2. Tính
3-(\(-\dfrac{6}{7}\))\(^0\) + (\(\dfrac{1}{2}\))\(^2\) : 2
Giúp mk vs
Những câu hỏi liên quan
1) thực hiện phép tính
d)\(\dfrac{4}{\sqrt{7}-\sqrt{3}}+\dfrac{6}{3+\sqrt{3}}+\dfrac{\sqrt{7}-7}{\sqrt{7}-1}\)
e) \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
giúp mk vs ạ mk cần gấp
B=\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{7}{6}}{2\dfrac{2}{5}}\)
Tính B.
Mn giúp tớ vs
\(B=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{7}{6}}{2\dfrac{2}{5}}\)
\(B=\dfrac{\dfrac{3}{6}+\dfrac{2}{6}+\dfrac{7}{6}}{\dfrac{12}{5}}\)
\(B=\dfrac{\dfrac{12}{6}}{\dfrac{12}{5}}\)
\(B=\dfrac{2}{\dfrac{12}{5}}\)
\(B=\dfrac{2\cdot5}{12}\)
\(B=\dfrac{10}{12}=\dfrac{5}{6}\)
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(1) tính
a) \(\sqrt{3}+2\sqrt{12}+4\sqrt{75}-\sqrt{300}\)
b) \(2\sqrt{20}+\sqrt{\left(1-\sqrt{5}\right)^2}-\dfrac{20}{\sqrt{5}+1}\)
c) \(\dfrac{6}{\sqrt{13}-1}+\dfrac{6}{\sqrt{13}+1}\)
d) \(\sin^238^0+\cot23^0+\sin^252^0-\tan67^0\)
giúp mk vs ạ mai mk hc rồi
\(a,=\sqrt{3}+4\sqrt{3}+20\sqrt{3}-10\sqrt{3}=15\sqrt{3}\\ b,=4\sqrt{5}+\sqrt{5}-1-\dfrac{20\left(\sqrt{5}-1\right)}{4}\\ =5\sqrt{5}-1-5\sqrt{5}+5=4\\ c,=\dfrac{6\sqrt{13}+6+6\sqrt{13}-6}{\left(\sqrt{13}-1\right)\left(\sqrt{13}+1\right)}=\dfrac{12\sqrt{13}}{12}=\sqrt{13}\\ d,=\left(\sin^238^0+\cos^238^0\right)+\left(\tan67^0-\tan67^0\right)=1+0=1\)
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(1) tính
a) \(\sqrt{3}+2\sqrt{12}+4\sqrt{75}-\sqrt{300}\)
b) \(2\sqrt{20}+\sqrt{\left(1-\sqrt{5}\right)^2}-\dfrac{20}{\sqrt{5}+1}\)
c) \(\dfrac{6}{\sqrt{13}-1}+\dfrac{6}{\sqrt{13}+1}\)
d) \(\sin^238^0+\cot23^0+\sin^252^0-\tan67^0\)
giúp mk vs ạ mai mk hc rồi
a: \(=\sqrt{3}+4\sqrt{3}+4\cdot5\sqrt{3}-10\sqrt{3}\)
\(=15\sqrt{3}\)
b: \(=2\cdot2\sqrt{5}+\sqrt{5}-1-5+5\sqrt{5}\)
=-6
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1) thực hiện phép tính:
a) \(\dfrac{4}{\sqrt{7}-\sqrt{3}}+\dfrac{6}{3+\sqrt{3}}+\dfrac{\sqrt{7}-7}{\sqrt{7}-1}\)
b) \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
giúp mk vs ah mk đang cần gấp lắm
a: Ta có: \(\dfrac{4}{\sqrt{7}-\sqrt{3}}+\dfrac{6}{3+\sqrt{3}}+\dfrac{\sqrt{7}-7}{\sqrt{7}-1}\)
\(=\sqrt{7}+\sqrt{3}+3-\sqrt{3}-\sqrt{7}\)
=3
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tìm x biết
\(\dfrac{7}{9}:\left(2+\dfrac{3}{4}.x\right)+\dfrac{5}{9}=\dfrac{23}{27}\)
|x|\(-\dfrac{3}{4}=\dfrac{5}{3}\)
\(\left|2.x-\dfrac{1}{3}\right|+\dfrac{5}{6}=1\)
giúp mk vs nhanh lên mình đang bận
đấy nháb) Ta có: \(\left|x\right|-\dfrac{3}{4}=\dfrac{5}{3}\)
\(\Leftrightarrow\left|x\right|=\dfrac{5}{3}+\dfrac{3}{4}=\dfrac{20}{12}+\dfrac{9}{12}=\dfrac{29}{12}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{12}\\x=-\dfrac{29}{12}\end{matrix}\right.\)
