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\(a,\)\(x^3-13x-12\)
\(=x^3-x-12x-12\)
\(=x\left(x^2-1\right)-12\left(x+1\right)\)
\(=x\left(x-1\right)\left(x+1\right)-12\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x-12\right)\)
\(=\left(x+1\right)\left(x^2-4x+3x-12\right)\)
\(=\left(x+1\right)\left[x\left(x-4\right)+3\left(x+4\right)\right]\)
\(=\left(x+1\right)\left(x-4\right)\left(x+3\right)\)
a) \(x^3-13x-12\)
\(=x^3+x^2-x^2-x-12x-12\)
\(=x^2\left(x+1\right)-x\left(x+1\right)-12\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x-12\right)\)
\(=\left(x+1\right)\left(x^2-4x+3x-12\right)\)
\(=\left(x+1\right)\left[x\left(x-4\right)+3\left(x-4\right)\right]\)
\(=\left(x+1\right)\left(x-4\right)\left(x+3\right)\)
b) \(2x^4+3x^3-9x^2-3x+2\)câu này hình như sai đề rồi, bạn xem lại nhen
c) \(x^4-3x^3-6x^2+3x+1\)câu này cx thế, bạn xem lại nha
/ (4x−2)(10x+4)(5x+7)(2x+1)+17=0(4x−2)(10x+4)(5x+7)(2x+1)+17=0
⇔(4x−2)(5x+7)(10x+4)(2x+1)+17=0⇔(4x−2)(5x+7)(10x+4)(2x+1)+17=0
⇔(20x2+18x−14)(20x2+18x+4)+17=0⇔(20x2+18x−14)(20x2+18x+4)+17=0
Đặt t= 20x2+18x+4(t≥0)20x2+18x+4(t≥0) ta có:
(t-18).t +17=0
⇔t2−18t+17=0⇔t2−18t+17=0
⇔(t−17)(t−1)=0⇔(t−17)(t−1)=0
⇔[t=17(tm)t=1(tm)⇔[t=17(tm)t=1(tm) ⇔[20x2+18x+4=1720x2+18x+4=1⇔[20x2+18x−13=020x2+18+3=0⇔[20x2+18x+4=1720x2+18x+4=1⇔[20x2+18x−13=020x2+18+3=0
⇔[(20x+9−341−−−√)(20x+9+341−−−√)=0(20x+9−21−−√)(20x+9+21−−√)=0⇔[(20x+9−341)(20x+9+341)=0(20x+9−21)(20x+9+21)=0
⇔⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢x=−9+341−−−√20x=−9−341−−−√20x=−9+21−−√20x=−9−21−−√20
\(a,\)\(\left(4x-2\right)\left(10x+4\right)\left(5x+7\right)\left(2x+1\right)+17\)
\(=\left(4x-2\right)\left(5x+7\right)\left(10x+4\right)\left(2x+1\right)+17\)
\(=\left(20x^2+18x-5\right)\left(20x^2+18x+4\right)+17\)
Đặt ....
