Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(5-7x^2=\left(\sqrt{5}\right)^2-\left(x\sqrt{7}\right)^2\)
\(=\left(\sqrt{5}-x\sqrt{7}\right)\left(\sqrt{5}+x\sqrt{7}\right)\)
\(3+4x=\left(\sqrt{3}\right)^2-\left(2\sqrt{x}\right)^2\) ( do x<0 )
\(=\left(\sqrt{3}-2\sqrt{x}\right)\left(3+2\sqrt{x}\right)\)
\(4x^4+625=\left(2x^2\right)^2+\left(5^2\right)^2=\left(2x^2\right)^2+2.2x^2.5^2+\left(5^2\right)^2-2.2x^2.5^2\)
\(=\left(2x^2+25\right)^2-100x^2=\left(2x^2+25-10x\right)\left(2x^2+25+10x\right)\)
\(4x^4+625\)
\(=4x^4+20x^3-20x^3+50x^2+50x^2-100x^2-250x+250x+625\)
\(=\left(4x^4+20x^3+50x^2\right)-\left(20x^3-100x^2-250x\right)+\left(50x^2+250x+625\right)\)
\(=2x^2\left(2x^2+10x+25\right)-10x\left(2x^2+10x+25\right)+25\left(2x^2+10x+25\right)\)
\(=\left(2x^2+10x+25\right)\left(2x^2-10x+25\right)\)
\(4x^4-37x^2+9=4x^4-36x^2-x^2+9=4x^2\left(x^2-9\right)-\left(x^2-9\right).\)
\(=\left(x^2-9\right)\left(4x^2-1\right)=\left(x-3\right)\left(x+3\right)\left(2x-1\right)\left(2x+1\right)\)
Đặt t = x2
đa thức trở thành 4t2 - 37t + 9
= 4t2 - t - 36t + 9
= ( 4t2 - 36t ) - ( t - 9 )
= 4t( t - 9 ) - ( t - 9 )
= ( t - 9 )( 4t - 1 )
= ( x2 - 9 )( 4x2 - 1 )
= ( x - 3 )( x + 3 )( 2x - 1 )( 2x + 1 )
a/ \(x^2-4x+3=\left(x^2-x\right)-\left(3x-3\right)=x\left(x-1\right)-3\left(x-1\right)=\left(x-1\right)\left(x-3\right)\)
b/ \(3x^2-5x+2=\left(3x^2-3x\right)-\left(2x-2\right)=3x\left(x-1\right)-2\left(x-1\right)=\left(x-1\right)\left(3x-2\right)\)
\(\left(x^2+4x+6\right)\left(x^2+6x+6\right)-3x^2\left(1\right)\)
Đặt \(x^2+5x+6=t\)Thay vào (1) ta được:
\(\left(t-x\right)\left(t+x\right)-3x^2\)
\(=t^2-x^2-3x^2\)
\(=t^2-4x^2\)
\(=\left(t-2x\right)\left(t+2x\right)\)Thay \(t=x^2+5x+6\)ta được:
\(\left(x^2+5x+6-2x\right)\left(x^2+5x+6+2x\right)\)
\(=\left(x^2+3x+6\right)\left(x^2+7x+6\right)\)
\(=\left(x^2+3x+6\right)\left(x^2+x+6x+6\right)\)
\(=\left(x^2+3x+6\right)\left[x\left(x+1\right)+6\left(x+1\right)\right]\)
\(=\left(x^2+3x+6\right)\left(x+1\right)\left(x+6\right)\)
\(x^4-2y^4-x^2y^2+x^2+y^2=\left(x^4-y^4\right)-\left(x^2y^2-x^2\right)+\left(y^2-y^4\right)=\left(x^2-y^2\right)\left(x^2+y^2\right)-x^2\left(y^2-1\right)-y^2\left(y^2-1\right)=\left(x^2+y^2\right)\left(x^2-y^2\right)-\left(y^2-1\right)\left(x^2+y^2\right)=\left(x^2+y^2\right)\left(x^2-y^2-y^2+1\right)=\left(x^2+y^2\right)\left(x^2-2y^2+1\right)\)
đề sai bét