ban da lam duoc chua vay, giup minh voi!

ban hoc  nha thay phuc nhom 1 ak

ko bít làm bn ạ !

Lời giải:

Đặt \(x-2=a(a\neq 0)\). PT trở thành:

\(\left(\frac{a-1}{a}\right)^3-(a-1)^3=16\)

\(\Leftrightarrow 16+(a-1)^3-\left(1-\frac{1}{a}\right)^3=0\)

\(\Leftrightarrow 16+a^3-3a^2+3a-1-\left(1-\frac{3}{a}+\frac{3}{a^2}-\frac{1}{a^3}\right)=0\)

\(\Leftrightarrow a^3+\frac{1}{a^3}-3(a^2+\frac{1}{a^2})+3(a+\frac{1}{a})+14=0\)

\(\Leftrightarrow (a+\frac{1}{a})^3-3(a^2+\frac{1}{a^2})+14=0\)

\(\Leftrightarrow (a+\frac{1}{a})^3-3(a+\frac{1}{a})^2+20=0\)

Đặt \(a+\frac{1}{a}=t\Rightarrow t^3-3t^2+20=0\)

\(\Leftrightarrow t^2(t+2)-5(t^2-4)=0\)

\(\Leftrightarrow (t+2)(t^2-5t+10)=0\)

Dễ thấy \(t^2-5t+10>0, \forall t\in\mathbb{R}\Rightarrow t+2=0\Leftrightarrow t=-2\)

Do đó: \(a+\frac{1}{a}=-2\Leftrightarrow \frac{(a+1)^2}{a}=0\Rightarrow a=-1\)

\(\Rightarrow x=2+a=1\)

Vậy $x=1$ là nghiệm của pt.

kho thay duoc t^2-5t+10>0. t€R.

5x (1/5x -2) + 3(6-1/3x^2) =12

x^2 - 10x + 18 -x^2 =12

-10x + 18 = 12

-10x = -6

x= 6/10

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

5x^2 - 15x + 5 + x - 5x^2 = x-2

-15x = -7

x= 7/15

\(\Leftrightarrow\dfrac{3}{x-2}>0\)

=>x-2>0

hay x>2

`P=((3+x)/(3-x)-(3-x)/(3+x)+(4x^2)/(x^2-9)):((2x+1)/(x+3)-1)`

`=((4x^2-(3-x)^2-(3+x)^2)/(x^2-9)):((2x+1-x-3)/(x+3))`

`=((4x^2-x^2+6x-9-x^2-6x-9)/(x^2-9)):((x-2)/(x+3))`

`=((2x^2-18)/(x^2-9))*(x+3)/(x-2)`

`=((2(x^2-9))/(x^2-9))*(x+3)/(x-2)`

`=(2x+6)/(x-2)`

ĐKXĐ: \(x\ne\pm3;x\ne-\dfrac{1}{2};x\ne2\)

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

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

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

\(=\dfrac{12x-4x^2}{3-x}\cdot\dfrac{1}{x-2}\)

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

Chắc mình làm tắt quá để mình làm lại bước biến đổi.

`P=((3+x)/(3-x)-(3-x)/(3+x)+(4x^2)/(x^2-9)):((2x+1)/(x+3)-1)`

`=((x-3)/(x+3)+(4x^2)/(x^2-9)-(x+3)/(x-3)):((2x+1-x-3)/(x+3))`

`=((x-3)^2/(x^2-9)+(4x^2)/(x^2-9)-(x+3)^2/(x^2-9)):((x-2)/(x+3))`

`=(((x-3)^2+4x^2-(x+3)^2)/(x^2-9))*(x+3)/(x-2)`

`=(x^2-6x+9+4x^2-x^2-6x-9)/(x^2-9)*(x+3)/(x-2)`

`=(4x^2-12x)/(x^2-9)*(x+3)/(x-2)`

`=(4x(x-3))/((x-3)(x+3))*(x+3)/(x-2)`

`=(4x)/(x+3)*(x+3)/(x-2)`

`=(4x)/(x-2)`

làm ơn giúp 🙏🙏🙏

a: =>1/3x+2/5x-2/5=0

=>11/15x-2/5=0

=>11/15x=2/5

=>x=2/5:11/15=2/5*15/11=30/55=6/11

b: =>-5x-1-1/2x+1/3=x

=>-11/2x-2/3-x=0

=>-13/2x=2/3

=>x=-2/3:13/2=-2/3*2/13=-4/39

c: (x+1/2)(2/3-2x)=0

=>x+1/2=0 hoặc 2/3-2x=0

=>x=1/3 hoặc x=-1/2

d: 9(3x+1)^2=16

=>(3x+1)^2=16/9

=>3x+1=4/3 hoặc 3x+1=-4/3

=>3x=1/3 hoặc 3x=-7/3

=>x=1/9 hoặc x=-7/9