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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 )
\(\)
ĐK: x>0,x\(\ne4\)
a) Ta thay x=\(\dfrac{1}{4}\) vào \(A=\dfrac{6}{x+2\sqrt{x}}=\dfrac{6}{\dfrac{1}{4}+2\sqrt{\dfrac{1}{4}}}=\dfrac{6}{\dfrac{1}{4}+2.\dfrac{1}{2}}=\dfrac{6}{\dfrac{1}{4}+1}=6:\left(\dfrac{1}{4}+1\right)=6:\dfrac{5}{4}=6.\dfrac{4}{5}=\dfrac{24}{5}=4,8\)B=\(\dfrac{\sqrt{x}}{x-4}+\dfrac{2}{2-\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}=\dfrac{\sqrt{x}}{x-4}-\dfrac{2}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}=\dfrac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-2\sqrt{x}-4+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{6}{\left(2-\sqrt{x}\right)\left(\sqrt{x}+2\right)}=\dfrac{6}{4-x}\)
b) Ta có M=\(\dfrac{A}{B}=A\div B=\dfrac{6}{x+2\sqrt{x}}\div\dfrac{6}{4-x}=\dfrac{6}{x+2\sqrt{x}}.\dfrac{4-x}{6}=\dfrac{4-x}{x+2\sqrt{x}}=\dfrac{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{2-\sqrt{x}}{\sqrt{x}}\)
Ta lại có M>1\(\Leftrightarrow\dfrac{2-\sqrt{x}}{\sqrt{x}}>1\Leftrightarrow2-\sqrt{x}>\sqrt{x}\Leftrightarrow2>2\sqrt{x}\Leftrightarrow\sqrt{x}< 1\Leftrightarrow x< 1\)
Kết hợp với ĐK
Vậy 0<x<1 thì M>1
c) Ta có M\(=\dfrac{2-\sqrt{x}}{\sqrt{x}}=\dfrac{2}{\sqrt{x}}-1\)
Vậy để \(M\in Z\) thì \(\sqrt{x}\inƯ\left(2\right)\in\left\{\pm1;\pm2\right\}\)
Vì \(\sqrt{x}>0\)
Nên \(\sqrt{x}\in\left\{1;2\right\}\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=2\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=1\left(tm\right)\\x=4\left(ktm\right)\end{matrix}\right.\)
Vậy x=1 thì M\(\in Z\)
Nguyễn Việt LâmTrầNguyễn Thị Khánh Như Trương NgọcThảo Vyn Trung NguyênBonkingsaint suppapong udomkaewkanjanaPhạm TiếnKHUÊ VŨMysterious PersonThiên Hàn
a: Để P là số nguyên thì \(\sqrt{x}-2+2⋮\sqrt{x}-2\)
=>\(\sqrt{x}-2\in\left\{1;-1;2;-2\right\}\)
hay \(x\in\left\{9;1;16;0\right\}\)
b: Để P là só nguyên thì \(2\sqrt{x}+6-7⋮\sqrt{x}+3\)
=>\(\sqrt{x}+3\in\left\{1;-1;7;-7\right\}\)
=>căn x+3=7
=>căn x=4
=>x=16
c: Để P là số nguyên thì \(3\sqrt{x}-1⋮2\sqrt{x}+1\)
\(\Leftrightarrow6\sqrt{x}-2⋮2\sqrt{x}+1\)
=>\(6\sqrt{x}+3-5⋮2\sqrt{x}+1\)
=>\(2\sqrt{x}+1\in\left\{1;5\right\}\)
=>x=0 hoặc x=4
a: Để P là số nguyên thì \(\sqrt{x}-2+2⋮\sqrt{x}-2\)
=>\(\sqrt{x}-2\in\left\{1;-1;2;-2\right\}\)
hay \(x\in\left\{9;1;16;0\right\}\)
b: Để P là só nguyên thì \(2\sqrt{x}+6-7⋮\sqrt{x}+3\)
=>\(\sqrt{x}+3\in\left\{1;-1;7;-7\right\}\)
=>căn x+3=7
=>căn x=4
=>x=16
c: Để P là số nguyên thì \(3\sqrt{x}-1⋮2\sqrt{x}+1\)
\(\Leftrightarrow6\sqrt{x}-2⋮2\sqrt{x}+1\)
=>\(6\sqrt{x}+3-5⋮2\sqrt{x}+1\)
=>\(2\sqrt{x}+1\in\left\{1;5\right\}\)
