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1, A=\(\left(1-\dfrac{2\sqrt{a}}{a+1}\right):\left(\dfrac{1}{\sqrt{a}+1}-\dfrac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}+a+1}\right)\)
ĐKXĐ: a≥0
A=\(\left(1-\dfrac{2\sqrt{a}}{a+1}\right):\left(\dfrac{1}{\sqrt{a}+1}-\dfrac{2\sqrt{a}}{\sqrt{a}\left(a+1\right)+1\left(a+1\right)}\right)\)
A=\(\left(\dfrac{a+1}{a+1}-\dfrac{2\sqrt{a}}{a+1}\right):\left(\dfrac{a+1}{\left(\sqrt{a}+1\right)\left(a+1\right)}-\dfrac{2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(a+1\right)}\right)\)
A=\(\left(\dfrac{a+1-2\sqrt{a}}{a+1}\right):\left(\dfrac{a+1-2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(a+1\right)}\right)\)
A=\(\left(\dfrac{a+1-2\sqrt{a}}{a+1}\right).\left(\dfrac{\left(a+1\right)\left(\sqrt{a}+1\right)}{a+1-2\sqrt{a}}\right)\)
A=\(\sqrt{a}+1\)
Vậy A=\(\sqrt{a}+1\)
2, a=1996-2\(\sqrt{1995}\)
a=\(1995-2\sqrt{1995}+1\)
a=\(\left(\sqrt{1995}-1\right)^2\) (TMĐKXĐ)
thay a=\(\left(\sqrt{1995}-1\right)^2\) vào A ta có:
A=\(\sqrt{\left(\sqrt{1995}-1\right)^2}+1\)
A=\(\sqrt{1995}\)
Vậy a=1996-2\(\sqrt{1995}\) thì A=\(\sqrt{1995}\)
1: \(A=\dfrac{a+1-2\sqrt{a}}{a+1}:\dfrac{a+1-2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(a+1\right)}\)
\(=\sqrt{a}+1\)
2: Khi \(a=2010-2\sqrt{2009}\) thì \(A=\sqrt{2009}-1+1=\sqrt{2009}\)
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)\(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right)\):\(\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a-1}\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a-1}\right)}\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{\left(a-1\right)\left(a-4\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{a-1-a+4}\)
\(=\dfrac{1}{\sqrt{a}}.\dfrac{\sqrt{a}-2}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\) ĐKXĐ: \(x>0\) \(a\ne4\) \(a\ne1\)
b) \(Q>0\)
\(\Leftrightarrow\dfrac{\sqrt{a}-2}{3\sqrt{a}}>0\)
mà \(3\sqrt{a}>0\) (Kết hợp ĐKXĐ \(a>0\))
\(\Leftrightarrow\sqrt{a}-2>0\)
\(\Leftrightarrow\sqrt{a}>2\)
\(\Leftrightarrow a>4\) (Thỏa mãn ĐKXĐ)
Vậy \(a>4\) thì \(Q>0\)
____♫ Chúc bạn học tốt ♫____
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)
\(P=\left(\dfrac{1}{\sqrt{a}+1}-\dfrac{1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right).\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}-1}\)
\(P=\left(\dfrac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right).\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)}\)
\(P=\dfrac{\sqrt{a}+1}{\sqrt{a}}\)
b/
\(a=2\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\)
\(a=\sqrt{3-\sqrt{5}}\left(6+2\sqrt{5}\right)\sqrt{2}\left(\sqrt{5}-1\right)\)
\(a=\sqrt{6-2\sqrt{5}}\left(6+2\sqrt{5}\right)\left(\sqrt{5}-1\right)=\sqrt{\left(\sqrt{5}-1\right)^2}\left(\sqrt{5}+1\right)^2\left(\sqrt{5}-1\right)\)
\(a=\left(\sqrt{5}+1\right)^2.\left(\sqrt{5}-1\right)^2\)
\(a=\left[\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)\right]^2=4^2=16\)
\(\Rightarrow P=\dfrac{\sqrt{a}+1}{\sqrt{a}}=\dfrac{\sqrt{16}+1}{\sqrt{16}}=\dfrac{4+1}{4}=\dfrac{5}{4}\)
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
ĐKXĐ: a ≥ 0
a) Ta có:
P = \(\left(1-\dfrac{2\sqrt{a}}{a+1}\right):\left(\dfrac{1}{\sqrt{a}+1}-\dfrac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}+a+1}\right)\)
= \(\dfrac{a-2\sqrt{a}+1}{a+1}:\left(\dfrac{1}{\sqrt{a}+1}-\dfrac{2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(a+1\right)}\right)\)
= \(\dfrac{\left(\sqrt{a}-1\right)^2}{a+1}:\dfrac{a-2\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(a+1\right)}\)
= \(\dfrac{\left(\sqrt{a}-1\right)^2}{a+1}.\dfrac{\left(\sqrt{a}+1\right)\left(a+1\right)}{\left(\sqrt{a}-1\right)^2}\)
Vậy P = \(\sqrt{a}+1\) với a ≥ 0
b) Ta có: a = \(1996-2\sqrt{1995}\) = \(\left(\sqrt{1995}-1\right)^2\) (TMĐK)
⇒ \(\sqrt{a}=\sqrt{1995}-1\). Thay vào P ta được
P = \(\sqrt{1995}-1+1=\sqrt{1995}\)
Vậy P = \(\sqrt{1995}\) khi a = \(1996-2\sqrt{1995}\)