H
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
31 tháng 10 2020
a)ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Ta có: \(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\frac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{15\sqrt{x}-11-\left(3x+9\sqrt{x}-2\sqrt{x}-6\right)-\left(2x-2\sqrt{x}+3\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{-5x+5\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{-5\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(-5\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
Ta có: \(A-\frac{2}{3}=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}-\frac{2}{3}\)
\(=\frac{3\left(-5\sqrt{x}+2\right)}{3\left(\sqrt{x}+3\right)}-\frac{2\left(\sqrt{x}+3\right)}{3\left(\sqrt{x}+3\right)}\)
\(=\frac{-15\sqrt{x}+6-2\sqrt{x}-6}{3\left(\sqrt{x}+3\right)}\)
\(=\frac{-17\sqrt{x}}{3\left(\sqrt{x}+3\right)}\)
\(=\frac{-17\sqrt{x}-51+51}{3\left(\sqrt{x}+3\right)}\)
\(=\frac{-17}{3}+\frac{17}{\sqrt{x}+3}\)
Ta có: \(\sqrt{x}+3\ge3\forall x\) thỏa mãn ĐKXĐ
\(\Rightarrow\frac{17}{\sqrt{x}+3}\le\frac{17}{3}\forall x\) thỏa mãn ĐKXĐ
\(\Rightarrow\frac{17}{\sqrt{x}+3}-\frac{17}{3}\le\frac{17}{3}-\frac{17}{3}=0\forall x\) thỏa mãn ĐKXĐ
\(\Rightarrow A-\frac{2}{3}\le0\forall x\) thỏa mãn ĐKXĐ
nên \(A\le\frac{2}{3}\)(đpcm)
c) Ta có: \(C=\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right):\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\left(\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right):\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{a-2\sqrt{ab}+b}{a-b}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{\sqrt{a}-\sqrt{b}+2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}+\sqrt{b}}=1\)
Vậy: Giá trị của C không phụ thuộc vào a,b(đpcm)
30 tháng 5 2022
Bài 2:
a: \(A=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-\left(5\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
b: Thay \(x=5-2\sqrt{6}\) vào A, ta được:
\(A=\dfrac{-5\left(\sqrt{3}-\sqrt{2}\right)+2}{\sqrt{3}-\sqrt{2}+3}=\dfrac{-5\sqrt{3}+5\sqrt{2}+2}{\sqrt{3}-\sqrt{2}+3}\simeq0,124\)
d: Để A=1/2 thì \(\sqrt{x}+3=-10\sqrt{x}+4\)
\(\Leftrightarrow11\sqrt{x}=1\)
hay x=1/121
19 tháng 8 2018
a) điều kiện xác định : \(a>2;a\ne11\)
ta có : \(P=\left(\dfrac{\sqrt{a-2}+2}{3}\right)\left(\dfrac{\sqrt{a-2}}{3+\sqrt{a-2}}+\dfrac{a+7}{11-a}\right):\left(\dfrac{3\sqrt{a-2}+1}{a-3\sqrt{a-2}-2}-\dfrac{1}{\sqrt{a-2}}\right)\)
\(\Leftrightarrow P=\left(\dfrac{\sqrt{a-2}+2}{3}\right)\left(\dfrac{\sqrt{a-2}}{3+\sqrt{a-2}}+\dfrac{a+7}{\left(3+\sqrt{a-2}\right)\left(3-\sqrt{a-2}\right)}\right):\left(\dfrac{3\sqrt{a-2}+1}{\sqrt{a-2}\left(\sqrt{a-2}-3\right)}-\dfrac{1}{\sqrt{a-2}}\right)\) \(\Leftrightarrow P=\left(\dfrac{\sqrt{a-2}+2}{3}\right)\left(\dfrac{\sqrt{a-2}\left(3-\sqrt{a-2}\right)+a+7}{\left(3+\sqrt{a-2}\right)\left(3-\sqrt{a-2}\right)}\right):\left(\dfrac{3\sqrt{a-2}+1-\sqrt{a-2}+3}{\sqrt{a-2}\left(\sqrt{a-2}-3\right)}\right)\) \(\Leftrightarrow P=\left(\dfrac{\sqrt{a-2}+2}{3}\right)\left(\dfrac{3\left(\sqrt{a-2}+3\right)}{\left(3+\sqrt{a-2}\right)\left(3-\sqrt{a-2}\right)}\right):\left(\dfrac{2\sqrt{a-2}+4}{\sqrt{a-2}\left(\sqrt{a-2}-3\right)}\right)\) \(\Leftrightarrow P=\left(\dfrac{\sqrt{a-2}+2}{3}\right)\left(\dfrac{3}{\left(3-\sqrt{a-2}\right)}\right)\left(\dfrac{\sqrt{a-2}\left(\sqrt{a-2}-3\right)}{2\left(\sqrt{a-2}+2\right)}\right)\) \(\Leftrightarrow P=\dfrac{-\sqrt{a-2}}{2}\)
