Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(Q=\frac{\sqrt{x}\cdot\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\cdot\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\frac{2\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(Q=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(Q=x+1\)
Không thể tìm được GTLN hay GTNN của Q.
b)
\(\frac{3x+3}{\sqrt{x}}=3\sqrt{x}+\frac{3}{\sqrt{x}}\)
Để \(\frac{3Q}{\sqrt{x}}\) nguyên thì \(\frac{3}{\sqrt{x}}\)nguyên hay \(\sqrt{x}\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
Vì \(\sqrt{x}\)dương nên \(\sqrt{x}\in\left\{1;3\right\}\)
Vậy x=1, x=9 là các giá trị cần tìm
1) +) ta có : \(C-\dfrac{1}{3}\Leftrightarrow\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{3}=\dfrac{3\sqrt{x}-x+\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{-\left(x-4\sqrt{x}+4\right)+3}{3\left(x+\sqrt{x}+1\right)}=\dfrac{-\left(\sqrt{x}-2\right)^2+3}{3\left(x+\sqrt{x}+1\right)}\)
không thể cm được đâu bn --> xem lại đề
2) +) ta có : \(D=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}=1-\dfrac{3}{\sqrt{x}+2}\)
--> để \(D\in Z\Leftrightarrow\sqrt{x}+2\) là ước của 3 \(\Leftrightarrow\sqrt{x}+2\in\left\{\pm1;\pm3\right\}\)
\(\Leftrightarrow x=1\) vậy \(x=1\)
3) +) tương tự 2)
4) a) +) điều kiện xác định : \(x>0;x\ne4\)
ta có : \(A=\left(\dfrac{2}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}}\right):\dfrac{\sqrt{x}-2}{x+3\sqrt{x}}\)
\(\Leftrightarrow A=\left(\dfrac{2\sqrt{x}-\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}+3\right)}\right):\dfrac{x+3\sqrt{x}}{\sqrt{x}-2}=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\)
b) ta có : \(A=3\Leftrightarrow\dfrac{\sqrt{x}-3}{\sqrt{x}-2}=3\Leftrightarrow\sqrt{x}-3=3\sqrt{x}-6\)
\(\Leftrightarrow2\sqrt{x}=3\Leftrightarrow\sqrt{x}=\dfrac{3}{2}\Leftrightarrow x=\dfrac{9}{4}\) vậy \(x=\dfrac{9}{4}\)
c) ta có : \(B=A.\dfrac{\sqrt{x}+3}{\sqrt{x}+2}=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}.\dfrac{\sqrt{x}+3}{\sqrt{x}+2}=\dfrac{x-9}{x-4}=1-\dfrac{5}{x-4}\)
tương tự 2 )
\(\)
Mình bị nhầm
b) \(P=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1+2}{\sqrt{x}-1}=1+\dfrac{2}{\sqrt{x}-1}\)
Để P\(\in Z\) thì \(\sqrt{x}-1\inƯ\left(2\right)\in\left\{\pm1;\pm2\right\}\)
Vì \(\sqrt{x}-1\ge-1\)
Vậy \(\sqrt{x}-1\in\left\{\pm1;2\right\}\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\sqrt{x}-1=-1\\\sqrt{x}-1=2\\\sqrt{x}-1=1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=0\left(tm\right)\\x=4\left(tm\right)\\x=9\left(tm\right)\end{matrix}\right.\)
Vậy x=0, x=4,x=9 thì P\(\in Z\)
a)
\(=\dfrac{3x+3\sqrt{x}-3-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{3x+3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\) với \(x\ge0;x\ne1\)
b)
P = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=1+\dfrac{2}{\sqrt{x}-1}\)
Vì 1 \(\in Z\) nên
Để P \(\in\) Z thì \(2⋮\sqrt{x}-1=>\sqrt{x}-1\in\) Ư(2) = { -2;-1;1;2 }
=> \(\sqrt{x}\) = { -1;0;2;3 }
=> x ={0;4;9} thỏa mãn đkxđ
Vậy, ...............
