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Câu 1 :
Đk: \(x\ge1\)
\(\sqrt{x-1}+\sqrt{2x-1}=5\\ \Leftrightarrow x-1+2\sqrt{\left(x-1\right)\left(2x-1\right)}+2x-1=25\\ \Leftrightarrow2\sqrt{2x^2-3x+1}=27-3x\\ \)
\(\Leftrightarrow\begin{cases}27-3x\ge0\\4\left(2x^2-3x+1\right)=9x^2-162x+729\end{cases}\) \(\Leftrightarrow\begin{cases}x\le9\\x^2-150x+725=0\end{cases}\)
\(\Leftrightarrow\begin{cases}x\le9\\x=145hoặcx=5\end{cases}\)
với x= 5 thoản mãn điều kiện, x=145 loại
Vậy \(S=\left\{5\right\}\)
a) ĐK: \(x\ge0,x\ne1,x\ne\frac{1}{4}\)
\(A=1+\left(\frac{2x+\sqrt{x}-1}{1-x}-\frac{2x\sqrt{x}-\sqrt{x}+x}{1-x\sqrt{x}}\right)\frac{x-\sqrt{x}}{2\sqrt{x}-1}\)
\(A=1+\left[\frac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(1-\sqrt{x}\right)}-\frac{\sqrt{x}\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(1-\sqrt{x}\right)\left(x+\sqrt{x}+1\right)}\right]\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}-1}\)
\(A=1+\left[\frac{2\sqrt{x}-1}{1-\sqrt{x}}-\frac{\sqrt{x}\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(1-\sqrt{x}\right)\left(x+\sqrt{x}+1\right)}\right]\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}-1}\)
\(A=1-\sqrt{x}+\frac{x\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}\)
\(A=\frac{x+1}{x+\sqrt{x}+1}\)
Để \(A=\frac{6-\sqrt{6}}{5}\Rightarrow\frac{x+1}{x+\sqrt{x}+1}=\frac{6-\sqrt{6}}{5}\)
\(\Rightarrow5x+5=\left(6-\sqrt{6}\right)x+\left(6-\sqrt{6}\right)\sqrt{x}+6-\sqrt{6}\)
\(\Rightarrow\left(1-\sqrt{6}\right)x+\left(6-\sqrt{6}\right)\sqrt{x}+1-\sqrt{6}=0\)
\(\Rightarrow x-\sqrt{6}.\sqrt{x}+1=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=\frac{\sqrt{2}+\sqrt{6}}{2}\\\sqrt{x}=\frac{-\sqrt{2}+\sqrt{6}}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=2+\sqrt{3}\\x=2-\sqrt{3}\end{cases}}\left(tmđk\right)\)
b) Xét \(A-\frac{2}{3}=\frac{x+1}{x+\sqrt{x}+1}-\frac{2}{3}=\frac{3x+3-2x-2\sqrt{x}-2}{3\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x-2\sqrt{x}+1}{3\left(x+\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)^2}{3\left(x+\sqrt{x}+1\right)}\)
Do \(x\ge0,x\ne1,x\ne\frac{1}{4}\Rightarrow\left(\sqrt{x}-1\right)^2>0\)
Lại có \(x+\sqrt{x}+1=\left(\sqrt{x}+\frac{1}{2}\right)+\frac{3}{4}>0\)
Nên \(A-\frac{2}{3}>0\Rightarrow A>\frac{2}{3}\).
