CM BIỂU THỨC KHÔNG PHỤ THUỘC VÀO BIẾN X
\(A=\sqrt{x-1-\sqrt{2x-3}}+\sqrt{x+11+5\sqrt{2x-3}}\) với \(\dfrac{3}{2}\le x\le2\)
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Những câu hỏi liên quan
10 tháng 9 2017
\(\frac{2x}{x+3\sqrt{x}+2}+\frac{5\sqrt{x}+1}{x+4\sqrt{x}+3}+\frac{\sqrt{x}+10}{x+5\sqrt{x}+6}\)
\(=\frac{2x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}+\frac{5\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}+10}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{2x\left(\sqrt{x}+3\right)+\left(5\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+\left(\sqrt{x}+10\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{2\sqrt{x^3}+6x+5x+11\sqrt{x}+2+x+11\sqrt{x}+10}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{12x+22\sqrt{x}+2\sqrt{x^3}+12}{6x+11\sqrt{x}+\sqrt{x^3}+6}\)
\(=\frac{2\left(6x+11\sqrt{x}+\sqrt{x^3}+6\right)}{6x+11\sqrt{x}+\sqrt{x^3}+6}\)
\(=2\) (ko phụ thuộc vào biến ) (đpcm)
6 tháng 9 2022
\(=\dfrac{2x\left(\sqrt{x}+3\right)+\left(5\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+\left(\sqrt{x}+10\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{2x\sqrt{x}+6x+5x+11\sqrt{x}+2+x+11\sqrt{x}+11}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{2x\sqrt{x}+12x+22\sqrt{x}+13}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
biểu thức này có phụ thuộc vào biến nha bạn
18 tháng 7 2019
\(R=\frac{2x}{x+3\sqrt{x}+2}+\frac{5\sqrt{x}}{x+4\sqrt{x}+3}+\frac{\sqrt{x}+10}{x+5\sqrt{x}+6}\)
\(=\frac{2x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}+\frac{5\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}+10}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
=\(\frac{2x\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}+\frac{5\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}+\frac{\left(\sqrt{x}+10\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{2x\sqrt{x}+6x+5x+10\sqrt{x}+x+\sqrt{x}+10\sqrt{x}+10}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{2x\sqrt{x}+12x+21\sqrt{x}+10}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
@@@@@@@@@@@ Đề sai hay mình sai??@@@@@@@@@@
28 tháng 7 2021
A = \(\left(\dfrac{\sqrt{x}+1}{2\sqrt{x}-2}+\dfrac{3}{x-1}-\dfrac{\sqrt{x}+3}{2\sqrt{x}+2}\right)\cdot\dfrac{4x-4}{5}\) (ĐK: x \(\ge\) 0; x \(\ne\) 1)
A = \(\left(\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}+\dfrac{3}{x-1}-\dfrac{\sqrt{x}+3}{2\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{4\left(x-1\right)}{5}\)
A = \(\left(\dfrac{\left(\sqrt{x}+1\right)^2}{2\left(x-1\right)}+\dfrac{6}{2\left(x-1\right)}-\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{2\left(x-1\right)}\right)\cdot\dfrac{4\left(x-1\right)}{5}\)
A = \(\left(\dfrac{x+2\sqrt{x}+1+6-x-3\sqrt{x}+\sqrt{x}+3}{2\left(x-1\right)}\right)\cdot\dfrac{4\left(x-1\right)}{5}\)
A = \(\dfrac{10}{2\left(x-1\right)}\cdot\dfrac{4\left(x-1\right)}{5}\)
A = 4
Vậy A không phụ thuộc vào x
Chúc bn học tốt!
29 tháng 7 2021
Ta có: \(A=\left(\dfrac{\sqrt{x}+1}{2\sqrt{x}-2}+\dfrac{3}{x-1}-\dfrac{\sqrt{x}+3}{2\sqrt{x}+2}\right)\cdot\dfrac{4x-4}{5}\)
\(=\dfrac{x+2\sqrt{x}+1+6-\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{4\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{5}\)
\(=\dfrac{x+2\sqrt{x}+7-x-2\sqrt{x}+3}{1}\cdot\dfrac{2}{5}\)
\(=10\cdot\dfrac{2}{5}=4\)
AH
Akai Haruma
Giáo viên
22 tháng 6 2023
1.
$x+3+\sqrt{x^2-6x+9}=x+3+\sqrt{(x-3)^2}=x+3+|x-3|$
$=x+3+(3-x)=6$
2.
$\sqrt{x^2+4x+4}-\sqrt{x^2}=\sqrt{(x+2)^2}-\sqrt{x^2}$
$=|x+2|-|x|=x+2-(-x)=2x+2$
3.
$\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}$
$=\sqrt{(\sqrt{x^2-1}+1)^2}-\sqrt{(\sqrt{x^2-1}-1)^2}$
$=|\sqrt{x^2-1}+1|+|\sqrt{x^2-1}-1|$
$=\sqrt{x^2-1}+1+|\sqrt{x^2-1}-1|$
AH
Akai Haruma
Giáo viên
22 tháng 6 2023
4.
$\frac{\sqrt{x^2-2x+1}}{x-1}=\frac{\sqrt{(x-1)^2}}{x-1}$
$=\frac{|x-1|}{x-1}=\frac{x-1}{x-1}=1$
5.
$|x-2|+\frac{\sqrt{x^2-4x+4}}{x-2}=2-x+\frac{\sqrt{(x-2)^2}}{x-2}$
$=2-x+\frac{|x-2|}{x-2}|=2-x+\frac{2-x}{x-2}=2-x+(-1)=1-x$
6.
$2x-1-\frac{\sqrt{x^2-10x+25}}{x-5}=2x-1-\frac{\sqrt{(x-5)^2}}{x-5}$
$=2x-1-\frac{|x-5|}{x-5}$