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8 tháng 6 2022
Sửa đề: A>-4
\(A=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{x-1+1-\sqrt{x}}{x+\sqrt{x}}\)
\(=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)
\(A+4=\dfrac{x-2\sqrt{x}+1+4\sqrt{x}}{\sqrt{x}}=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}>0\)
=>A>-4
7 tháng 12 2017
\(=\dfrac{x-1}{\sqrt{x}}:\left(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+\left(1-\sqrt{x}\right)}{\sqrt{x}+x}\right)\)
\(=\dfrac{x-1}{\sqrt{x}}:\left(\dfrac{\left(\sqrt{x}-1\right)\sqrt{x}}{\left(\sqrt{x}+1\right)\sqrt{x}}\right)\)
\(=\dfrac{x-1}{\sqrt{x}}.\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\dfrac{\left(1+\sqrt{x}\right)^2}{\sqrt{x}}\)
ma \(\left(1+\sqrt{x}\right)^2>4\) voi moi x
\(\Rightarrow A>4\)
29 tháng 7 2022
a: \(T=1:\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right)\)
\(=1:\dfrac{x+2+x-1-x-\sqrt{x}-1}{x\sqrt{x}-1}\)
\(=\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
b: \(T-3=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\)
=>T>3
19 tháng 3 2021
a) - Với \(x>0,x\ne1\), ta có:
\(A=\left(\frac{1}{x-1}+\frac{3\sqrt{x}+5}{x\sqrt{x}-x-\sqrt{x}+1}\right)\left[\frac{\left(\sqrt{x}+1\right)^2}{4\sqrt{x}}-1\right]\)
\(A=\left[\frac{1}{x-1}+\frac{3\sqrt{x}+5}{\sqrt{x}\left(x-1\right)-\left(x-1\right)}\right]\left[\frac{x+2\sqrt{x}+1}{4\sqrt{x}}-\frac{4\sqrt{x}}{4\sqrt{x}}\right]\)
\(A=\left[\frac{1}{x-1}+\frac{3\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(x-1\right)}\right]\left[\frac{x+2\sqrt{x}-4\sqrt{x}+1}{4\sqrt{x}}\right]\)
\(A=\left[\frac{\sqrt{x}-1}{\left(x-1\right)\left(\sqrt{x}-1\right)}+\frac{3\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(x-1\right)}\right]\left[\frac{x^2-2\sqrt{x}+1}{4\sqrt{x}}\right]\)
\(A=\frac{\sqrt{x}+3\sqrt{x}-1+5}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(A=\frac{4+4\sqrt{x}}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(A=\frac{4\left(\sqrt{x}+1\right)}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(A=\frac{4\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{4\left(x-1\right)\left(\sqrt{x}-1\right).\sqrt{x}}\)
\(A=\frac{4\left(x-1\right)\left(\sqrt{x}-1\right)}{4\left(x-1\right)\left(\sqrt{x}-1\right).\sqrt{x}}=\frac{1}{\sqrt{x}}\)
Vậy với \(x>0,x\ne1\)thì \(A=\frac{1}{\sqrt{x}}\)
19 tháng 3 2021
\(A=\left(\frac{1}{x-1}+\frac{3\sqrt{x}+5}{x\sqrt{x}-x-\sqrt{x}+1}\right)\left[\frac{\left(\sqrt{x}+1\right)^2}{4\sqrt{x}}-1\right]\)
\(=\left[\frac{1}{x-1}+\frac{3\sqrt{x}+5}{\sqrt{x}\left(x-1\right)-\left(x-1\right)}\right]\left[\frac{x+2\sqrt{x}+1}{4\sqrt{x}}-\frac{4\sqrt{x}}{4\sqrt{x}}\right]\)
\(=\left[\frac{1}{x-1}+\frac{3\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(x-1\right)}\right]\left[\frac{x+2\sqrt{x}-4\sqrt{x}+1}{4\sqrt{x}}\right]\)
\(=\left[\frac{\sqrt{x}-1}{\left(x-1\right)\left(\sqrt{x}-1\right)}+\frac{3\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(x-1\right)}\right]\left[\frac{x^2-2\sqrt{x}+1}{4\sqrt{x}}\right]\)
\(=\frac{\sqrt{x}+3\sqrt{x}-1+5}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(=\frac{4+4\sqrt{x}}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(=\frac{4\left(\sqrt{x}+1\right)}{\left(x-1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}\)
\(=\frac{4\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{4\left(x-1\right)\left(\sqrt{x}-1\right).\sqrt{x}}\)
\(=\frac{4\left(x-1\right)\left(\sqrt{x}-1\right)}{4\left(x-1\right)\left(\sqrt{x}-1\right).\sqrt{x}}=\frac{1}{\sqrt{x}}\)
b) \(B=\left(x-\sqrt{x}+1\right)\cdot A=\frac{1}{\sqrt{x}}\left(x-\sqrt{x}+1\right)=\frac{x}{\sqrt{x}}-\frac{\sqrt{x}}{\sqrt{x}}+\frac{1}{\sqrt{x}}=\frac{1}{\sqrt{x}}+\sqrt{x}-1\)
Xét hiệu B - 1 ta có : \(B-1=\frac{1}{\sqrt{x}}+\sqrt{x}-2=\frac{1}{\sqrt{x}}+\frac{x}{\sqrt{x}}-\frac{2\sqrt{x}}{\sqrt{x}}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)
Dễ thấy \(\hept{\begin{cases}\sqrt{x}>0\forall x>0\\\left(\sqrt{x}-1\right)^2\ge0\forall x\ge0\end{cases}}\Rightarrow\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\ge0\forall x>0\)
Đẳng thức xảy ra <=> x = 1 ( ktm ĐKXĐ )
Vậy đẳng thức không xảy ra , hay chỉ có B - 1 > 0 <=> B > 1 ( đpcm )