a) Để biểu thức P xác định thì \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)
Vậy ĐKXĐ:x\(\ge0\),x\(\ne9\)
\(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)=\left[\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}=\left[\dfrac{2x-6\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{\left(-3\sqrt{x}-3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}=\dfrac{-3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}=\dfrac{-3}{\sqrt{x}+3}\)
b) Ta có \(P< \dfrac{1}{2}\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}< \dfrac{1}{2}\Leftrightarrow-6< \sqrt{x}+3\Leftrightarrow\sqrt{x}>-9\)
Vì \(\sqrt{x}\ge0\) và 0>-9
Vậy \(x\ge0\)
Kết hợp với ĐKXĐ, Vậy \(x\ge0\) và \(x\ne9\) thì P<\(\dfrac{1}{2}\)
Bài 3:
a: \(=\left(4\sqrt{2}-6\sqrt{2}\right)\cdot\dfrac{\sqrt{2}}{2}=-2\sqrt{2}\cdot\dfrac{\sqrt{2}}{2}=-2\)
b: \(=\dfrac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-2\left(\sqrt{6}-1\right)\)
\(=\sqrt{6}-2\sqrt{6}+2=2-\sqrt{6}\)
a) ĐKXĐ: x\(\ne1,x\ge0\)
\(P=\left(\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}+\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+2}\right):\dfrac{1}{x-1}=\left(\dfrac{3x+3\sqrt{x}-3+\sqrt{x}+2+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=\left(\dfrac{\left(3x+5\sqrt{x}-2\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right)=\dfrac{\left(3\sqrt{x}-1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+2}=3x+3\sqrt{x}-\sqrt{x}-1=3x+2\sqrt{x}-1\)b)Ta có \(\dfrac{1}{P}=\dfrac{1}{3x+2\sqrt{x}-1}=\dfrac{1}{\left(3\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
Để \(\dfrac{1}{P}\in N\) thì \(\left(3\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\inƯ\left(1\right)\)\(\in\left(\pm1\right)\)
Ta có \(\sqrt{x}+1>0\Rightarrow3\sqrt{x}-1>0\)
Mà 1 là số nguyên tố \(\Rightarrow\left\{{}\begin{matrix}3\sqrt{x}-1=1\\\sqrt{x}+1=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{4}{9}\\x=0\end{matrix}\right.\)
(loại)
Vậy không có giá trị x nguyên
Câu 2:
a, ĐKXĐ: x\(\ge\)0; x\(\ne\)\(\pm\)1
B=
\(\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}}{\sqrt{x}+1}\\ =\dfrac{\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}}{\sqrt{x}+1}\\ =\dfrac{-2.2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}}{\sqrt{x}+1}\\ =\dfrac{-4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}}{\sqrt{x}+1}\\ =\dfrac{4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\\ =-\dfrac{4}{\sqrt{x}-1}\)
1. \(\left(1+\sqrt{2}+\sqrt{3}\right)\left(1+\sqrt{2}-\sqrt{3}\right)\)
\(=\left(1+\sqrt{2}\right)^2-\sqrt{3}^2\)
\(=1+2\sqrt{2}+2-3\)
\(=2\sqrt{2}\)
3. \(A=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)\cdot\left(1+\dfrac{1}{\sqrt{x}}\right)\)(1)
ĐKXĐ \(x>0,x\ne1\)
pt (1) <=> \(\left(\dfrac{\sqrt{x}+1+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\right)\cdot\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}\right)\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right)\cdot\left(\sqrt{x}+1+\sqrt{x}-1\right)}{\sqrt{x}\cdot\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow\dfrac{2\sqrt{x}}{x-\sqrt{x}}\)
\(\Leftrightarrow\dfrac{\sqrt{x}\cdot2}{\sqrt{x}\cdot\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}\)
b) Để \(\sqrt{A}>A\Leftrightarrow\sqrt{\dfrac{2}{\sqrt{x}-1}}>\dfrac{2}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}>\dfrac{4}{x-2\sqrt{x}+1}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}-\dfrac{4}{x-2\sqrt{x}+1}>0\)
\(\Leftrightarrow\dfrac{2\cdot\left(\sqrt{x}-1\right)-4}{x-2\sqrt{x}+1}>0\)
\(\Leftrightarrow\dfrac{2\sqrt{2}-2-4}{x-2\sqrt{x}+1}>0\)
\(\Leftrightarrow\dfrac{2\sqrt{2}-6}{x-2\sqrt{x}+1}>0\)
Vì \(2\sqrt{2}-6< 0\Rightarrow x-2\sqrt{x}+1< 0\)
mà \(x-2\sqrt{x}+1=\left(\sqrt{x}-1\right)^2\ge0\forall x\)
Vậy không có giá trị nào của x thỏa mãn \(\sqrt{A}>A\)
(P/s Đề câu b bị sai hay sao vậy, chả có số nào mà \(\sqrt{A}>A\) cả, check lại đề giùm với nhé)
\(1.a.A=\left(1-\dfrac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{1}{\sqrt{x}+1}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\left(x\ge0;x\ne4;x\ne9\right)\)
\(b.A< 0\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\)
\(\Leftrightarrow\sqrt{x}-2< 0\)
\(\Leftrightarrow x< 4\)
Kết hợp với ĐKXĐ , ta có : \(0\le x< 4\)
KL............
\(2.\) Tương tự bài 1.
\(3a.A=\dfrac{1}{x-\sqrt{x}+1}=\dfrac{1}{x-2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{4}{3}\)
\(\Rightarrow A_{Max}=\dfrac{4}{3}."="\Leftrightarrow x=\dfrac{1}{4}\)
a/ đkxđ: x >=0; x khác 1
\(Q=\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}=\dfrac{3x+\sqrt{9x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{3x+3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b/ Q là số nguyên <=> \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\in Z\)
Có: \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1+2}{\sqrt{x}-1}=1+\dfrac{2}{\sqrt{x}-1}\)
=> \(\sqrt{x}-1\inƯ\left(2\right)\)
\(\Leftrightarrow\sqrt{x}-1=\left\{-2;-1;1;2\right\}\)
\(\Leftrightarrow\sqrt{x}=\left\{0;2;3\right\}\)
\(\Leftrightarrow x=\left\{0;4;9\right\}\) (TM)
Vậy.........