Rút gọn: \(A=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\)
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N
1
15 tháng 7 2018
\(A=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}=\sqrt{x-2+2\sqrt{2}.\sqrt{x-2}+2}+\sqrt{x-2-2\sqrt{2}.\sqrt{x-2}+2}=\text{| }\sqrt{x-2}+\sqrt{2}\text{| }+\text{| }\sqrt{x-2}-\sqrt{2}\text{| }\) +) \(A=\sqrt{x-2}+\sqrt{2}+\sqrt{x-2}-\sqrt{2}=2\sqrt{x-2}\left(\text{ }\sqrt{x-2}\text{≥}\sqrt{2}\right)\)
+) \(A=\sqrt{x-2}+\sqrt{2}+\sqrt{2}-\sqrt{x-2}=2\sqrt{2}\left(\sqrt{x-2}< \sqrt{2}\right)\)
PH
2
PD
16 tháng 8 2020
\(\sqrt{2x-2\sqrt{x^2-4}}+\sqrt{x-2}=\sqrt{\left(\sqrt{x-2}-\sqrt{x+2}\right)^2}+\sqrt{x-2}\)
\(=\left|\sqrt{x-2}-\sqrt{x+2}\right|+\sqrt{x-2}\)
\(=-\sqrt{x-2}+\sqrt{x+2}+\sqrt{x-2}\)
\(=\sqrt{x+2}\)
15 tháng 11 2018
1, ĐKXĐ: x\(\ge0\);x\(\ne1\)
Rút gọn P với \(x\ge0;x\ne1\)ta có
P=\(\dfrac{-\sqrt{x}+\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\left(\dfrac{-\left(\sqrt{x}-0,5\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-0,5\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right)\)
\(=\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\left(\dfrac{-\sqrt{x}+0,5}{\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}-0,5\right)}{x-\sqrt{x}+1}\right)\)
=\(\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\left(\dfrac{-x\sqrt{x}+x-\sqrt{x}+0,5x-0,5\sqrt{x}+0,5+x\sqrt{x}-x-0,5x+0,5\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\right)\)
=\(\dfrac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\dfrac{-1}{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\)
=\(\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)
15 tháng 11 2018
2, Thay x=7-4\(\sqrt{3}\)thỏa mãn đk vào P ta có:
P\(=\dfrac{7-4\sqrt{3}-\sqrt{7-4\sqrt{3}}+1}{\sqrt{7-4\sqrt{3}}}\)
=\(\dfrac{7-4\sqrt{3}-\sqrt{\left(\sqrt{3}-2\right)^2}+1}{\sqrt{\left(\sqrt{3}-2\right)^2}}\)
=\(\dfrac{7-4\sqrt{3}-2+\sqrt{3}+1}{2-\sqrt{3}}\)
\(=\dfrac{6-3\sqrt{3}}{2-\sqrt{3}}=12+6\sqrt{3}-6\sqrt{3}-9\)=3
19 tháng 6 2015
\(P=\frac{\sqrt{x}\left(\sqrt{x^3}+1\right)}{\left(x-\sqrt{x}+1\right)}+1-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}\)
\(P=\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+1-2\sqrt{x}-1\)
\(P=\sqrt{x}\left(\sqrt{x}+1\right)-2\sqrt{x}=x+\sqrt{x}-2\sqrt{x}=x-\sqrt{x}\)
8 tháng 10 2022
\(F=\left(\dfrac{\sqrt{x}\left(2x+\sqrt{x}-1\right)}{x\sqrt{x}-1}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\cdot\dfrac{x-1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\dfrac{2x\sqrt{x}+x-\sqrt{x}-x\sqrt{x}-x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)}{2\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\dfrac{x\sqrt{x}-2\sqrt{x}}{x+\sqrt{x}+1}\cdot\dfrac{1}{2\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\dfrac{x\sqrt{x}-2\sqrt{x}+x\sqrt{x}+x+\sqrt{x}}{\left(2\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{2x\sqrt{x}+x-\sqrt{x}}{\left(2\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}{\left(2\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}\)
NA
5 tháng 10 2018
\(\left(x-1\right)-\frac{2x-2\sqrt{x}}{\sqrt{x}-1}+\frac{x\sqrt{x}+1}{x-\sqrt{x}+1}\)
\(=\left(x-1\right)-\frac{2\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}\)
\(=x-1-2\sqrt{x}+\sqrt{x}+1\)
\(=x-\sqrt{x}\)
Ta có \(A\sqrt{2}=\sqrt{2x+4\sqrt{2x-4}}+\sqrt{2x-4\sqrt{2x-4}}=\sqrt{2x-4+4\sqrt{2x-4}+4}\)
+\(\sqrt{2x-4-4\sqrt{2x-4}+4}=\sqrt{2x-4}+2+\sqrt{2x-4}-2=2\sqrt{2x-4}\)
=> A=\(2\sqrt{x-2}\)
A = \(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\)
\(\sqrt{2}A=\sqrt{2}\left(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\right)\)
\(\sqrt{2}A=\sqrt{2\left(x+2\sqrt{2x-4}\right)}+\sqrt{2\left(x-2\sqrt{2x-4}\right)}\)
\(\sqrt{2}A=\sqrt{2x+4\sqrt{2x-4}}+\sqrt{2x-4\sqrt{2x-4}}\)
\(\sqrt{2}A=\sqrt{\left(2x-4\right)+2.2\sqrt{2x-4}+4}+\sqrt{\left(2x-4\right)-2.2\sqrt{2x-4}+4}\)
\(\sqrt{2}A=\sqrt{\left(\sqrt{2x-4}+2\right)^2}+\sqrt{\left(\sqrt{2x-4}-2\right)^2}\)
\(\sqrt{2}A=|\sqrt{2x-4}+2|+|\sqrt{2x-4}+2|\)
\(\sqrt{2}A=\sqrt{2x-4}+2+|\sqrt{2x-4}-2|\)
Xét 2 trường hợp:
+)\(\sqrt{2x-4}\ge2\)
\(\sqrt{2}A=\sqrt{2x-4}+2+\sqrt{2x-4}-2\)
\(\sqrt{2}A=2\sqrt{2x-4}\)
\(A=\sqrt{2}\sqrt{2x-4}=\sqrt{4x-8}\)
+)\(\sqrt{2x-4}< 2\)
\(\sqrt{2}A=\sqrt{2x-4}+2+2-\sqrt{2x-4}=4\)
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