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PQ
4
25 tháng 1 2017
6x^5+15x^4+20x^3+15x^2+6x+1
=3x^4(2x+1)+6x^3(2x+1)+7x^2(2x+1)+4x(2x+1)+(2x+1)
=(2x+1)(3x^4+6x^3+7x^2+4x+1)
=(2x+1)(3x^2(x^2+x+1)+3x(x^2+x+1)+(x^2+x+1)
=(2x+1)(x^2+x+1)(3x^2+3x+1)
PT
0
YB
1
TN
3 tháng 1 2017
a)\(\left(x^2-x+2\right)^2+\left(x-2\right)^2\)
\(=x^4-2x^3+6x^2-8x+8\)
\(=x^4-2x^3+2x^2+4x^2-8x+8\)
\(=x^2\left(x^2-2x+2\right)+4\left(x^2-2x+2\right)\)
\(=\left(x^2+4\right)\left(x^2-2x+2\right)\)
b)\(6x^5+15x^4+20x^3+15x^2+6x+1\)
\(=6x^5+3x^4+12x^4+6x^3+14x^3+7x^2+8x^2+4x+2x+1\)
\(=3x^4\left(2x+1\right)+6x^3\left(2x+1\right)+7x^2\left(2x+1\right)+4x\left(2x+1\right)+\left(2x+1\right)\)
\(=\left(3x^4+6x^3+7x^2+4x+1\right)\left(2x+1\right)\)
\(=\left[3x^4+3x^3+x^2+3x^3+3x^2+x+3x^2+3x+1\right]\left(2x+1\right)\)
\(=\left[x^2\left(3x^2+3x+1\right)+x\left(3x^2+3x+1\right)+\left(3x^2+3x+1\right)\right]\left(2x+1\right)\)
\(=\left(x^2+x+1\right)\left(3x^2+3x+1\right)\left(2x+1\right)\)
PT
0
DV
1
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 3 2019
a/ \(\left(x^2-x+2\right)^2+\left(x-2\right)^2=\left(x^2-x+2\right)^2-x^2+x^2+\left(x-2\right)^2\)
\(=\left(x^2-2x+2\right)\left(x^2+2\right)+2x^2-4x+4\)
\(=\left(x^2-2x+2\right)\left(x^2+2\right)+2\left(x^2-2x+2\right)\)
\(=\left(x^2-2x+2\right)\left(x^2+4\right)\)
b/ \(6x^5+15x^4+20x^3+15x^2+6x+1\)
\(=6x^5+6x^4+2x^3+9x^4+9x^3+3x^2+9x^3+9x^2+3x+3x^2+3x+1\)
\(=2x^3\left(3x^2+3x+1\right)+3x^2\left(3x^2+3x+1\right)+3x\left(3x^2+3x+1\right)+3x^2+3x+1\)
\(=\left(3x^2+3x+1\right)\left(2x^3+3x^2+3x+1\right)\)
\(=\left(3x^2+3x+1\right)\left(x^3+\left(x+1\right)^3\right)\)
\(=\left(3x^2+3x+1\right)\left(2x+1\right)\left(x^2-\left(x+1\right)x+\left(x+1\right)^2\right)\)
\(=\left(3x^2+3x+1\right)\left(3x+1\right)\left(x^2+x+1\right)\)
RC
2
PA
26 tháng 9 2016
\(8x^3+12x^2+6x+7-3\left(2x+1\right)^2=6\)
\(\left(2x\right)^3+3\times\left(2x\right)^2\times1+3\times2x\times1^2+1^3+6-3\left(2x+1\right)^2=6\)
\(\left(2x+1\right)^3-3\left(2x+1\right)^2=6-6\)
\(\left(2x+1\right)^2\left(2x+1-3\right)=0\)
\(\left(2x+1\right)^2\left(2x-2\right)=0\)
\(2\left(2x+1\right)^2\left(x-1\right)=0\)
\(\left[\begin{array}{nghiempt}2x+1=0\\x-1=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=1\end{array}\right.\)
PA
26 tháng 9 2016
\(20x^3-15x^2+7x=45x^2-38x\)
\(20x^3-15x^2-45x^2+7x+38x=0\)
\(20x^3-60x^2+45x=0\)
\(5x\left(4x^2-12x+9\right)=0\)
\(5x\left(2x-3\right)^2=0\)
\(\left[\begin{array}{nghiempt}x=0\\2x-3=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=0\\x=\frac{3}{2}\end{array}\right.\)
ND
28 tháng 2 2018
b. sửa đề
\(6x^4+25x^3+12x-25x^2+6=0\)
\(\Leftrightarrow6x^4+12x^3+13x^3+26x^2-14x^2-28x+3x+6=0\)
\(\Leftrightarrow6x^3\left(x+2\right)+13x^2\left(x+2\right)-14x\left(x+2\right)+3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(6x^3+13x^2-14x+3\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)\left(2x-1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=-3\\x=\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy........
