Cho :
\(P=\dfrac{\sqrt{a}\cdot\left(16-\sqrt{a}\right)}{a-4}+\dfrac{3+2\cdot\sqrt{a}}{2-\sqrt{a}}-\dfrac{2-3\cdot\sqrt{a}}{\sqrt{a}+2}\)
a) Rút gọn P
b) Tính giá trị của P tại \(a=4+2\cdot\sqrt{3}\)
c) Tìm min \(Q=P+\sqrt{a}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1. \(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right).\left(\sqrt{a}.\dfrac{4}{\sqrt{a}}\right)=\dfrac{\left(\sqrt{a}-2\right)^2-\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}.4=\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}.4=\dfrac{-64\sqrt{a}}{a-4}\)Nếu nhân tu thứ 2 của phép tính là \(\sqrt{a}-\dfrac{4}{\sqrt{a}}\) thì kết quả của phép tính là -16 nha bạn
2.\(\left(\dfrac{1}{1-\sqrt{a}}-\dfrac{1}{1+\sqrt{a}}\right).\left(1-\dfrac{1}{\sqrt{a}}\right)=\dfrac{1+\sqrt{a}-1+\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}.\dfrac{-\left(1-\sqrt{a}\right)}{\sqrt{a}}=\dfrac{-2\sqrt{a}}{\left(1+\sqrt{a}\right)\sqrt{a}}=\dfrac{-2}{1+\sqrt{a}}\)\(\left(a>0,a\ne1\right)\)
a: ĐKXĐ: x>1; x<>2
b: \(P=\left(\dfrac{\sqrt{x}+\sqrt{x-1}}{x-x+1}-\sqrt{x-1}-\sqrt{2}\right)\cdot\dfrac{2\sqrt{x}-\sqrt{x}-\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\)
\(=\left(\sqrt{x}-\sqrt{2}\right)\cdot\dfrac{\sqrt{x}-\sqrt{2}}{\sqrt{x}\left(2-\sqrt{x}\right)}=\dfrac{-\sqrt{x}+\sqrt{2}}{\sqrt{x}}\)
c: Khi x=3+2căn 2 thì
P=(-căn 2-1+căn 2)/(căn 2+1)=căn 2-1
a) \(A=\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right).\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\left(đk:a>0,x\ne1\right)\)
\(=\dfrac{a-1}{2\sqrt{a}}.\dfrac{\left(a-\sqrt{a}\right)\left(\sqrt{a}-1\right)-\left(a+\sqrt{a}\right)\left(\sqrt{a}+1\right)}{a-1}\)
\(=\dfrac{a\sqrt{a}-2a+\sqrt{a}-a\sqrt{a}-2a-\sqrt{a}}{2\sqrt{a}}\)
\(=\dfrac{-4a}{2\sqrt{a}}=-2\sqrt{a}\)
b) \(A=-2\sqrt{a}>-6\)
\(\Leftrightarrow\sqrt{a}< 3\Leftrightarrow0\le a< 9\) và \(a\ne1\)
c) \(a^2-3=0\Leftrightarrow a^2=3\Leftrightarrow\sqrt{a}=\sqrt[4]{3}\)
\(\Rightarrow A=-2\sqrt{a}=-2\sqrt[4]{3}\)
a: \(=ab+2\cdot\sqrt{\dfrac{b}{a}\cdot ab}-\sqrt{ab\cdot\left(\dfrac{a}{b}+\dfrac{1}{\sqrt{ab}}\right)}\)
\(=ab+2b-\sqrt{ab\cdot\dfrac{a\sqrt{a}+\sqrt{b}}{b\sqrt{a}}}\)
\(=ab+2b-\sqrt{\sqrt{a}\cdot\left(a\sqrt{a}+\sqrt{b}\right)}\)
b: \(=\left(\sqrt{\dfrac{a^2m^2\cdot n}{b^2\cdot m}}-\sqrt{mn\cdot\dfrac{a^2b^2}{n^2}}+\sqrt{\dfrac{a^4}{b^4}\cdot\dfrac{m}{n}}\right)\cdot a^2b^2\cdot\sqrt{\dfrac{n}{m}}\)
