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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,ĐKXĐ \(x\ge0;x\ne1\)
Ta có A=\(\dfrac{x+2\sqrt{x}+1+2\sqrt{x}-x-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{x+2\sqrt{x}+1-x+\sqrt{x}-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
A=\(\dfrac{4\sqrt{x}}{x-1}.\dfrac{x-1}{2\sqrt{x+1}}\)
A=\(\dfrac{4\sqrt{x}}{2\sqrt{x}+1}\)
b, Thay x=\(1-\dfrac{\sqrt{3}}{2}\) vào biểu thức A ta có
A=\(\dfrac{4\sqrt{1-\dfrac{\sqrt{3}}{2}}}{2\sqrt{1-\dfrac{\sqrt{3}}{2}}+1}=\dfrac{\sqrt{16-8\sqrt{3}}}{\sqrt{4-2\sqrt{3}}+1}=\dfrac{6-2\sqrt{3}}{3}\)
ĐKXĐ : \(x>0\) và \(x\ne1\)
Câu a : \(P=\left(\dfrac{2-x}{x-\sqrt{x}}-\dfrac{1}{1-\sqrt{x}}+\dfrac{\sqrt{x}+1}{\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
\(=\left(\dfrac{2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(=\dfrac{2-x+\sqrt{x}+x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
Câu b : Thay \(x=\dfrac{9}{16}\) vào P ta được :
\(P=\dfrac{\sqrt{\dfrac{9}{16}}-1}{\sqrt{\dfrac{9}{16}}}=\dfrac{\dfrac{3}{4}-1}{\dfrac{3}{4}}=\dfrac{\dfrac{-1}{4}}{\dfrac{3}{4}}=-\dfrac{1}{3}\)
Câu c : Để \(P< \dfrac{1}{2}\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}< \dfrac{1}{2}\)
\(\Leftrightarrow2\sqrt{x}-2< \sqrt{x}\)
\(\Leftrightarrow\sqrt{x}< 2\Leftrightarrow x< 4\)
a)đkxđ:\(x\ge0,x\ne4\)
P=\(\left(\dfrac{2}{x-4}+\dfrac{1}{\sqrt{x}+2}\right):\dfrac{1}{\sqrt{x+2}}\)
\(=\left(\dfrac{2}{\left(\sqrt{x-2}\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)}\right):\dfrac{1}{\sqrt{x}+2}\)
\(=\dfrac{2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\sqrt{x}+2\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
Mọi ngươi giúp em với ạ chứ em làm câu a Bài 1 và 2 ra kết quả dài quá :(
Bài 1:
a: \(P=\dfrac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}\)
\(=\dfrac{a-\sqrt{a}-12}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+3\right)}=\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\)
b: Để P<1 thì P-1<0
\(\Leftrightarrow\dfrac{\sqrt{a}-4-\sqrt{a}+2}{\sqrt{a}-2}< 0\)
=>căn a-2>0
=>a>4
a: \(A=\dfrac{-x+\sqrt{x}-3-\sqrt{x}-1}{x-1}:\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1-8\sqrt{x}}{x-1}\)
\(=\dfrac{-x-4}{x-1}\cdot\dfrac{x-1}{-4\sqrt{x}}=\dfrac{x+4}{4\sqrt{x}}\)
b: Khi x=4-2 căn 3 thì \(A=\dfrac{4-2\sqrt{3}+4}{4\left(\sqrt{3}-1\right)}=\dfrac{8-2\sqrt{3}}{4\left(\sqrt{3}-1\right)}=\dfrac{3\sqrt{3}+1}{4}\)
c: \(A-1=\dfrac{x+4-4\sqrt{x}}{4\sqrt{x}}=\dfrac{\left(\sqrt{x}-2\right)^2}{4\sqrt{x}}>0\)
=>A>1
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne1\end{matrix}\right.\)
\(A=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right):\dfrac{x+1}{x-1}\)
\(=\dfrac{\sqrt{x}+1+\sqrt{x}-1}{x-1}\cdot\dfrac{x-1}{x+1}\)
\(=\dfrac{2\sqrt{x}}{x+1}\)
b: Thay x=9 vào A, ta được:
\(A=\dfrac{2\cdot\sqrt{9}}{9+1}=\dfrac{2\cdot3}{10}=\dfrac{6}{10}=\dfrac{3}{5}\)
c: Để A=1 thì \(\dfrac{2\sqrt{x}}{x+1}=1\)
=>\(x+1=2\sqrt{x}\)
=>\(x-2\sqrt{x}+1=0\)
=>\(\left(\sqrt{x}-1\right)^2=0\)
=>\(\sqrt{x}-1=0\)
=>\(\sqrt{x}=1\)
=>x=1(loại)