vẽ góc vuông xay vẽ góc xay' đối đỉnh với góc xay hãy tính số đo các góc còn lại trong hình
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\(\left(2x-3\right)\left(x-\dfrac{1}{2}\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-3=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=3\\x=\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\) (Thêm KL cuối dòng: Vậy \(x\in\left\{\dfrac{3}{2};\dfrac{1}{2}\right\}\))
\(a,\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{1999\cdot2000}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{1999}-\dfrac{1}{2000}\\ =1-\dfrac{1}{2000}\\ =\dfrac{1999}{2000}\\ b,\dfrac{1}{1\cdot4}+\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+...+\dfrac{1}{100\cdot103}\\ =\dfrac{1}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{100\cdot103}\right)\\ =\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\\ =\dfrac{1}{3}\cdot\left(1-\dfrac{1}{103}\right)\\ =\dfrac{102}{309}\)
\(c,\dfrac{8}{9}-\dfrac{1}{2}-\dfrac{1}{6}-...-\dfrac{1}{72}\\ =\dfrac{8}{9}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{8\cdot9}\right)\\ =\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\right)\\ =\dfrac{8}{9}-\left(1-\dfrac{1}{9}\right)\\ =\dfrac{8}{9}-\dfrac{8}{9}\\ =0\)
\(\left(5-x\right)\left(7-x\right)< 0\)
\(5-x=0\Rightarrow x=5\)
\(7-x=0\Rightarrow x=7\)
Lập bảng xét dấu
\(x\) | 5 7 |
\(5-x\) | + 0 - \(|\) - |
\(7-x\) | + \(|\) + 0 - |
\(\left(5-x\right)\left(7-x\right)\) | + 0 - 0 + |
\(\Rightarrow5< x< 7\)
\(A=\dfrac{2}{4.7}-\dfrac{3}{5.9}+\dfrac{2}{7.10}-\dfrac{3}{9.13}+...+\dfrac{2}{301.304}-\dfrac{3}{401.405}\)
\(A=\dfrac{2}{4.7}+\dfrac{2}{7.10}+\dfrac{2}{301.304}...-\left(\dfrac{3}{5.9}+\dfrac{3}{9.13}+...+\dfrac{3}{401.405}\right)\)
\(A=2\left(\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{301.304}\right)...-3\left(\dfrac{1}{5.9}+\dfrac{1}{9.13}+...+\dfrac{1}{401.405}\right)\)
\(A=2\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{301}-\dfrac{1}{304}\right)...-3\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{401}-\dfrac{1}{405}\right)\)
\(A=2\left(\dfrac{1}{4}-\dfrac{1}{304}\right)-3\left(\dfrac{1}{5}-\dfrac{1}{405}\right)\)
\(A=2\left(\dfrac{76}{304}-\dfrac{1}{304}\right)-3\left(\dfrac{81}{5}-\dfrac{1}{405}\right)\)
\(A=2.\dfrac{75}{304}-3.\dfrac{80}{405}=\dfrac{75}{152}-\dfrac{80}{135}=\dfrac{10125-12160}{152.135}=-\dfrac{2035}{152.135}=-\dfrac{407}{4104}\)
\(125.5^2.\dfrac{1}{625}.5^3=5^3.5^2.\dfrac{1}{5^4}.5^3=5^{3+2-4+3}=5^4\\ 8.32.\left(2^4.\dfrac{1}{32}\right)=2^3.2^5.2^4.\dfrac{1}{2^5}=2^{3+5+4-5}=2^7\\ 6^3.5^2.\left(\dfrac{5}{6}\right)^3=6^3.5^2.5^3:6^3=5^{2+3}.6^{3-3}=5^5.6^0=5^5.1=5^5\\ Bài.5A\)
\(Bài.5B\\ a,2401.\left(\dfrac{1}{7}\right)^2.\dfrac{1}{7}.49^2=7^4.\left(\dfrac{1}{7}\right)^3.\left(7^2\right)^2=7^4.\dfrac{1}{7^3}.7^4=7^{4-3+4}=7^5\\ b,9.81:\left(3^5.\dfrac{1}{27}\right)=3^2.3^4:\left(3^5.\dfrac{1}{3^3}\right)=3^{2+4}:\left(3^{5-3}\right)=3^6:3^2=3^{6-2}=3^4\\ c,3^4.7^2.\left(\dfrac{7}{3}\right)^4=3^4.7^2.7^4:3^4=\left(3^4:3^4\right).\left(7^2.7^4\right)=1.7^6=7^6\)
a) Số đo \(\widehat{xAy}\) là: 90o vì có kí hiệu vuông góc.
b) Số đo \(\widehat{x'Ay}\):
Vì \(\widehat{x'Ay}\) và \(\widehat{xAy}\) là hai góc kề bù nên
nên \(\widehat{x'Ax}\) = \(\widehat{x'Ay}\) + \(\widehat{xAy}\)
180o = \(\widehat{x'Ay}\) + 90o
\(\widehat{x'Ay}\) = 180o - 90o
\(\widehat{x'Ay}\) = 90o
c) Số đo \(\widehat{x'Ay'}\):
Vì \(\widehat{xAy}\) và \(\widehat{x'Ay'}\) là hai góc đối đỉnh
nên: \(\widehat{x'Ay'}\) = \(\widehat{xAy}\) = 90o
d) Số đo \(\widehat{xAy'}\):
Vì \(\widehat{xAy'}\) và \(\widehat{x'Ay}\) là hai góc đối đỉnh
nên \(\widehat{xAy'}\) = \(\widehat{x'Ay}\) = 90o
(2^10*3^10.2^10*3^9):(2^9*3^10)