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a) x2 + 5x + 6
= x2 + 2x + 3x + 6
= (x2 + 2x) + (3x + 6)
= x(x + 2) + 3 (x + 2)
= (x + 2) (x + 3)
b) x2 + 6x + 8
= x2 + 2x + 4x + 8
= (x2 + 2x) + (4x + 8)
= x(x + 2) + 4(x + 2)
= (x + 2)(x + 4)
c) x2 - 5x - 14
= x2 + 2x - 7x - 14
= (x2 + 2x) - (7x + 14)
= x(x + 2) - 7(x + 2)
= (x + 2)(x - 7)
d) x2 - 9x + 18
= x2 - 3x - 6x + 18
= (x2 - 3x) - (6x + 18)
= x(x - 3) - 6 (x - 3)
= (x - 3)(x - 6)
e) x2 - 7x + 12
= x2 -3x - 4x + 12
= (x2 - 3x) - (4x + 12)
= x(x - 3) - 4(x - 3)
= (x - 3)(x - 4)
f) 3x2 + 9x - 30
= 3(x2 + 3x - 10)
= 3\(\left[\left(x^2+5x-2x-10\right)\right]\)
= 3\(\left[\left(x^2+5x\right)-\left(2x-10\right)\right]\)
= 3\(\left[x\left(x+5\right)-2\left(x+5\right)\right]\)
= 3(x + 5)(x - 2)
Chuc ban hoc tot
a) \(x^2+5x+6=\left(x+2\right)\left(x+3\right)\)
b) \(x^2+6x+8=\left(x+2\right)\left(x+4\right)\)
c) \(x^2-5x-14=\left(x-7\right)\left(x+2\right)\)
d) \(x^2-9x+18=\left(x-3\right)\left(x-6\right)\)
e) \(x^2-7x+12=\left(x-3\right)\left(x-4\right)\)
f) \(3x^2+9x-30=3\left(x^2+3x-10\right)=3\left(x+5\right)\left(x-2\right)\)
f)\(x^2-5x-14=x^2-7x+2x-14=x\left(x-7\right)+2\left(x-7\right)=\left(x-7\right)\left(x+2\right)\)
i)\(x^2-7x+10=x^2-2x-5x+10=x\left(x-2\right)-5\left(x-2\right)=\left(x-5\right)\left(x-2\right)\)
h)\(x^2-7x+12=x^2-3x-4x+12=x\left(x-3\right)-4\left(x-3\right)=\left(x-4\right)\left(x-3\right)\)
g)\(x^2+6x+5=x^2+x+5x+5=x\left(x+1\right)+5\left(x+1\right)=\left(x+1\right)\left(x+5\right)\)
f)\(x^2-5x-14=x^2-7x+2x-14\)
\(=\left(x+2\right)\left(x-7\right)\)
i)\(x^2-7x+10=x^2-5x-2x+10\)
\(=\left(x-2\right)\left(x-5\right)\)
h)\(x^2-7x+12=x^2-4x-3x+12\)
\(=\left(x-3\right)\left(x-4\right)\)
g)\(x^2+6x+5=x^2+x+5x+5\)
\(=\left(x+5\right)\left(x+1\right)\)
a) \(x^2-5x+6=x^2-2x-3x+6=\left(x-2\right)\left(x-3\right)\)
b)\(3x^2+9x-30=3x^2-6x+15x-30=3\left(x-2\right)\left(x+5\right)\)
c)\(x^2-7x+12=x^2-3x-4x+12=\left(x-3\right)\left(x-4\right)\)
d)\(x^2-7x+10=x^2-2x-5x+10=\left(x-2\right)\left(x-5\right)\)
a) \(x^2-5x+6=x^2-2x-3x+6=\left(x^2-2x\right)-\left(3x-6\right)\)
\(=x\left(x-2\right)-3\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)
b) \(3x^2+9x-30=3\left(x^2+3x-10\right)=3\left(x^2-2x+5x-10\right)\)
\(=3\left[\left(x^2-2x\right)+\left(5x-10\right)\right]=3\left[x\left(x-2\right)+5\left(x-2\right)\right]\)
\(=3\left(x-2\right)\left(x+5\right)\)
c) \(x^2-7x+12=x^2-3x-4x+12=\left(x^2-3x\right)-\left(4x-12\right)\)
\(=x\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x-4\right)\)
d) \(x^2-7x+10=x^2-2x-5x+10=\left(x^2-2x\right)-\left(5x-10\right)\)
