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a) Ta có: \(x^2+9x+20\)
\(=x^2+4x+5x+20\)
\(=x\left(x+4\right)+5\left(x+4\right)\)
\(=\left(x+4\right)\left(x+5\right)\)
b) Ta có: \(x^2+x-12\)
\(=x^2+4x-3x-12\)
\(=x\left(x+4\right)-3\left(x+4\right)\)
\(=\left(x+4\right)\left(x-3\right)\)
c) Ta có: \(6x^2-11x-16\)
\(=6\left(x^2-\frac{11}{6}x-\frac{16}{6}\right)\)
\(=6\left(x^2-2\cdot x\cdot\frac{11}{12}+\frac{121}{144}-\frac{505}{144}\right)\)
\(=6\left[\left(x-\frac{11}{12}\right)^2-\frac{505}{144}\right]\)
\(=6\left(x-\frac{11+\sqrt{505}}{12}\right)\left(x-\frac{11-\sqrt{505}}{12}\right)\)
d) Ta có: \(4x^2-8x-5\)
\(=4x^2-10x+2x-5\)
\(=2x\left(2x-5\right)+\left(2x-5\right)\)
\(=\left(2x-5\right)\left(2x+1\right)\)
e) Ta có: \(x^3-6x^2-x+30\)
\(=x^3+2x^2-8x^2-16x+15x+30\)
\(=x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-8x+15\right)\)
\(=\left(x+2\right)\left(x^2-3x-5x+15\right)\)
\(=\left(x+2\right)\left[x\left(x-3\right)-5\left(x-3\right)\right]\)
\(=\left(x+2\right)\left(x-3\right)\left(x-5\right)\)
g) Ta có: \(x^3+9x^2+23x+15\)
\(=x^3+x^2+8x^2+8x+15x+15\)
\(=x^2\left(x+1\right)+8x\left(x+1\right)+15\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+8x+15\right)\)
\(=\left(x+1\right)\left(x^2+3x+5x+15\right)\)
\(=\left(x+1\right)\left[x\left(x+3\right)+5\left(x+3\right)\right]\)
\(=\left(x+1\right)\left(x+3\right)\left(x+5\right)\)
h) Ta có: \(2x^4-x^3-9x^2+13x\)
\(=x\left(2x^3-x^2-9x+13\right)\)
i) Ta có: \(x^4+2x^3-16x^2-2x+15\)
\(=x^4-3x^3+5x^3-15x^2-x^2+3x-5x+15\)
\(=x^3\left(x-3\right)+5x^2\left(x-3\right)-x\left(x-3\right)-5\left(x-3\right)\)
\(=\left(x-3\right)\left(x^3+5x^2-x-5\right)\)
\(=\left(x-3\right)\left[x^2\left(x+5\right)-\left(x+5\right)\right]\)
\(=\left(x-3\right)\left(x+5\right)\left(x^2-1\right)\)
\(=\left(x-3\right)\left(x+5\right)\left(x-1\right)\left(x+1\right)\)
a) \(x^2-6x+8\)
\(=x^2-2\cdot x\cdot3+3^2-1\)
\(=\left(x-3\right)^2-1^2\)
\(=\left(x-3-1\right)\left(x-3+1\right)\)
\(=\left(x-4\right)\left(x-2\right)\)
Còn lại tương tự
a) \(x^2-6x+8=x^2-2x-4x+8\)
\(=\left(x^2-2x\right)-\left(4x-8\right)\)
=x(x-2)-4(x-2) = (x-2)(x-4)
a) \(x^2+4x+3=\left(x^2+4x+4\right)-1=\left(x+2\right)^2-1^2=\left(x+1\right)\left(x+3\right)\) (mình sửa lại)
b) \(x^2+8x-9=\left(x^2+8x+16\right)-25=\left(x+4\right)^2-5^2=\left(x-1\right)\left(x+9\right)\)
c) \(3x^2+6x-9=3\left[\left(x^2+2x+1\right)-4\right]=3\left[\left(x+1\right)^2-2^2\right]=3\left(x-1\right)\left(x+3\right)\)
d) \(2x^2+x-3=2x^2-4x+2+5x-5=2\left(x^2-2x+1\right)+5\left(x-1\right)=2\left(x-1\right)^2+5\left(x-1\right)=\left(x-1\right)\left(2x+3\right)\)
x3 + 7x - 6=x2 . x + 7x - 22 + 2 = (x2 - 22) + (x+7x)+2=(x-2) . (x+2) + 8x + 2
x3 - 5x + 8x - 4=x2 . x -5x + 8x -22 = (x2 - 22) . (x -5x + 8x )=(x-2) . (x+2) . 4x
x3 - 9x2 + 6x + 16=x2 . x - 9x2 + 6x + 16 = (x2 - 9x2) . (x+6x) + 16=(x-9x) . (x+9x) . 7x + 16
k mk nha
