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b) \(B=\dfrac{x-\sqrt{x}}{1-\sqrt{x}}-\dfrac{x\sqrt{x}}{\sqrt{x}}=\dfrac{\sqrt{x}\left(x-\sqrt{x}\right)-x\sqrt{x}\left(1-\sqrt{x}\right)}{\sqrt{x}\left(1-\sqrt{x}\right)}\) = \(\dfrac{x\sqrt{x}-x-x\sqrt{x}+x^2}{\sqrt{x}-x}=\dfrac{x^2-x}{\sqrt{x}-x}\)
c) \(C=\dfrac{x+2\sqrt{x}}{\sqrt{x}-x}-\dfrac{x\sqrt{x}}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}+1\right)\left(x+2\sqrt{x}\right)-x\sqrt{x}\left(\sqrt{x}-x\right)}{\left(\sqrt{x}-x\right)\left(\sqrt{x}+1\right)}=x+2\sqrt{x}-x\sqrt{x}\)
\(d,D=\dfrac{x+2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{x-4}=\dfrac{x+2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\) \(\dfrac{\left(x+2\sqrt{x}\right)\left(\sqrt{x}-2\right)+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{x+7\sqrt{x}-2}{\sqrt{x}+2}\)
e) \(E=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{\sqrt{x}-24}{x-9}=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)+\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\) = \(\dfrac{2\sqrt{x}-24}{\sqrt{x}+3}\)
F) F = \(\dfrac{3}{\sqrt{x}+5}+\dfrac{20-2\sqrt{x}}{x-25}=\dfrac{3\left(\sqrt{x}-5\right)+20-2\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\dfrac{23-2\sqrt{x}}{\sqrt{x}+5}\)
Tìm đc mỗi GTNN, cách tìm GTLN chưa chắc chắn lắm nên mk ko lm nha :D
1/ \(A=\sqrt{\left(x-1\right)^2}+\sqrt{\left(3-x\right)^2}=\left|x-1\right|+\left|3-x\right|\ge\left|x-1+3-x\right|=2\)
2/ \(B=\sqrt{x-1-2\sqrt{x-1}+1}+\sqrt{x-1+2\sqrt{x-1}+1}=\sqrt{\left(1-\sqrt{x-1}\right)^2}+\sqrt{\left(\sqrt{x-1}+1\right)^2}\)
\(=\left|1-\sqrt{x-1}\right|+\left|\sqrt{x-1}+1\right|\ge\left|1-\sqrt{x-1}+\sqrt{x-1}+1\right|=2\)
M= \(\sqrt{2}+1-\) \(\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}+1-\sqrt{2}+1=2\)
N=\(\sqrt{1+2\sqrt{\left(\sqrt{2}+1\right)^2}}=\sqrt{1+2\left(\sqrt{2}+1\right)}=\) \(\sqrt{1+2\sqrt{2}+2}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)
P= \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}+\frac{2\sqrt{x}.\sqrt{x}}{\sqrt{x}}\) (dk \(x>0\))
=\(\sqrt{x}+1+2\sqrt{x}=3\sqrt{x}+1\)
Q= \(\sqrt{\left(\sqrt{x}+1\right)^2}+\sqrt{\left(\sqrt{x}-1\right)^2}\) (dk \(x\ge0\) )
=\(\left|\sqrt{x}+1\right|+\left|\sqrt{x}-1\right|\)
th1 \(\sqrt{x}\ge1\Leftrightarrow x\ge1\) Q=\(\sqrt{x}+1+\sqrt{x}-1=2\sqrt{x}\)
th2 \(0\le x< 1\) Q=\(\sqrt{x}+1+1-\sqrt{x}=2\)
a) \(M=\sqrt{2}+1-\sqrt{1,5.2-2.\sqrt{2}}\)
\(=\sqrt{2}+1-\sqrt{2.\left(1,5-\sqrt{2}\right)}\)\(=\sqrt{2}+1-\sqrt{2}.\sqrt{1,5-\sqrt{2}}\)
\(=\sqrt{2}.\left(1+1,5-\sqrt{2}\right)+1=\sqrt{2}.\left(2,5-\sqrt{2}\right)+1\)
\(=\sqrt{2}.2,5-2+1=\sqrt{2}.2,5-1\)
P/s: Theo em thì em nghĩ là đúng '-' Khoảng 90% :)
\(a,Đkxđ:\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
\(P=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x+1}\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x^3}-1\right)}{x+\sqrt{x}+1}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}=\sqrt{x}\left(\sqrt{x}-1\right)\)
\(=x-\sqrt{x}\)
\(b,P=x-\sqrt{x}=x-\sqrt{x}+\frac{1}{4}-\frac{1}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\)
Ta có: \(\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\forall x\ge0\)
\(\Leftrightarrow\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\forall x\ge0\)
Dấu " = " xảy ra \(\Leftrightarrow x=\frac{1}{4}\)
\(Min_P=-\frac{1}{4}\Leftrightarrow x=\frac{1}{4}\)
c, Đề thiếu không bạn?
Bài 1:
a: ĐKXĐ: 2x+3>=0 và x-3>0
=>x>3
b: ĐKXĐ:(2x+3)/(x-3)>=0
=>x>3 hoặc x<-3/2
c: ĐKXĐ: x+2<0
hay x<-2
d: ĐKXĐ: -x>=0 và x+3<>0
=>x<=0 và x<>-3
\(\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
\(=\frac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{x-1-\left(x-4\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{3}\)
\(=\frac{\sqrt{x}-2}{3\sqrt{x}}\)
\(\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)Đkxđ : x>2
=(\(\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)) \(:\left(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(x-1\right)-\left(x-4\right)}\)
\(=\frac{1}{\sqrt{x}}.\frac{\sqrt{x}-2}{3}=\frac{\sqrt{x}-2}{3\sqrt{x}}\)
b) \(< =>\sqrt{x+1}\left(\sqrt{x^2-x+1}+\sqrt{x+3}\right)=0\)
<=> x=-1
hoặc \(x^2-x+1=x+3\) => \(x^2-2x-2=0...\)
bạn lộn dấu rồi phải là \(|a|+|b|\ge|a+b|\)
phải là GTln chớ bạn