`A=(2\sqrtx-9)(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)-(2sqrtx+1)(3-sqrtx)(x>=0,x ne 4, x ne 9)`

`=(2\sqrtx-9)(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)+(2sqrtx+1)(sqrtx-3)`

`=(2sqrtx-9-x+9+2x-3sqrtx-2)/(x-5sqrtx+6)`
`=(x-sqrtx-2)/(x-5sqrtx+6)`
`=((\sqrtx+1)(sqrtx-2))/((sqrtx-2)(sqrtx-3))`
`=(sqrtx+1)/(sqrtx-3)`

`A=(2\sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)-(2sqrtx+1)/(3-sqrtx)(x>=0,x ne 4, x ne 9)`

`=(2\sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)+(2sqrtx+1)/(sqrtx-3)`

`=(2sqrtx-9-x+9+2x-3sqrtx-2)/(x-5sqrtx+6)`
`=(x-sqrtx-2)/(x-5sqrtx+6)`
`=((\sqrtx+1)(sqrtx-2))/((sqrtx-2)(sqrtx-3))`
`=(sqrtx+1)/(sqrtx-3)`

a: (x-3)(x-2)<0

=>x-2>0 và x-3<0

=>2<x<3

b: \(\left(x+3\right)\left(x+4\right)\left(x^2+2\right)\ge0\)

=>(x+3)(x+4)>=0

=>x+3>=0 hoặc x+4<=0

=>x>=-3 hoặc x<=-4

c: \(\dfrac{x-1}{x-2}\ge0\)

=>x-2>0 hoặc x-1<=0

=>x>2 hoặc x<=1

d: \(\dfrac{x+3}{2-x}>=0\)

=>\(\dfrac{x+3}{x-2}< =0\)

=>x+3>=0 và x-2<0

=>-3<=x<2

a: 

Sửa đề: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{3x+3}{9-x}\right)\cdot\left(\dfrac{\sqrt{x}-7}{\sqrt{x}+1}+1\right)\)

\(P=\left(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\right)\cdot\dfrac{\sqrt{x}-7+\sqrt{x}+1}{\sqrt{x}+1}\)

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\cdot\dfrac{2\sqrt{x}-6}{\sqrt{x}+1}\)

\(=\dfrac{-3\sqrt{x}-3}{\sqrt{x}+3}\cdot\dfrac{2}{\sqrt{x}+1}=\dfrac{-6}{\sqrt{x}+3}\)

b: P>=1/2

=>P-1/2>=0

=>\(\dfrac{-6}{\sqrt{x}+3}-\dfrac{1}{2}>=0\)

=>\(\dfrac{-12-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>=0\)

=>\(-\sqrt{x}-15>=0\)

=>\(-\sqrt{x}>=15\)

=>căn x<=-15

=>\(x\in\varnothing\)

c: căn x+3>=3

=>6/căn x+3<=6/3=2

=>P>=-2

Dấu = xảy ra khi x=0

viết kiểu gì khó hiểu quá

Ta có : (x - 3)(x - 2) < 0

Nên sảy ra 2 trường hợp : D

Th1 : \(\hept{\begin{cases}x-3< 0\\x-2>0\end{cases}\Rightarrow\hept{\begin{cases}x< 3\\x>2\end{cases}\Rightarrow}2< x< 3}\)

Th2 : \(\hept{\begin{cases}x-3>0\\x-2< 0\end{cases}\Rightarrow\hept{\begin{cases}x>3\\x< 2\end{cases}\left(loại\right)}}\)

Vậy 2 < x < 3

1.Ta có :\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\)

\(=x^2-xy+y^2\) (do x+y=1)

\(=\dfrac{3}{4}\left(x-y\right)^2+\dfrac{1}{4}\left(x+y\right)^2\ge\dfrac{1}{4}\left(x+y\right)^2\)\(=\dfrac{1}{4}.1=\dfrac{1}{4}\)

Dấu "=" xảy ra khi :\(x=y=\dfrac{1}{2}\)

Vậy \(x^3+y^3\ge\dfrac{1}{4}\)

2.

a) Sửa đề: \(a^3+b^3\ge ab\left(a+b\right)\)

\(\Leftrightarrow\left(a^3-a^2b\right)+\left(b^3-ab^2\right)\ge0\)

\(\Leftrightarrow a^2\left(a-b\right)+b^2\left(b-a\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)\left(a^2-b^2\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0\) (luôn đúng vì \(a,b\ge0\))

Đẳng thức xảy ra \(\Leftrightarrow a=b\)

b) Lần trước mk giải rồi nhá

3.

a) Áp dụng BĐT Cauchy-Schwarz dạng Engel\(P=\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}\ge\dfrac{\left(1+1+1\right)^2}{\left(x+y+z\right)+3}=\dfrac{9}{3+3}=\dfrac{3}{2}\)

Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}=\dfrac{1}{y+1}=\dfrac{1}{z+1}\\x+y+z=3\end{matrix}\right.\Leftrightarrow x=y=z=1\)

b) \(Q=\dfrac{x}{x^2+1}+\dfrac{y}{y^2+1}+\dfrac{z}{z^2+1}\le\dfrac{x}{2\sqrt{x^2.1}}+\dfrac{y}{2\sqrt{y^2.1}}+\dfrac{z}{2\sqrt{z^2.1}}\)

\(=\dfrac{x}{2x}+\dfrac{y}{2y}+\dfrac{z}{2z}=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{3}{2}\)

Đẳng thức xảy ra \(\Leftrightarrow x^2=y^2=z^2=1\Leftrightarrow x=y=z=1\)

a: =>4(2x-1)-12x=3(x+3)+24

=>8x-4-12x=3x+9+24

=>-4x-4=3x+33

=>-7x=37

=>x=-37/7

b: =>(x-2)(x+2+x-9)=0

=>(2x-7)(x-2)=0

=>x=2 hoặc x=7/2

c: =>(x-1)(x+3)-x+3=3x+3

=>x^2+2x-3-x+3=3x+3

=>x^2+x-3x-3=0

=>x^2-2x-3=0

=>(x-3)(x+1)=0

=>x=-1

d: Ta có: \(\dfrac{x}{x+3}-\dfrac{2x}{x-3}-\dfrac{3x}{9-x^2}=0\)

\(\Leftrightarrow x^2-3x-2x^2-6x+3x=0\)

\(\Leftrightarrow-x^2-6x=0\)

\(\Leftrightarrow-x\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-6\left(nhận\right)\end{matrix}\right.\)

a: Ta có: \(3\left(x-1\right)-3=2\left(x+3\right)\)

\(\Leftrightarrow3x-3-3=2x+6\)

\(\Leftrightarrow x=12\)

b: Ta có: \(\dfrac{x+4}{4}-\dfrac{x+3}{3}=\dfrac{x+6}{6}\)

\(\Leftrightarrow3x+12-4x-12=2x+12\)

\(\Leftrightarrow-3x=12\)

hay x=-4

c: Ta có: \(\left(2x-1\right)^2-x^2=0\)

\(\Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)