b: \(A=\dfrac{2-1}{3\cdot2}=\dfrac{1}{6}\)

nhanh giùm mình được không

Bài 1: 

a) Ta có: \(P=1+\dfrac{3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)\)

\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}-\dfrac{3x}{3\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)

\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{4}{x-2}-\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)

\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\dfrac{4\left(x+2\right)-x-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{4x+8-x-x+2}\)

\(=1+3\cdot\dfrac{\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)

\(=1+\dfrac{3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)

\(=\dfrac{\left(x+3\right)\left(2x+10\right)+3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)

\(=\dfrac{2x^2+10x+6x+30+3x-6}{\left(x+3\right)\left(2x+10\right)}\)

\(=\dfrac{2x^2+19x-6}{\left(x+3\right)\left(2x+10\right)}\)

Bài 1:

\(a,6x^2-15x^3y\\ b,=-\dfrac{2}{3}x^2y^3+\dfrac{2}{3}x^4y-\dfrac{8}{3}xy\)

Bài 2:

\(a,=20x^3-10x^2+5x-20x^3+10x^2+4x=9x\\ b,=3x^2-6x-5x+5x^2-8x^2+24=24-11x\\ c,=x^5+x^3-2x^3-2x=x^5-x^3-2x\)

câu d của bài 2 là của bài 1 nha mình để nhầm chỗ huhu

a, ĐKXĐ : x khác -4;4;-2

P =[ 8+x-4/(x-4).(x+4) ] : 1/(x+2).(x-4)

   = x+4/(x+4).(x-4)   . (x+2).(x-4)

   = x+2

b, x^2-9x+20 = 0

<=> (x^2-4x)-(5x-20)=0

<=> (x-4).(x-5)=0

<=> x-4=0 hoặc x-5=0

<=> x=4 hoặc x=5

+, Với x=4 thì P = 4+2 = 6

+, Với x=5 thì P = 5+2 = 7

k mk nha

cứuuuuuuuuuuu

`@` `\text {Ans}`

`\downarrow`

Gửi c!

Bài 1: 

a) \(3x^2\left(2x^3-x+5\right)-6x^5-3x^3+10x^2\)

\(=6x^5-3x^3+10x^2-6x^5-3x^3+10x^2\)

\(=10x^2+10x^2\)

\(=20x^2\)

b) \(-2x\left(x^3-3x^2-x+11\right)-2x^4+3x^3+2x^2-22x\)

\(=-2x^4+6x^3+2x^2-22x-2x^4+3x^3+2x^2-22x\)

\(=-4x^4+9x^3+4x^2-44x\)

4:

a: =>1/4x^2-1/4x^2+2x=-14

=>2x=-14

=>x=-7

b: =>2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20

=>3x^2-12x-2=3x^2-17x+20

=>5x=22

=>x=22/5

a) \(A=\left(x+2\right)\left(x^2-2x+4\right)-x^3+2\)

\(A=x^3+8-x^3+2\)

\(A=10\)

b) \(B=\left(x-1\right)\left(x^2+x+1\right)-\left(x+1\right)\left(x^2-x+1\right)\)

\(B=x^3-1-\left(x^3+1\right)\)

\(B=x^3-1-x^3-1\)

\(B=-2\)

c) \(C=\left(2x-y\right)\left(4x^2+2xy+y^2\right)+\left(y-3x\right)\left(y^2+3xy+9x^2\right)\)

\(C=\left(2x\right)^3-y^3+y^3-\left(3x\right)^3\)

\(C=8x^3-y^3+y^3-27x^3\)

\(C=-19x^3\)

a)

\(A=\left(x+2\right)\left(x-2\right)\left(x-2\right)-x^3+2\\ =\left(x^2-4\right)\left(x-2\right)-x^3+2\\ =x^3-2x^2-4x+8-x^3+2\\ =-2x^2-4x+10\)

b)

\(B=x^3-1-\left(x^3+1\right)\\ =x^3-1-x^3-1\\ =-2\)

c)

\(C=\left(2x\right)^3-y^3+\left(y\right)^3-\left(3x\right)^3\\ =8x^3-y^3+y^3-27x^3\\ =-19x^3\)

a, \(A=x^2\left(2x-1\right)+x\left(x+8\right)=2x^3-x^2+x^2+8x=2x^3+8x\)

Thay x = -2, ta có:

\(2\cdot\left(-2\right)^3+8\cdot\left(-2\right)=-32\)

b, \(A=2x^3+8x=0\\ \Leftrightarrow2x\left(x^2+4\right)=0\\ \Leftrightarrow x=0\)

Vậy A=0 khi x=0

a,A = \(x^2\).( 2\(x\) - 1) + \(x\)(\(x+8\))

A = 2\(x^3\) - \(x^2\) + \(x^2\) + 8\(x\)

A = 2\(x^3\) + 8\(x\)

b, \(x=-2\) ⇒ A = 2.(-2)³ + 8.(-2) = - 32 

A = 0 ⇔ 2\(x^3\) + 8\(x\) = 0

             2\(x\left(x^2+4\right)\) = 0 

              vì \(x^2\) + 4 > 0 ∀ \(x\) ⇒ \(x\)  =0 

\(a,A=x^2\left(2x-1\right)+x\left(x+8\right)\\ =2x^3-x^2+x^2+8x=2x^3+8x\\ =2x.\left(x^2+8\right)\\ Thế.x=-2.vào.A:A=2x.\left(x^2+8\right)=2.\left(-2\right).\left[\left(-2\right)^2+8\right]=-4.12=-48\\ b,A=0\\ \Leftrightarrow2x\left(x^2+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+8=0\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=0\\ Vậy:A=0\Leftrightarrow x=0\)

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