Ta có: \(x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)

\(B\in Z\Rightarrow\frac{7}{x^2-x+1}\in Z\Rightarrow7⋮\left(x^2-x+1\right)\Rightarrow x^2-x+1\in\left\{1;7\right\}\left(x^2-x+1>0\right)\)

TH1: \(x^2-x+1=1\Rightarrow x\left(x-1\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\) (thỏa mãn)

TH2: \(x^2-x+1=7\Rightarrow x^2-x-6=0\Rightarrow\left(x+2\right)\left(x-3\right)=0\Rightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)(thỏa mãn) 

Vậy \(x\in\left\{0;1;-2;3\right\}\)

Ta có: \(\hept{\begin{cases}x^2+2ax+b=\left(x-1\right)A\left(1\right)\\x^2+2ax+b=\left(x+2\right)B+4\left(2\right)\end{cases}}\)

Thay x=1 vào (1) rồi thay x=-2 vào (2) ta được:

\(\hept{\begin{cases}1+2a+b=0\\4-4a+b=4\end{cases}\Leftrightarrow\hept{\begin{cases}2a+b=-1\\-4a+b=0\end{cases}\Leftrightarrow}\hept{\begin{cases}a=\frac{-1}{6}\\b=-\frac{4}{6}\end{cases}}}\)

a) \(A=y^2+2y+1\)

\(A=\left(y+1\right)^2\)

Thay y = 99 vào A ta có :

\(A=\left(99+1\right)^2\)

\(A=100^2=10000\)

b) \(B=x^2-6x+9\)

\(B=x^2-2\cdot x\cdot3+3^2\)

\(B=\left(x-3\right)^2\)

Thay x = 103 vào B ta có :

\(B=\left(103-3\right)^2\)

\(B=100^2=10000\)

c) \(C=x^2+4x+4\)

\(C=x^2+2\cdot x\cdot2+2^2\)

\(C=\left(x+2\right)^2\)

Thay x = 98 vào C ta có :

\(C=\left(98+2\right)^2\)

\(C=100^2=10000\)

d) \(D=y^2-2xy+x^2\)

\(D=\left(y-x\right)^2\)

Thay y = 109, x = 9 vào D ta có :

\(D=\left(109-9\right)^2\)

\(D=100^2=10000\)

a) x ^ 2 + 2x + 1 = ( x + 1 ) ^ 2 = ( 99 + 1 ) ^ 2 = 100 ^ 2 = 10000

b) x ^ 2 - 6x + 9 = ( x - 3 ) ^ 2 = ( 103 - 3 ) ^ 2 = 100 ^ 2 = 10000

c) x ^ 2 + 4x + 4 = ( x + 2 ) ^ 2 = ( 98 + 2 ) ^ 2 = 100 ^ 2 = 10000

d) y ^ 2 - 2xy + x ^ 2 = ( y - x ) ^ 2 = ( 109 - 9 ) ^ 2 = 100 ^ 2 = 10000

a) x⁴ + 1997x² + 1996x +1997

= x⁴ + 1997x² + 1997x - x +1997

=(x⁴-x) + (1997x² +1997x+1997)

=x(x³-1) + 1997(x²+x+1)

=x(x-1)(x²+x+1) + 1997(x²+x+1)

=(x²+x+1)(x²-x) + 1997(x²+x+1)

=(x²+x+1)(x²-x+1997)

b) x² -x -2001.2002

=x² - x -2002² +2002

=(x²-2002²)-(x-2002)

=(x-2002)(x+2002) - (x-2002)

=(x-2002)(x+2002+1)

=(x-2002)(x+2003)

c)x⁸ + 98x⁴ +1

= (x⁸+2x⁴+1) + 96x⁴

= (x⁴+1)² + 96x⁴

=[(x⁴+1)² + 2.(x⁴+1).8 + 64x⁴ ]+[32x⁴ - 16x²(x⁴+1)]

=(x⁴+1+8x²)-16x²(-2x²+x⁴+1)

=(x⁴+8x²+1)²- 16x²(x²-1)²

=(x⁴ + 8x² +1)²- [4x(x²-1)]²

=(x⁴+8x²+1)² - (4x³-4x)²

=(x⁴-4x³+8x²+4x+1)(x⁴+4x³+8x²-4x+1)

2

k nha Cầu...........xin.

lp 8 mak ko bít 1+1 = ?

x³ - 19x - 30 

= x³ - 4x - 15x - 30

= x(x² - 4) - 15(x + 2)

= x(x - 2)(x + 2) - 15(x + 2)

= (x² - 2x) (x + 2) - 15(x + 2)

= (x + 2)(x² - 2x - 15)

= (x + 2)(x² - 5x + 3x - 15)

= (x + 2)(x - 5)(x + 3)

2bc(b + 2c) + 2ac(c - 2a) - 2ab(a + 2b) - 7abc

= 2b²c + 4bc² + 2ac² - 4a²c - 2ab(a + 2b) - 7abc

= 2b²c + abc + 4bc² + 2ac² - 4a²c - 8abc - 2ab(a + 2b)

= bc(2b + a) + 2c²(2b + a) - 4ac(a + 2b) - 2ab(a + 2b)

= (a + 2b)(bc + 2c² - 4ac - 2ab)

= (a + 2b)[c(b + 2c) - 2a(2c + b)]

= (a + 2b)(b + 2c)(c - 2a)