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a) Ta có:
\(P=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\left(\frac{2\sqrt{x}\left(\sqrt{x-3}\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x-3}\right)}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{3x+3}{x-9}\right):\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\left(\frac{2x-6}{x-9}+\frac{x+3\sqrt{x}}{x-9}-\frac{3x+3}{x-9}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{2x-6+x+3\sqrt{x}-3x-3}{x-9}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+3}\)
\(=\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+3}\)
\(=\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)^2}\)
b) \(P< \frac{-1}{2}\Rightarrow\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)^2}< \frac{-1}{2}\)
.....Chưa nghĩ ra....
c) Ta có: \(\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)^2}\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}-3=0\Rightarrow x=9\)
Vậy Min P = 0 khi x =9.
k - kb với tớ nhia mn!
\(P=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x+1\right)}+\frac{1}{x+1}\right).\frac{x+1}{\sqrt{x}-1}\)ĐK x>=0 x khác -1
=\(\frac{\sqrt{x}+1}{x+1}.\frac{x+1}{\sqrt{x}-1}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
b/ x =\(\frac{2+\sqrt{3}}{2}=\frac{4+2\sqrt{3}}{4}=\frac{3+2\sqrt{3}+1}{4}=\frac{\left(\sqrt{3}+1\right)^2}{4}\)
\(\Rightarrow\sqrt{x}=\frac{\sqrt{3}+1}{2}\)
Em thay vào tính nhé!
c) với x>1
A=\(\frac{\sqrt{x}+1}{\sqrt{x}-1}.\sqrt{x}=\frac{x+\sqrt{x}}{\sqrt{x}-1}=\sqrt{x}+2+\frac{2}{\sqrt{x}-1}=\sqrt{x}-1+\frac{2}{\sqrt{x}-1}+3\)
Áp dụng bất đẳng thức Cosi
A\(\ge2\sqrt{2}+3\)
Xét dấu bằng xảy ra ....
\(Q=\frac{\sqrt{x}\cdot\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\cdot\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\frac{2\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(Q=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(Q=x+1\)
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
Lời giải:
a) ĐKXĐ: \(x\geq 0, x\neq 1\)
Ta có:
\(A=\left(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{x(\sqrt{x}-1)+(\sqrt{x}-1)}\right):\left(\frac{x+\sqrt{x}}{x(\sqrt{x}+1)+(\sqrt{x}+1)}+\frac{1}{x+1}\right)\)
\(=\frac{x+1-2\sqrt{x}}{(x+1)(\sqrt{x}-1)}:\left(\frac{\sqrt{x}(\sqrt{x}+1)}{(x+1)(\sqrt{x}+1)}+\frac{1}{x+1}\right)\)
\(=\frac{(\sqrt{x}-1)^2}{(x+1)(\sqrt{x}-1)}:\left(\frac{\sqrt{x}}{x+1}+\frac{1}{x+1}\right)\)
\(=\frac{\sqrt{x}-1}{x+1}.\frac{x+1}{\sqrt{x}+1}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
b) \(A=\sqrt{x}-2\)
\(\Leftrightarrow \frac{\sqrt{x}-1}{\sqrt{x}+1}=\sqrt{x}-2\)
\(\Rightarrow \sqrt{x}-1=(\sqrt{x}-2)(\sqrt{x}+1)=x-\sqrt{x}-2\)
\(\Rightarrow x-2\sqrt{x}-1=0\)
\(\Leftrightarrow (\sqrt{x}-1)^2=2\Rightarrow \left[\begin{matrix} \sqrt{x}-1=\sqrt{2}\rightarrow x=3+2\sqrt{2}\\ \sqrt{x}-1=-\sqrt{2}\rightarrow \sqrt{x}=1-\sqrt{2}< 0(\text{vô lý})\end{matrix}\right.\)
Vậy \(x=3+2\sqrt{2}\)
c)
\(A=\frac{\sqrt{x}-1}{\sqrt{x}+1}=\frac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\frac{2}{\sqrt{x}+1}\)
Vì \(\sqrt{x}\geq 0\Rightarrow \frac{2}{\sqrt{x}+1}\leq \frac{2}{0+1}=2\)
\(\Rightarrow A=1-\frac{2}{\sqrt{x}+1}\geq 1-2=-1\)
Vậy $A$ min bằng $-1$. Dấu bằng xảy ra khi $x=0$
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
thử xem lại đề coi
\(A=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{x\sqrt{x}-\sqrt{x}+x+1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\right)\)
\(=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{x\sqrt{x}-\sqrt{x}+x+1}\right)\cdot\left(\sqrt{x}+1\right)\)
\(=1-\dfrac{2\left(x-1\right)}{x\sqrt{x}-\sqrt{x}+x+1}\)
a: \(B=\dfrac{2x+\sqrt{x}-1}{1-x}+\dfrac{2x\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}\)
\(=\left(2x+\sqrt{x}-1\right)\left(\dfrac{-1}{x-1}+\dfrac{\sqrt{x}}{x\sqrt{x}+1}\right)\)
\(=\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)\cdot\left(\dfrac{-x+\sqrt{x}-1+x-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\right)\)
\(=-\dfrac{\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\)
\(A=\dfrac{-\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}{2\sqrt{x}-1}\)
\(=\dfrac{-x+\sqrt{x}-1}{\sqrt{x}}\)
b: Khi \(x=17-12\sqrt{2}=\left(3-2\sqrt{2}\right)^2\) thì
\(A=\dfrac{-17+12\sqrt{2}+3-2\sqrt{2}-1}{3-2\sqrt{2}}=-5\)
c: \(A=\dfrac{-\left(x-\sqrt{x}+1\right)}{\sqrt{x}}=\dfrac{-\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{3}{4}}{\sqrt{x}}< 0\)
=>căn A không tồn tại