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a: \(A=\dfrac{\sqrt{x}-1+\sqrt{x}}{\sqrt{x}\left(1-\sqrt{x}\right)}:\left(\dfrac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)}+\dfrac{\sqrt{x}\left(2x+\sqrt{x}-1\right)}{1+x\sqrt{x}}\right)\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(1-\sqrt{x}\right)}:\left(2x+\sqrt{x}-1\right)\cdot\left(\dfrac{1}{1-x}+\dfrac{\sqrt{x}}{1+x\sqrt{x}}\right)\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(1-\sqrt{x}\right)}:\left[\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)\cdot\dfrac{1+x\sqrt{x}+\sqrt{x}-x\sqrt{x}}{\left(1-x\right)\left(1+x\sqrt{x}\right)}\right]\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(1-\sqrt{x}\right)}:\left[\dfrac{\left(2\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}{\left(1-\sqrt{x}\right)\left(1+x\sqrt{x}\right)}\right]\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(1-\sqrt{x}\right)}\cdot\dfrac{\left(1-\sqrt{x}\right)\left(x+\sqrt{x}+1\right)}{\left(2\sqrt{x}-1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
b: Khi x=17-12 căn 2 thì \(A=\dfrac{17-12\sqrt{2}+3-2\sqrt{2}+1}{3-2\sqrt{2}}=7\)
a: \(A=\dfrac{\sqrt{x}-1+\sqrt{x}}{\sqrt{x}\left(1-\sqrt{x}\right)}:\left(\dfrac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)}+\dfrac{\sqrt{x}\left(2x+\sqrt{x}-1\right)}{1+x\sqrt{x}}\right)\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(1-\sqrt{x}\right)}:\left(2x+\sqrt{x}-1\right)\cdot\left(\dfrac{1}{1-x}+\dfrac{\sqrt{x}}{1+x\sqrt{x}}\right)\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(1-\sqrt{x}\right)}:\left[\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)\cdot\dfrac{1+x\sqrt{x}+\sqrt{x}-x\sqrt{x}}{\left(1-x\right)\left(1+x\sqrt{x}\right)}\right]\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(1-\sqrt{x}\right)}:\left[\dfrac{\left(2\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}{\left(1-\sqrt{x}\right)\left(1+x\sqrt{x}\right)}\right]\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(1-\sqrt{x}\right)}\cdot\dfrac{\left(1-\sqrt{x}\right)\left(x+\sqrt{x}+1\right)}{\left(2\sqrt{x}-1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
b: Khi x=17-12 căn 2 thì \(A=\dfrac{17-12\sqrt{2}+3-2\sqrt{2}+1}{3-2\sqrt{2}}=7\)
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
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
Bài 3:
a: \(=\left(4\sqrt{2}-6\sqrt{2}\right)\cdot\dfrac{\sqrt{2}}{2}=-2\sqrt{2}\cdot\dfrac{\sqrt{2}}{2}=-2\)
b: \(=\dfrac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-2\left(\sqrt{6}-1\right)\)
\(=\sqrt{6}-2\sqrt{6}+2=2-\sqrt{6}\)
1. \(\left(1+\sqrt{2}+\sqrt{3}\right)\left(1+\sqrt{2}-\sqrt{3}\right)\)
\(=\left(1+\sqrt{2}\right)^2-\sqrt{3}^2\)
\(=1+2\sqrt{2}+2-3\)
\(=2\sqrt{2}\)
3. \(A=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)\cdot\left(1+\dfrac{1}{\sqrt{x}}\right)\)(1)
ĐKXĐ \(x>0,x\ne1\)
pt (1) <=> \(\left(\dfrac{\sqrt{x}+1+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\right)\cdot\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}\right)\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right)\cdot\left(\sqrt{x}+1+\sqrt{x}-1\right)}{\sqrt{x}\cdot\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow\dfrac{2\sqrt{x}}{x-\sqrt{x}}\)
\(\Leftrightarrow\dfrac{\sqrt{x}\cdot2}{\sqrt{x}\cdot\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}\)
b) Để \(\sqrt{A}>A\Leftrightarrow\sqrt{\dfrac{2}{\sqrt{x}-1}}>\dfrac{2}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}>\dfrac{4}{x-2\sqrt{x}+1}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}-\dfrac{4}{x-2\sqrt{x}+1}>0\)
