Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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)
Sửa đề: \(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}-1}{\sqrt{a}+2}\right)\)
a) ĐKXĐ: \(\left\{{}\begin{matrix}a\ge0\\a\notin\left\{1;4\right\}\end{matrix}\right.\)
Ta có: \(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}-1}{\sqrt{a}+2}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a+3\sqrt{a}+2-a+3\sqrt{a}-2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{6\sqrt{a}}\)
\(=\dfrac{a-4}{6a\left(\sqrt{a}-1\right)}\)
c) Thay \(a=9-4\sqrt{5}\) vào Q, ta được:
\(Q=\dfrac{5-4\sqrt{5}}{6\left(9-4\sqrt{5}\right)\left(\sqrt{5}-3\right)}\)
\(=\dfrac{5-4\sqrt{5}}{6\left(9\sqrt{5}-27-20+12\sqrt{5}\right)}\)
\(=\dfrac{5-4\sqrt{5}}{6\left(21\sqrt{5}-47\right)}\)
\(=\dfrac{\left(5-4\sqrt{5}\right)\left(21\sqrt{5}+47\right)}{-24}\)
\(=\dfrac{105\sqrt{5}+235-420-188\sqrt{5}}{-24}\)
\(=\dfrac{-83\sqrt{5}-185}{-24}=\dfrac{83\sqrt{5}+185}{24}\)
\(ĐKXĐ:a>0,a\ne1\)
\(A=\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\)
\(A=\dfrac{a+2\sqrt{a}+1-a+2\sqrt{a}-1+4a\sqrt{a}-4\sqrt{a}}{a-1}.\dfrac{a\sqrt{a}-\sqrt{a}}{a}\)
\(A=\dfrac{4a\sqrt{a}}{a-1}.\dfrac{\sqrt{a}\left(a-1\right)}{a}=\dfrac{4a^2}{a}=4a\)
c. \(A=4a=4.\dfrac{\sqrt{6}}{2+\sqrt{6}}=\dfrac{4\sqrt{6}}{2+\sqrt{6}}=12-4\sqrt{6}\)
a) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}=\sqrt{16}-6+\sqrt{20}-\sqrt{5}=4-6+2\sqrt{5}-\sqrt{5}=\sqrt{5}-2\)
b) \(0,2\sqrt{\left(-10\right)^3.3}+2\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}=0,2\left|-10\right|\sqrt{3}+2\left|\sqrt{3}-\sqrt{5}\right|=0,2.10.\sqrt{3}+2\left(\sqrt{5}-\sqrt{3}\right)=2\sqrt{3}+2\sqrt{5}-2\sqrt{3}=2\sqrt{5}\)
c) \(\left(\dfrac{1}{2}\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}\sqrt{2}+\dfrac{4}{5}\sqrt{200}\right):\dfrac{1}{8}=\left(\dfrac{1}{2}\sqrt{\dfrac{2}{4}}-\dfrac{3}{2}\sqrt{2}+8\sqrt{2}\right):\dfrac{1}{8}=\left(\dfrac{1}{4}\sqrt{2}-\dfrac{2}{3}\sqrt{2}+8\sqrt{2}\right):\dfrac{1}{8}=\dfrac{27}{4}\sqrt{2}.8=54\sqrt{2}\)
d) \(2\sqrt{\left(\sqrt{2}-3\right)^2}+\sqrt{2.\left(-3\right)^2}-5\sqrt{\left(-1\right)^4}=2\left(3-\sqrt{2}\right)+3\sqrt{2}-5=6-2\sqrt{2}+3\sqrt{2}-5=1+\sqrt{2}\)
\(A=\dfrac{a-\sqrt{a}-6}{4-a}-\dfrac{1}{\sqrt{a}-2}=\dfrac{a+2\sqrt{a}-3\sqrt{a}-6}{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}-\dfrac{1}{\sqrt{a}-2}=\dfrac{\sqrt{a}-3}{2-\sqrt{a}}+\dfrac{1}{2-\sqrt{a}}=\dfrac{\sqrt{a}-2}{2-\sqrt{a}}=-1\) \(F=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right).\dfrac{1-\sqrt{a}}{1-a}.\dfrac{1-\sqrt{a}}{1-a}=\left(a+2\sqrt{a}+1\right).\dfrac{\left(1-\sqrt{a}\right)^2}{\left(\sqrt{a}+1\right)^2\left(1-\sqrt{a}\right)^2}=1\)
Sửa đề: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right):\dfrac{2\sqrt{x}}{x-4}\)
ĐKXĐ: x>0; x<>4
\(A=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)+\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{2\sqrt{x}}\)
\(=\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{2\sqrt{x}}=\dfrac{2x}{2\sqrt{x}}=\sqrt{x}\)
Điều kiện: x>2, \(x\ne4\)
\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x-2}}+\dfrac{\sqrt{x}}{\sqrt{x+2}}\right):\dfrac{2\sqrt{x}}{x-4}\\ \Rightarrow A=\sqrt{x}\cdot\dfrac{\sqrt{x+2}+\sqrt{x-2}}{\sqrt{x^2-4}}\cdot\dfrac{x-4}{2\sqrt{x}}\\ \Rightarrow A=\dfrac{\left(x-4\right)\left(\sqrt{x+2}+\sqrt{x-2}\right)}{2\sqrt{x^2-4}}\)
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.
ĐKXĐ :
\(\left\{{}\begin{matrix}\sqrt{a}+2\ne0\\\sqrt{a}-2\ne0\\\sqrt{a}\ne0\\\sqrt{a}x\text{đ}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{a}\ne2\\a\ne0\\a\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a\ne4\\a>0\end{matrix}\right.\)
Rút gọn :
\(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right).\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\)
\(=\dfrac{\left(\sqrt{a}-2\right)^2-\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}.\dfrac{a-4}{\sqrt{a}}\)
\(=\dfrac{\left(\sqrt{a}-2+\sqrt{a}+2\right)\left(\sqrt{a}-2-\sqrt{a}-2\right)}{\left(\sqrt{a}\right)^2-2^2}.\dfrac{a-4}{\sqrt{a}}\)
\(=\dfrac{2\sqrt{a}.\left(-4\right)}{\sqrt{a}}\)
\(=2.\left(-4\right)=-8\)