\(A=\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}+\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}+\dfrac{x+1}{\sqrt{x}}\)
Thu gọn và tìm x để A-|A|=0
1) Cho biểu thức A= (\(\dfrac{3\sqrt{x}}{\sqrt{x}-1}\)-\(\dfrac{1}{\sqrt{x}+1}\)- 3) . \(\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\) với x≥0 và x≠1
a) rút gọn A
b) tìm x để A<0
\(a,A=\left(\dfrac{3\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}-3\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\left(đk:x\ge0;x\ne1\right)\)
\(=\left[\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{3\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
\(=\dfrac{3x+3\sqrt{x}-\sqrt{x}+1-3\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
\(=\dfrac{3x+2\sqrt{x}+1-3x+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2}{\sqrt{x}-1}\)
\(---\)
\(b,A< 0\Leftrightarrow\dfrac{2}{\sqrt{x}-1}< 0\)
\(\Leftrightarrow\sqrt{x}-1< 0\)
\(\Leftrightarrow\sqrt{x}< 1\)
\(\Leftrightarrow x< 1\)
Kết hợp với điều kiện của \(x\), ta được:
\(0\le x< 1\)
Vậy: ...
\(Toru\)
a) \(A=\left(\dfrac{3\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}-3\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
\(A=\left[\dfrac{3\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}\right]\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
\(A=\dfrac{3x+3\sqrt{x}-\sqrt{x}+1-3x+3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
\(A=\dfrac{2\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
\(A=\dfrac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
\(A=\dfrac{2}{\sqrt{x}-1}\)
b) \(A< 0\) khi
\(\dfrac{2}{\sqrt{x}-1}< 0\Leftrightarrow\sqrt{x}-1< 0\)
\(\Leftrightarrow\sqrt{x}< 1\)
\(\Leftrightarrow x< 1\)
Kết hợp với đk:
\(0\le x< 1\)
Cho P = (\(\dfrac{\sqrt{x}}{\sqrt{x}-1 }\) - \(\dfrac{1}{x-\sqrt{x}}\))(\(\dfrac{1}{1+\sqrt{x}}\) + \(\dfrac{2}{x-1}\))
a. Tìm đkxđ và rút gọn P
b. Tìm x để P>0
a) ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right)\cdot\left(\dfrac{1}{1+\sqrt{x}}+\dfrac{2}{x-1}\right)\)
\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\cdot\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
b) Để P>0 thì \(\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}>0\)
mà \(\sqrt{x}+1>0\forall x\) thỏa mãn ĐKXĐ
nên \(\sqrt{x}\left(\sqrt{x}-1\right)>0\)
mà \(\sqrt{x}>0\forall x\) thỏa mãn ĐKXĐ
nên \(\sqrt{x}-1>0\)
\(\Leftrightarrow\sqrt{x}>1\)
hay x>1
Kết hợp ĐKXĐ,ta được: x>1
Vậy: Để P>0 thì x>1
Cho A= \(\dfrac{\sqrt{x}+4}{{}\sqrt{x}-1}\) và B= \(\dfrac{x+2\sqrt{x}}{\sqrt{x}(\sqrt{x}+1)} -\dfrac{3\sqrt{x}-3}{x-1}\) (đk: x>0,x≠1)
a) Rút gọn P=A.B
b) Tìm x để P(\(\sqrt{x}+1\)) ≤ 6-x
c) Tìm x để P nhận giá trị nguyên
Cho biểu thức P = \(\left(\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\right):\dfrac{\sqrt{x}}{2-\sqrt{x}}\) (với x>0; x\(\ne\)0)
a,Rút gọn biểu thức P và tìm x để P = \(\dfrac{-3}{5}\)
b,Tìm GTNN của biểu thức A=P . \(\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
\(a,P=\dfrac{\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{2-\sqrt{x}}{\sqrt{x}}=\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{-2}{\sqrt{x}+2}\\ P=-\dfrac{3}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\\ \Leftrightarrow3\sqrt{x}+6=10\Leftrightarrow\sqrt{x}=\dfrac{4}{3}\Leftrightarrow x=\dfrac{16}{9}\left(tm\right)\)
Cho biểu thức: A = \(\dfrac{x-1}{\sqrt{x}}\) : (\(\dfrac{\sqrt{x}-1}{\sqrt{x}}\) + \(\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\)) với x > 0; x ≠ 1
1) Rút gọn A
2) Tìm x để A . \(\sqrt{x}\)= 25
3) Chứng minh A > 4
1) \(A=\dfrac{x-1}{\sqrt{x}}:\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\right)\)
\(A=\dfrac{x-1}{\sqrt{x}}:\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)
\(A=\dfrac{x-1}{\sqrt{x}}:\left(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)
