tính: \(\dfrac{\sqrt{18}}{\sqrt{2}}\cdot\sqrt{2}\)
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Tính:
\(\left(\dfrac{3\sqrt{3}-2\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\dfrac{3\sqrt{2}+2\sqrt{3}}{\sqrt{3}+\sqrt{2}}\right)\cdot\dfrac{5-2\sqrt{6}}{4}\)
Ta có: \(\left(\dfrac{3\sqrt{3}-2\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\dfrac{3\sqrt{2}+2\sqrt{3}}{\sqrt{3}+\sqrt{2}}\right)\cdot\dfrac{5-2\sqrt{6}}{4}\)
\(=\left(\dfrac{\left(\sqrt{3}-\sqrt{2}\right)\left(3+\sqrt{6}+2\right)}{\sqrt{3}-\sqrt{2}}+\dfrac{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}\right)\cdot\dfrac{5-2\sqrt{6}}{4}\)
\(=\left(5+\sqrt{6}+\sqrt{6}\right)\cdot\dfrac{5-2\sqrt{6}}{4}\)
\(=\dfrac{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)}{4}\)
\(=\dfrac{25-24}{4}=\dfrac{1}{4}\)
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\(\sqrt{2a}-\sqrt{18^3}+4\sqrt{\dfrac{a}{2}}\)
\(\sqrt{\dfrac{a}{1+2b+b^2}}\cdot\sqrt{\dfrac{4a+8ab+4ab^2}{225}}\)
\(\sqrt{2a}-\sqrt{18^3}+4\sqrt{\dfrac{a}{2}}=\sqrt{2}.\sqrt{a}-54\sqrt{2}+2\sqrt{2}.\sqrt{a}=3\sqrt{2}.\sqrt{a}-54\sqrt{2}\)
\(\sqrt{\dfrac{a}{1+2b+b^2}}.\sqrt{\dfrac{4a+8ab+4ab^2}{225}}=\sqrt{\dfrac{a}{\left(b+1\right)^2}}.\sqrt{\dfrac{4a\left(1+2b+b^2\right)}{225}}=\dfrac{\sqrt{a}}{\left|b+1\right|}.\dfrac{\sqrt{4a\left(b+1\right)^2}}{15}=\dfrac{\sqrt{a}}{\left|b+1\right|}.\dfrac{2\sqrt{a}\left|b+1\right|}{15}=\dfrac{2a}{15}\)
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Tính:
a)\(\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}}\cdot\sqrt{18}-\sqrt{128}}}\)
b)\(\sqrt{6+2\cdot\sqrt{5-\sqrt{13+4\sqrt{3}}}}\)
thực hiện phép tính
A=\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)
B=\(\sqrt{\dfrac{3-\sqrt{5}}{\sqrt{10}+\sqrt{2}}}\cdot\left(3+\sqrt{5}\right)\)
a) \(A=\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)
\(=\dfrac{2-\sqrt{3}}{1}+\dfrac{2+\sqrt{3}}{1}\)
=4
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\(\dfrac{\sqrt{8-4\sqrt{3}}}{\sqrt{2}}=\dfrac{\sqrt{4\cdot2-4\sqrt{3}}}{\sqrt{2}}=\dfrac{\sqrt{4}\cdot\sqrt{2-\sqrt{3}}}{\sqrt{2}}=\sqrt{2}\cdot\sqrt{2-\sqrt{3}}\)
15 tháng 12 2021 lúc 21:58
Thực hiện phép tính (tính nhanh nếu có thể):
4) \(4\cdot\left(\dfrac{-1}{2}\right)^3+\left|-1\dfrac{1}{2}+\sqrt{\dfrac{9}{4}}\right|:\sqrt{25}\)
5) \(\left[6-3\cdot\left(\dfrac{-1}{3}\right)^2+\sqrt{\dfrac{1}{4}}\right]:\sqrt{0,\left(9\right)}\)
Bài 58 (trang 32 SGK Toán 9 Tập 1)
Rút gọn các biểu thức sau:
a) $5 \sqrt{\dfrac{1}{5}}+\dfrac{1}{2} \sqrt{20}+\sqrt{5}$ ; b) $\sqrt{\dfrac{1}{2}}+\sqrt{4,5}+\sqrt{12,5}$ ;
c) $\sqrt{20}-\sqrt{45}+3 \sqrt{18}+\sqrt{72}$ ; d) $0,1 . \sqrt{200}+2 \cdot \sqrt{0,08}+0,4 \cdot \sqrt{50}$.
