Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
\begin{array}{l} a){\left( {ab - 1} \right)^2} + {\left( {a + b} \right)^2}\\ = {a^2}{b^2} - 2ab + 1 + {a^2} + 2ab + {b^2}\\ = {a^2}{b^2} + 1 + {a^2} + {b^2}\\ = {a^2}\left( {{b^2} + 1} \right) + \left( {{b^2} + 1} \right)\\ = \left( {{a^2} + 1} \right)\left( {{b^2} + 1} \right)\\ c){x^3} - 4{x^2} + 12x - 27\\ = {x^3} - 27 + \left( { - 4{x^2} + 12x} \right)\\ = \left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right) - 4x\left( {x - 3} \right)\\ = \left( {x - 3} \right)\left( {{x^2} + 3x + 9 - 4x} \right)\\ = \left( {x - 3} \right)\left( {{x^2} - x + 9} \right)\\ b){x^3} + 2{x^2} + 2x + 1\\ = {x^3} + 2{x^2} + x + x + 1\\ = x\left( {{x^2} + 2x + 1} \right) + \left( {x + 1} \right)\\ = x{\left( {x + 1} \right)^2} + \left( {x + 1} \right)\\ = \left( {x + 1} \right)\left( {x\left( {x + 1} \right) + 1} \right)\\ = \left( {x + 1} \right)\left( {{x^2} + x + 1} \right)\\ d){x^4} - 2{x^3} + 2x - 1\\ = {x^4} - 2{x^3} + {x^2} - {x^2} + 2x - 1\\ = {x^2}\left( {{x^2} - 2x + 1} \right) - \left( {{x^2} - 2x + 1} \right)\\ = \left( {{x^2} - 2x + 1} \right)\left( {{x^2} - 1} \right)\\ = {\left( {x - 1} \right)^2}\left( {x - 1} \right)\left( {x + 1} \right)\\ = {\left( {x - 1} \right)^3}\left( {x + 1} \right)\\ e){x^4} + 2{x^3} + 2{x^2} + 2x + 1\\ = {x^4} + 2{x^3} + {x^2} + {x^2} + 2x + 1\\ = {x^2}\left( {{x^2} + 2x + 1} \right) + \left( {{x^2} + 2x + 1} \right)\\ = \left( {{x^2} + 2x + 1} \right)\left( {{x^2} + 1} \right)\\ = {\left( {x + 1} \right)^2}\left( {{x^2} + 1} \right) \end{array} |
a) 3x2 - 7x + 2
= 3x2 - 6x - x + 2
= (3x2 - 6x) - (x - 2)
= 3x (x - 2) - (x - 2)
= (3x - 1) (x - 2)
a) \(=2xy^2\left(x^2+8x+15\right)\)
\(=2xy^2\left[\left(x^2+8x+16\right)-1\right]\)
\(=2xy^2\left[\left(x+4\right)^2-1\right]\)
\(=2xy^2\left(x+4+1\right)\left(x+4-1\right)\)
\(=2xy^2\left(x+5\right)\left(x-3\right)\)
mấy câu sau tự làm nha :*
b,=(x^2-10x+25)-4
=(x-5)^2-2^2
=(x-5-2)(x-5+2)
=(x-7)(x-3)
a) Ta có : x2 - 4x + 3
= x2 - x - 3x + 3
= x(x - 1) - (3x - 3)
= x(x - 1) - 3(x - 1)
= (x - 1) (x - 3)
a) \(x^2-4x+3\)
\(=x^2-x-3x+3\)
\(=x\left(x-1\right)-3\left(x-1\right)\)
\(=\left(x-1\right)\left(x-3\right)\)
b) \(x^2+5x+4\)
\(=x^2+x+4x+4\)
\(=x\left(x+1\right)+4\left(x+1\right)\)
\(=\left(x+1\right)\left(x+4\right)\)
c) \(x^2-x-6\)
\(=x^2-3x+2x-6\)
\(=x\left(x-3\right)+2\left(x-3\right)\)
\(=\left(x+2\right)\left(x-3\right)\)
d) \(x^4+1997x^2+1996x+1997\)
\(=x^4+x^2+1996x^2+1996x+1996+1\)
\(=\left(x^4+x^2+1\right)+\left(1996x^2+1996x+1996\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+1\right)+1996\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+1997\right)\)
