Question

giúp e giải bài này với ạ. e xin cảm ơn

Answer 1

a: Thay x=9 vào P, ta được:

\(P=\dfrac{3+3}{3-2}=6\)

b: \(Q=\dfrac{x-3\sqrt{x}+2+3\sqrt{x}-2}{x-4}=\dfrac{x}{x-4}\)

Answer 2

`a)` Thế `x=9` vào `P` ta được:

\(P=\dfrac{9+3}{\sqrt{9}-2}=\dfrac{12}{1}=12\)

`b)`\(Q=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{x-4}\)

\(Q=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(Q=\dfrac{x-2\sqrt{x}-\sqrt{x}+2+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

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

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

\(Q=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

Answer 3

`a)` Thay `x=9` (t/m đk) vào `P` có: `P=[9+3]/[\sqrt{9}-2]=12`

`b)` Với `x > 0,x \ne 4` có:

`Q=[\sqrt{x}-1]/[\sqrt{x}+2]+[5\sqrt{x}-2]/[x-4]`

`Q=[(\sqrt{x}-1)(\sqrt{x}-2)+5\sqrt{x}-2]/[(\sqrt{x}-2)(\sqrt{x}+2)]`

`Q=[x-2\sqrt{x}-\sqrt{x}+2+5\sqrt{x}-2]/[(\sqrt{x}-2)(\sqrt{x}+2)]`

`Q=[x+2\sqrt{x}]/[(\sqrt{x}-2)(\sqrt{x}+2)]`

`Q=[\sqrt{x}(\sqrt{x}+2)]/[(\sqrt{x}-2)(\sqrt{x}+2)]=\sqrt{x}/[\sqrt{x}-2]`

Answer 4

\(a,Thayx=9vàoPtađc\\ P=\dfrac{9+3}{\sqrt{9}-2}=\dfrac{12}{1}=12\)

b, Với  \(Q=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{x-3\sqrt{x}+2+5\sqrt{x}-2}{\left(x-4\right)}=\dfrac{x+2\sqrt{x}}{\left(x-4\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{x-4}=\dfrac{\sqrt{x}}{\left(\sqrt{x}-2\right)}\)