c) Ta có: \(\left|2x-\dfrac{1}{3}\right|+\dfrac{5}{6}=1\)
\(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{1}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{1}{3}=\dfrac{1}{6}\\2x-\dfrac{1}{3}=\dfrac{-1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{1}{6}+\dfrac{1}{3}=\dfrac{1}{2}\\2x=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{1}{12}\end{matrix}\right.\)
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tính
\(\dfrac{2}{\sqrt{3}+1}-\dfrac{1}{\sqrt{3}-2}+\dfrac{12}{\sqrt{3}+3}\)
giúp mk vs ạ mk cần gấp
BT=\(\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}+\dfrac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+\dfrac{12\left(3-\sqrt{3}\right)}{\left(\sqrt{3}+3\right)\left(3-\sqrt{3}\right)}\)
\(=\dfrac{2\left(\sqrt{3}-1\right)}{2}+\dfrac{2+\sqrt{3}}{4-3}+\dfrac{12\left(3-\sqrt{3}\right)}{9-3}\)
\(=\sqrt{3}-1+2+\sqrt{3}+2\left(3-\sqrt{3}\right)\)
\(=\sqrt{3}-1+2+\sqrt{3}+6-2\sqrt{3}=7\)
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1) rút gọn
A= \(3\sqrt{2}+5\sqrt{8}-2\sqrt{50}\)
B= \(\dfrac{1}{3+\sqrt{5}}+\dfrac{1}{3-\sqrt{5}}\)
C= \(\sqrt{7-4\sqrt{3}}+\sqrt{12+6\sqrt{3}}\)
Giúp mk vs ạ mk cần gấp
\(A=3\sqrt{2}+5\sqrt{8}-2\sqrt{50}\)
\(=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}\)
\(=3\sqrt{2}\)
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\(B=\dfrac{1}{3+\sqrt{5}}+\dfrac{1}{3-\sqrt{5}}\)
\(=\dfrac{3-\sqrt{5}}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}+\dfrac{3+\sqrt{5}}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}\)
\(=\dfrac{3-\sqrt{5}+3+\sqrt{5}}{9-5}\)
\(=\dfrac{3}{2}\)
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\(A=3\sqrt{2}+5\sqrt{8}-2\sqrt{50}\)
\(A=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}=3\sqrt{2}\)
\(B=\dfrac{1}{3+\sqrt{5}}+\dfrac{1}{3-\sqrt{5}}\)
\(B=\dfrac{3-\sqrt{5}+3+\sqrt{5}}{9-5}=\dfrac{6}{4}=\dfrac{3}{2}\)
\(C=\sqrt{7-4\sqrt{3}}+\sqrt{12+6\sqrt{3}}\)
\(C=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(3+\sqrt{3}\right)^2}\)
\(C=2-\sqrt{3}+3+\sqrt{3}=5\)
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Xem thêm câu trả lời
1) sắp xếp theo thứ tự tăng dần
\(23;2\sqrt{7};5\sqrt{6};-8\sqrt{2};-\sqrt{127}\)
2) sắp xếp theo thứ tự giảm dần
\(6\sqrt{\dfrac{1}{4}};4\sqrt{\dfrac{1}{2}};-\sqrt{132};2\sqrt{3};\sqrt{\dfrac{15}{5}}\)
giúp mk vs ah
tính:
a) \(\left(2^{-1}+3^{-1}\right):\left(2^{-1}-3^{-1}\right)+\left(2^{-1}.2^0\right):2^3\)
b) \(\left(-\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)
c) \(\left[\left(0,1\right)^2\right]^0+\left[\left(\dfrac{1}{7}\right)^{-1}\right]^2.\dfrac{1}{49}.\left[\left(2^2\right)^3:2^5\right]\)
giúp mk vs các bn mk dg cần gấp lm
