a√a +1 /√a +1 ( với a>=0) a,rút gọn A b, chứng minh a>= √a
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CM
11 tháng 4 2018
a) Với a > 0 và a ≠ 1 ta có:
P = a − 1 ( a − 1 ) ( a − 1 ) + 3 a + 5 ( a − 1 ) ( a − 1 ) . ( a + 2 a + 1 ) − 4 a 4 a = 4 a + 4 ( a − 1 ) 2 ( a + 1 ) . a − 2 a + 1 4 a = 4 ( a − 1 ) 2 . ( a − 1 ) 2 4 a = 1 a
b, Có Q = a − a + 1 a
Xét Q − 1 = a − 2 a + 1 a = ( a − 1 ) 2 a
Vì ( a − 1 ) 2 > 0 , a > 0 , ∀ a > 0 , a ≠ 1 ⇒ Q − 1 > 0 ⇒ Q > 1
14 tháng 7 2016
a) \(A=\left(\frac{1}{a-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)
\(=\left[\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right].\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)
\(=\frac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\frac{\sqrt{a}-1}{\sqrt{a}}=1-\frac{1}{\sqrt{a}}\)
b) Ta có : \(\frac{1}{\sqrt{a}}>0\Leftrightarrow-\frac{1}{\sqrt{a}}< 0\Rightarrow\) \(A=1-\frac{1}{\sqrt{a}}< 1\)
13 tháng 7 2016
a) \(A=\left(\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\)\(=\frac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}=\frac{\sqrt{a}-1}{\sqrt{a}}\)
b) Do \(\sqrt{a}\ge0\) => \(\sqrt{a}-1< \sqrt{a}\)=> \(\frac{\sqrt{a}-1}{\sqrt{a}}< 1\)
7 tháng 2 2022
a) \(B=\dfrac{x-\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{9-\sqrt{9}+1}{\sqrt{9}-1}=\dfrac{9-3+1}{3-1}=\dfrac{7}{2}\)
b) \(A=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)+2\left(\sqrt{x}-2\right)-9\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+3}\)
c) \(A=\dfrac{\sqrt{x}-1}{\sqrt{x}+3}>0\Leftrightarrow\sqrt{x}-1>0\left(do.\sqrt{x}+3>0\right)\)
\(\Leftrightarrow\sqrt{x}>1\Leftrightarrow x>1\)
\(B=\dfrac{x-\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+1}{\sqrt{x}-1}=\sqrt{x}+\dfrac{1}{\sqrt{x}-1}\)
Do \(\sqrt{x}>1\Leftrightarrow\sqrt{x}-1>0\)
Áp dụng BĐT Cauchy cho 2 số k âm:
\(B=\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}+1\ge2\sqrt{\left(\sqrt{x}-1\right).\dfrac{1}{\sqrt{x}-1}}+1=2+1=3\)
Dấu "=" xảy ra \(\Leftrightarrow\left(\sqrt{x}-1\right)^2=1\Leftrightarrow x=4\)
4 tháng 12 2018
a, ĐK: \(a\ne0,b\ne0,a+b\ne0\)
\(A=\left[\frac{1}{a^2}+\left(\frac{1}{a}+\frac{1}{b}\right):\frac{a+b}{2}+\frac{1}{b^2}\right].\frac{a^2b^2}{a^3+b^3}:\left(a+b\right)\)
\(=\left[\frac{1}{a^2}+\frac{a+b}{ab}:\frac{a+b}{2}+\frac{1}{b^2}\right].\frac{a^2b^2}{a^3+b^3}:\left(a+b\right)\)
\(=\left[\frac{1}{a^2}+\frac{2}{ab}+\frac{1}{b^2}\right].\frac{a^2b^2}{a^3+b^3}:\left(a+b\right)\)
\(=\frac{\left(a+b\right)^2}{a^2b^2}.\frac{a^2b^2}{\left(a+b\right)\left(a^2-ab+b^2\right)}.\frac{1}{a+b}\)
\(=\frac{1}{a^2-ab+b^2}\)
b, \(a^2-ab+b^2=\left(a-\frac{1}{2}b\right)^2+\frac{3}{4}b^2>0\left(a,b\ne0\right)\)
\(\Rightarrow A=\frac{1}{a^2-ab+b^2}>0\forall a;b\)
3 tháng 6 2021
`A=sqrt{3-sqrt5}-sqrt{3+sqrt5}`
`<=>sqrt2A=sqrt{6-2sqrt5}-sqrt{6+2sqrt5}`
`<=>sqrt2A=sqrt{(sqrt5-1)^2}-sqrt{(sqrt5+1)^2}`
`<=>sqrt2A=sqrt5-1-sqrt5-1=-2`
`<=>A=-sqrt2`
Câu b đề sai sai kiểu gì ý `sqrt{a+1}/a` là sao ;-;?
30 tháng 5 2023
2:
\(VT=\dfrac{a^2b}{a-b}\cdot\dfrac{2\sqrt{2}\left(a-b\right)}{5\sqrt{3}\cdot a^2\sqrt{b}}=\dfrac{2}{15}\cdot\sqrt{6b}=VP\)
1: \(=9\sqrt{ab}+\dfrac{7\sqrt{ab}}{b}-\dfrac{5\sqrt{ab}}{a}-3\sqrt{ab}=\)6căn ab+căn ab(7/b-5/a)
=căn ab(6+7/b-5/a)
H9
HT.Phong (9A5)
CTVHS
3 tháng 8 2023
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)
3 tháng 8 2023
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