Tính giá trị các biểu thức sau:
a) \({\left( { - 5} \right)^{ - 1}}\);
b) \({2^0}.{\left( {\frac{1}{2}} \right)^{ - 5}}\);
c) \({6^{ - 2}}.{\left( {\frac{1}{3}} \right)^{ - 3}}:{2^{ - 2}}\).
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H9
HT.Phong (9A5)
CTVHS
18 tháng 8 2023
a) \(\left(\dfrac{3}{4}\right)^{-2}\cdot3^2\cdot12^0=16\)
b) \(\left(\dfrac{1}{12}\right)^{-1}\cdot\left(\dfrac{2}{3}\right)^{-2}=27\)
c) \(\left(2^{-2}\cdot5^2\right)^{-2}:\left(5\cdot5^{-5}\right)=16\)
H
18 tháng 8 2023
a)
$16^{\alpha }+16^{-\alpha } = (4^2)^{\alpha }+(4^2)^{-\alpha } = 4^{2\alpha }+4^{-2\alpha }$
$4^{2\alpha }+4^{-2\alpha } = 4^{2\log_4{\frac{1}{5}}}+4^{-2\log_4{\frac{1}{5}}} = \left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^{-2} = \frac{1}{25}+25 = \frac{26}{25}$
H
18 tháng 8 2023
b)
$\left(2^{\alpha }+2^{-\alpha }\right)^2 = \left(\sqrt{4}\right)^{\alpha }+\left(\sqrt{4}\right)^{-\alpha } = 4^{\frac{\alpha}{2}}+4^{-\frac{\alpha}{2}}$
$4^{\frac{\alpha}{2}}+4^{-\frac{\alpha}{2}} = 4^{\frac{\log_4{\frac{1}{5}}}{2}}+4^{-\frac{\log_4{\frac{1}{5}}}{2}} = \left(\frac{1}{5}\right)^{\frac{1}{2}}+\left(\frac{1}{5}\right)^{-\frac{1}{2}} = \sqrt{\frac{1}{5}}+\frac{1}{\sqrt{5}} = \frac{2}{\sqrt{5}}$
HQ
Hà Quang Minh
Giáo viên
16 tháng 9 2023
a)
\(\begin{array}{l}\left( {0,25 - \frac{5}{6}} \right).1,6 + \frac{{ - 1}}{3}\\ =(\frac{25}{100}-\frac{5}{6}).\frac{16}{10}+\frac{-1}{3}\\= \left( {\frac{1}{4} - \frac{5}{6}} \right).\frac{8}{5} + \frac{{ - 1}}{3}\\ = \left( {\frac{6}{{24}} - \frac{{20}}{{24}}} \right).\frac{8}{5} + \frac{{ - 1}}{3}\\ = \frac{{ - 14}}{{24}}.\frac{8}{5} + \frac{{ - 1}}{3}\\ = \frac{{ - 14}}{{15}} + \frac{{ - 1}}{3}\\ = \frac{{ - 14}}{{15}} + \frac{{ - 5}}{{15}}\\ = \frac{{ - 19}}{{15}}\end{array}\)
b)
\(\begin{array}{l}3 - 2.\left[ {0,5 + \left( {0,25 - \frac{1}{6}} \right)} \right]\\ = 3 - 2.\left[ {\frac{1}{2} + \left( {\frac{1}{4} - \frac{1}{6}} \right)} \right]\\ = 3 - 2.\left( {\frac{1}{2} + \frac{1}{{12}}} \right)\\ =3-2.(\frac{6}{12}+\frac{1}{12})\\= 3 - 2.\frac{7}{{12}}\\ = 3 - \frac{7}{6}\\=\frac{18}{6}-\frac{7}{6}\\ = \frac{{11}}{6}\end{array}\)
4 tháng 7 2021
Bài 1:
a) Ta có: \(A=-1.7\cdot2.3+1.7\cdot\left(-3.7\right)-1.7\cdot3-0.17:0.1\)
\(=1.7\cdot\left(-2.3\right)+1.7\cdot\left(-3.7\right)+1.7\cdot\left(-3\right)+1.7\cdot\left(-1\right)\)
\(=1.7\cdot\left(-2.3-3.7-3-1\right)\)
\(=-10\cdot1.7=-17\)
b) Ta có: \(B=2\dfrac{3}{4}\cdot\left(-0.4\right)-1\dfrac{2}{3}\cdot2.75+\left(-1.2\right):\dfrac{4}{11}\)
\(=\dfrac{11}{4}\cdot\left(-0.4\right)-\dfrac{5}{3}\cdot\dfrac{11}{4}+\left(-1.2\right)\cdot\dfrac{11}{4}\)
\(=\dfrac{11}{4}\left(-0.4-\dfrac{5}{3}-1.2\right)\)
\(=-\dfrac{539}{60}\)
c) Ta có: \(C=\dfrac{\left(2^3\cdot5\cdot7\right)\cdot\left(5^2\cdot7^3\right)}{\left(2\cdot5\cdot7^2\right)^2}\)
\(=\dfrac{2^3\cdot5^3\cdot7^4}{2^2\cdot5^2\cdot7^4}\)
\(=10\)
H9
HT.Phong (9A5)
CTVHS
18 tháng 8 2023
a) \(25^{\dfrac{1}{2}}=5\)