=>x=0 hoặc x=4
a: Để P là số nguyên thì \(\sqrt{x}-2+2⋮\sqrt{x}-2\)
=>\(\sqrt{x}-2\in\left\{1;-1;2;-2\right\}\)
hay \(x\in\left\{9;1;16;0\right\}\)
b: Để P là só nguyên thì \(2\sqrt{x}+6-7⋮\sqrt{x}+3\)
=>\(\sqrt{x}+3\in\left\{1;-1;7;-7\right\}\)
=>căn x+3=7
=>căn x=4
=>x=16
c: Để P là số nguyên thì \(3\sqrt{x}-1⋮2\sqrt{x}+1\)
\(\Leftrightarrow6\sqrt{x}-2⋮2\sqrt{x}+1\)
=>\(6\sqrt{x}+3-5⋮2\sqrt{x}+1\)
=>\(2\sqrt{x}+1\in\left\{1;5\right\}\)
=>x=0 hoặc x=4
Câu 3:
\(C=\dfrac{3\sqrt{x}-x+x+9}{9-x}:\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-3\left(\sqrt{x}+3\right)}{x-9}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}\)
\(=\dfrac{-3\sqrt{x}}{2\sqrt{x}+4}\)
Để C<-1 thì C+1<0
=>-3 căn x+2 căn x+4<0
=>-căn x<-4
=>x>16
Bài 1:
a: ĐKXĐ: 2x+3>=0 và x-3>0
=>x>3
b: ĐKXĐ:(2x+3)/(x-3)>=0
=>x>3 hoặc x<-3/2
c: ĐKXĐ: x+2<0
hay x<-2
d: ĐKXĐ: -x>=0 và x+3<>0
=>x<=0 và x<>-3
1: \(=3\left(x+\dfrac{2}{3}\sqrt{x}+\dfrac{1}{3}\right)\)
\(=3\left(x+2\cdot\sqrt{x}\cdot\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{2}{9}\right)\)
\(=3\left(\sqrt{x}+\dfrac{1}{3}\right)^2+\dfrac{2}{3}>=3\cdot\dfrac{1}{9}+\dfrac{2}{3}=1\)
Dấu '=' xảy ra khi x=0
2: \(=x+3\sqrt{x}+\dfrac{9}{4}-\dfrac{21}{4}=\left(\sqrt{x}+\dfrac{3}{2}\right)^2-\dfrac{21}{4}>=-3\)
Dấu '=' xảy ra khi x=0
3: \(A=-2x-3\sqrt{x}+2< =2\)
Dấu '=' xảy ra khi x=0
5: \(=x-2\sqrt{x}+1+1=\left(\sqrt{x}-1\right)^2+1>=1\)
Dấu '=' xảy ra khi x=1
Bài 6:
a: \(\Leftrightarrow\sqrt{x^2+4}=\sqrt{12}\)
=>x^2+4=12
=>x^2=8
=>\(x=\pm2\sqrt{2}\)
b: \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>x+1=1
=>x=0
c: \(\Leftrightarrow3\sqrt{2x}+10\sqrt{2x}-3\sqrt{2x}-20=0\)
=>\(\sqrt{2x}=2\)
=>2x=4
=>x=2
d: \(\Leftrightarrow2\left|x+2\right|=8\)
=>x+2=4 hoặcx+2=-4
=>x=-6 hoặc x=2
a)ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
\(\Rightarrow A=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow A=\dfrac{x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow A=\dfrac{x+\sqrt{x}-2\sqrt{x}+2-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow A=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow A=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow A=\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
b) \(x=9\Rightarrow A=\dfrac{3}{3+1}=\dfrac{3}{4}\)
\(x=7-4\sqrt{3}\Rightarrow A=\dfrac{\sqrt{7-4\sqrt{3}}}{\sqrt{7-4\sqrt{3}}+1}=\dfrac{\sqrt{7-2\sqrt{12}}}{\sqrt{7-2\sqrt{12}}+1}=\dfrac{\sqrt{4-2\sqrt{3}\sqrt{4}+3}}{\sqrt{4-2\sqrt{3}\sqrt{4}+3}+1}=\dfrac{2-\sqrt{3}}{2-\sqrt{3}+1}=\dfrac{2-\sqrt{3}}{3-\sqrt{3}}=\dfrac{\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)}=\dfrac{3-\sqrt{3}}{6}\)