TRẢ LỜI :
\(=\sqrt{5}+\sqrt{5}+\sqrt{5}=3\sqrt{5}\)
c) √20 - √45 + 3√18 + √72
= √4.5 - √9.5 + 3√9.2 + √36.2
= 2√5 - 3√5 + 9√2 + 6√2
= -√5 + 15√2
a) 3√5 b) 9√2 / 2
c) -√5 + 15√2 d)
3,4√2
a) .
b) hay .
c) .
d) .
Xem thêm câu trả lời
tính: \(\sqrt{1-\dfrac{1}{2^2}}\cdot\sqrt{1-\dfrac{1}{3^2}}\cdot\cdot\cdot\sqrt{1-\dfrac{1}{2006^2}}\)
\(P^2=\dfrac{\left(2^2-1\right)\left(3^2-1\right)\left(4^2-1\right)....\left(2005^2-1\right)\left(2006^2-1\right)}{2^2.3^2...2006^2}\\ \)
\(P^2=\dfrac{1.3.2.4.3.5...2006.2007}{2^2.3^2....2006^2}=\dfrac{1.2007}{2.2006}\)
\(P=\dfrac{1}{4}\sqrt{\dfrac{2007}{1003}}\)
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Thực hiện phép tính:
a) \(\left(\sqrt{ab}+2\sqrt{\dfrac{b}{a}}-\sqrt{\dfrac{a}{b}+\sqrt{\dfrac{1}{ab}}}\right)\cdot\sqrt{ab}\)
b) \(\left(\dfrac{am}{b}\sqrt{\dfrac{n}{m}}-\dfrac{ab}{n}\sqrt{mn}+\dfrac{a^2}{b^2}\sqrt{\dfrac{m}{n}}\right)\cdot a^2b^2\cdot\sqrt{\dfrac{n}{m}}\)
a: \(=ab+2\cdot\sqrt{\dfrac{b}{a}\cdot ab}-\sqrt{ab\cdot\left(\dfrac{a}{b}+\dfrac{1}{\sqrt{ab}}\right)}\)
\(=ab+2b-\sqrt{ab\cdot\dfrac{a\sqrt{a}+\sqrt{b}}{b\sqrt{a}}}\)
\(=ab+2b-\sqrt{\sqrt{a}\cdot\left(a\sqrt{a}+\sqrt{b}\right)}\)
b: \(=\left(\sqrt{\dfrac{a^2m^2\cdot n}{b^2\cdot m}}-\sqrt{mn\cdot\dfrac{a^2b^2}{n^2}}+\sqrt{\dfrac{a^4}{b^4}\cdot\dfrac{m}{n}}\right)\cdot a^2b^2\cdot\sqrt{\dfrac{n}{m}}\)
\(=\left(\dfrac{a\sqrt{mn}}{b}-\sqrt{a^2b^2\cdot\dfrac{m}{n}}+\dfrac{a^2}{b^2}\cdot\sqrt{\dfrac{m}{n}}\right)\cdot\sqrt{\dfrac{n}{m}}\cdot a^2b^2\)
\(=\left(\dfrac{an}{b}-ab+\dfrac{a^2}{b^2}\right)\cdot a^2b^2\)
\(=a^3nb-a^3b^3+a^4\)
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