2x^2+4x+4=0\Leftrightarrow\left(\sqrt{2}x\right)^2+2.\sqrt{2}.\sqrt{2}x+\left(\sqrt{2}\right)^2+2=0\) \(\Leftrightarrow\left(\sqrt{2}x+\sqrt{2}\right)^2+2=0\)
ta có : \(\left(\sqrt{2}x+\sqrt{2}\right)^2\ge0\Rightarrow\left(\sqrt{2}x+\sqrt{2}\right)^2+2\ge2>0\forall x\)
\(\Rightarrow\) phương trình vô nghiệm
vậy phương trình vô nghiệm
b) \(\left(4x-3\right)\left(4x+3\right)-2\left(x+2\right)^2=14x^2\)
\(\Leftrightarrow16x^2-9-2\left(x^2+4x+4\right)=14x^2\)
\(\Leftrightarrow16x^2-9-2x^2-8x-8=14x^2\)
\(\Leftrightarrow16x^2-9-2x^2-8x-8-14x^2=0\)
\(\Leftrightarrow-8x-17=0\Leftrightarrow-8x=17\Leftrightarrow x=\dfrac{-17}{8}\)
vậy \(x=\dfrac{-17}{8}\)
c) \(\left(2x-1\right)\left(x+1\right)-x^2+1=\dfrac{1}{2}\left(x-1\right)^2\)
\(\Leftrightarrow2x^2+2x-x-1-x^2+1=\dfrac{1}{2}\left(x^2-2x+1\right)\)
\(\Leftrightarrow2x^2+2x-x-1-x^2+1=\dfrac{1}{2}x^2-x+\dfrac{1}{2}\)
\(\Leftrightarrow2x^2+2x-x-1-x^2+1-\dfrac{1}{2}x^2+x-\dfrac{1}{2}=0\)
\(\Leftrightarrow\dfrac{1}{2}x^2+2x-\dfrac{1}{2}=0\Leftrightarrow\left(\dfrac{\sqrt{2}}{2}x\right)^2+2.\sqrt{2}.\dfrac{\sqrt{2}}{2}x+\left(\sqrt{2}\right)^2-\dfrac{5}{2}=0\)
\(\Leftrightarrow\left(\dfrac{\sqrt{2}}{2}x+\sqrt{2}\right)^2-\dfrac{5}{2}=0\Leftrightarrow\left(\dfrac{\sqrt{2}}{2}x+\sqrt{2}\right)^2=\dfrac{5}{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{\sqrt{2}}{2}x+\sqrt{2}=\sqrt{\dfrac{5}{2}}\\\dfrac{\sqrt{2}}{2}x+\sqrt{2}=-\sqrt{\dfrac{5}{2}}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{\sqrt{2}}{2}x=\sqrt{\dfrac{5}{2}}-\sqrt{2}=\dfrac{\sqrt{10}-2\sqrt{2}}{2}\\\dfrac{\sqrt{2}}{2}x=-\sqrt{\dfrac{5}{2}}-\sqrt{2}=-\dfrac{\sqrt{10}+2\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-2+\sqrt{5}\\x=-2-\sqrt{5}\end{matrix}\right.\)
vậy \(x=-2+\sqrt{5};x=-2-\sqrt{5}\)