Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A = \(\dfrac{4\sqrt{x}+9}{2\sqrt{x}+1}\)
Mà \(4\sqrt{x}+9>0\)
\(2\sqrt{x}+1>0\)
=> A > 0
A = \(\dfrac{2\left(2\sqrt{x}+1\right)+7}{2\sqrt{x}+1}\) = \(2+\dfrac{7}{2\sqrt{x}+1}\)
Mà \(2\sqrt{x}+1\ge1< =>\dfrac{7}{2\sqrt{x}+1}\le7\)
<=> \(A\le9\)
<=> 0 < A \(\le9\)
Mà A thuộc Z
<=> A \(\in\){1;2;3;4;5;6;7;8;9}
Đến đây bn thay A vào để tìm x nhé
A = \(\dfrac{2\left(2\sqrt{x}+1\right)+7}{2\sqrt{x}+1}=2+\dfrac{7}{2\sqrt{x}+1}\)
Mà \(2\sqrt{x}+1>0< =>\dfrac{7}{2\sqrt{x}+1}>0\)
<=> A > 2
Có \(2\sqrt{x}+1\ge1< =>\dfrac{7}{2\sqrt{x}+1}\le7\)
<=> \(A\le9\)
<=> 2 < A \(\le9\)
Mà A thuộc Z
<=> \(A\in\left\{3;4;5;6;7;8;9\right\}\)
Đến đây bn thay A vào để tìm x nhé
A = \(\dfrac{6\sqrt{x}+8}{3\sqrt{x}+2}=2+\dfrac{4}{3\sqrt{x}+2}\)
Có \(3\sqrt{x}+2>0< =>\dfrac{4}{3\sqrt{x}+2}>0\) <=> A > 2
Có: \(3\sqrt{x}+2\ge2< =>\dfrac{4}{3\sqrt{x}+2}\le2\) <=> A \(\le4\)
<=> 2 < A \(\le4\)
Mà A nguyên
<=> \(\left[{}\begin{matrix}A=3\\A=4\end{matrix}\right.\)
TH1: A = 3
<=> \(\dfrac{4}{3\sqrt{x}+2}=1\)
<=> \(3\sqrt{x}+2=4< =>x=\dfrac{4}{9}\)
TH2: A = 4
<=> \(\dfrac{4}{3\sqrt{x}+2}=2< =>3\sqrt{x}+2=2< =>x=0\)
1.2 với \(x\ge0,x\in Z\)
A=\(\dfrac{2\sqrt{x}+7}{\sqrt{x}+2}=2+\dfrac{3}{\sqrt{x}+2}\in Z< =>\sqrt{x}+2\inƯ\left(3\right)=\left(\pm1;\pm3\right)\)
*\(\sqrt{x}+2=1=>\sqrt{x}=-1\)(vô lí)
*\(\sqrt{x}+2=-1=>\sqrt{x}=-3\)(vô lí
*\(\sqrt{x}+2=3=>x=1\)(TM)
*\(\sqrt{x}+2=-3=\sqrt{x}=-5\)(vô lí)
vậy x=1 thì A\(\in Z\)
Bài 1.2
\(A=\dfrac{2\sqrt{x}+7}{\sqrt{x}+2}=2+\dfrac{3}{\sqrt{x}+2}\)
C1:Bạn dùng pp chặn như bài 2.2
C2: (Gợi ý)\(\sqrt{x}+2\ge2\) và \(\sqrt{x}+2\inƯ\left(3\right)\)\(\Rightarrow\sqrt{x}+2=3\Leftrightarrow x=1\)
Vậy x=1 thì A nguyên
Bài 2.2
\(A=\dfrac{\sqrt{x}+7}{\sqrt{x}+2}=1+\dfrac{5}{\sqrt{x}+2}\)
Do \(\sqrt{x}\ge0;\forall x\)\(\Rightarrow\sqrt{x}+2\ge2\) \(\Rightarrow\dfrac{5}{\sqrt{x}+2}\le\dfrac{5}{2}\)\(\Rightarrow A\le\dfrac{7}{2}\) (1)
mà \(\dfrac{5}{\sqrt{x}+2}>0;\forall x\Rightarrow A>1\) (2)
Từ (1) (2) \(\Rightarrow1< A\le\dfrac{7}{2}\) mà A nguyên
\(\Rightarrow\left[{}\begin{matrix}A=2\\A=3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}1+\dfrac{5}{\sqrt{x}+2}=2\\1+\dfrac{5}{\sqrt{x}+2}=3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+2=5\\\sqrt{x}+2=\dfrac{5}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy...