b) \(\left(\dfrac{36}{49}\right)^{-\dfrac{1}{2}}=\dfrac{7}{6}\)
c) \(100^{1,5}=1000\)
7 tháng 3 2022
a: \(A=49+\dfrac{8}{23}-14-\dfrac{8}{23}-5-\dfrac{7}{32}=30-\dfrac{7}{32}=\dfrac{953}{32}\)
b:
Sửa đề: \(B=71\dfrac{38}{45}-\left(43\dfrac{8}{45}-1\dfrac{17}{51}\right)\)
\(B=71+\dfrac{38}{45}-43-\dfrac{8}{45}+1+\dfrac{17}{51}\)
\(=71-43+1+1\)
=28+2=30
H9
HT.Phong (9A5)
CTVHS
18 tháng 8 2023
a) \(\sqrt[4]{\dfrac{1}{16}}=\dfrac{1}{2}\)
b) \(\left(\sqrt[6]{8}\right)^2=\sqrt[\dfrac{6}{2}]{8}=\sqrt[3]{8}=2\)
c) \(\sqrt[4]{3}\cdot\sqrt[4]{27}=\sqrt[4]{3\cdot27}=\sqrt[4]{81}=3\)
1 tháng 7 2021
a, đặt \(\sqrt{2-\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)\)
\(=\sqrt{2-\sqrt{3}}.\sqrt{2}.\left(\sqrt{3}+1\right)\)
\(=\sqrt{4-2\sqrt{3}}\left(\sqrt{3}+1\right)\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)=3-1=2\)
\(b,\)
\(\left(\sqrt{21}+7\right)\sqrt{10-2\sqrt{21}}=\left[\sqrt{7}\left(\sqrt{7}+\sqrt{3}\right)\right].\sqrt{10-2\sqrt{21}}\)
\(=\sqrt{7}\left(\sqrt{7}+\sqrt{3}\right)\sqrt{\left(\sqrt{7}\right)^2-2\sqrt{7.3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{7}\left(\sqrt{7}+\sqrt{3}\right)\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)
\(=\sqrt{7}\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)\)
\(=\sqrt{7}\left(7-3\right)=4\sqrt{7}\)
1 tháng 7 2021
a) Ta có: \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}+\sqrt{2}\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
=3-1=2
b) Ta có: \(\left(\sqrt{21}+7\right)\cdot\sqrt{10-2\sqrt{21}}\)
\(=\sqrt{7}\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)\)
\(=4\sqrt{7}\)
HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
Ta có:
a) \(\sin \left( {\alpha + \frac{\pi }{6}} \right) = \sin \alpha \cos \frac{\pi }{6} + \cos \alpha \sin \frac{\pi }{6} = \frac{{\sqrt 6 }}{3}.\frac{{\sqrt 3 }}{2} + \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{1}{2} = \frac{{ - \sqrt 3 + 3\sqrt 2 }}{6}\)
b) \(\cos \left( {\alpha + \frac{\pi }{6}} \right) = \cos \alpha .\cos \frac{\pi }{6} - \sin \alpha \sin \frac{\pi }{6} = \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{{\sqrt 3 }}{2} - \frac{{\sqrt 6 }}{3}.\frac{1}{2} = - \frac{{3 + \sqrt 6 }}{6}\)
c) \(\sin \left( {\alpha - \frac{\pi }{3}} \right) = \sin \alpha \cos \frac{\pi }{3} - \cos \alpha \sin \frac{\pi }{3} = \frac{{\sqrt 6 }}{3}.\frac{1}{2} - \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{{\sqrt 3 }}{2} = \frac{{3 + \sqrt 6 }}{6}\)
d) \(\cos \left( {\alpha - \frac{\pi }{6}} \right) = \cos \alpha \cos \frac{\pi }{6} + \sin \alpha \sin \frac{\pi }{6} = \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{{\sqrt 3 }}{2} + \frac{{\sqrt 6 }}{3}.\frac{1}{2} = \frac{{ - 3 + \sqrt 6 }}{6}\)
a) \(\left(-5\right)^{-1}=-\dfrac{1}{5}\)
b) \(2^0\cdot\left(\dfrac{1}{2}\right)^{-5}=1\cdot32=32\)
c) \(6^{-2}\cdot\left(\dfrac{1}{3}\right)^{-3}:2^{-2}\)
\(=\dfrac{1}{36}\cdot27:\dfrac{1}{4}\)
\(=\dfrac{27\cdot4}{36}=3\)