a) \(2x^3-5x^2+8x-3\)
\(=\left(2x^3-4x^2+6x\right)-\left(x^2-2x+3\right)\)
\(=2x\left(x^2-2x+3\right)-\left(x^2-2x+3\right)\)
\(=\left(2x-1\right)\left(x^2-2x+3\right)\)
b) bn ktra lại đề
c) \(12x^2+5x-12y^2+12y-10xy-3\)
\(=\left(12x^2-18xy+9x\right)+\left(8xy-12y^2+6y\right)-\left(4x-6y+3\right)\)
\(=3x\left(4x-6y+3\right)+2y\left(4x-6y+3\right)-\left(4x-6y+3\right)\)
\(=\left(3x+2y-1\right)\left(4x-6y+3\right)\)
a﴿ 2x − 5x + 8x − 3 = 2x − 4x + 6x − x − 2x + 3 = 2x x − 2x + 3 − x − 2x + 3 = 2x − 1 x − 2x + 3
c﴿ 12x + 5x − 12y + 12y − 10xy − 3
= 12x − 18xy + 9x + 8xy − 12y + 6y − 4x − 6y + 3
= 3x 4x − 6y + 3 + 2y 4x − 6y + 3 − 4x − 6y + 3
= 3x + 2y − 1 4x − 6y + 3 3 2
Câu 1:
\(\dfrac{x^2-10x+21}{x^3-7x^2+x-7}=\dfrac{\left(x-7\right)\left(x-3\right)}{\left(x-7\right)\left(x^2+1\right)}=\dfrac{x-3}{x^2+1}\)
\(\dfrac{2x^2-x-15}{2x^3+5x^2+2x+5}=\dfrac{2x^2-6x+5x-15}{\left(2x+5\right)\left(x^2+1\right)}=\dfrac{\left(2x+5\right)\left(x-3\right)}{\left(2x+5\right)\left(x^2+1\right)}=\dfrac{x-3}{x^2+1}\)
Do đó: \(\dfrac{x^2-10x+21}{x^3-7x^2+x-7}=\dfrac{2x^2-x-15}{2x^3+5x^2+2x+5}\)
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 )
\(x^5+y^5-\left(x+y\right)^5\)
\(=x^5+y^5-\left(x^5+5x^4y+10x^3y^2+10x^2y^3+8xy^4+y^5\right)\)
\(=-5xy\left(x^3+2x^2y+2xy^2+y^3\right)\)
\(=-5xy\left[\left(x+y\right)\left(x^2-xy+y^2\right)+2xy\left(x+y\right)\right]\)
\(=-5xy\left(x+y\right)\left(x^2+xy+y^2\right)\)
Bài 1:
a) 3x2 - 3y2
= 3(x2 - y2)
= 3(x - y)(x + y)
b) x2 - xy + 7x - 7y
= x(x - y) + 7(x - y)
= (x - y)(x + 7)
c) x2 - 3x + 2
= x2 - x - 2x + 2
= x(x - 1) - 2(x - 1)
= (x - 1)(x - 2)
Trần Hoàng NghĩaSiêu sao bóng đáRibi Nkok NgokNguyễn Thanh HằngVũ ElsaNguyễn Ngô Minh TríAkai Harumalê thị hương giangPhạm Hoàng Giang với các bạn khác giải hộ mình nhé! Vô trang cá nhân giải hộ mình mấy bài khác nữa.
a: \(\Leftrightarrow2x^2-8x+3x-12+x^2-7x+10=3x^2-12x-5x+20\)
\(\Leftrightarrow3x^2-12x-2=3x^2-17x+20\)
=>5x=22
hay x=22/5
b: \(\Leftrightarrow24x^2+16x-9x-6-4x^2-16x-7x-28=10x^2-2x+5x-1\)
\(\Leftrightarrow20x^2-16x-34=10x^2+3x-1\)
\(\Leftrightarrow10x^2-19x-33=0\)
hay \(x\in\left\{3;-\dfrac{11}{10}\right\}\)
c: \(\Leftrightarrow x^3+2x^2-5x-10+5x=2x^2+17\)
\(\Leftrightarrow x^3+2x^2-10-2x^2-17=0\)
=>x3=27
=>x=3
d: \(\Leftrightarrow x^3+1-x^3+3x=15\)
=>3x=14
hay x=14/3
a, \(3x^3-4x^2+5x-4\)
\(=3x^3-3x^2-x^2+x+4x-4\)
\(=3x^2\left(x-1\right)-x\left(x-1\right)+4\left(x-1\right)\)
\(=\left(3x^2-x+4\right)\left(x-1\right)\)
b, \(4x^3-3x^2+5x-21\)
\(=4x^3-7x^2+4x^2-7x+12x-21\)
\(=x^2\left(4x-7\right)+x\left(4x-7\right)+3\left(4x-7\right)\)
\(=\left(x^2+x+3\right)\left(4x-7\right)\)
c, \(3x^3+8x^2+14x+15\)
\(=3x^3+5x^2+3x^2+5x+9x+15\)
\(=x^2\left(3x+5\right)+x\left(3x+5\right)+3\left(3x+5\right)\)
\(=\left(x^2+x+3\right)\left(3x+5\right)\)
Bài này dùng phương pháp nhẩm nghiệm (tối ưu nhất với đa thức bậc ba)
Chúc bạn học tốt.