phân tích đa thức thành nhân tử
A=(x2+3x)2-2(x2+3x)-8
B=(x2+4x+10)2-7(x2+4x+11)+7
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31 tháng 7 2021
a) x3+4x-5 = x3-x2+x2+4x-5=(x3-x2)+(x2-x)+(5x-5)=x2(x-1)+x(x-1)+5(x-1)=(x2+x+5)(x-1)
b) x3-3x2+4=x3-2x2-x2+4=(x3-2x2)-(x2-4)=x2(x-2)-(x-2)(x+2)=(x2-x+2)(x-2)
c) x3+2x2+3x+2=x3+x2+x2+x+2x+2=(x3+x2)+(x2+x)+(2x+2)=x2(x+1)+x(x+1)+2(x+1)=(x2+x+2)(x+1)
d) bạn xem lại đề đúng ko
e) (x2+3x)2-2(x2+3x)-8=x4+6x3+9x2-2x2-6x-8=x4+6x3+7x2-6x-8=x4-x3+7x3-7x2+14x2-14x+8x-8=(x4-x3)+(7x3-7x2)+(14x2-14x)+(8x-8)=x3(x-1)+7x2(x-1)+14x(x-1)+8(x-1)=(x3+7x2+14x+8)(x-1)=(x3+x2+6x2+6x+8x+8)(x-1)=\(\left[\left(x^3+x^2\right)+\left(6x^2+6x\right)+\left(8x+8\right)\right]\left(x-1\right)\)\(=\left[x^2\left(x+1\right)+6x\left(x+1\right)+8\left(x+1\right)\right]\left(x-1\right)\)\(=\left(x^2+6x+8\right)\left(x+1\right)\left(x-1\right)\)\(=\left(x^2+2x+4x+8\right)\left(x+1\right)\left(x-1\right)\)\(=\left[\left(x^2+2x\right)+\left(4x+8\right)\right]\left(x+1\right)\left(x-1\right)\)\(=\left[x\left(x+2\right)+4\left(x+2\right)\right]\left(x+1\right)\left(x-1\right)\)=\(\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x+4\right)\)
f) (x2+4x+10)2-7(x2+4x+11)+7=(x2+4x+10)2-\(\left[7\left(x^2+4x+11\right)-7\right]\)\(=\left(x^2+4x+10\right)^2-7\left(x^2+4x+10\right)\)\(=\left(x^2+4x+10\right)\left(x^2+4x+3\right)\)
31 tháng 7 2021
a) Ta có: \(x^3+4x-5\)
\(=x^3-x+5x-5\)
\(=x\left(x-1\right)\left(x+1\right)+5\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+5\right)\)
b) Ta có: \(x^3-3x^2+4\)
\(=x^3+x^2-4x^2+4\)
\(=x^2\left(x+1\right)-4\left(x-1\right)\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-4x+4\right)\)
\(=\left(x+1\right)\cdot\left(x-2\right)^2\)
c) Ta có: \(x^3+2x^2+3x+2\)
\(=x^3+x^2+x^2+x+2x+2\)
\(=x^2\left(x+1\right)+x\left(x+1\right)+2\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+x+2\right)\)
d) Ta có: \(x^2+2xy+y^2+2x+2y-3\)
\(=\left(x+y\right)^2+2\left(x+y\right)-3\)
\(=\left(x+y\right)^2+3\left(x+y\right)-\left(x+y\right)-3\)
\(=\left(x+y\right)\left(x+y+3\right)-\left(x+y+3\right)\)
\(=\left(x+y+3\right)\left(x+y-1\right)\)
CM
31 tháng 1 2017
a) (x - 1)(x - 2). b) 4(x - 2)(x - 7).
c) (x + 2)(2x +1). d) (x - l)(2x - 7).
e) (2x + 3y - 3)(2x - 3y +1). g) (x - 3)( x 3 + x 2 - x +1).
h) (x + y)(x + y-l)(x + y + l).
