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Tính giá trị biểu thức M= sin²42⁰+sin²43⁰+sin²44⁰+sin²45⁰+sin²46⁰+sin²47⁰+sin²48⁰ Giúp em với mn ;-;
![](https://rs.olm.vn/images/avt/0.png?1311)
Các góc dưới đây đều là độ:
Sử dụng đẳng thức \(sina=cos\left(90^0-a\right)\) và \(sin^2a+cos^2a=1\):
\(M=sin^242+sin^243+sin^244+sin^245+cos^2\left(90-46\right)+cos^2\left(90-47\right)+cos^2\left(90-48\right)\)
\(=sin^242+sin^243+sin^244+sin^245+cos^244+cos^243+cos^242\)
\(=\left(sin^242+cos^242\right)+\left(sin^243+cos^243\right)+\left(sin^244+cos^244\right)+\left(\dfrac{\sqrt{2}}{2}\right)^2\)
\(=1+1+1+\dfrac{1}{2}=\dfrac{7}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(D=\left(4\sqrt{2}-\dfrac{4}{3}\sqrt{10}+\dfrac{9}{7}\sqrt{2}\right)\cdot\dfrac{\sqrt{2}}{2}\)
\(=\dfrac{37}{7}-\dfrac{4}{3}\sqrt{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\left(\sqrt{9.5}-\sqrt{4.5}+\sqrt{5}\right):\sqrt{6}\\ =\left(3\sqrt{5}-2\sqrt{5}+\sqrt{5}\right):\sqrt{6}\\ =\dfrac{2\sqrt{5}}{\sqrt{6}}\\ =\dfrac{\sqrt{2}.\sqrt{2}.\sqrt{5}}{\sqrt{2}.\sqrt{3}}\\ =\dfrac{\sqrt{10}}{\sqrt{3}}=\dfrac{\sqrt{30}}{3}\)
\(\left(\sqrt{45}-\sqrt{20}+\sqrt{5}\right):\sqrt{6}\\ =\left(\sqrt{9.5}-\sqrt{4.5}+\sqrt{5}\right):\sqrt{6}\\ =\left(3\sqrt{5}-2\sqrt{5}+\sqrt{5}\right):\sqrt{6}\\ =2\sqrt{5}:\sqrt{6}\\ =2\sqrt{5}.\dfrac{1}{\sqrt{6}}\\ =\dfrac{2\sqrt{5}}{\sqrt{6}}\\ =\dfrac{2\sqrt{5}.\sqrt{6}}{6}\\ =\dfrac{1}{3}.\sqrt{30}\\ =\dfrac{\sqrt{30}}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1+(2+58)+(45+345)+(45+345)+(373+43+4)+(574+84+4)
=1+60+390+390+420+662
=1923
NẾU ĐÚNG THÌ K NHA
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2\sqrt{18}-4\sqrt{27}+3\sqrt{43}-6\sqrt{32}+5\sqrt{48}-3\sqrt{20}=-5.343529954\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)
\(=6\sqrt{5}-3\sqrt{5}+6\sqrt{2}+6\sqrt{2}\)
\(=3\sqrt{5}+12\sqrt{2}\)
b) Ta có: \(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)
\(=\dfrac{12\left(3+\sqrt{5}\right)}{4}-\dfrac{16\left(\sqrt{5}-1\right)}{4}\)
\(=3\left(3+\sqrt{5}\right)-4\left(\sqrt{5}-1\right)\)
\(=9+3\sqrt{5}-4\sqrt{5}+4\)
\(=13-\sqrt{5}\)
c) Ta có: \(C=10\sqrt{\dfrac{1}{5}}+\dfrac{1}{5}\sqrt{125}-2\sqrt{20}\)
\(=\dfrac{10}{\sqrt{5}}+\dfrac{1}{5}\cdot5\sqrt{5}-2\cdot2\sqrt{5}\)
\(=2\sqrt{5}+\sqrt{5}-4\sqrt{5}\)
\(=-\sqrt{5}\)
e) Ta có: \(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\sqrt{3}+1-2+\sqrt{3}\)
\(=2\sqrt{3}-1\)
f) Ta có: \(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{5}+1-\sqrt{5}+2\)
=3
![](https://rs.olm.vn/images/avt/0.png?1311)
e) Ta có: \(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\sqrt{3}+1-2+\sqrt{3}\)
\(=2\sqrt{3}-1\)
f) Ta có: \(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{5}+1-\sqrt{5}+2\)
=3
a) Ta có: \(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)
\(=6\sqrt{5}-3\sqrt{5}+6\sqrt{2}+6\sqrt{2}\)
\(=3\sqrt{5}+12\sqrt{2}\)
b) Ta có: \(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)
\(=\dfrac{12\left(3+\sqrt{5}\right)}{4}-\dfrac{16\left(\sqrt{5}-1\right)}{4}\)
\(=3\left(3+\sqrt{5}\right)-4\left(\sqrt{5}-1\right)\)
\(=9+3\sqrt{5}-4\sqrt{5}+4\)
\(=13-\sqrt{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Vì 75 ° + 15 ° = 90 ° nên sin 75 ° = cos 15 °
Vì 53 ° + 37 ° = 90 ° nên cos 53 ° = sin 37 °
Vì 47 ° 20 ' + 42 ° 40 ' = 90 ° nên sin 47 ° 20 ' = cos 42 ° 40 '
Vì 62 ° + 28 ° = 90 ° nên tg 62 ° = cotg 28 °
Vì 82 ° 45 ' + 7 ° 15 ' = 90 ° nên cotg 82 ° 45 ' = tg 7 ° 15 '
![](https://rs.olm.vn/images/avt/0.png?1311)
\(ĐK:x\ge5\\ \Leftrightarrow\sqrt{x-5}+\dfrac{1}{3}\cdot3\sqrt{x-5}=\dfrac{1}{5}\sqrt{25x-119}\\ \Leftrightarrow2\sqrt{x-5}=\dfrac{1}{5}\sqrt{25x-119}\\ \Leftrightarrow4\left(x-5\right)=\dfrac{1}{25}\left(25x-119\right)\\ \Leftrightarrow4x-20=x-\dfrac{119}{25}\\ \Leftrightarrow3x=\dfrac{381}{25}\Leftrightarrow x=\dfrac{127}{25}\)
Ta có: \(\sqrt{x-5}+\dfrac{1}{3}\sqrt{9x-45}=\dfrac{1}{5}\sqrt{25x-125}+6\)
\(\Leftrightarrow x-5=36\)
hay x=41
45,6=2352,534231
82,1=78,79324245
43,5=(22)3,5=27=128