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a: ĐKXĐ: a>=0; a<>1
b: \(A=\left(\dfrac{a+3\sqrt{a}+2}{3\sqrt{a}-2}-\dfrac{\sqrt{a}}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}-1+\sqrt{a}+1}{a-1}\)
\(=\left(\dfrac{\left(a-1\right)\left(\sqrt{a}+2\right)-3a+2\sqrt{a}}{\left(3\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\cdot\dfrac{a-1}{2\sqrt{a}}\)
\(=\dfrac{a\sqrt{a}+2a-\sqrt{a}-2-3a+2\sqrt{a}}{\left(3\sqrt{a}-2\right)}\cdot\dfrac{\sqrt{a}+1}{2\sqrt{a}}\)
\(=\dfrac{\left(a\sqrt{a}-a+\sqrt{a}-2\right)}{3\sqrt{a}-2}\cdot\dfrac{\sqrt{a}+1}{2\sqrt{a}}\)
\(B=\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{a-4}\cdot\dfrac{a-4}{\sqrt{a}}\)
\(=\dfrac{-8\sqrt{a}}{\sqrt{a}}=-8\)
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)
1/ đkxđ: a > 0; a khác 1
a/ A= (\(\dfrac{\sqrt{a}}{2\sqrt{a}}-\dfrac{1}{2\sqrt{a}}\))\(\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-1}{2\sqrt{a}}\cdot\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)^2-\sqrt{a}\left(\sqrt{a}+1\right)^2}{a-1}\)
\(=\dfrac{1}{2\sqrt{a}}\cdot\dfrac{a\sqrt{a}-2a+\sqrt{a}-a\sqrt{a}-2a-\sqrt{a}}{a-1}\)
\(=\dfrac{1}{2\sqrt{a}}\cdot\dfrac{-4a}{a-1}=-\dfrac{2\sqrt{a}}{a-1}=\dfrac{2\sqrt{a}}{a+1}\)
b/+) A = 4
\(\Leftrightarrow\dfrac{2\sqrt{a}}{a+1}=4\)\(\Leftrightarrow2\sqrt{a}=4a+4\)
=> Không có gt a nào t/m
+) \(A>-6\)
\(\Leftrightarrow\dfrac{2\sqrt{a}}{a+1}>-6\)
\(\Leftrightarrow2\sqrt{a}>-6a-6\)
\(\Leftrightarrow6a+2\sqrt{a}+6>0\) (luôn đúng vì a > 0)
=> bpt có nghiệm với mọi a > 0
vậy........
c/ \(a^2-3=0\Leftrightarrow\left[{}\begin{matrix}a=\sqrt{3}\left(tm\right)\\a=-\sqrt{3}\left(ktmđkxđ\right)\end{matrix}\right.\)
Với a = \(\sqrt{3}\) ta có:
\(A=\dfrac{2\sqrt{3}}{\sqrt{3}+1}=\dfrac{2\sqrt{3}\left(\sqrt{3}-1\right)}{3-1}=\dfrac{2\sqrt{3}\left(\sqrt{3}-1\right)}{2}=\sqrt{3}\left(\sqrt{3}-1\right)=3-\sqrt{3}\)
P/s gọi a = x cho dễ viết nhé
a, Với \(x\ge0;x\ne1;x\ne4\)
\(P=\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
\(=\left(\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}-2}{3\sqrt{x}}\)
chỗ này mình nghĩ ko phải trục căn thức đâu ha :D
b, Ta có P > 1/6 hay \(\frac{\sqrt{x}-2}{3\sqrt{x}}>\frac{1}{6}\Leftrightarrow\frac{\sqrt[]{x}-2}{3\sqrt{x}}-\frac{1}{6}>0\)
\(\Leftrightarrow\frac{6\sqrt{x}-12-3\sqrt{x}}{18\sqrt{x}}>0\Leftrightarrow\frac{3\sqrt{x}-12}{18\sqrt{x}}>0\)
\(\Leftrightarrow3\sqrt{x}-12>0\)( vì \(18\sqrt{x}>0\))
\(\Leftrightarrow3\sqrt{x}>12\Leftrightarrow\sqrt{x}>4\Leftrightarrow x>16\)
Vậy \(x>16\)
cho mình hỏi đề có sai ko ? \(P< \frac{1}{6}\)mình nghĩ sẽ hợp lí hơn
んuﾘ ｲ hãy thuận theo ý thầy :)) và nhớ chú ý đến ĐKXĐ
\(P=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right)\div\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
