Tìm A và B
`4A+2B``=``x^2-2x+9`
`4A+3B``=``x^2-2x+1`
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
H9
HT.Phong (9A5)
CTVHS
26 tháng 6 2023
a) \(\left(2a-b\right)\left(b+4a\right)+2a\left(b-3a\right)\)
\(=2ab+8a^2-b^2-4ab+2ab-6a^2\)
\(=\left(2ab+2ab-4ab\right)+\left(8a^2-6a^2\right)-b^2\)
\(=2a^2-b^2\)
b) \(\left(3a-2b\right).\left(2a-3b\right)-6a\left(a-b\right)\)
\(=6a^2-9ab-4ab+6b^2-6a^2+6ab\)
\(=\left(6a^2-6a^2\right)-\left(9ab+4ab-6ab\right)+6b^2\)
\(=-7ab+b^2\)
c) \(5b\left(2x-b\right)-\left(8b-x\right)\left(2x-b\right)\)
\(=10bx-5b^2-\left(16bx-8b^2-2x^2+bx\right)\)
\(=10bx-5b^2-16bx+8b^2+2x^2-bx\)
\(=\left(10bx-16bx-bx\right)-\left(5b^2-8b^2\right)+2x^2\)
\(=-7bx+3b^2+2x^2\)
d) \(2x\left(a+15x\right)+\left(x-6a\right)\left(5a+2x\right)\)
\(=2ax+30x^2+5ax+2x^2-30a^2-12ax\)
\(=\left(2ax+5ax-12ax\right)+\left(30x^2+2x^2\right)-30a^2\)
\(=-5ax+32x^2-30a^2\)
26 tháng 6 2023
a: =2ab+8a^2-b^2-4ab+2ab-6a^2
=2a^2-b^2
b: =6a^2-9ab-4ab+6b^2-6a^2+6ab
=-7ab+6b^2
c: =10bx-5b^2-16bx+8b^2+2x^2-xb
=3b^2+2x^2-7xb
d: =2xa+30x^2+5ax+2x^2-30a^2-12ax
=32x^2-30a^2-5ax
A
19 tháng 7 2018
a) 4x^2 - 12xy + 9y^2
=(2x)^2 - 2.2.3xy + (3y)^2
=(2x+3y)^2
b) 27a^3 - 64b^3
=(3a)^3 - (4b)^3
=(3a - 4b) [(3a)^2 +3a.4b +(4B)^2]
d) (2x - 6y)^2 - (3xy - 4)^2
=[ (2x - 6y)+ (3xy - 4) ] [ (2x - 6y)- (3xy - 4) ]
19 tháng 7 2018
\(1,a,4x^2-12xy+9y^2\)
\(=\left(2x\right)^2-2.3.2xy+\left(3y\right)^2\)
\(=\left(2x-3y\right)^2\)
\(b,27a^3-64b^3\)
\(=\left(3a\right)^3-\left(4b\right)^3\)
\(\left(3a-4b\right)\left(9a^2+12ab+16b^2\right)\)
Y
22 tháng 6 2023
\(a,a=-\dfrac{3}{2}\)
\(\Rightarrow3\left[2\left(-\dfrac{3}{2}\right)-1\right]+5\left(3+\dfrac{3}{2}\right)=3.\left(-3-1\right)+5.\dfrac{9}{2}=-12+\dfrac{45}{2}=\dfrac{21}{2}\)
\(b,x=2,1\)
\(\Rightarrow25.2,1-4\left(3.2,1-1\right)+7\left(5-2.2,1\right)=52,5-4.5,3+7.0,8=36,9\)
\(c,b=\dfrac{1}{2}\)
\(\Rightarrow12\left(2-3.\dfrac{1}{2}\right)+35.\dfrac{1}{2}-9\left(\dfrac{1}{2}+1\right)=12.\dfrac{1}{2}+\dfrac{35}{2}-9.\dfrac{3}{2}=6+\dfrac{35}{2}-\dfrac{27}{2}=10\)
\(d,a=-0,2\)
\(\Rightarrow4.\left(-0,2\right)^2-2\left(10.\left(-0,2\right)-1\right)+4.\left(-0,2\right)\left(2-\left(-0,2\right)^2\right)\)
