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Bài 1:
a: \(=\dfrac{3x+5-5}{2x}=\dfrac{3x}{2x}=\dfrac{3}{2}\)
b: \(=\dfrac{2x}{x+3}\cdot\dfrac{\left(x+3\right)\left(x-3\right)}{x}=2\left(x-3\right)\)
Bài 2:
=>x^3+x+2x^2+2+a-2 chia hết cho x^2+1
=>a-2=0
=>a=2
Mk xin lỗi nha, câu c sai đề
c) (x+6)4 + (x+8)4 = 272
Vì dài quá nên mình chỉ có thể trả lời được mấy câu thôi
Bài 1:
27x3 - 8 : (6x + 9x2 +4)
= (3x - 2) (9x2 + 6x + 4) : (9x2 + 6x + 4)
= 3x - 2
Bài 3:
a, 81x4 + 4 = (9x2)2 + 36x2 + 4 - 36x2
= (9x2 + 2)2 - (6x)2
= (9x2 + 6x + 2)(9x2 - 6x + 2)
b, x2 + 8x + 15 = x2 + 3x + 5x + 15
= x(x + 3) + 5(x + 3)
= (x + 3)(x + 5)
c, x2 - x - 12 = x2 + 3x - 4x - 12
= x(x + 3) - 4(x + 3)
= (x + 3) (x - 4)
Câu 1:
(27x3 - 8) : (6x + 9x2 + 4)
= (3x - 2)(9x2 + 6x + 4) : (6x + 9x2 + 4)
= 3x - 2
Câu 2:
a) (3x - 5)(2x+ 11) - (2x + 3)(3x + 7)
= 6x2 + 33x - 10x - 55 - 6x2 - 14x - 9x - 21
= -76
⇒ đccm
b) (2x + 3)(4x2 - 6x + 9) - 2(4x3 - 1)
= 8x3 + 27 - 8x3 + 2
= 29
⇒ đccm
Câu 3:
a) 81x4 + 4
= (9x2)2 + 22
= (9x2 + 2)2 - (6x)2
= (9x2 - 6x + 2)(9x2 + 6x + 2)
b) x2 + 8x + 15
= x2 + 3x + 5x + 15
= x(x + 3) + 5(x + 3)
= (x + 3)(x + 5)
c) x2 - x - 12
= x2 - 4x + 3x - 12
= x(x - 4) + 3(x - 4)
= (x - 4)(x + 3)
a.
\(\left(2x-1\right)^3+6\left(3x-1\right)^3=2\left(x+1\right)^3+6\left(x+2\right)^3\)
\(\Leftrightarrow\left(2x\right)^3-3.\left(2x\right)^2.1+3.2x.1+1^3+6.\left[\left(3x\right)^3-3.\left(3x\right)^2.1+3.3x.1+1^3\right]=2\left(x^3+3x^2+3x+1\right)+6\left(x^2+3.x^2.2+3.x.2^2+2^3\right)\)
a ) \(\left(5x+2y\right)^2=25x^2+20xy+4y^2\)
b ) \(\left(-3x+2\right)^2=9x^2-12x+4\)
c ) \(\left(\dfrac{2}{3}x+\dfrac{1}{3}y\right)^2=\dfrac{4}{9}x^2+\dfrac{4}{9}xy+\dfrac{1}{9}y^2\)
d ) \(\left(2x-\dfrac{5}{2}y\right)^2=4x^2-10xy+\dfrac{25}{4}y^2\)
e ) \(\left(x+\dfrac{4}{3}y^2\right)^2=x^2+\dfrac{8}{3}xy^2+\dfrac{16}{9}y^4\)
f ) \(\left(2x^2+\dfrac{5}{3}y\right)^2=4x^4+\dfrac{20}{3}x^2y+\dfrac{25}{9}y^2\)
a) (x + 5)2 - (x - 3)2 = 2x - 7
(x + 5 - x + 3)(x + 5 + x - 3) = 2x - 7
8(2x + 2)= 2x - 7
16x + 16 = 2x - 7
16x - 2x = - 7 - 16
14x = - 23
x = - 23/14
b) (2x - 3)(4x2 + 6x + 9) = 98
(2x)3 - 33 = 98
8x3 - 27 = 98
8x3 = 125
x3 = 125/8
x3 = (5/2)3
x = 5/2
Bài 2:
a, \(A=3x\left(2x-5y\right)+\left(3x-y\right)\left(-2x\right)-\dfrac{1}{2}\left(2-26xy\right)\)
\(=6x^2-15xy-6x^2+2xy-1+13xy\)
\(=-1\)
\(\Rightarrowđpcm\)
b, \(B=\left(2x-3\right)\left(4x+1\right)-4\left(x-1\right)\left(2x-1\right)-2x+5\)
\(=8x^2+2x-12x-3-4\left(2x^2-x-2x+1\right)-2x+5\)
\(=8x^2-10x+2-8x^2+4x+8x-4-2x\)
\(=2-4=-2\)
\(\Rightarrowđpcm\)
Bài 2:
\(=\dfrac{3x^4+3x^2+x^3+x-3x^2-3+5x-5}{x^2+1}\)
\(=3x^2+x-3+\dfrac{5x-5}{x^2+1}\)
Bài 3:
\(\dfrac{A}{B}=\dfrac{2x^3-x^2-x+1}{x^2-2x}\)
\(=\dfrac{2x^3-4x^2+3x^2-6x+5x+1}{x^2-2x}\)
\(=2x^2+3+\dfrac{5x+1}{x^2-2x}\)
=>\(2x^3-x^2-x+1=\left(x^2-2x\right)\left(2x^2+3\right)+5x+1\)