`a,`

\(B=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}}{\sqrt{x}-1}\\ =\left(\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}\\ =\dfrac{x+2\sqrt{x}+1-\left(x-2\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}\\ =\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}\\ =\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}\)

\(=\dfrac{4}{\sqrt{x}+1}\)

`b,` Để `A *B<0` ta có :

\(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\cdot\dfrac{4}{\sqrt{x}+1}< 0\\ \Leftrightarrow\dfrac{4}{\sqrt{x}-1}< 0\\ \Leftrightarrow\sqrt{x}-1< 0\left(vì.4>0\right)\\ \Leftrightarrow\sqrt{x}< 1\\ \Leftrightarrow0\le x< 1\)

Kết hợp với đkxđ ta có : \(0< x< 1\)

okeoke:vv tưởng chia thành nhân mà đó nhân tử r=)))))))

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

`=>(x-3/2)^2=9/2-1/2`

`=>(x-3/2)^2= 8/2`

`=>(x-3/2)^2=4`

`=>(x-3/2)^2=(+-2)^2`

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=4\\x-\dfrac{3}{2}=-4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4+\dfrac{3}{2}\\x=-4+\dfrac{3}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{11}{2};-\dfrac{5}{2}\right\}\)

\(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{2\left(x-2\sqrt{x}+1\right)}{x-1}\left(đkxđ:x>0;x\ne1\right)\\ =\left(\dfrac{\left(\sqrt{x}\right)^3-1^3}{\sqrt{x}\left(\sqrt{x-1}\right)}-\dfrac{\left(\sqrt{x}\right)^3+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{x-1}{2\left(\sqrt{x}-1\right)^2}\\ =\left(\dfrac{\left(\sqrt{x}-1\right)\left(x+2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\\ =\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\\ =\dfrac{4\sqrt{x}}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\\ =\dfrac{2\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)

`b,` Để `A<0` thì :

\(\dfrac{2\left(\sqrt{x}+1\right)}{\sqrt{x}-1}< 0\\ \Leftrightarrow\sqrt{x}-1< 0\left(vì.2\left(\sqrt{x}+1\right)>0\right)\\ \Leftrightarrow\sqrt{x}< 1\\ \Leftrightarrow0\le x< 1\)

Kết hợp với điều kiện xác định ta có : \(0< x< 1\)

`a, x^2-4x-5=0`

`<=>x^2+x-5x-5=0`

`<=> (x^2+x)-(5x+5)=0`

`<=> x(x+1)-5(x+1)=0`

`<=>(x+1)(x-5)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)

`b, 5x^2-9x-2=0`

`<=> 5x^2+x-10x-2=0`

`<=> (5x^2+x)-(10x+2)=0`

`<=> x(5x+1)-2(5x+1)=0`

`<=>(5x+1)(x-2)=0`

\(\Leftrightarrow\left[{}\begin{matrix}5x+1=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}5x=-1\\x=2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=2\end{matrix}\right.\)

`c, x^3+2x-12=0`

`<=> x^3+2x-8-4=0`

`<=> (x^3-8)+(2x-4)=0`

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

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

`<=>(x-2)(x^2+2x+6)=0`

`<=>x-2=0`

`<=>x=2`

`d, -2x^3 +x^2-3=0`

`<=> -(2x^3-x^2+3)=0`

`<=> -(2x^3 -3x^2 +3x+2x^2 -3x+3)=0`

`<=>-[(2x^3 -3x^2 +3x) +(2x^2-3x+3)]=0`

`<=> - [x(2x^3 -3x^2 +3x) +(2x^2-3x+3)]=0`

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

`<=>x+1=0`

`<=>x=-1`

`a,` Khoảng cách : `2`

Số số hạng : \(\dfrac{44-4}{2}+1=21\)

Tổng dãy là : \(\dfrac{\left(44+4\right)\cdot21}{2}=504\)

`b,` Khoảng cách : `3`

Số số hạng là : \(\dfrac{135-3}{3}+1=45\)

Tổng dãy là : \(\dfrac{\left(135+3\right)\cdot45}{2}=3105\)

`c,` Khoảng cách là : `5`

Số số hạng là : \(\dfrac{250-5}{5}+1=50\)

Tổng dãy là : \(\dfrac{\left(250+5\right)\cdot50}{2}=6375\)

có e ạ ☕