27 tháng 8 2021
\(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2=\left(x^2+4x+8+\dfrac{3}{2}x\right)^2-\dfrac{1}{4}x^2=\left(x^2+\dfrac{11}{2}x+8\right)^2-\left(\dfrac{1}{2}x\right)^2=\left(x^2+\dfrac{11}{2}x+8-\dfrac{1}{2}x\right)\left(x^2+\dfrac{11}{2}x+8+\dfrac{1}{2}x\right)=\left(x^2+5x+8\right)\left(x^2+6x+8\right)=\left(x+2\right)\left(x+4\right)\left(x^2+5x+8\right)\)
27 tháng 8 2021
\(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\)
\(=\left(x^2+4x+8\right)^2+x\left(x^2+4x+8\right)+2x\left(x^2+4x+8\right)+2x^2\)
\(=\left(x^2+4x+8\right)\left(x^2+5x+8\right)+2x\left(x^2+5x+8\right)\)
\(=\left(x^2+5x+8\right)\left(x+2\right)\left(x+4\right)\)
KV
9 tháng 12 2023
a) x² - 9
= x² - 3²
= (x - 3)(x + 3)
b) 4x² - 1
= (2x)² - 1²
= (2x - 1)(2x + 1)
c) x⁴ - 16
= (x²)² - 4²
= (x² - 4)(x² + 4)
= (x² - 2²)(x² + 4)
= (x - 2)(x + 2)(x + 4)
d) x² - 4x + 4
= x² - 2.x.2 + 2²
= (x - 2)²
e) x³ - 8
= x³ - 2³
= (x - 2)(x² + 2x + 4)
f) x³ + 3x² + 3x + 1
= x³ + 3.x².1 + 3.x.1² + 1³
= (x + 1)³
HN
2
5 tháng 9 2021
\(a,\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\) (sửa \(2x\rightarrow2x^2\)
Đặt \(x^2+4x+8=a\)
\(=a^2+3ax+2x=a^2+ax+2ax+2x^2=\left(a+x\right)\left(a+2x\right)\\ =\left(x^2+5x+8\right)\left(x^2+6x+8\right)=\left(x^2+5x+8\right)\left(x+2\right)\left(x+4\right)\)
5 tháng 9 2021
\(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x\)
\(=\left(x^2+6x+8\right)\left(x^2+5x+8\right)\)
\(=\left(x+2\right)\left(x+4\right)\left(x^2+5x+8\right)\)
27 tháng 10 2021
\(a,=\left(x+1\right)\left(x+3\right)\\ b,=-5x^2+15x+x-3=\left(x-3\right)\left(1-5x\right)\\ c,=2x^2+2x+5x+5=\left(2x+5\right)\left(x+1\right)\\ d,=2x^2-2x+5x-5=\left(x-1\right)\left(2x+5\right)\\ e,=x^3+x^2-4x^2-4x+x+1=\left(x+1\right)\left(x^2-4x+1\right)\\ f,=x^2+x-5x-5=\left(x+1\right)\left(x-5\right)\)
29 tháng 8 2021
\(A=x^2+3x+2=\left(x+1\right)\left(x+2\right)\)
\(B=x^2-4x-5=\left(x-5\right)\left(x+1\right)\)
\(C=3x^2+7x+4=\left(x+1\right)\left(3x+4\right)\)
29 tháng 8 2021
\(A=x^2+3x+2=\left(x+1\right)\left(x+2\right)\)
\(B=x^2-4x-5=\left(x-5\right)\left(x+1\right)\)
\(C=3x^2+7x+4=\left(x+1\right)\left(3x+4\right)\)
\(A=\left(x^2+3x\right)^2-2\left(x^2+3x\right)-8\)
\(=\left(x^2+3x\right)^2-2\left(x^2+3x\right)+1-9\)
\(=\left(x^2+3x-1\right)^2-9\)
\(=\left(x^2+3x-1-3\right)\left(x^2+3x-1+3\right)\)
\(=\left(x^2+3x-4\right)\left(x^2+3x+2\right)\)
\(=\left[x^2+4x-x-4\right]\left[x^2+2x+x+2\right]\)
\(=\left[x\left(x+4\right)-\left(x+4\right)\right]\left[x\left(x+2\right)+x+2\right]\)
\(=\left(x-1\right)\left(x+4\right)\left(x+1\right)\left(x+2\right)\)
\(B=\left(x^2+4x+10\right)^2-7\left(x^2+4x+11\right)+7\)
Đặt \(x^2+4x+10=a\) , ta có :
\(a^2-7\left(a+1\right)+7\)
\(=a^2-7a-7+7\)
\(=a^2-7a\)
\(=a\left(a-7\right)\)
\(=\left(x^2+4x+10\right)\left(x^2+4x+10-7\right)\)
\(=\left(x^2+4x+10\right)\left(x^2+4x+3\right)\)
\(=\left(x^2+4x+10\right)\left[x\left(x+3\right)+x+3\right]\)
\(=\left(x^2+4x+10\right)\left(x+1\right)\left(x+3\right)\)
\(A=\left(x^2+3x\right)^2-2\left(x^2+3x\right)-8\)
\(\Leftrightarrow A=\left(x^2+3x\right)-2\left(x^2+3x\right)+1-3^2\)
\(\Leftrightarrow A=\left(x^2+3x-1\right)^2-3^2\)
\(\Leftrightarrow A=\left(x^2+3x-1-3\right)\left(x^2+3x-1+3\right)\)
\(\Leftrightarrow A=\left(x^2+3x-4\right)\left(x^2+3x+2\right)\)
\(\Leftrightarrow A=\left(x^2+4x-x-4\right)\left(x^2+x+2x+2\right)\)
\(\Leftrightarrow A=\left[\left(x^2-x\right)+\left(4x-4\right)\right].\left[\left(x^2+x\right)+\left(2x+2\right)\right]\)
\(\Leftrightarrow A=\left[x\left(x-1\right)+4\left(x-1\right)\right].\left[x\left(x+1\right)+2\left(x+1\right)\right]\)
\(\Leftrightarrow A=\left(x+4\right)\left(x-1\right)\left(x+2\right)\left(x+1\right)\)
Vậy .............................................................
\(B=\left(x^2+4x+10\right)^2-7\left(x^2+4x+11\right)+7\)
Đặt \(x^2+4x+10=y\) , Ta có:
\(y^2-7\left(y+1\right)+7\)
\(=y^2-7y-7+7\)
\(=y^2-7y\)
\(=y\left(y-7\right)\)
Thay \(y=x^2+4x+10\), ta có:
\(\left(x^2+4x+10\right).\left(x^2+4x+10-7\right)\)
\(=\left(x^2+4x+10\right).\left(x^2+4x+3\right)\)
\(=\left(x^2+4x+10\right).\left(x^2+3x+x+3\right)\)
\(=\left(x^2+4x+10\right).\left[\left(x^2+x\right)+\left(3x+3\right)\right]\)
\(=\left(x^2+4x+10\right).\left[x\left(x+1\right)+3\left(x+1\right)\right]\)
\(=\left(x^2+4x+10\right)\left(x+3\right)\left(x+1\right)\)
Vậy ................................................................
Chúc bn hok tốt!!!