19 tháng 8 2018
ta có : \(a+b=\sqrt{2017-a^2}+\sqrt{2017-b^2}\)
\(\Leftrightarrow\left(a+b\right)\left(\sqrt{2017-a^2}-\sqrt{2017-b^2}\right)=b^2-a^2\)
\(\Leftrightarrow b-a=\sqrt{2017-a^2}-\sqrt{2017-b^2}\)
\(\Leftrightarrow2b=2\sqrt{2017-a^2}\Leftrightarrow b^2=2017-a^2\Rightarrow\left(đpcm\right)\)
3 tháng 8 2017
1. ĐK \(\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
a. Ta có \(R=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right).\left(\frac{1}{\sqrt{x}+2}+\frac{4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}-2+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}+2}{\sqrt{x}}.\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
b. Với \(x=4+2\sqrt{3}\Rightarrow R=\frac{\sqrt{4+2\sqrt{3}}+2}{\sqrt{4+2\sqrt{3}}\left(\sqrt{4+2\sqrt{3}}-2\right)}=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}+2}{\sqrt{\left(\sqrt{3}+1\right)^2}\left(\sqrt{\left(\sqrt{3}+1\right)^2}-2\right)}\)
\(=\frac{\sqrt{3}+1+2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\frac{\sqrt{3}+3}{3-1}=\frac{\sqrt{3}+3}{2}\)
c. Để \(R>0\Rightarrow\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}>0\Rightarrow\sqrt{x}-2>0\Rightarrow x>4\)
Vậy \(x>4\)thì \(R>0\)
2. Ta có \(A=6+2\sqrt{2}=6+\sqrt{8};B=9=6+3=6+\sqrt{9}\)
Vì \(\sqrt{8}< \sqrt{9}\Rightarrow A< B\)
3. a. \(VT=\frac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}.\left(\sqrt{a}+\sqrt{b}\right)\)
\(=\left(\sqrt{a}-\sqrt{b}\right).\left(\sqrt{a}+\sqrt{b}\right)=a-b=VP\left(đpcm\right)\)
b. Ta có \(VT=\left(2+\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right).\left(2-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\)
\(=\left(2+\sqrt{a}\right)\left(2-\sqrt{a}\right)=4-a=VP\left(đpcm\right)\)
\(\left(a+\sqrt{a^2+3}\right)\left(b+\sqrt{b^2+3}\right)=3\left(1\right)\)
Nhân cả 2 vế của (1) với \(a-\sqrt{a^2+3}\) ,ta được:
\(\left(a-\sqrt{a^2+3}\right)\left(a+\sqrt{a^2+3}\right)\left(b+\sqrt{b^2+3}\right)=3\left(a-\sqrt{a^2+3}\right)\)
\(\left(a^2-a^2-3\right)\left(b+\sqrt{b^2+3}\right)=3\left(a-\sqrt{a^2+3}\right)\)
\(\Leftrightarrow-3\left(b+\sqrt{b^2+3}\right)=3\left(a-\sqrt{a^2+3}\right)\)
\(\Leftrightarrow-\left(b+\sqrt{b^2+3}\right)=a-\sqrt{a^2+3}\)(2)
Nhân cả 2 vế của (1) với \(b-\sqrt{b^2+3}\) ,ta được:
\(\left(a+\sqrt{a^2+3}\right)\left(b+\sqrt{b^2+3}\right)\left(b-\sqrt{b^2+3}\right)=3\left(b-\sqrt{b^2+3}\right)\)
\(\Leftrightarrow\left(a+\sqrt{a^2+3}\right)\left(b^2-b^2-3\right)=3\left(b-\sqrt{b^2+3}\right)\)
\(\Leftrightarrow-3\left(a+\sqrt{a^2+3}\right)=3\left(b-\sqrt{b^2+3}\right)\)
\(\Leftrightarrow-\left(a+\sqrt{a^2+3}\right)=b-\sqrt{b^2+3}\) (3)
Cộng 2 vế của (2) và (3), ta được:
\(-b-\sqrt{b^2+3}-a-\sqrt{a^2+3}=a-\sqrt{a^2+3}+b-\sqrt{b^2+3}\)
\(\Leftrightarrow-2a-2b=0\)
\(\Leftrightarrow-2\left(a+b\right)=0\)
\(\Leftrightarrow a+b=0\)
Vậy: a + b = 0
=.= hk tốt!!