\(=4.0,04-2.\left(-3\right)-0,8.1,96\)
\(=0,16+6-1,568\)
\(=4,592\)
22 tháng 6 2023
a: A=6a-3+15-5a=a+12
Khi a=-3/2 thì A=-3/2+12=10,5
b: B=25x-12x+4+35-8x=5x+39
Khi x=2,1 thì B=10,5+39=49,5
c: C=24-6b+35b-9b-9=20b+15
Khi b=0,5 thì C=10+15=25
d: D=4a^2-20a+2+8a-4a^3=-4a^3+4a^2-12a+2
Khi a=-0,2 thì
D=-4*(-1/5)^3+4*(-1/5)^2-12*(-1/5)+2=4,592
3 tháng 7 2019
a) (2a - b)(b + 4a) + 2a(b - 3a)
= 2a(b + 4a) - b(b + 4a) + 2ab - 6a^2
= 2ab + 8a^2 - b^2 - 4ab + 2ab - 6a^2
= (8a^2 - 6a^2) + (2ab + 2ab - 4ab) - b^2
= 2a^2 - b^2
b) .(3a - 2b)(2a - 3b) - 6a(a - b)
= 3a(2a - 3b) - 2b(2a - 3b) - (6a^2 - 6ab)
= 6a^2 - 9ab - (4ab - 6b^2) - (6a^2 - 6ab)
= 6a^2 - 9ab - 4ab + 6b^2 - 6a^2 + 6ab
= 6b^2 + (6a^2 - 6a^2) + (6ab - 4ab - 9ab)
= 6b^2 - 7ab
c. 5b(2x - b) - (8b - x)(2x - b)
= 10bx - 5b^2 - 8b(2x - b) + x(2x - b)
= 10bx - 5b^2 - 16bx + 8b^2 + 2x^2 - bx
= (10bx - 16bx - bx) + 2x^2 + (8b^2 - 5b^2)
= -7bx + 2x^2 + 3b^2
d. 2x(a + 15x) + (x - 6a)(5a + 2x)
= 2ax + 30x^2 + x(5a + 2x) - 6a(5a + 2x)
= 2ax + 30x^2 + 5ax + 2x^2 - 30a^2 - 12ax
= (30x^2 + 2x^2) + (2ax + 5ax - 12ax) - 30a^2
= 32x^2 - 5ax - 30a^2
Chúc bạn hok tốt !!!
NH
2
21 tháng 7 2017
a2-b2-4a+4b
=(a-b)(a+b)-4(a-b)
=(a-b)(a+b-4)
b,
x3-3x2-3x+1
=(x+1)(x2-x+1)-3x(x+1)
=(x+1)(x2-4x+1)
c,sai đề
22 tháng 7 2017
mình trả lời câu a,b đã mình đang bận
a, a^2-b^2-4a+4b
=(a-b)(a+b)-4(a-b)
=(a-b)(a+b-4)
b, x^3-3x^2-3x+1
=x^3 +x^2-4x^2-4x+x+1
=x(x+1)-4x(x+1)+(x+1)
=(x+1)(x-4x+1)
LC
0
Ta có:
\(4A+3B-\left(4A+2B\right)=x^2-2x+1-\left(x^2-2x+9\right)\)
\(\Rightarrow B=-8\)
Thay B vào \(4A+2B=x^2-2x+9\) được:
\(4A+2.\left(-8\right)=x^2-2x+9\)
\(\Rightarrow4A=x^2-2x+9+16\)
\(\Rightarrow4A=x^2-2x+25\)
\(\Rightarrow A=\dfrac{x^2-2x+25}{4}\)
Vậy...
\(\left\{{}\begin{matrix}4A+2B=x^2-2x+9\\4A+3B=x^2-2x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}B=-8\\4A+3B=x^2-2x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}B=-8\\4A+3.\left(-8\right)=x^2-2x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}B=-8\\4A-24=x^2-2x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}B=-8\\4A=x^2-2x+25\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}B=-8\\A=\dfrac{x^2-2x+25}{4}\end{matrix}\right.\)