Đáp án:
Giải thích các bước giải:
BT 1:
`a)M=({2x+3\sqrt{x}}/{x\sqrt{x}+1}+1/{x-\sqrt{x}+1}-1/{\sqrt{x}+1}).{x-\sqrt{x}+1}/\sqrt{x}`
`ĐK: x>=0`
`M={2x+3\sqrt{x}}/{(\sqrt{x}+1)(x-\sqrt{x}+1)}+{\sqrt{x}+1}/{(\sqrt{x}+1)(x-\sqrt{x}+1)}-{x-\sqrt{x}+1}/{(\sqrt{x}+1)(x-\sqrt{x}+1)}).{x-\sqrt{x}+1}/\sqrt{x}`
`M={x+5\sqrt{x}}/{(\sqrt{x}+1)(x-\sqrt{x}+1)}.{x-\sqrt{x}+1}/\sqrt{x}`
`M={\sqrt{x}(\sqrt{x}+5)}/{(\sqrt{x}+1)(x-\sqrt{x}+1)}.{x-\sqrt{x}+1}/\sqrt{x}`
`M={\sqrt{x}+5}/{\sqrt{x}+1}`
Vậy `M={\sqrt{x}+5}/{\sqrt{x}+1}`với `x>=0`
`b)` Với `x>=0` ta có `M={\sqrt{x}+5}/{\sqrt{x}+1}={\sqrt{x}+1+4}/{\sqrt{x}+1}=1+4/{\sqrt{x}+1}>1`
`=>M>1`
`b)` Với `x>=0` ta có `M={\sqrt{x}+5}/{\sqrt{x}+1}={\sqrt{x}+1+4}/{\sqrt{x}+1}=1+4/{\sqrt{x}+1} in Z ` khi
`\sqrt{x}+1 in Ư_(4)`
`<=>\sqrt{x}+1 in {+-1;+-2;+-4}`
`<=>x in {0;1;9}`
Vậy `x in {0;1;9}` để `M in Z`
BT 2:
`a) P=(1/{\sqrt{x}-1}+\sqrt{x}/{x-1}): {2\sqrt{x}+1}/{x+\sqrt{x}-2`
`ĐK : x>=0,x ne 1`
`P=({\sqrt{x}+1}/{(\sqrt{x}-1)(\sqrt{x}+1)}+\sqrt{x}/{( \sqrt{x}+1)(\sqrt{x}-1)}).{x+\sqrt{x}-2}/{2\sqrt{x}+1}`
`P={2\sqrt{x}+1}/{(\sqrt{x}-1)(\sqrt{x}+1)}.{(\sqrt{x}-1)(\sqrt{x}+2)}/{2\sqrt{x}+1}`
`P={\sqrt{x}+2}/{\sqrt{x}+1}`
Vậy `P={\sqrt{x}+2}/{\sqrt{x}+1}` với `x>=0,x ne 1`
`b)`với `x>=0,x ne 1` để `P in Z<=>{\sqrt{x}+2}/{\sqrt{x}+1} in Z`
`<=>{\sqrt{x}+1+1}/{\sqrt{x}+1} in Z`
`<=>1+1/{\sqrt{x}+1} in Z`
Biểu thức `P` nguyên khi `\sqrt{x}+1 in Ư _ (1)`
`<=> \sqrt{x}+ 1 in {+-1}`
`<=> x in {0}`
Vậy `x=0` nguyên đề `P` nguyên
`c)` Xét `P={\sqrt{x}+2}/{\sqrt{x}+1}` với `x>=0,x ne 1` có
`\sqrt{x}+2 > 0 forall x`(Vì `x>=0`)
`\sqrt{x}+1> 0 forall x`(Vì `x>=0`)
`P_{Max}` khi `\sqrt{x}+1` đạt GTNN
Ta có : `\sqrt{x}>=0 forall x`
`<=>\sqrt{x}+1>=1 forall x`
Dấu `"="` xảy ra khi `x=0`
Vậy `P` đạt GTLN là `2` khi `x=0`