\(=x\left(x-2\right)-5\left(x-2\right)=\left(x-2\right)\left(x-5\right)\)
a) \(x^3+4x^2-21x\)
\(=x\left(x^2+4x-21\right)\)
\(=x\left(x^2-3x+7x-21\right)\)
\(=x\left[x\left(x-3\right)+7\left(x-3\right)\right]\)
\(=x\left(x-3\right)\left(x+7\right)\)
b) \(5x^3+6x^2+x\)
\(=x\left(5x^2+6x+1\right)\)
\(=x\left(5x^2+5x+x+1\right)\)
\(=x\left[5x\left(x+1\right)+\left(x+1\right)\right]\)
\(=x\left(x+1\right)\left(5x+1\right)\)
c) \(x^3-7x+6\)
\(=x^3+2x^2-3x-2x^2-4x+6\)
\(=x\left(x^2+2x-3\right)-2\left(x^2+2x-3\right)\)
\(=\left(x-2\right)\left(x^2+2x-3\right)\)
\(=\left(x-2\right)\left(x-1\right)\left(x+3\right)\)
d) \(3x^3+2x-5\)
\(=3x^3+3x^2+5x-3x^2-3x-5\)
\(=x\left(3x^2+3x+5\right)-\left(3x^2+3x+5\right)\)
\(=\left(x-1\right)\left(3x^2+3x+5\right)\)
a) \(x^2-3x+2\)
\(\Leftrightarrow x^2-2x-x+2\)
\(\Leftrightarrow x\left(x-2\right)-\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\)
b) \(x^2-6x+8\)
\(\Leftrightarrow x^2-4x-2x+8\)
\(\Leftrightarrow x\left(x-4\right)-2\left(x-4\right)\)
\(\Leftrightarrow\left(x-4\right)\left(x-2\right)\)
c) \(3x^2+9x-30\)
\(\Leftrightarrow3\left(x^2+3x-10\right)\)
\(\Leftrightarrow3\left[\left(x^2+2\cdot\frac{3x}{2}+\frac{9}{4}\right)-\frac{49}{4}\right]\)
\(\Leftrightarrow3\left[\left(x+\frac{3}{2}\right)^2-\left(\frac{7}{2}\right)^2\right]\)
\(\Leftrightarrow3\left(x+\frac{3}{2}+\frac{7}{2}\right)\left(x+\frac{3}{2}-\frac{7}{2}\right)\)
\(\Leftrightarrow3\left(x-2\right)\left(x+5\right)\)
d) \(x^2-9x+18\)
\(\Leftrightarrow x^2-3x-6x+18\)
\(\Leftrightarrow x\left(x-3\right)-6\left(x-3\right)\)
\(\Leftrightarrow\left(x-3\right)\left(x-6\right)\)
TK MK NKA !!!! TH@NK !!!
a) x2 - 3x + 2 ( như này mới phân tích được ạ :) )
= x2 - x - 2x + 2
= x( x - 1 ) - 2( x - 1 )
= ( x - 2 )( x - 1 )
b) x2 - 6x + 8
= x2 - 2x - 4x + 8
= x( x - 2 ) - 4( x - 2 )
= ( x - 4 )( x - 2 )
c) 3x2 + 9x - 30
= 3( x2 + 3x - 10 )
= 3( x2 - 2x + 5x - 10 )
= 3[ x( x - 2 ) + 5( x - 2 )]
= 3( x + 5 )( x - 2 )
d) x2 - 9x + 18
= x2 - 3x - 6x + 18
= x( x - 3 ) - 6( x - 3 )
= ( x - 6 )( x - 3 )
a/\(x^2-5x+6=x^2-2x-3x+6=x\left(x-2\right)-3\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)b/
\(3x^2+9x-30=3\left(x^2+3x-10\right)\)
c/
\(x^2-3x+2=x^2-x-2x+2=x\left(x-1\right)-2\left(x-1\right)=\left(x-1\right)\left(x-2\right)\)
d/\(x^2-9x+18=x^2-3x-6x+18=x\left(x-3\right)-6\left(x-3\right)=\left(x-3\right)\left(x-6\right)\)e/
\(x^2-6x+8=x^2-2x-4x+8=x\left(x-2\right)-4\left(x-2\right)=\left(x-2\right)\left(x-4\right)\)f/\(x^2-5x-14=x^2+2x-7x-14=x\left(x+2\right)-7\left(x+2\right)=\left(x+2\right)\left(x-7\right)\)
g/