K
28 tháng 2 2018
Bài 1 : Giải phương trình
a) (x + 3)4 + (x + 5)4 = 16
Đặt : x + 3 = t
=> x + 5 = x + 3 + 2 = t + 2
Thay x + 3 = t và x + 5 = t + 2 vào phương trình, ta có :
t4 + (t + 2)4 = 16
<=> 2t4 + 8t3 + 24t2 + 32t + 16 = 16
<=> 2(t4 + 4t3 + 12t2 + 16t) = 0
<=> t4 + 4t3 + 12t2 + 16t = 0
<=> (t + 2) . t . (t2 + 2y + 4) = 0
TH1 : t = 0
TH2 : t + 2 = 0 <=> t = -2
TH3 : t2 + 2y + 4 = 0 (vô nghiệm => loại)
Nên t = 0 hoặc t = -2
hay x + 3 = -2 hoặc x + 3 = 0
<=> x = -5 hoặc x = -3
\(S=\left\{-5;-3\right\}\)
b) 6x4 + 25x3 + 12x2 - 25x + 6 = 0
<=> 6x4 + 12x3 + 13x3 + 26x2 - 14x2 - 28x + 3x + 6 = 0
<=> 6x3 (x + 2) + 13x2 (x + 2) - 14x (x + 2) + 3(x + 2) = 0
<=> (x + 2)(6x3 + 13x2 - 14x + 3) = 0
<=> (x + 2)(6x3 + 18x2 - 5x2 - 15x + x + 3) = 0
\(\Leftrightarrow\left(x+2\right)[6x^2\left(x+3\right)-5x\left(x+3\right)+\left(x+3\right)]=0\)
<=> (x + 2)(x + 3) (6x2 - 5x + 1) = 0
<=> (x + 2)(x + 3)(2x - 1)(3x - 1) = 0
TH1 : x + 2 = 0 <=> x = -2
TH2 : x + 3 = 0 <=> x = -3
TH3 : 2x - 1 = 0 <=> 2x = 1 <=> x = \(\dfrac{1}{2}\)
TH4 : 3x - 1 = 0 <=> 3x = 1 <=> 3x = \(\dfrac{1}{3}\)
\(S=\left\{-2;-3;\dfrac{1}{2};\dfrac{1}{3}\right\}\)
MA
1 tháng 10 2016
1) \(6x^2-15x+6\)
\(=6x^2-3x-12x+6\)
\(=3x\left(2x-1\right)-6\left(2x-1\right)\)
\(=\left(3x-6\right)\left(2x-1\right)\)
\(=3\left(x-2\right)\left(2x-1\right)\)
b) \(6x^2-20x+6\)
\(=6x^2-2x-18x+6\)
\(=2x\left(3x-1\right)-6\left(3x-1\right)\)
\(=\left(2x-6\right)\left(3x-1\right)\)
\(=2\left(x-3\right)\left(3x-1\right)\)
c) \(-10x^2-7x+6\)
\(=5x-10x^2+6-12x\)
\(=5x\left(1-2x\right)+6\left(1-2x\right)\)
\(=\left(5x+6\right)\left(1-2x\right)\)
d) \(10x^2-7x-12\)
\(=10x^2-15x+8x-12\)
\(=5x\left(2x-3\right)+4\left(2x-3\right)\)
\(=\left(5x+4\right)\left(2x-3\right)\)
QD
1 tháng 10 2016
1) 6x^2 - 15x + 6
= 6x^2 - 12x - 3x +6
= 6x (x - 2) - 3 (x - 2)
= (x - 2) (6x - 3)
= 3 (x - 2) (x - 1)
2) 6x^2 - 20x + 6
= 6x^2 - 18x - 2x + 6
= 6x (x - 3) - 2 (x - 3)
= (x - 3) (6x - 2)
= 2 (x - 3) (3x - 1)
3) -10x^2 - 7x + 6
= -10x^2 + 5x - 12x + 6
= 5x (1 - 2x) + 6 (1 - 2x)
= (1 - 2x) (5x + 6)
4) 10x^2 - 7x - 12
= 10x^2 - 15x + 8x - 12
= 5x (2x - 3) + 4 (2x - 3)
= (2x - 3) (5x + 4)
good luck !
28 tháng 1 2020
Bài 1:
a)\(5x^2y^3-25x^3y^4+10x^3y^3=5x^2y^3\left(1-5xy+2x\right)\)
b)\(x^3-2xy-x^2y+2y^2=\left(x^3-x^2y\right)-\left(2xy-2y^2\right)=x^2\left(x-y\right)-2y\left(x-y\right)=\left(x-y\right)\left(x^2-2y\right)\)
c)Đề sai hoàn toàn
d) \(2x^2+4xy+2y^2-8z^2=2\left(x^2+2xy+y^2-4z^2\right)=2\left[\left(x+y\right)^2-\left(2z\right)^2\right]=2\left(x+y-2z\right)\left(x+y+2z\right)\)e) \(3x-3a+yx-ya=3\left(x-a\right)+y\left(x-a\right)=\left(x-a\right)\left(3+y\right)\)
f)\(\left(x^2+y^2\right)^2-4x^2y^2=\left(x-y\right)^2\left(x+y\right)^2\)
g)\(2x^2-5x+2=2x^2-x-4x+2=x\left(2x-1\right)-2\left(2x-1\right)=\left(2x-1\right)\left(x-2\right)\)
i)\(14x\left(x-y\right)-21y\left(y-x\right)+28z\left(x-y\right)=14x\left(x-y\right)+21y\left(x-y\right)+28z\left(x-y\right)=7\left(x-y\right)\left(2x+3y+4z\right)\)
hình như là hệ số đối xứng
phân tích thành nhân tử ak