\(=\left(\dfrac{a\sqrt{mn}}{b}-\sqrt{a^2b^2\cdot\dfrac{m}{n}}+\dfrac{a^2}{b^2}\cdot\sqrt{\dfrac{m}{n}}\right)\cdot\sqrt{\dfrac{n}{m}}\cdot a^2b^2\)
\(=\left(\dfrac{an}{b}-ab+\dfrac{a^2}{b^2}\right)\cdot a^2b^2\)
\(=a^3nb-a^3b^3+a^4\)
\(A=\left[1:\left(1-\frac{\sqrt{a}}{1+\sqrt{a}}\right)\right]\left[\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{a\sqrt{a}-a+\sqrt{a}-1}\right]\)
\(=\left[1:\left(\frac{1+\sqrt{a}-\sqrt{a}}{1+\sqrt{a}}\right)\right]\left[\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}-1\right)}\right]\)
\(=\left(1:\frac{1}{1+\sqrt{a}}\right).\frac{a+1-2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+1\right)}\)
\(=\left(\sqrt{a}+1\right).\frac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(a+1\right)}\)
\(=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{a+1}=\frac{a-1}{a+1}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\\\sqrt{y}=b\end{matrix}\right.\), ta có:
\(A=\left[\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\times\dfrac{2}{a+b}+\dfrac{1}{a^2}+\dfrac{1}{b^2}\right]\)\(\times\dfrac{a^3+ab^2+a^2b+b^3}{ab^3+a^3b}\)
\(=\left(\dfrac{b+a}{ab}\times\dfrac{2}{a+b}+\dfrac{b^2+a^2}{a^2b^2}\right)\)\(\times\dfrac{a^2\left(a+b\right)+b^2\left(a+b\right)}{ab\left(a^2+b^2\right)}\)
\(=\dfrac{2ab+b^2+a^2}{a^2b^2}\times\dfrac{\left(a+b\right)\left(a^2+b^2\right)}{ab\left(b^2+a^2\right)}\)
\(=\dfrac{\left(a+b\right)^3}{a^3b^3}\)
\(=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^3}{\sqrt{\left(xy\right)^3}}\)
\(A=\sqrt{8}-\sqrt{7}+5\sqrt{7}+2\sqrt{2}\\ =2\sqrt{2}-\sqrt{7}+5\sqrt{7}+2\sqrt{2}\\ =4\sqrt{2}+4\sqrt{7}\)
\(B=\left(3+2\sqrt{6}+2\right)\left(25-20\sqrt{6}+24\right)\sqrt{3-2\sqrt{6}+2}\\ =\left(\sqrt{3}+\sqrt{2}\right)^2\left(5-2\sqrt{6}\right)^2\left(\sqrt{3}-\sqrt{2}\right)\\ =\left(\sqrt{3}+\sqrt{2}\right)\left(3-2\sqrt{6}+2\right)^2\\ =\left(\sqrt{3}-\sqrt{2}\right)^3\\ =9\sqrt{3}-11\sqrt{2}\)
a: \(P=\dfrac{16\sqrt{a}-a-\left(2\sqrt{a}+3\right)\left(\sqrt{a}+2\right)+\left(3\sqrt{a}-2\right)\left(\sqrt{a}-2\right)}{a-4}\)
\(=\dfrac{16\sqrt{a}-a-2a-7\sqrt{a}-6+3a-8\sqrt{a}+4}{a-4}\)
\(=\dfrac{\sqrt{a}-2}{a-4}=\dfrac{1}{\sqrt{a}+2}\)
b: Thay \(a=4+2\sqrt{3}\) vào P, ta được:
\(P=\dfrac{1}{\sqrt{3}+1+2}=\dfrac{1}{3+\sqrt{3}}=\dfrac{3-\sqrt{3}}{6}\)