`b,2(x+1)=5x-7`

`=>2x+2=5x-7`

`=>3x=9`

`=>x=3`

`c,3-4x(25-2x)=8x^2+x-300`

`<=>3-100x+8x^2=8x^2+x-300`

`<=>101x=303`

`<=>x=3`

`d,(10x+3)/12=1+(6+8x)/9`

`<=>(10x+3)/12=(8x+15)/9`

`<=>30x+9=32x+60`

`<=>2x=-51`

`<=>x=-51/2`

nhanh giùm mình được không

Bài 1: 

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

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

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

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

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

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

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

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

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

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

x bằng bao nhiêu

e: =>x(x^3-4x^2-8x+8)=0

=>x[(x^3+8)-4x(x+2)]=0

=>x(x+2)(x^2-2x+4-4x)=0

=>x(x+2)(x^2-6x+4)=0

=>\(x\in\left\{0;-2;3+\sqrt{5};3-\sqrt{5}\right\}\)

g: =>2x^4+5x^3-6x^3-15x^2+6x^2+15x-2x-5=0

=>(2x+5)(x^3-3x^2+3x-1)=0

=>(2x+5)(x-1)^3=0

=>x=1 hoặc x=-5/2

h: =>(x^2+8x+7)(x^2+8x+15)+15=0

=>(x^2+8x)^2+22(x^2+8x)+120=0

=>(x^2+8x+10)(x^2+8x+12)=0

=>(x^2+8x+10)(x+2)(x+6)=0

=>\(x\in\left\{-2;-6;-4+\sqrt{6};-4-\sqrt{6}\right\}\)

\(\dfrac{8x^2-8x+2}{\left(4x-2\right)\left(15-x\right)}=\dfrac{2\left(4x^2-4x+1\right)}{\left(4x-2\right)\left(15-x\right)}=\dfrac{2\left(2x-1\right)^2}{2\left(2x-1\right)\left(15-x\right)}=\dfrac{2x-1}{15-x}=\dfrac{1-2x}{x-15}\Rightarrow?=x-15\)

B = x15 - 8x14 + 8x13 - 8x2 + ... - 8x2 + 8x - 5

B = x^15 - 7x^14 -x^14+7x^13+x^13-7x^12-...-x^2+7x+x-5

B = x^14(x-7) - x^14(x-7) +...+x^2(x-7)-x(x-7)+x-5

B = 7-5=2

`B = x^15 - 7x^14 - x^14 + 7x^13 + x^13 - .... +7x + x - 7 + 2`

`<=> x^14(x-7) - x^13(x-7) + ... + x - 7 + 2`

`<=> (x^14-x^13 + ... + 1)(x-7) + 2`

Thay `x = 7 <=> (x^14 - x^13 + ... + 1) xx 0 + 2 = 2`.

B = \(x^{15}-8x^{14}+8x^{13}+8x^2+...-8x^2+8x-5\)

B = \(x^{15}-7x^{14}+x^{14}+7x^{13}+x^{13}+7x^{12}-...-x^2+7x+x-5\)

B = \(x^{14}\left(x-7\right)-x^{14}\left(x-7\right)+...+x^2\left(x-7\right)-x\left(x-7\right)-5\)

B = \(7-5=2\)

ta có: 8=7+1=x+1

\(B=x^{15}-8x^{14}+8x^{13}-...-8x^2+8x-5\)

\(\Rightarrow B=x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-...-\left(x+1\right)x^2+\left(x+1\right)x-5\)

\(\Rightarrow B=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-...-x^3-x^2+x^2+x-5\)

\(\Rightarrow B=x-5\)

\(\Rightarrow B=7-5\)

\(\Rightarrow B=2\)

B = x15 - 8x14 + 8x13 - 8x2 + ... - 8x2 + 8x - 5

B = x^15 - 7x^14 -x^14+7x^13+x^13-7x^12-...-x^2+7x+x-5

B = x^14(x-7) - x^14(x-7) +...+x^2(x-7)-x(x-7)+x-5

B = 7-5=2

B = x15 - 8x14 + 8x13 - 8x2 + ... - 8x2 + 8x - 5

B = x^15 - 7x^14 -x^14+7x^13+x^13-7x^12-...-x^2+7x+x-5

B = x^14(x-7) - x^14(x-7) +...+x^2(x-7)-x(x-7)+x-5

B = 7-5=2

Tham khảo cách này nhoá~