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a, \(x^4-x^3-x^3+x^2-x^2+x+x-1\)\(1\)
=\(x^3\left(x-1\right)+x^2\left(x-1\right)-x\left(x-1\right)+\left(x-1\right)\)
=\(\left(x-1\right)\left(x^3+x^2-x+1\right)\)
b, \(\left(ab-1\right)^2+\left(a+b\right)^2\)
=\(a^2b^2-2ab+1+a^2+2ab+b^2\)
=\(a^2b^2+a^2+b^2+1\)
=\(a^2\left(b^2+1\right)+\left(b^2+1\right)\)
=\(\left(b^2+1\right)\left(a^2+1\right)\)
c,\(x^4+2x^3+2x^2+2x+1\)
=\(x^4+x^3+x^3+x^2+x^2+x+x+1\)
=\(x^3\left(x+1\right)+x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)\)
=\(\left(x+1\right)\left(x^3+x^2+x+1\right)\)
=\(\left(x+1\right)^2\left(x^2+1\right)\)
1. \(B=\left(x-2\right)\left(x+2\right)\left(x+3\right)-\left(x+1\right)^3\)
\(=\left(x^2-4\right)\left(x+3\right)-\left(x^3+3x^2+3x+1\right)\)
\(=x^3+3x^2-4x-12-x^3-3x^2-3x-1\)
\(=-7x-13\)
2. \(64-x^2-y^2+2xy=64-\left(x^2+y^2-2xy\right)\)
\(=64-\left(x-y\right)^2=\left(8+x-y\right)\left(8-x+y\right)\)
3. \(2x^3-x^2+2x-1=0\)
\(\Leftrightarrow x^2.\left(2x-1\right)+\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x^2+1\right)=0\)
Vì \(x^2\ge0\)\(\Rightarrow x^2+1>0\)
\(\Rightarrow2x-1=0\)\(\Rightarrow2x=1\)\(\Rightarrow x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
Bài 1.
B = ( x - 2 )( x + 2 )( x + 3 ) - ( x + 1 )3
= ( x2 - 4 )( x + 3 ) - ( x3 + 3x2 + 3x + 1 )
= x3 + 3x2 - 4x - 12 - x3 - 3x2 - 3x - 1
= -7x - 13
Bài 2.
64 - x2 - y2 + 2xy
= 64 - ( x2 - 2xy + y2 )
= 82 - ( x - y )2
= ( 8 - x + y )( 8 + x - y )
Bài 3.
2x3 - x2 + 2x - 1 = 0
<=> ( 2x3 - x2 ) + ( 2x - 1 ) = 0
<=> x2( 2x - 1 ) + 1( 2x - 1 ) = 0
<=> ( 2x - 1 )( x2 + 1 ) = 0
<=> \(\orbr{\begin{cases}2x-1=0\\x^2+1=0\end{cases}}\Leftrightarrow x=\frac{1}{2}\)( vì x2 + 1 ≥ 1 > 0 ∀ x )
Ta có ; x2 - 11x + 24
= x2 - 3x - 8x + 24
= x(x - 3) - (8x - 24)
= x(x - 3) - 8(x - 3)
= (x - 3)(x - 8)
a, =x4(x+2)-x3(x+2)+x2(x+2)-x(x+2)+(x+2)
=(x+2)(x4-x3+x2-x+1)
Câu 2 nha
\(a,x^4+2x^3+x^2\)
\(=x^2\left(x^2+2x+1\right)\)
\(=x^2\left(x+1\right)^2\)
\(c,x^2-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)
a) \(x^4+2x^2-3=x^4-x^2+3x^2-3=\left(x^4-x^2\right)+\left(3x^2-3\right)=x^2\left(x^2-1\right)+3\left(x^2-1\right)=\left(x^2-1\right)\left(x^2+3\right)=\left(x-1\right)\left(x+1\right)\left(x^2+3\right)\)
b) Đặt \(x^2+2x=a\)
\(\Rightarrow B=a\left(a+4\right)+3\) ( Đặt biểu thức trên là B)
= \(a^2+4a+3\)
=\(a^2+a+3a+3\)
=\(\left(a^2+a\right)+\left(3a+3\right)\)
=\(a\left(a+1\right)+3\left(a+1\right)\)
=\(\left(a+1\right)\left(a+3\right)\)
Thay \(a=x^2+2x\)
\(\Rightarrow\left(x^2+2x+1\right)\left(x^2+2x+3\right)\)
a) x4 + 2x2 - 3
= x4 + 2x2 + 1 - 4
= (x2 + 1)2 - 22
= (x2 + 1 - 2)(x2 + 1 + 2)
= (x2 - 1)(x2 + 3)
= (x - 1)(x + 1)(x2 + 3)
c) 2xy2 + 4xy + 2x - 2xz2 + 4xzt - 2xt2
= 2x(y2 + 2y + 1 - z2 + 2zt - t2)
= \(2x\left [ (y + 1)^{2} - (z -t)^{2} \right ]\)
= 2x(y - 1 - z + t)(y + 1 + z - t)
= 2x(y - z + t - 1)(y + z - t + 1)
câu b mk đang suy nghĩ
c) \(2xy^2+4xy+2x-2xz^2+4xzt-2xt^2=2x\left(y^2+2y+1-z^2+2zt-t^2\right)=2x\left[\left(y^2+2y+1\right)-\left(z^2-2zt+t^2\right)\right]=2x\left[\left(y+1\right)^2-\left(z-t\right)^2\right]=2x\left(y+1-z+t\right)\left(y+1-z-t\right)\)
a, \(x^4+2x^2-3\)\(=\left[\left(x^2\right)^2+2.x^2.1+1\right]-\left(1-3\right)\)
\(=\left(x^2+1\right)^2-\left(\sqrt{2}\right)^2\)
\(=\left(x^2+1-\sqrt{1}\right)\left(x^2+1+\sqrt{2}\right)\)
b,\(\left(x^2+2x\right)\left(x^2+2x+4\right)+3=x^4+2x^3+4x^2+2x^3+4x^2+8x\)
\(=\left(x^4+4x^3\right)+\left(8x^2+8x\right)=x\left(x^3+4x^2\right)+x\left(8x+8\right)\)
\(=x\left(x^3+4x^2+8x+8\right)\)
c,\(2xy^2+4xy+2x-2xz^2+4xzt-2xt^2\)
\(=\left(2xy^2+4xy\right)+\left(2x-2xz^2\right)+\left(4xzt-2xt^2\right)\)
\(=2x\left(y^2+2y\right)+2x\left(1-z^2\right)+2x\left(2zt-t^2\right)\)
\(=2x\left(y^2+2y+1-z^2+2zt-t^2\right)\)
\(=2x\left[\left(y^2+2y+1\right)-\left(z^2-2zt+t^2\right)\right]\)
\(=2x\left[\left(y+1\right)^2-\left(z-t\right)^2+\left(2y-t^2\right)\right]\)
\(=2x\left[\left(y+1-z-t\right)\left(y+1+z-t\right)\right]\)
Ở ha đúng rồi quên mất thanks bạn nhiều lắm còn mấy câu kia bn giúp mk nữa nhá
Mik làm xong rồi ạ! Các bn ko cần giúp mik nữa đâu