\(A=\dfrac{x-1}{\sqrt{x}}:\left(\dfrac{x-1+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)
\(A=\dfrac{x-1}{\sqrt{x}}:\dfrac{x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-\sqrt{x}}\)
\(A=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\cdot\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
b) Ta có:
\(A\cdot\sqrt{x}=25\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\cdot\sqrt{x}=25\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)^2=25\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)^2=5^2\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+1=5\\\sqrt{x}+1=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=16\\\sqrt{x}=-6\text{(vô lý)}\end{matrix}\right.\)
c) Ta xét hiệu:
\(A-4=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}-4\)
\(A-4=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}-\dfrac{4\sqrt{x}}{\sqrt{x}}\)
\(A-4=\dfrac{x+2\sqrt{x}+1-4\sqrt{x}}{\sqrt{x}}\)
\(A-4=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}}\)
\(A-4=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)
Với \(x>0\) thì \(\left(\sqrt{x}-1\right)>0\) và \(\sqrt{x}>0\)
\(\Rightarrow\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\)
Nên A > 4 (đpcm)
1: \(A=\dfrac{x-1}{\sqrt{x}}:\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1+1-\sqrt{x}}\)
\(=\dfrac{\left(x-1\right)\cdot\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
2: A*căn x=25
=>(căn x+1)^2=25
=>căn x+1=5
=>x=16
3: \(A-4=\dfrac{\left(\sqrt{x}+1\right)^2-4\sqrt{x}}{\sqrt{x}}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\)
=>A>4
A=\(\dfrac{3+2\sqrt{3}}{\sqrt{3}}\)-\(\dfrac{1}{\sqrt{3}-\sqrt{2}}\)+\(\dfrac{2+\sqrt{2}}{\sqrt{2}+1}\)
B=\(\dfrac{\sqrt{x}}{\sqrt{x}+3}\)+\(\dfrac{2\sqrt{x}}{\sqrt{x}-3}\)-\(\dfrac{3x+9}{x-9}\)với x≥0;x≠9
a. Rút gọn biểu thức A và B
b. Tìm x để một phần ba giá trị của A bằng giá trị của biểu thức B
a) Ta có: \(A=\dfrac{3+2\sqrt{3}}{\sqrt{3}}-\dfrac{1}{\sqrt{3}-\sqrt{2}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}\)
\(=2+\sqrt{3}-\sqrt{3}-\sqrt{2}+\sqrt{2}\)
=2
Ta có: \(B=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)
\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{3}{\sqrt{x}+3}\)
cho biểu thứ A=\(\left(1+\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}-x}\right)+\dfrac{5}{\sqrt{x}}\) với x>0 ,x≠1
a)rút gọn A
b)tìm x để A=5
c)tìm x để A>4
a: \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}}:\dfrac{\sqrt{x}-1+1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{5}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}}=\dfrac{x-1}{\sqrt{x}}+\dfrac{5}{\sqrt{x}}=\dfrac{x+4}{\sqrt{x}}\)
b: Để A=5 thì \(x+4=5\sqrt{x}\)
=>x=1(loại) hoặc x=16(nhận)
1. Thu gọn A = \(\left(\dfrac{x}{x\sqrt{x}-4\sqrt{x}}-\dfrac{6}{3\sqrt{x}-6}+\dfrac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\right)\)
2. Tìm x để A<2
ĐKXĐ: \(x>0;x\ne4\)
\(A=\left(\dfrac{x}{\sqrt{x}\left(x-4\right)}-\dfrac{6}{3\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}+2}\right):\left(\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+10-x}{\sqrt{x}+2}\right)\)
\(=\left(\dfrac{\sqrt{x}-2\left(\sqrt{x}+2\right)+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{6}{\sqrt{x}+2}\right)\)
\(=\dfrac{-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+2}{6}=\dfrac{6}{2-\sqrt{x}}\)
Để \(A< 2\Rightarrow\dfrac{6}{2-\sqrt{x}}< 2\)
\(\Rightarrow\dfrac{3}{2-\sqrt{x}}-1< 0\Rightarrow\dfrac{\sqrt{x}+1}{2-\sqrt{x}}< 0\)
\(\Rightarrow2-\sqrt{x}< 0\) (do \(\sqrt{x}+1>0;\forall x\in TXĐ\))
\(\Rightarrow x>4\)
cho biểu thức A=\(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\)với x≥0,x≠1
a)rút gọn A
b)tìm x nguyên để M =A.\(\dfrac{\sqrt{x}+2}{2\sqrt{x}+1}+\dfrac{x-\sqrt{x}-5}{\sqrt{x}+3}\)có giá trị nguyên
a: \(A=\dfrac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)