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HT
1
rút gọn giúp mình vs ạ
Rút gọn giúp mình với ạ
HN
0
DT
1
7 tháng 10 2021
a) \(\sqrt{0,64.a^2}\left(a>0\right)=0,8.\left|a\right|=0,8a\)
b) \(\sqrt{a^2\left(a-2\right)^2}\left(a>2\right)=\left|a\left(a-2\right)\right|=a\left(a-2\right)=a^2-2a\)
c) \(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}\left(a\ge0,a\ne1\right)=\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}=1+\sqrt{a}+a\)
MO
2
AT
13 tháng 7 2021
\(P=\left(\dfrac{x-1}{\sqrt{x}+1}-\dfrac{x-2\sqrt{x}+1}{x-\sqrt{x}}+1\right).\dfrac{1}{x\sqrt{x}+1}\)
\(=\left(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+1}-\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}+1\right).\dfrac{1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\left(\sqrt{x}-1-\dfrac{\sqrt{x}-1}{\sqrt{x}}+1\right).\dfrac{1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)+\sqrt{x}}{\sqrt{x}}.\dfrac{1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{x-\sqrt{x}+1}{\sqrt{x}}.\dfrac{1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
13 tháng 7 2021
Bài 2:
Ta có: \(P=\left(\dfrac{x-1}{\sqrt{x}+1}-\dfrac{x-2\sqrt{x}+1}{x-\sqrt{x}}+1\right)\cdot\dfrac{1}{x\sqrt{x}+1}\)
\(=\left(\sqrt{x}-1-\dfrac{\sqrt{x}-1}{\sqrt{x}}+1\right)\cdot\dfrac{1}{x\sqrt{x}+1}\)
\(=\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\cdot\dfrac{1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{1}{x+\sqrt{x}}\)
N
1
24 tháng 10 2021
d: \(\dfrac{-\left(\sqrt{3}-\sqrt{6}\right)}{1-\sqrt{2}}+\dfrac{6\sqrt{3}+3}{\sqrt{3}}-\dfrac{13}{4+\sqrt{3}}\)
\(=-\sqrt{3}+6+\sqrt{3}-4+\sqrt{3}\)
\(=2+\sqrt{3}\)
TM
25 tháng 4 2019
ĐKXĐ: x khác 1
\(M=\frac{x-\sqrt[3]{x}}{x-1}+\frac{1}{\sqrt[3]{x}-1}+\frac{1}{\sqrt[3]{x^2}+\sqrt[3]{x}+1}\)
\(=\frac{x-\sqrt[3]{x}}{x-1}+\frac{\sqrt[3]{x^2}+\sqrt[3]{x}+1+\sqrt[3]{x}-1}{\left(\sqrt[3]{x}-1\right)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}=\frac{x-\sqrt[3]{x}}{x-1}+\frac{\sqrt[3]{x^2}+2\sqrt[3]{x}}{\left(\sqrt[3]{x}\right)^3-1}\)
\(=\frac{x-\sqrt[3]{x}}{x-1}+\frac{\sqrt[3]{x^2}+2\sqrt[3]{x}}{x-1}=\frac{x+\sqrt[3]{x^2}+\sqrt[3]{x}}{x-1}=\frac{\sqrt[3]{x}\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}{\left(\sqrt[3]{x}-1\right)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}\)
\(=\frac{\sqrt[3]{x}}{\sqrt[3]{x}-1}\)
bạn nhớ kiểm tra lại nhé
\(\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\dfrac{6+2\sqrt{6}}{\sqrt{3}+\sqrt{2}}\\ =\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{4}\right)}{\sqrt{5}-\sqrt{4}}+\dfrac{\sqrt{2}.\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}\\ =\sqrt{3}+\sqrt{12}\\ =\sqrt{3}+\sqrt{2^2.3}\\ =\sqrt{3}+2\sqrt{3}\\ =3\sqrt{3}\)
\(\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\dfrac{6+2\sqrt{6}}{\sqrt{3}+\sqrt{2}}\)
\(=\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{4}\right)}{\sqrt{5}-2}+\dfrac{\sqrt{12}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}\\ =\sqrt{3}+\sqrt{12}\\ =\sqrt{3}+2\sqrt{3}=3\sqrt{3}\)