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\(\left(x^2-2x+3\right)\left(\dfrac{1}{2}x-5\right)\)
\(=\dfrac{1}{2}x^3-x^2+\dfrac{3}{2}x-5x^2+10x-15\)
\(=\dfrac{1}{2}x^3-6x^2+\dfrac{23}{2}x-15\)
\(\left(x^2-2x+3\right)\left(\dfrac{1}{2}x-5\right)\)
\(=\dfrac{1}{2}x^3-5x^2-x^2+10x+\dfrac{3}{2}x-15\)
\(=\dfrac{1}{2}x^3-6x^2+\dfrac{23}{2}x-15\)
\(\left(x^2-2x+3\right)\left(1212x-5\right)\)
\(=1212x^3-5x^2-2424x^2+10x+3636x-15\)
\(=1212x^3-2429x^2+3646x-15\)
\(=1212x^3-5x^2-2424x^2+10x+3636x-15\\ =1212x^3-2429x^2+3646x-15\)
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 ) \(\left(2x^2+\dfrac{5}{3}y\right)^2=4x^4+\dfrac{20}{3}x^2y+\dfrac{25}{9}y^2\)
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:
\(\dfrac{A}{B}=\dfrac{2x^3-x^2-x+1}{x^2-2x}\)
\(=\dfrac{2x^3-4x^2+3x^2-6x+5x+1}{x^2-2x}\)
\(=2x^2+3+\dfrac{5x+1}{x^2-2x}\)
=>\(2x^3-x^2-x+1=\left(x^2-2x\right)\left(2x^2+3\right)+5x+1\)
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
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.
\(\frac{x^2}{4}+x+3=\frac{x^2}{4}+x+1+2=\left(\frac{x}{2}+1\right)^2+2>0;\forall x\)
b.
\(A=-3x^2+2x-5=-3\left(x^2-2.\frac{1}{3}x+\frac{1}{9}\right)-\frac{14}{3}=-3\left(x-\frac{1}{3}\right)^2-\frac{14}{3}\le-\frac{14}{3}\)
\(A_{max}=-\frac{14}{3}\) khi \(x=\frac{1}{3}\)
c.
Đề thiếu (để ý 2 số hạng cuối)
\(A=x^4-2x^3+x^2+3x^2-6x+3-1\)
\(=\left(x^2-x\right)^2+3\left(x-1\right)^2-1\ge-1\)
\(A_{min}=-1\) khi \(x=1\)
d.
\(27x^2-\frac{9}{2}x+\frac{3}{16}=3\left(9x^2-\frac{3}{2}x+\frac{1}{16}\right)=3\left(3x-\frac{1}{4}\right)^2\)
e.
\(=\left[\left(b+c\right)+a\right]^2+\left[\left(b+c\right)-a\right]^2+\left[a-\left(b-c\right)\right]^2+\left[a+\left(b-c\right)\right]^2\)
\(=2\left(b+c\right)^2+2a^2+2a^2+2\left(b-c\right)^2\)
\(=4a^2+2b^2+4bc+2c^2+2b^2-4bc+2c^2\)
\(=4\left(a^2+b^2+c^2\right)\)
f.
\(\left(a^2+b^2\right)\left(c^2+d^2\right)=a^2c^2+b^2d^2+a^2d^2+b^2c^2\)
\(=\left(a^2c^2+b^2d^2+2ac.bd\right)+\left(a^2d^2+b^2c^2-2ad.bc\right)\)
\(=\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
\(=\dfrac{1}{2}x^3-5x^2-x^2+10x+\dfrac{3}{2}x-15=\dfrac{1}{2}x^3-6x^2+\dfrac{23}{2}x-15\)
(x^2−2x+3)(12x−5)
=12x^3−x^2+32x−5x^2+10x−15
=12x^3−6x^2+23/2x−15