a) | \(-\sqrt{2}\)- x | = \(\sqrt{2}\)
b) | x - 1 | = \(\sqrt{3}\)- 2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
AH
Akai Haruma
Giáo viên
26 tháng 6 2021
Bài 1: Bạn đã post 1 lần
Bài 2:
\(C=\sqrt{(x-3)-2\sqrt{x-3}+1}-\sqrt{(x-3)-4\sqrt{x-3}+4}\)
\(=\sqrt{(\sqrt{x-3}-1)^2}-\sqrt{(\sqrt{x-3}-2)^2}\)
\(=|\sqrt{x-3}-1|-|\sqrt{x-3}-2|\)
Áp dụng BĐT dạng $|a|-|b|\leq |a-b|(*)$ thì:
$C\leq |\sqrt{x-3}-1-(\sqrt{x-3}-2)|$ hay $C\leq 1$
Vậy $C_{\max}=1$
Mặt khác, vẫn áp dụng BĐT $(*)$:
\(|\sqrt{x-3}-1|=|(\sqrt{x-3}-2-(-1)|\geq |\sqrt{x-3}-2|-|-1|\)
\(=|\sqrt{x-3}-2|-1\Rightarrow C\geq -1\)
Vậy $C_{\min}=-1$
28 tháng 8 2021
a: Ta có: \(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\dfrac{15\sqrt{x}-11-\left(3x+7\sqrt{x}-6\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
b: Ta có: \(\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\)
\(=\sqrt{a}\left(\sqrt{a}+1\right)-\left(2\sqrt{a}-1\right)+1\)
\(=a+\sqrt{a}-2\sqrt{a}+1+1\)
\(=a-\sqrt{a}+2\)
28 tháng 8 2021
a,ĐKXĐ: tự tìm :v
\(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)
\(=\dfrac{15\sqrt{x}-11}{\left(x+2\sqrt{x}+1\right)-4}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)
\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}+1\right)^2-4}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)
\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}+\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)
\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+7\sqrt{x}-6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{2x+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6+2x+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{9\sqrt{x}-x-8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\left(9\sqrt{x}-9\right)-\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{9\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(10-\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(\dfrac{10-\sqrt{x}}{\sqrt{x}+3}\)
15 tháng 1 2022
a: \(A=\dfrac{2x-6\sqrt{x}+\sqrt{x}-3-2x+4\sqrt{x}+\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{3x-3\sqrt{x}-\sqrt{x}-4}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}-1}{3x-4\sqrt{x}-4}\)
\(=\dfrac{1}{\sqrt{x}-2}\cdot\dfrac{3x-6\sqrt{x}+2\sqrt{x}-4}{\sqrt{x}-1}=\dfrac{3\sqrt{x}+2}{\sqrt{x}-1}\)
b: Để A<2 thì \(\dfrac{3\sqrt{x}+2-2\sqrt{x}+2}{\left(\sqrt{x}-1\right)}< 0\)
=>x<1
20 tháng 8 2021
b: Ta có: \(B=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\cdot\left(\dfrac{x\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\left(x+\sqrt{x}+1+\sqrt{x}\right)\)
\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\sqrt{x}-1}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}-1}\)
8 tháng 9 2022
Bài 3:
a: \(=\dfrac{3+2\sqrt{2}}{1}-\dfrac{\sqrt{2}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}\)
\(=3+2\sqrt{2}-\sqrt{2}=3+\sqrt{2}\)
b: \(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{a-b}\cdot\sqrt{\dfrac{ab+b^2-2b\sqrt{ab}}{a^2+2a\sqrt{b}+b}}\)
\(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{\left(\sqrt{ab}-b\right)}{\left(a+\sqrt{b}\right)^2}\)
\(=\dfrac{\sqrt{b}}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{a+\sqrt{b}}=\dfrac{b}{a+\sqrt{b}}\)
c: \(=x+\sqrt{x}-2\sqrt{x}-1+1=x-\sqrt{x}\)
b) | x - 1 | = \(\sqrt{3}-2\)
\(\Rightarrow\orbr{\begin{cases}x-1=\sqrt{3}-2\\x-1=-\left(\sqrt{3}-2\right)\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\sqrt{3}-2+1\\x=-\sqrt{3}+2+1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\sqrt{3}-1\\x=-\sqrt{3}+3\end{cases}}\)
a) | \(-\sqrt{2}\)- x | = \(\sqrt{2}\)
\(\Rightarrow\orbr{\begin{cases}-\sqrt{2}-x=\sqrt{2}\\-\sqrt{2}-x=-\sqrt{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\sqrt{2}-\sqrt{2}\\x=-\sqrt{2}+\sqrt{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-2\sqrt{2}\\x=0\end{cases}}\)