Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a,ĐK:x≠4;x>0
b,A=(\(\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}\))*\(\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
=\(\dfrac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)*\(\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
=\(\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
Để A>\(\dfrac{1}{2}\)thì\(\dfrac{\sqrt{x}-2}{\sqrt{x}}\)>\(\dfrac{1}{2}\)
⇔\(\dfrac{\sqrt{x}-4}{2\sqrt{x}}\)>0
⇔\(\sqrt{x}-4>0\left(2\sqrt{x}>0\right)\)
⇔ x>16(tm)
Để A>\(\dfrac{1}{2}\)thì 0<x>16và x≠4
Mọi ngươi giúp em với ạ chứ em làm câu a Bài 1 và 2 ra kết quả dài quá :(
Bài 1:
a: \(P=\dfrac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)
\(=\dfrac{a-\sqrt{a}-12}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}=\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\)
b: Để P<1 thì P-1<0
\(\Leftrightarrow\dfrac{\sqrt{a}-4-\sqrt{a}+2}{\sqrt{a}-2}< 0\)
=>căn a-2>0
=>a>4
ĐKXĐ : \(x>0\) và \(x\ne1\)
Câu a : \(P=\left(\dfrac{2-x}{x-\sqrt{x}}-\dfrac{1}{1-\sqrt{x}}+\dfrac{\sqrt{x}+1}{\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
\(=\left(\dfrac{2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(=\dfrac{2-x+\sqrt{x}+x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
Câu b : Thay \(x=\dfrac{9}{16}\) vào P ta được :
\(P=\dfrac{\sqrt{\dfrac{9}{16}}-1}{\sqrt{\dfrac{9}{16}}}=\dfrac{\dfrac{3}{4}-1}{\dfrac{3}{4}}=\dfrac{\dfrac{-1}{4}}{\dfrac{3}{4}}=-\dfrac{1}{3}\)
Câu c : Để \(P< \dfrac{1}{2}\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}< \dfrac{1}{2}\)
\(\Leftrightarrow2\sqrt{x}-2< \sqrt{x}\)
\(\Leftrightarrow\sqrt{x}< 2\Leftrightarrow x< 4\)
a: ĐKXĐ: x>1
b: \(B=\sqrt{x-1}+\sqrt{x}+\sqrt{x-1}-\sqrt{x}+x=x+2\sqrt{x-1}\)
c: Khi B=4 thì \(\left(\sqrt{x-1}+1\right)^2=4\)
\(\Leftrightarrow\sqrt{x-1}+1=2\)
=>x-1=1
=>x=2
1) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
\(P=\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}-\frac{4x}{x-4}\)
\(\Leftrightarrow P=\frac{\left(2+\sqrt{x}\right)^2-\left(2-\sqrt{x}\right)^2+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
\(\Leftrightarrow P=\frac{4+4\sqrt{x}+x-4+4\sqrt{x}-x+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
\(\Leftrightarrow P=\frac{4x+8\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
\(\Leftrightarrow P=\frac{4\sqrt{x}}{2-\sqrt{x}}\)
2) Để \(P=2\)
\(\Leftrightarrow\frac{4\sqrt{x}}{2-\sqrt{x}}=2\)
\(\Leftrightarrow4\sqrt{x}=4-2\sqrt{x}\)
\(\Leftrightarrow6\sqrt{x}=4\)
\(\Leftrightarrow\sqrt{x}=\frac{2}{3}\)
\(\Leftrightarrow x=\frac{4}{9}\)
Vậy để \(P=2\Leftrightarrow x=\frac{4}{9}\)
3) Khi \(\left(\sqrt{x}-2\right)\left(2\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-2=0\\2\sqrt{x}-1==0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=2\\\sqrt{x}=\frac{1}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\left(ktm\right)\\x=\frac{1}{4}\left(tm\right)\end{cases}}\)