TN
10 tháng 8 2018
a. \(B=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\\ =\left(\dfrac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-1\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}-1}{2}\\ =\dfrac{\left(\sqrt{x}-1\right)^2.2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{2}{x+\sqrt{x}+1}\)
b.Ta có:
\(B=\dfrac{2}{x+\sqrt{x}+1}\). Mà \(\left[{}\begin{matrix}2>0\\x+\sqrt{x}+1=\left[\left(\sqrt{x}\right)^2+2\sqrt{x}.\dfrac{1}{2}+\dfrac{1}{4}\right]+\dfrac{3}{4}>0\end{matrix}\right.\)
Vậy B>0 \(\forall x\)
T
6 tháng 12 2018
a,\(P=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)
\(P=\left[\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right].\dfrac{2}{\sqrt{x}-1}\)
\(P=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(P=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(P=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}=\dfrac{2}{x+\sqrt{x}+1}\)
Vậy \(P=\dfrac{2}{x+\sqrt{x}+1}\)
b, Ta có \(x+\sqrt{x}+1=\left(x+2\sqrt{x}.\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\)Suy ra \(\dfrac{2}{x+\sqrt{x}+1}>0\forall x>0,x\ne1\)
hay \(P>0\forall x>0,x\ne1\)(đpcm)
19 tháng 3 2021
a) Với \(x>0;x\ne1\), ta có:
\(P=\left(\frac{x-2}{x+2\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right).\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(P=\left[\frac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\frac{1}{\sqrt{x}+2}\right].\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(P=\left[\frac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\right].\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(P=\frac{x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\sqrt{x}-1}{\sqrt{x}}.\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(P=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}}\)
Vậy với \(x>0,x\ne1\)thì \(P=\frac{\sqrt{x}+1}{\sqrt{x}}\)
19 tháng 3 2021
\(P=\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(\Rightarrow2P=\frac{2\sqrt{x}+2}{\sqrt{x}}\)
\(2P=2\sqrt{x}+5\Leftrightarrow\frac{2\sqrt{x}+2}{\sqrt{x}}=2\sqrt{x}+5\left(ĐKXĐ:x\ne0\right)\left(1\right)\)
Mà theo đề bài : \(x>0\)nên phương trình luôn được xác định.
\(\left(1\right)\Leftrightarrow\frac{2\sqrt{x}+2}{\sqrt{x}}=\frac{\sqrt{x}\left(2\sqrt{x}+5\right)}{\sqrt{x}}\)
\(\Rightarrow2\sqrt{x}+2=\sqrt{x}\left(2\sqrt{x}+5\right)\)
\(\Leftrightarrow2\sqrt{x}+2=2x+5\sqrt{x}\)
\(\Leftrightarrow2\sqrt{x}+2-2x-5\sqrt{x}\)
\(\Leftrightarrow-2x-3\sqrt{x}+2=0\Leftrightarrow2x+3\sqrt{x}-2=0\)
\(\Leftrightarrow\left(2\sqrt{x}-1\right)\left(\sqrt{x}+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2\sqrt{x}-1=0\\\sqrt{x}+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2\sqrt{x}=1\\\sqrt{x}=-2\left(vn\right)\end{cases}}\Leftrightarrow2\sqrt{x}=1\)
\(\Leftrightarrow\sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\left(TMĐK:x>0;x\ne1\right)\)
Vậy \(2P=2\sqrt{x}+5\Leftrightarrow x=\frac{1}{4}\)
30 tháng 7 2022
a: \(T=1:\left(\dfrac{x+2+x-1}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right)\)
\(=1:\left(\dfrac{2x+1-x-\sqrt{x}-1}{x\sqrt{x}-1}\right)\)
\(=\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
b: \(T-3=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\)
=>T>3
27 tháng 5 2018
x>0
y=[√x.(√x+1).(x-√x+1)]/(x-√x+1)-1-[√x.(2√x+1)]
=√x.(√x+1)-2√x-2
=x-√x-2
b.
y=(√x-1/2)^2-9/4≥-9/4
x=1/4
c.
x≥4=>(√x-1/2)^2≥9/4=>y≥0
=>y≥0=>|y|=y
=>y-|y|=y-y=0
25 tháng 10 2018
Bạn nào làm được bài này thì giúp mình với ạ ! mình đang cần gấp
29 tháng 10 2022
Bài 4:
\(AH=\sqrt{9\cdot16}=12\left(cm\right)\)
\(AB=\sqrt{9\cdot25}=15\left(cm\right)\)
AC=căn(25^2-15^2)=20(cm)
Xét ΔABC vuông tại A có sin ABC=AC/BC=4/5
nên góc ABC=53 độ
Ta thấy \(A-4=\dfrac{x+2\sqrt{x}+1}{\sqrt{x}}-4\)
\(=\dfrac{x+2\sqrt{x}+1-4\sqrt{x}}{\sqrt{x}}\)
\(=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)
Do \(\left(\sqrt{x}-1\right)^2\ge0\) và \(\sqrt{x}>0\) nên \(\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\ge0\). ĐTXR \(\Leftrightarrow x=1\).
Như vậy \(A-4\ge0\) \(\Leftrightarrow A\ge4\)
(không phải là \(A>4\) như trong đề đâu nhé, dấu "=" vẫn có thể xảy ra nếu \(x=1\))