ĐKXĐ : \(\hept{\begin{cases}x>0\\x\ne1\\x\ne4\end{cases}}\)
\(=\left(\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right)\div\left(\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\div\left(\frac{a-1}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\frac{a-4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\div\frac{3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\times\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}=\frac{\sqrt{a}-2}{3\sqrt{a}}\)
Để P > 1/6 thì \(\frac{\sqrt{a}-2}{3\sqrt{a}}>\frac{1}{6}\)
<=> \(\frac{\sqrt{a}-2}{3\sqrt{a}}-\frac{1}{6}>0\)
<=> \(\frac{2\sqrt{a}-4}{6\sqrt{a}}-\frac{\sqrt{a}}{6\sqrt{a}}>0\)
<=> \(\frac{\sqrt{a}-4}{6\sqrt{a}}>0\)
Dễ thấy \(6\sqrt{a}>0\forall x>0\)
=> \(\sqrt{a}-4>0\)<=> \(\sqrt{a}>4\)<=> \(a>16\)
Vậy với a > 16 thì P > 1/6
Bài 2:
\(P=\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right)^2.\left(\dfrac{\sqrt{a}-1}{\sqrt{a}+1}-\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
\(P=\left(\dfrac{a-1}{2\sqrt{a}}\right)^2.\left(\dfrac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
\(P=\left[\dfrac{\left(a-1\right)^2}{4a}\right].\left(\dfrac{\left(\sqrt{a}-1+\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\sqrt{a}-1}{a-1}\right)\)
\(P=\dfrac{\left(a-1\right)^2}{4a}.\dfrac{2\sqrt{a}.\left(-2\right)}{a-1}\)
\(P=\dfrac{\left(a-1\right)^2\left(-4\sqrt{a}\right)}{4a.\left(a-1\right)}\)
\(P=\dfrac{\left(a-1\right).\left(-\sqrt{a}\right)}{a}=\dfrac{-a\sqrt{a}+\sqrt{a}}{a}\)
Bài 1:
\(A=\dfrac{2}{\sqrt{2}}-\dfrac{1}{\sqrt{3}-\sqrt{2}}+\dfrac{2}{\sqrt{3}-1}\)\(A=\dfrac{2\sqrt{2}}{2}-\dfrac{1\left(\sqrt{3}+\sqrt{2}\right)}{3-2}+\dfrac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}\right)^2-1}\)
\(A=\sqrt{2}-\dfrac{\sqrt{3}+\sqrt{2}}{1}+\dfrac{2\left(\sqrt{3}+1\right)}{3-1}\)
\(A=\sqrt{2}-\sqrt{3}-\sqrt{2}+\sqrt{3}+1\)
\(A=1\)
\(A=\dfrac{a-\sqrt{a}-6}{4-a}-\dfrac{1}{\sqrt{a}-2}=\dfrac{a+2\sqrt{a}-3\sqrt{a}-6}{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}-\dfrac{1}{\sqrt{a}-2}=\dfrac{\sqrt{a}-3}{2-\sqrt{a}}+\dfrac{1}{2-\sqrt{a}}=\dfrac{\sqrt{a}-2}{2-\sqrt{a}}=-1\) \(F=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right).\dfrac{1-\sqrt{a}}{1-a}.\dfrac{1-\sqrt{a}}{1-a}=\left(a+2\sqrt{a}+1\right).\dfrac{\left(1-\sqrt{a}\right)^2}{\left(\sqrt{a}+1\right)^2\left(1-\sqrt{a}\right)^2}=1\)
a: ĐKXĐ: a>0; a<>1
b: \(A=\dfrac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\sqrt{a}-1+2}{a-1}\)
\(=\dfrac{\sqrt{a}+1}{\sqrt{a}}\cdot\dfrac{a-1}{\sqrt{a}+1}=\dfrac{a-1}{\sqrt{a}}\)
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: \(=\dfrac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-2\left(\sqrt{6}-1\right)\)
