a) \(x-\sqrt{x}+1>0\)mà \(\sqrt{x}\)>0 => biểu thức > 0

b) \(\sqrt{x}\)\(\le x-\sqrt{x}+1\)<=> \(x-2\sqrt{x}+1\ge0\)(nhân lên do không âm)

<=> \(\left(\sqrt{x}-1\right)^2\ge0\)=> đpcm ^^

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

\(=\dfrac{2}{x+\sqrt{x}+1}\)

Với \(0\le x;y\le1\) ta có:

\(\frac{x}{\sqrt{y+3}}+\frac{y}{\sqrt{x+3}}\ge\frac{x}{\sqrt{1+3}}+\frac{y}{\sqrt{1+3}}=\frac{x+y}{2}\)

Dấu "=" xảy ra <=> x = y = 1

Có: \(0\le x;y\le1\)

=> \(0\le x^2\le x\le1;0\le y^2\le y\le1\)

\(\left(\frac{x}{\sqrt{y+3}}+\frac{y}{\sqrt{x+3}}\right)^2\le2\left(\frac{x^2}{y+3}+\frac{y^2}{x+3}\right)\le2\left(\frac{x}{x+y+2}+\frac{y}{x+y+2}\right)\)

\(=2\left(\frac{x+y+2}{x+y+2}-\frac{2}{x+y+2}\right)\le2\left(1-\frac{2}{1+1+2}\right)=1\)

=> \(\sqrt{\frac{x}{\sqrt{y+3}}+\frac{y}{\sqrt{x+3}}}\le1\)

Dấu "=" xảy ra x<=>  = y =1

Câu b bạn sửa lại đề

\(a,VT=\left[1+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\\ =\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x=VP\\ b,VT=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}+\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\\ =\sqrt{a}-\sqrt{b}+\sqrt{a}+\sqrt{b}=2\sqrt{a}=VP\)

a: \(=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x\)

a,\(P=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)

\(P=\left[\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right].\dfrac{2}{\sqrt{x}-1}\)

\(P=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)

\(P=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)

\(P=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}=\dfrac{2}{x+\sqrt{x}+1}\)

Vậy \(P=\dfrac{2}{x+\sqrt{x}+1}\)

b, Ta có \(x+\sqrt{x}+1=\left(x+2\sqrt{x}.\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\)Suy ra \(\dfrac{2}{x+\sqrt{x}+1}>0\forall x>0,x\ne1\)

hay \(P>0\forall x>0,x\ne1\)(đpcm)

ĐKXĐ: \(x\ge1\)

\(3\sqrt[]{x-1}+m\sqrt[]{x+1}=2\sqrt[4]{\left(x-1\right)\left(x+1\right)}\)

\(\Leftrightarrow3\sqrt[]{\dfrac{x-1}{x+1}}+m=2\sqrt[4]{\dfrac{x-1}{x+1}}\)

Đặt \(\sqrt[4]{\dfrac{x-1}{x+1}}=t\Rightarrow0\le t< 1\)

\(\Rightarrow3t^2+m=2t\Leftrightarrow-3t^2+2t=m\)

Xét \(f\left(t\right)=-3t^2+2t\) trên \([0;1)\)

\(f'\left(t\right)=-6t+2=0\Rightarrow t=\dfrac{1}{3}\)

\(f\left(0\right)=0;f\left(\dfrac{1}{3}\right)=\dfrac{1}{3};f\left(1\right)=-1\)

\(\Rightarrow-1< f\left(t\right)\le\dfrac{1}{3}\)

\(\Rightarrow-1< m\le\dfrac{1}{3}\)

Chọn C

ĐK:`x>=0`

Nhân hai vế với `sqrt{x+1}`

`2sqrt2+sqrt{x^2+x}<=sqrt{x^2+10x+9}`

BP 2 vế ta có:

`8+x^2+x+4\sqrt{2x^2+2x}<=x^2+10x+9`

`<=>4\sqrt{2x^2+2x}<=9x-1`

ĐK:`x>=1/9`

`<=>16(2x^2+2x)<=81x^2-18x+1`

`<=>32x^2+32x<=81x^2-18x+1`

`<=>49x^2-50x+1>=0`

`<=>(x-1)(49x-1)>=0`

Vì `x>=1/9=>49x-1>0`

`=>x-1>=0<=>x>=1`

Vậy bpt có nghiệm `S={x|x>=1}`