b, \(\left(x^2+x\right)^2+4x^2+4x-12=x^4+2x^3+x^2+4x^2+4x-12\)
\(=x^4+2x^3+5x^2+4x-12\)
\(=\left(x^4-x^3\right)+\left(3x^3-3x^2\right)+\left(8x^2-8x\right)+\left(12x-12\right)\)
\(=x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)\)
\(=\left(x^3+3x^2+8x+12\right)\left(x-1\right)\)
\(=\left[\left(x^3+2x^2\right)+\left(x^2+2x\right)+\left(6x+12\right)\right]\left(x-1\right)\)
\(=\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]\left(x-1\right)\)
\(=\left(x^2+x+6\right)\left(x+2\right)\left(x-1\right)\)
c, \(x^3+3x^2-4=\left(x^3+2x^2\right)+\left(x^2+2x\right)-\left(2x+4\right)\)
\(=x^2\left(x+2\right)+x\left(x+2\right)-2\left(x+2\right)\)
= \(\left(x^2+x-2\right)\left(x+2\right)\)
a)\(x^5+x^4+1=x^5-\left(-x^3+x^3\right)+x^4+\left(x^2-x^2\right)+\left(x-x\right)+1\)
\(=x^5-x^3+x^2+x^4-x^2+x+x^3-x+1\)
\(=x^2\left(x^3-x+1\right)+x\left(x^3-x+1\right)+\left(x^3-x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^3-x+1\right)\)
b,c có ng lm rồi
d)\(2x^4-3x^3-7x^2+6x+8\)
Ta thấy x=-1 là nghiệm của đa thức
=>đa thức có 1 hạng tử là x+1
\(\Rightarrow\left(x+1\right)\left(2x^3-5x^2-2x+8\right)\)
\(\Rightarrow\left(x+1\right)\left[2x^3-x^2-4x-4x^2+2x+8\right]\)
\(\Rightarrow\left(x+1\right)\left[x\left(2x^2-x-4\right)-2\left(2x^2-x-4\right)\right]\)
\(\Rightarrow\left(x+1\right)\left(x-2\right)\left(2x^2-x-4\right)\)
phần còn lại bạn tự lo nhé
a) \(x^2-6x+9-9y^2=x^2-2\cdot3+3^2-\left(3y\right)^2=\left(x-3\right)^2-\left(3y\right)^2=\left(x-3-3y\right)\cdot\left(x-3+3y\right)\)
b) \(x^3-3x^2+2x-1+2\cdot\left(x^2-x\right)=\left(x-1\right)^3+2x\cdot\left(x-1\right)=\left(x-1\right)\cdot\left[\left(x-1\right)^2+2x\right]\)
Bài làm:
a) \(x^2-6x+4=\left(x^2-6x+9\right)-5=\left(x-3\right)^2-\left(\sqrt{5}\right)^2\)
\(=\left(x-3-\sqrt{5}\right)\left(x-3+\sqrt{5}\right)\)
b) \(x^2-4x+3=x^2-x-3x+3=\left(x-1\right)\left(x-3\right)\)
c) \(6x^2-5x+1=6x^2-3x-2x+1=\left(2x-1\right)\left(3x-1\right)\)
d) \(3x^2+13x-10=3x^2+15x-2x-10=\left(x-5\right)\left(3x-2\right)\)
a, 8x2+10x =2x.(4x+5)
b, 4x2-8x+4 =4.(x2 -2x+1)=4.(x-1)2
c, 3x2 -3xy -5x +5y =(3x2-5x) - (3xy-5y) = x.(3x-5)- y.(3x-5)= (x-y).(3x-5)
d, x2+ 4x- 45=x2+ 9x- 5x- 45= x.(x+9)- 5.(x+9)=(x-5).(x+9)
a , 8 x 2 + 10 x
= 2 x ( 4 x + 5 )
b , 4 x 2 - 8 x + 4
= ( 2x ) 2 - 2 . 2 x . 2 + 2 2
= ( 2x + 2 ) 2
c ) 3 x 2 - 3 x y - 5 x + 5 y
= 3 x ( x - y ) - 5 ( x - y )
= ( 3x - 5 ) ( x - y )
d ) x 2 + 4x - 45
= x 2 + 2 x . 2 + 4 - 49
= ( x + 2 ) 2 - 49
= ( x + 2 ) 2 - 7 2
= ( x + 2 - 7 ) ( x + 2 + 7)
= ( x - 5 ) ( x + 9 )
a)\(x^2+9x+20\)
\(\Leftrightarrow x^2+4x+5x+20\)
\(\Leftrightarrow x\left(x+4\right)+5\left(x+4\right)\)
\(\Leftrightarrow\left(x+5\right)\left(x+4\right)\)
b)\(x^2+x-12\)
\(\Leftrightarrow x-3x+4x-12\)
\(\Leftrightarrow x\left(x-3\right)+4\left(x-3\right)\)
\(\Leftrightarrow\left(x+4\right)\left(x-3\right)\)
Vừa vừa phải phải thôi người ta mất công gửi lên còn chửi người ta đó điên mất lịch sự