\(=\sqrt{6}-2\sqrt{6}+2=2-\sqrt{6}\)
bài 2 ) a) đk : \(a>0;b>0\)
b) P = \(\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}.\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}\)
P = \(\dfrac{a-2\sqrt{ab}+b+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}.\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)
P = \(\dfrac{a+2\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}.\sqrt{a}-\sqrt{b}\) = \(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}.\sqrt{a}-\sqrt{b}\) = \(\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)\) = \(a-b\)
c) ta có P = \(a-b\) thay \(a=2\sqrt{3};b=\sqrt{3}\) vào ta có
P = \(2\sqrt{3}-\sqrt{3}=\sqrt{3}\) vậy khi \(a=2\sqrt{3};b=\sqrt{3}\) thì P = \(\sqrt{3}\)
bài 1) a) P = \(\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}+\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}+\dfrac{\sqrt{a}-1}{\sqrt{a}+1}\right)\)
P = \(\dfrac{\left(a\sqrt{a}-1\right)\left(a+\sqrt{a}\right)-\left(a\sqrt{a}+1\right)\left(a-\sqrt{a}\right)}{\left(a+\sqrt{a}\right)\left(a-\sqrt{a}\right)}+\dfrac{a-1}{\sqrt{a}}.\dfrac{\left(\sqrt{a}+1\right)^2+\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
P = \(\dfrac{a^2\sqrt{a}+a^2-a-\sqrt{a}-\left(a^2\sqrt{a}-a^2+a-\sqrt{a}\right)}{\left(a+\sqrt{a}\right)\left(a-\sqrt{a}\right)}+\dfrac{a-1}{\sqrt{a}}.\dfrac{a+2\sqrt{a}+1+a-2\sqrt{a}+1}{a-1}\)
P = \(\dfrac{a^2\sqrt{a}+a^2-a-\sqrt{a}-a^2\sqrt{a}+a^2-a+\sqrt{a}}{\left(a+\sqrt{a}\right)\left(a-\sqrt{a}\right)}+\dfrac{2a+2}{\sqrt{a}}\)
P = \(\dfrac{2a^2-2a}{a^2-a}+\dfrac{2a+1}{\sqrt{a}}\) = \(\dfrac{2\left(a^2-a\right)}{a^2-a}+\dfrac{2a+2}{\sqrt{a}}\)
P = \(2+\dfrac{2a+2}{\sqrt{a}}\) = \(\dfrac{2a+2\sqrt{a}+2}{\sqrt{a}}\)
b) ta có P = 7 \(\Leftrightarrow\) \(\dfrac{2a+2\sqrt{a}+2}{\sqrt{a}}=7\) \(\Leftrightarrow\) \(2a+2\sqrt{a}+2=7\sqrt{a}\)
\(\Leftrightarrow\) \(2a-5\sqrt{a}+2=0\) (1)
đặc \(\sqrt{a}=u\) \(\left(u\ge0\right)\) (1) \(\Leftrightarrow\) \(2u^2-5u+2\)
\(\Delta=\left(-5\right)^2-4.2.2\) = \(25-16=9>0\)
\(\Rightarrow\) phương trình có 2 nghiệm phân biệt
\(u_1=\dfrac{5+3}{4}=\dfrac{8}{4}=2\left(tmđk\right)\)
\(u_2=\dfrac{5-3}{4}=\dfrac{2}{4}=\dfrac{1}{2}\left(tmđk\right)\)
ta có : \(u=\sqrt{a}=2\Leftrightarrow x=4\)
\(u=\sqrt{a}=\dfrac{1}{2}\Leftrightarrow a=\dfrac{1}{4}\)
vậy \(a=4;a=\dfrac{1}{4}\) thì P = 7
\(ĐKXĐ:a>0,a\ne1\)
\(A=\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\)
\(A=\dfrac{a+2\sqrt{a}+1-a+2\sqrt{a}-1+4a\sqrt{a}-4\sqrt{a}}{a-1}.\dfrac{a\sqrt{a}-\sqrt{a}}{a}\)
\(A=\dfrac{4a\sqrt{a}}{a-1}.\dfrac{\sqrt{a}\left(a-1\right)}{a}=\dfrac{4a^2}{a}=4a\)
c. \(A=4a=4.\dfrac{\sqrt{6}}{2+\sqrt{6}}=\dfrac{4\sqrt{6}}{2+\sqrt{6}}=12-4\sqrt{6}\)