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4 tháng 8 2020

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!

2 tháng 7 2021

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) x- 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) \(3x^2+9x-30=3\left(x^2+3x-10\right)=3\left(x+5\right)\left(x-2\right)\)

15 tháng 9 2017

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)\(x^2+6x+5=x^2+x+5x+5=x\left(x+1\right)+5\left(x+1\right)=\left(x+1\right)\left(x+5\right)\)

15 tháng 9 2017

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)\(x^2+6x+5=x^2+x+5x+5\)

                           \(=\left(x+5\right)\left(x+1\right)\)

                             

6 tháng 9 2020

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)\(x^2-7x+10=x^2-2x-5x+10=\left(x-2\right)\left(x-5\right)\)

6 tháng 9 2020

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) \(x^2-7x+10=x^2-2x-5x+10=\left(x^2-2x\right)-\left(5x-10\right)\)

\(=x\left(x-2\right)-5\left(x-2\right)=\left(x-2\right)\left(x-5\right)\)

22 tháng 10 2023

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) \(3x^3+2x-5\)

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

\(=x\left(3x^2+3x+5\right)-\left(3x^2+3x+5\right)\)

\(=\left(x-1\right)\left(3x^2+3x+5\right)\)

13 tháng 5 2017

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 !!! 

3 tháng 8 2020

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 )

13 tháng 8 2020

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: \(x^2-7x+10\)

\(=x^2-2x-5x+10\)

\(=x\left(x-2\right)-5\left(x-2\right)\)

\(=\left(x-2\right)\left(x-5\right)\)

22 tháng 9 2019

Bạn tải ứng dụng PhotoMath về nha. Ứng dụng này sẽ giải toán số chi tiết

22 tháng 9 2019

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) \(6x-9-x^2=-\left(x-3\right)^2\)