e) \(x^2-2001\cdot2002\)( hình như sai sai)
a/ x4 +5x3 +10x-4
=(x4- 4)+(5x3 + 10x)
=(x2+2) (x2-2) + 5x(x2 +2 )
=(x2+2)(x2 -2 +5x)
b/x5 - x4 +x3 -x2 +x-1
=x4(x-1)+x3(x-1)+(x-1)
=(x-1)(x4+x3+1)
Bài 14:Tìm x
a,\(x-3=\left(3-x\right)^2\)
\(\Rightarrow\left(x-3\right)-\left(3-x\right)^2=0\)
\(\Rightarrow\left(x-3\right)+\left(x-3\right)^2=0\)
\(\Rightarrow\left(x-3\right)\left(1+x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
b,\(\left(2x-5\right)-\left(5+2x\right)^2=0\)
\(\Rightarrow\left(2x-5\right)+\left(2x-5\right)^2=0\)
\(\Rightarrow\left(2x-5\right)\left(1+2x-5\right)=0\)
\(\Rightarrow\left(2x-5\right)\left(2x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-5=0\\2x-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=5\\2x=4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=2\end{matrix}\right.\)
a) (x^2+x)^2-14(x^2+x)+24
=(x^2+x)^2-2(x^2+x)-12(x^2+x)24
=(x^2+x)(x^2+x-2)-12(x^2+x-2)
=(x^2+x-12)(x^2+x-2)
a) \(x^4+4y^4\)
\(=\left(x^2\right)+\left(2y^2\right)^2\)
\(=\left(x^2\right)^2+2.x^2.2y^2+\left(2y^2\right)^2-2.x^2.2y^2\)
\(=\left(x^2+2y^2\right)^2-4x^2y^2\)
\(=\left(x^2+2y^2\right)^2-\left(2xy\right)^2\)
\(=\left(x^2+2y^2-2xy\right)\left(x^2+2y^2+2xy\right)\)
b) \(x^4+y^4\)
\(=\left(x^2\right)^2+2x^2y^2+\left(y^2\right)^2-2x^2y^2\)
\(=\left(x^2+y^2\right)^2-\left(\sqrt{2}xy\right)^2\)
\(=\left(x^2+y^2-\sqrt{2}xy\right)\left(x^2+y^2+\sqrt{2}xy\right)\)
c) \(4x^4+y^4\)
\(=\left(2x^2\right)^2+\left(y^2\right)^2\)
\(=\left(2x^2\right)^2++2.2x^2.y^2+\left(y^2\right)^2-2.2x^2.y^2\)
\(=\left(2x^2+y^2\right)^2-4x^2y^2\)
\(=\left(2x^2+y^2\right)^2-\left(2xy\right)^2\)
\(=\left(2x^2+y^2\right)^2-\left(\sqrt{2}xy\right)^2\)
\(=\left(2x^2+y^2-\sqrt{2}xy\right)\left(2x^2+y^2+\sqrt{2}xy\right)\)
d) \(x^4+324\)
\(=\left(x^2\right)^2+18^2\)
\(=\left(x^2\right)^2+2.x^2.18+18^2-2.x^2.18\)
\(=\left(x^2+18\right)^2-36x^2\)
\(=\left(x^2+18\right)^2-\left(6x\right)^2\)
\(=\left(x^2+18-6x\right)\left(x^2+18+6x\right)\)
e) \(x^5+x+1\)
\(=x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1\)
\(=\left(x^5+x^4+x^3\right)-\left(x^4+x^3+x^2\right)+\left(x^2+x+1\right)\)
\(=x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
f) \(x^5-x^4-1\)
\(=x^5-x^3-x^2-x^4+x^2+x+x^3-x-1\)
\(=\left(x^5-x^3-x^2\right)-\left(x^4-x^2-x\right)+\left(x^3-x-1\right)\)
\(=x^2\left(x^3-x-1\right)-x\left(x^3-x-1\right)+\left(x^3-x-1\right)\)
\(=\left(x^3-x-1\right)\left(x^2-x+1\right)\)
\(a)x^4+4y^4\)
\(=\left(x^2\right)^2+\left(2y^2\right)^2\)
\(=\left(x^2\right)^2+4x^2y^2+\left(2y^2\right)^2-4x^2y^2\)
\(=\left(x^2+y^2\right)^2-\left(2xy\right)^2\)
\(=\left(x^2+2y^2-2xy\right)\left(x^2+2y^2+2xy\right)\)