\(x^2-6x+5=x^2-x-5x+5=x\left(x-1\right)-5\left(x-1\right)=\left(x-1\right)\left(x-5\right)\)
h/
\(x^2-7x+12=x^2-4x-3x+12=x\left(x-4\right)-3\left(x-4\right)=\left(x-4\right)\left(x-3\right)\)i/\(x^2-7x+10=x^2-2x-5x+10=x\left(x-2\right)-5\left(x-2\right)=\left(x-2\right)\left(x-5\right)\)
a) Ta có: \(x^2-5x+6\)
\(=x^2-2x-3x+6\)
\(=x\left(x-2\right)-3\left(x-2\right)\)
\(=\left(x-2\right)\left(x-3\right)\)
b) Ta có: \(3x^2+9x-30\)
\(=3\left(x^2+3x-10\right)\)
\(=3\left(x^2+5x-2x-10\right)\)
\(=3\left[x\left(x+5\right)-2\left(x+5\right)\right]\)
\(=3\left(x+5\right)\left(x-2\right)\)
c) Ta có: \(x^2-3x+2\)
\(=x^2-x-2x+2\)
\(=x\left(x-1\right)-2\left(x-1\right)\)
\(=\left(x-1\right)\left(x-2\right)\)
d) Ta có: \(x^2-9x+18\)
\(=x^2-3x-6x+18\)
\(=x\left(x-3\right)-6\left(x-3\right)\)
\(=\left(x-3\right)\left(x-6\right)\)
e) Ta có: \(x^2-6x+8\)
\(=x^2-4x-2x+8\)
\(=x\left(x-4\right)-2\left(x-4\right)\)
\(=\left(x-4\right)\left(x-2\right)\)
f) Ta có: \(x^2-5x-14\)
\(=x^2-7x+2x-14\)
\(=x\left(x-7\right)+2\left(x-7\right)\)
\(=\left(x-7\right)\left(x+2\right)\)
g) Ta có: \(x^2-6x+5\)
\(=x^2-x-5x+5\)
\(=x\left(x-1\right)-5\left(x-1\right)\)
\(=\left(x-1\right)\left(x-5\right)\)
h) Ta có: \(x^2-7x+12\)
\(=x^2-3x-4x+12\)
\(=x\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x-4\right)\)
i) Ta có: \(x^2-7x+10\)
\(=x^2-2x-5x+10\)
\(=x\left(x-2\right)-5\left(x-2\right)\)
\(=\left(x-2\right)\left(x-5\right)\)
Bạn tải ứng dụng PhotoMath về nha. Ứng dụng này sẽ giải toán số chi tiết
a) \(x^3-4x^2-12x+27\)
\(=\left(x^3+27\right)-\left(4x^2+12x\right)\)
\(=\left(x+3\right)\left(x^2-3x+9\right)-4x\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2-7x+9\right)\)
b) \(x^3-3x^2-4x+12\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x^2-4\right)\left(x-3\right)\)
\(=\left(x+2\right)\left(x-2\right)\left(x-3\right)\)
a) \(9x^2+6xy+y^2=\left(3x+y\right)^2\)
b) \(6x-9-x^2=-\left(x-3\right)^2\)
Bài 1 : Phân tích các đa thức sau thành nhân tử : ( tách một hạn tử thành nhiều hạng tử )
a, 3x2 + 9x - 30
= 3(x2 + 3x - 10)
= 3(x2 + 5x - 2x - 10)
= 3[x(x + 5) - 2(x + 5)]
= 3(x + 5)(x - 2)
b, x2 - 3x + 2
= x2 - x - 2x + 2
= x(x - 1) - 2(x - 1)
= (x - 1)(x - 2)
c, x2 - 9x + 18
= x2 - 6x - 3x + 18
= x(x - 6) - 3(x - 6)
= (x - 6)(x - 3)
d, x2 - 6x + 8
= x2 - 4x - 2x + 8
= x(x - 4) - 2(x - 4)
= (x - 4)(x - 2)
e, x2 - 5x - 14
= x2 + 2x - 7x - 14
= x(x + 2) - 7(x + 2)
= (x + 2)(x - 7)
f, x2 + 6x + 5
= x2 + x + 5x + 5
= x(x + 1) + 5(x + 1)
= (x + 1)(x + 5)
h, x2 - 7x + 12
= x2 - 3x - 4x + 12
= x(x - 3) - 4(x - 3)
= (x - 3)(x - 4)
i, x2 - 7x + 10
= x2 - 2x - 5x + 10
= x(x - 2) - 5(x - 2)
= (x - 2)(x - 5)
#Học tốt!