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Sửa đề; \(A=\left(\dfrac{a\sqrt{a}-3}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-3\right)}-\dfrac{2\left(\sqrt{a}-3\right)}{\sqrt{a}+1}-\dfrac{\sqrt{a}+3}{\sqrt{a}-3}\right):\dfrac{a+8}{a-1}\)
\(A=\dfrac{a\sqrt{a}-3-2a+12\sqrt{a}-18-a-4\sqrt{a}-3}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-3\right)}:\dfrac{a+8}{a-1}\)
\(=\dfrac{a\sqrt{a}-3a+8\sqrt{a}-24}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-3\right)}\cdot\dfrac{a-1}{a+8}\)
\(=\sqrt{a}-1\)
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)
Bài 1 : Rút gọn biểu thức :
\(\left(2-\sqrt{2}\right)\left(-5\sqrt{2}\right)-\left(3\sqrt{2}-5\right)^2\)
\(=\left(-10\sqrt{2}+10\right)-\left(18-30\sqrt{2}+25\right)\)
\(=\left(-10\sqrt{2}+10\right)-\left(7-30\sqrt{2}\right)\)
\(=-10\sqrt{2}+10-7+30\sqrt{2}\)
\(=20\sqrt{2}+3\)
Bài 2:
a) ĐKXĐ : x # 4 ; x # - 4
P = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)
P =\(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2+5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{x+2\sqrt{x}+\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
b ) Để P = 2 \(\Leftrightarrow\dfrac{3\sqrt{x}}{\sqrt{x}+2}\) = 2
\(\Leftrightarrow3\sqrt{x}=2\sqrt{x}+4\)
\(\Leftrightarrow\sqrt{x}=4\)
\(\Leftrightarrow x=16\)
Vậy, để P = 2 thì x = 16.
a: Sửa đề: \(P=\left(\dfrac{2a+1}{a\sqrt{a}+1}-\dfrac{\sqrt{a}}{a-\sqrt{a}+1}\right)\cdot\left(\dfrac{1+\sqrt{a^3}}{1+\sqrt{a}}-\sqrt{a}\right)\)
\(=\dfrac{2a+1-a-\sqrt{a}}{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}\cdot\left(1-2\sqrt{a}+a\right)\)
\(=\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)
b: \(P\cdot\sqrt{1-a}=\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\cdot\sqrt{1-a}>0\)
Câu 1:
Sửa đề: \(P=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
a: \(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
b: Để P>0 thì căn a-2>0
=>a>4
a: \(P=\dfrac{\sqrt{a}\left(1-a\right)^2}{1+a}:\left[\left(a+2\sqrt{a}+1\right)\left(a-2\sqrt{a}+1\right)\right]\)
\(=\dfrac{\sqrt{a}\left(a-1\right)^2}{a+1}\cdot\dfrac{1}{\left(a-1\right)^2}=\dfrac{\sqrt{a}}{a+1}\)
b: \(P-\dfrac{1}{2}=\dfrac{\sqrt{a}}{a+1}-\dfrac{1}{2}=\dfrac{2\sqrt{a}-a-1}{2\left(a+1\right)}=\dfrac{-\left(\sqrt{a}-1\right)^2}{2\left(a+1\right)}\)
\(\Leftrightarrow a\left(P-\dfrac{1}{2}\right)< 0\)