x . \(\frac{1}{2}\)- x.\(\frac{2}{3}\) + x.\(\frac{3}{4}\)- x. \(\frac{5}{6}\) = \(\frac{5}{6}\) -\(\frac{3}{4}\) + \(\frac{2}{3}\) -\(\frac{1}{2}\)
x . \(\frac{1}{2}\)- x.\(\frac{2}{3}\) + x.\(\frac{3}{4}\)- x. \(\frac{5}{6}\) = \(\frac{10}{12}\)-\(\frac{9}{12}\)+\(\frac{8}{12}\)-\(\frac{6}{12}\)
x . \(\frac{1}{2}\)- x.\(\frac{2}{3}\) + x.\(\frac{3}{4}\)- x. \(\frac{5}{6}\)= \(\frac{1}{4}\)=> x. (\(\frac{1}{2}\)- \(\frac{2}{3}\) + \(\frac{3}{4}\)- \(\frac{5}{6}\)) = \(\frac{1}{4}\)=> x.( \(\frac{6}{12}\)- \(\frac{8}{12}\)+\(\frac{9}{12}\)-\(\frac{10}{12}\))= \(\frac{1}{4}\)=> x. \(\frac{-1}{4}\)=\(\frac{1}{4}\)=> x = \(\frac{1}{4}\): \(\frac{-1}{4}\)=> x = -1=>x.(1/2-2/3+3/4)=1/4
=>x.7/12=1/4
=>x=1/4:7/12
=>x=1/4.12/7
=>x=3/7
\(\begin{cases}\sqrt{xy}+\frac{1}{\sqrt{xy}}=\frac{5}{2}\\\sqrt{x}+\sqrt{y}+\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}=\frac{9}{2}\end{cases}\)
<=>\(\begin{cases}xy+1=\frac{5\sqrt{xy}}{2}\\\sqrt{xy}.\left(\sqrt{x}+\sqrt{y}\right)+\sqrt{x}+\sqrt{y}=\frac{9\sqrt{xy}}{2}\end{cases}\)
Đặt P=\(\sqrt{xy}\);S=\(\sqrt{x}+\sqrt{y}\)(S2\(\ge\)4P)
Ta có HPT: \(\begin{cases}P^2+1=\frac{5P}{2}\\S.P+P=\frac{9P}{2}\end{cases}\)
Tới đây dễ tự làm
Ta có :
\(K=\frac{2\sqrt{x}+3}{\sqrt{x}-5}=\frac{2\sqrt{x}-10}{\sqrt{x}-5}+\frac{13}{\sqrt{x}-5}=2+\frac{13}{\sqrt{x}-5}\)là số nguyên dương
<=> 13 chia hết cho \(\sqrt{x}-5\)
<=> \(\sqrt{x}-5\inƯ\left(13\right)=\left\{-13;-1;1;13\right\}\)
<=> \(\sqrt{x}\in\left\{-12;4;6;18\right\}\)
<=> \(x\in\left\{16;36;324\right\}\) (vì \(\sqrt{x}\ge0\))
Do x nguyên và x có GTLN nên x = 324
tìm số nguyên x để A có giá trị là 1 số nguyên \(A=\frac{\sqrt{x}+1}{\sqrt{x}-3}\left(x\ge0\right)\)
\(A=\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\left(\sqrt{x}-3\right)+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\) E Z
<=>4 chia hết cho \(\sqrt{x}-3\)
<=>\(\sqrt{x}-3\) E Ư(4)={-4;-2;-1;1;2;4}
+)\(\sqrt{x}-3=-4=>\sqrt{x}=-1\) (loại vì \(\sqrt{x}\) >= 0)
+)\(\sqrt{x}-3=-2=>\sqrt{x}=1=>x=1\)
+)\(\sqrt{x}-3=-1=>\sqrt{x}=2=>x=4\)
+)\(\sqrt{x}-3=1=>\sqrt{x}=4=>x=16\)
+)\(\sqrt{x}-3=2=>\sqrt{x}=5=>x=25\)
+)\(\sqrt{x}-3=4=>\sqrt{x}=7=>x=49\)
Vậy x E {1;4;16;25;49} thì thỏa mãn đề bài
A=\(\frac{\sqrt{x}+1}{\sqrt{x}-3}\)=\(\frac{\sqrt{x}-3+4}{\sqrt{x}-3}\)=1+\(\frac{4}{\sqrt{x}-3}\)
Để A \(\in\) Z\(\Leftrightarrow\)\(\frac{4}{\sqrt{x}-3}\)\(\in\) Z
\(\Leftrightarrow\)\(\sqrt{x}-3\) \(\in\) ư(4)=4;-4;1;-1;2;-
| \(\sqrt{x}-3\) | 1 | -1 | 2 | -2 | 4 | -4 |
| \(\sqrt{x}\) | 4 | 2 | 5 | 1 | 7 | -1 |
| \(x\) | 16 | 4 | 25 | 1 | 49 | loại |
Vậy x\(\in\)\(\left\{1;4;16;25;49\right\}\)thì A\(\in\)Z
\(\int\limits^1_{\sqrt{ }3}\)\(\sqrt{\left(1+x^2\right)}\)\(dx\)
\(A=\frac{\sqrt{x}-5}{\sqrt{x}+5}=\frac{\sqrt{x}+5-10}{\sqrt{x}+5}=1-\frac{10}{\sqrt{x}+5}\)
Vì \(A< \frac{1}{3}=>1-\frac{10}{\sqrt{x}+5}< \frac{1}{3}\)
\(=>1-\frac{1}{3}< \frac{10}{\sqrt{x}+5}=>\frac{2}{3}< \frac{10}{\sqrt{x}+5}\)
\(=>2.\left(\sqrt{x}+5\right)< 30=>2\sqrt{x}+10< 30=>2\sqrt{x}< 20\)
\(=>\sqrt{x}< 10=>\left(\sqrt{x}\right)^2< 10^2=>x< 100\)
Vậy x<100 thì A<1/3