\(\Leftrightarrow\dfrac{2\cdot\left(\sqrt{x}-1\right)-4}{x-2\sqrt{x}+1}>0\)
\(\Leftrightarrow\dfrac{2\sqrt{2}-2-4}{x-2\sqrt{x}+1}>0\)
\(\Leftrightarrow\dfrac{2\sqrt{2}-6}{x-2\sqrt{x}+1}>0\)
Vì \(2\sqrt{2}-6< 0\Rightarrow x-2\sqrt{x}+1< 0\)
mà \(x-2\sqrt{x}+1=\left(\sqrt{x}-1\right)^2\ge0\forall x\)
Vậy không có giá trị nào của x thỏa mãn \(\sqrt{A}>A\)
(P/s Đề câu b bị sai hay sao vậy, chả có số nào mà \(\sqrt{A}>A\) cả, check lại đề giùm với nhé)
A)
Đặt \(\sqrt{1+2x}=a; \sqrt{1-2x}=b\) (\(a,b>0\) )
\(\Rightarrow \left\{\begin{matrix} a^2+b^2=2\\ a^2-b^2=4x=\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} 2a^2=2+\sqrt{3}\rightarrow 4a^2=4+2\sqrt{3}=(\sqrt{3}+1)^2\\ 2b^2=2-\sqrt{3}\rightarrow 4b^2=4-2\sqrt{3}=(\sqrt{3}-1)^2\end{matrix}\right.\)
\(\Rightarrow a=\frac{\sqrt{3}+1}{2}; b=\frac{\sqrt{3}-1}{2}\)
\(\Rightarrow ab=\frac{(\sqrt{3}+1)(\sqrt{3}-1)}{4}=\frac{1}{2}; a-b=1\)
Có:
\(A=\frac{a^2}{1+a}+\frac{b^2}{1-b}=\frac{a^2-a^2b+b^2+ab^2}{(1+a)(1-b)}\)
\(=\frac{2-ab(a-b)}{1+(a-b)-ab}=\frac{2-\frac{1}{2}.1}{1+1-\frac{1}{2}}=1\)
B)
\(2x=\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\)
\(\Rightarrow 4x^2=\frac{a}{b}+\frac{b}{a}+2\)
\(\rightarrow 4(x^2-1)=\frac{a}{b}+\frac{b}{a}-2=\left(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\right)^2\)
\(\Rightarrow \sqrt{4(x^2-1)}=\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\) do $a>b$
T có: \(B=\frac{b\sqrt{4(x^2-1)}}{x-\sqrt{x^2-1}}=\frac{2b\sqrt{4(x^2-1)}}{2x-\sqrt{4(x^2-1)}}=\frac{2b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}-\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}\)
\(=\frac{2b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{2\sqrt{\frac{b}{a}}}=\frac{b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{\sqrt{\frac{b}{a}}}=\frac{\frac{b(a-b)}{\sqrt{ab}}}{\sqrt{\frac{b}{a}}}=a-b\)
ĐKXĐ: \(x>0;x\ne1\)
\(\dfrac{2x+\sqrt{x}-1}{1-x}+\dfrac{2x\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}=\dfrac{x-1+x+\sqrt{x}}{1-x}+\dfrac{x\sqrt{x}-\sqrt{x}+x\sqrt{x}+x}{1+x\sqrt{x}}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+\sqrt{x}\left(\sqrt{x}+1\right)}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+x\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{2\sqrt{x}-1}{1-\sqrt{x}}+\dfrac{\left(2\sqrt{x}-1\right)\sqrt{x}}{x-\sqrt{x}+1}=\left(2\sqrt{x}-1\right)\left(\dfrac{1}{1-\sqrt{x}}+\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\right)\)
\(=\dfrac{2\sqrt{x}-1}{\left(1-\sqrt{x}\right)\left(x-\sqrt{x}+1\right)}\)
Vậy \(A=\left(\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(1-\sqrt{x}\right)}\right):\left(\dfrac{2\sqrt{x}-1}{\left(1-\sqrt{x}\right)\left(x-\sqrt{x}+1\right)}\right)\)
\(A=\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)
b/ Dễ dàng nhận ra \(A>0\)\(A=\dfrac{x-\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}-1+\dfrac{1}{\sqrt{x}}=\sqrt{17-12\sqrt{2}}-1+\dfrac{1}{\sqrt{17-12\sqrt{2}}}\)
\(A=\sqrt{17-12\sqrt{2}}-1+\sqrt{17+12\sqrt{2}}=\sqrt{\left(3-2\sqrt{2}\right)^2}-1+\sqrt{\left(3+2\sqrt{2}\right)^2}\)
\(\Rightarrow A=3-2\sqrt{2}+3+2\sqrt{2}-1=6-1=5\)
c/ Ta có \(A=\sqrt{x}+\dfrac{1}{\sqrt{x}}-1>2\sqrt{\sqrt{x}.\dfrac{1}{\sqrt{x}}}-1=1\) (dấu "=" không xảy ra)
Mà \(A>0\Rightarrow\sqrt{A}>1\Rightarrow\sqrt{A}-1>0\)
Ta có \(A-\sqrt{A}=\sqrt{A}\left(\sqrt{A}-1\right)>0\) do \(\left\{{}\begin{matrix}\sqrt{A}>0\\\sqrt{A}-1>0\end{matrix}\right.\)
\(\Rightarrow A>\sqrt{A}\) \(\forall x\)