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\(-25x^2\sqrt{2}+10x+4\sqrt{2}=-\sqrt{2}\left(25x^2-\dfrac{10}{\sqrt{2}}-4\right)=-\sqrt{2}.\left(\left(25x\right)^2-2.5.\dfrac{1}{\sqrt{2}}+\dfrac{1}{2}-\dfrac{5}{2}\right)=-\sqrt{2}\left[\left(5x-\dfrac{1}{\sqrt{2}}\right)^2-\dfrac{5}{2}\right]=-\sqrt{2}.\left(5x-\dfrac{1}{\sqrt{2}}-\dfrac{\sqrt{5}}{\sqrt{2}}\right).\left(5x-\dfrac{1}{\sqrt{2}}+\dfrac{\sqrt{5}}{\sqrt{2}}\right)=-\sqrt{2}.\left(5x-\dfrac{1+\sqrt{5}}{\sqrt{2}}\right)\left(5x-\dfrac{1-\sqrt{5}}{\sqrt{2}}\right)\)
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\(25x^2-4y^2-4y-1\)
\(=25x^2-\left(2y+1\right)^2=\left(5x-2y-1\right)\left(5x+2y+1\right)\)
25x2 - 4y2 - 4y - 1
= 25x2 - (4y2 + 4y + 1)
= (5x)2 - (2y + 1)2
= [5x - (2y + 1)][5x + (2y + 1)]
= (5x - 2y - 1)(5x + 2y + 1)
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\(-\left(x+2y\right)^2\)
\(-\left(x-3\right)^2\)
\(\left(3-5x\right)^2\)
\(-x^2-4xy-4y^2=-\left(x+2y\right)^2\)
\(-x^2+6x-9=-\left(x-3\right)^2\)
\(25x^2-30x+9=\left(5x-3\right)^2\)
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Answer:
\(25x^2-10x+4y-4y^2\)
\(=25x^2-10x+1-4x^2+4y-1\)
\(=\left(25x^2-10x+1\right)-\left(4y^2-2y+1\right)\)
\(=[\left(5x\right)^2-2.5x.1+1]-[\left(2y\right)^2-2.2y.1+1]\)
\(=\left(5x-1\right)^2-\left(2y-1\right)^2\)
\(=\left(5x-1-2y+1\right).\left(5x-1+2y-1\right)\)
\(=\left(5x-2y\right).\left(5x+2y-2\right)\)
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A=12x2-16x-9x+12
A=4x(3x-4)-3(3x-4)
A=(3x-4)(4x-3)
dễ mà, tick nha
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\(25x^2-4y^2-4y-1=25x^2-\left(2y+1\right)^2\)
\(=\left(5x-2y-1\right)\left(5x+2y+1\right)\)
\(\left(25x^2-2\right)=\left(5x-\sqrt[]{2}\right)\left(5x+\sqrt[]{2}\right)\)