Thay \(x=\frac{1}{4}\)vào P, ta được :
\(\Leftrightarrow P=\frac{4\sqrt{\frac{1}{4}}}{2-\sqrt{\frac{1}{4}}}=\frac{4\cdot\frac{1}{2}}{2-\frac{1}{2}}=\frac{2}{\frac{3}{2}}=\frac{4}{3}\)
4) Để \(P=\frac{\sqrt{x}+3}{2\sqrt{x}-1}\)
\(\Leftrightarrow\frac{4\sqrt{x}}{2-\sqrt{x}}=\frac{\sqrt{x}+3}{2\sqrt{x}-1}\)
\(\Leftrightarrow8x-4\sqrt{x}=-x-\sqrt{x}+6\)
\(\Leftrightarrow9x-3\sqrt{x}-6=0\)
\(\Leftrightarrow3x-\sqrt{x}-2=0\)
\(\Leftrightarrow\sqrt{x}=3x-2\)
\(\Leftrightarrow x=9x^2-12x+4\)
\(\Leftrightarrow9x^2-13x+4=0\)
\(\Leftrightarrow\left(9x-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}9x-4=0\\x-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{4}{9}\\x=1\end{cases}}\)
Thử lại ta được kết quá : \(x=\frac{4}{9}\left(ktm\right)\); \(x=1\left(tm\right)\)
Vậy để \(P=\frac{\sqrt{x}+3}{2\sqrt{x}-1}\Leftrightarrow x=1\)
5) Để biểu thức nhận giá trị nguyên
\(\Leftrightarrow\frac{4\sqrt{x}}{2-\sqrt{x}}\inℤ\)
\(\Leftrightarrow4\sqrt{x}⋮2-\sqrt{x}\)
\(\Leftrightarrow-4\left(2-\sqrt{x}\right)+8⋮2-\sqrt{x}\)
\(\Leftrightarrow8⋮2-\sqrt{x}\)
\(\Leftrightarrow2-\sqrt{x}\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{1;3;0;4;-2;6;-6;10\right\}\)
Ta loại các giá trị < 0
\(\Leftrightarrow\sqrt{x}\in\left\{1;3;0;4;6;10\right\}\)
\(\Leftrightarrow x\in\left\{1;9;0;16;36;100\right\}\)
Vậy để \(P\inℤ\Leftrightarrow x\in\left\{1;9;0;16;36;100\right\}\)
\(\)
a: \(P=\dfrac{x+\sqrt{x}+1+11\sqrt{x}-11+34}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{x+\sqrt{x}+1-x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{x+12\sqrt{x}+24}{\sqrt{x}+2}\)
b: Thay \(x=3-2\sqrt{2}\) vào P, ta được:
\(P=\dfrac{3-2\sqrt{2}+12\left(\sqrt{2}-1\right)+24}{\sqrt{2}-1+2}\)
\(=\dfrac{27-2\sqrt{2}+12\sqrt{2}-12}{\sqrt{2}+1}=5+5\sqrt{2}\)
1: Khi x=64 thì \(A=\dfrac{8+2}{8}=\dfrac{5}{4}\)
2: \(B=\dfrac{x-1+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
a: \(P=\left(\dfrac{x-1}{2\sqrt{x}}\right)^2\cdot\dfrac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{x-1}\)
\(=\dfrac{\left(x-1\right)^2}{4x}\cdot\dfrac{-4\sqrt{x}}{x-1}=\dfrac{-\left(x-1\right)}{\sqrt{x}}\)
b: Khi \(x=4+2\sqrt{3}\) thì \(P=\dfrac{-\left(4+2\sqrt{3}-1\right)}{\sqrt{3}+1}=\dfrac{-\left(3+2\sqrt{3}\right)}{\sqrt{3}+1}=\dfrac{-3-\sqrt{3}}{2}\)
c: Để P<0 thì -(x-1)<0
=>x-1>0
=>x>1
\(Q=\frac{\sqrt{x}\cdot\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\cdot\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\frac{2\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(Q=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(Q=x+1\)
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
a, đk x >= 0 ; x khác 4
b, \(A=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{x-4}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
c, Ta có \(\dfrac{\sqrt{x}}{\sqrt{x}-2}=-\dfrac{1}{3}\Rightarrow3\sqrt{x}=2-\sqrt{x}\Leftrightarrow4\sqrt{x}=2\Leftrightarrow x=\dfrac{1}{16}\)(tm)