Bài 3.2
\(A=\dfrac{-x-2\sqrt{x}-5}{\sqrt{x}+2}\)\(=\dfrac{-\sqrt{x}\left(\sqrt{x}+2\right)-5}{\sqrt{x}+2}=-\sqrt{x}-\dfrac{5}{\sqrt{x}+2}\)
\(=2-\left(\sqrt{x}+2+\dfrac{5}{\sqrt{x}+2}\right)\)
Áp dụng bđt cosi: \(\sqrt{x}+2+\dfrac{5}{\sqrt{x}+2}\ge2\sqrt{\left(\sqrt{x}+2\right).\dfrac{5}{\sqrt{x}+2}}=2\sqrt{5}\)
\(\Rightarrow A\le2-2\sqrt{5}\)
Dấu = xảy ra \(\Leftrightarrow\sqrt{x}+2=\dfrac{5}{\sqrt{x}+2}\Leftrightarrow x=9-4\sqrt{5}\)
\(3.A=\dfrac{2\sqrt{x}+17}{\sqrt{x}+5}=\dfrac{2\left(\sqrt{x}+5\right)+7}{\sqrt{x}+5}\)\(=2+\dfrac{7}{\sqrt{x}+5}\)
\(\sqrt{x}+5\ge5=>2+\dfrac{7}{\sqrt{x}+5}\le2+\dfrac{7}{5}=3,4\)
dấu'=' xảy ra<=>x=0=>MaxA=3,4
Bài này ko phải tìm giá trị lớn hơn nhỏ hơn mà nó là tìm x để A thuộc Z bạn ơi
Ta có:
\(\dfrac{\sqrt{x}+5}{\sqrt{x}+2}=\dfrac{\left(\sqrt{x}+2\right)+3}{\sqrt{x}+2}=1+\dfrac{3}{\sqrt{x}+2}\)
Để \(A\in Z\Leftrightarrow\dfrac{3}{\sqrt{x}+2}\in Z\)
\(\Rightarrow\left(\sqrt{x}+2\right)\inƯ_{\left(3\right)}=\left\{1;-1;3;-3\right\}\)
\(\Rightarrow\sqrt{x}=\left\{-1;-3;1;-5\right\}\)
Mà \(\sqrt{x}\ge0\)
Nên \(\sqrt{x}=1\Rightarrow x=1\)
Vậy x=1 thì \(A\in Z\)
A = \(\dfrac{2\left(3\sqrt{x}+2\right)+4}{3\sqrt{x}+2}\)
= \(2+\dfrac{4}{3\sqrt{x}+2}\)
Để A nguyên
<=> \(\dfrac{4}{3\sqrt{x}+2}\) nguyên
<=> \(4⋮3\sqrt{x}+2\)
Ta có bảngg
| \(3\sqrt{x}+2\) | 1 | -1 | 2 | -2 | 4 | -4 |
| x | \(\varnothing\) | \(\varnothing\) | 0 | \(\varnothing\) | \(\dfrac{4}{9}\) | \(\varnothing\) |
| Thử lại | tm | loại |
KL: x = 0
A=\(\dfrac{6\sqrt{x}+8}{3\sqrt{x}+2}\)=\(\dfrac{2(3\sqrt{x}+4)}{3\sqrt{x}+2}\)=\(2\cdot\left(1+\dfrac{2}{3\sqrt{x}+2}\right)\)
Để A∈Z
Thì \(3\sqrt{x}+2\)∈Ư(2)
Tức là \(3\sqrt{x}+2\)∈\(\left\{1;-1;2;-2\right\}\)
\(3\sqrt{x}+2=1\)(vô lí);\(3\sqrt{x}+2=-1\)(vô lí);\(3\sqrt{x}+2=-2\)(vô lí)
\(3\sqrt{x}+2=2\)=>x=0
Vì 0∈Z
Vậy x=0 thì thỏa mãn đề bài
`A=(6sqrtx+8)/(3sqrtx+2)`
`=(6sqrtx+4+4)/(3sqrtx+2)`
`=2+4/(3sqrtx+2)>2AAx>=0(1)`
Vì `3sqrtx>=0`
`=>3sqrtx+2>=2`
`=>4/(3sqrtx+2)<=2`
`=>A<=4(2)`
`(1)(2)=>2<A<=4`
Mà `A in ZZ`
`=>A in {3,4}`
`**A=3`
`<=>4/(3sqrtx+2)=1`
`<=>4=3sqrtx+2`
`<=>3sqrtx=2`
`<=>x=4/9`
`**A=4`
`<=>4/(3sqrtx+2)=2`
`<=>6sqrtx+4=4`
`<=>6sqrtx=0`
`<=>sqrtx=0`
`<=>x=0`
đk: \(x\ge0\)
A = \(\dfrac{2\left(3\sqrt{x}+2\right)+4}{3\sqrt{x}+2}\)
= \(2+\dfrac{4}{3\sqrt{x}+2}\)
Để A \(\in Z\)
<=> \(4⋮3\sqrt{x}+2\)
Ta có bảng:
| \(3\sqrt{x}+2\) | 1 | -1 | 2 | -2 | 4 | -4 |
| x | \(\varnothing\) | \(\varnothing\) | 0 | \(\varnothing\) | \(\dfrac{4}{9}\) | \(\varnothing\) |
| tm | tm |
`A=(2sqrtx+17)/(sqrtx+5)`
`=(2sqrtx+10+7)/(sqrtx+5)`
`=(2(sqrtx+5)+7)/(sqrtx+5)`
`=2+7/(sqrtx+5)`
`A in ZZ`
`=>7/(sqrtx+5) in ZZ`
`=>sqrtx+5 in Ư(7)={+-1,+-7}`
Mà `sqrtx+5>=5`
`=>sqrtx+5=7`
`=>sqrtx=2`
`=>x=4`
Vậy `x=4` thì `A in ZZ`
làm pp chặn mà bạn