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\(\dfrac{\left(x+3\right)}{7}+\dfrac{\left(x+4\right)}{6}+\dfrac{\left(x+90\right)}{10}+3=0\\ \Leftrightarrow\dfrac{60\left(x+3\right)}{420}+\dfrac{70\left(x+4\right)}{420}+\dfrac{42\left(x+90\right)}{420}+\dfrac{1260}{420}=0\\ \Leftrightarrow60x+180+70x+280+42x+3780+1260=0\\ \Leftrightarrow172x=-5500\\ \Leftrightarrow x=-\dfrac{1375}{43}\)
\(\dfrac{x+23}{2021}+\dfrac{x+22}{2022}+\dfrac{x+21}{2023}+\dfrac{x+20}{2024}=-4\)
Vì \(\dfrac{x+23}{2021}+\dfrac{x+22}{2022}+\dfrac{x+21}{2023}+\dfrac{x+20}{2024}=-4\)
\(\Rightarrow\dfrac{x+23}{2021}+\dfrac{x+22}{2022}+\dfrac{x+21}{2023}+\dfrac{x+20}{2024}+4=0\)
\(\Rightarrow\left(\dfrac{x+23}{2021}+1\right)+\left(\dfrac{x+22}{2022}+1\right)+\left(\dfrac{x+21}{2023}+1\right)+\left(\dfrac{x+20}{2024}+1\right)=0\)
\(\Rightarrow\dfrac{x+2044}{2021}+\dfrac{x+2044}{2022}+\dfrac{x+2044}{2023}+\dfrac{x+2044}{2024}=0\)
\(\Rightarrow\left(x+2044\right)\left(\dfrac{1}{2021}+\dfrac{1}{2022}+\dfrac{1}{2023}+\dfrac{1}{2024}\right)=0\)
\(\Rightarrow x+2044=0\left(\dfrac{1}{2021}+\dfrac{1}{2022}+\dfrac{1}{2023}+\dfrac{1}{2024}\ne0\right)\)
\(\Rightarrow x=-2024\)
`@` `\text {Ans}`
`\downarrow`
`16,`
`@` Các cặp góc đồng vị:
`+`\(\widehat {M_4}\) và \(\widehat {N_4}\)
`+`\(\widehat {M_1}\) và \(\widehat {N_1}\)
`+`\(\widehat {M_2}\) và \(\widehat {N_2}\)
`+`\(\widehat {M_3}\) và \(\widehat {N_3}\)
`@` Các cặp góc sole trong:
`+`\(\widehat {M_3} \) và \(\widehat {N_1}\)
`+`\(\widehat {M_2}\) và \(\widehat {N_4}\)
`b,`
Ta có: \(\widehat {M_3} = \widehat {M_1} (\text {đối đỉnh})\)
`=>`\(\widehat {M_1}=50^0\)
\(\widehat {M_3}+\widehat {M_2}=180^0 (\text {kề bù})\)
`=>`\(50^0+\widehat {M_2}=180^0\)
`=>`\(\widehat {M_2}=180^0-50^0=130^0\)
\(\widehat {M_2}=\widehat {M_4} (\text {2 góc đối đỉnh})\)
`=>`\(\widehat {M_4} = 130^0\)
Vì \(\widehat {M_3}\) và \(\widehat {N_1}\) là `2` góc sole trong
`=>`\(\widehat {M_3}=\widehat {N_1}=50^0\)
\(\widehat {M_3}=\widehat {N_3}=50^0 (\text {2 góc đồng vị})\)
\(\widehat {M_2}=\widehat {N_2}=130^0 (\text {2 góc đồng vị})\)
\(\widehat {M_2}=\widehat {N_4}=130^0 (\text {2 góc slt})\)
`17,`
Vì \(\widehat {A_1}\) và \(\widehat {A_2}\) là `2` góc kề bù
`=>`\(\widehat {A_1}+\widehat {A_2}=180^0\)
\(3\widehat {A_1}=2\widehat {A_2}\) (gt)
`=>`\(\widehat{A_1}=\dfrac{2}{3}\cdot\widehat{A_2}\)
Thay \(\widehat{A_1}=\dfrac{2}{3}\widehat{A_2}\)
\(\dfrac{2}{3}\cdot\widehat{A_2}+\widehat{A_2}=180^0\)
`=>`\(\widehat{A_2}\left(\dfrac{2}{3}+1\right)=180^0\)
`=>`\(\widehat{A_2}\cdot\dfrac{5}{3}=180^0\)
`=>`\(\widehat{A_2}=180^0\div\dfrac{5}{3}\)
`=>`\(\widehat{A_2}=108^0\)
Vậy, số đo \(\widehat{A_2}=108^0\)
\(\widehat {A_1}+\widehat {A_2}=180^0 (\text {kề bù})\)
`=>`\(\widehat{A_1}+108^0=180^0\)
`=>`\(\widehat{A_1}=72^0\)
\(\widehat {A_1}=\widehat {A_3}=72^0 (\text {đối đỉnh})\)
\(\widehat {A_2}=\widehat {A_4}=108^0 (\text {đối đỉnh})\)
`@` Số đo các góc của đỉnh B:
`+`\(\widehat {A_4}=\widehat {B_4}=108^0 (\text {đồng vị})\)
`+`\(\widehat {A_2}=\widehat {B_2}=108^0 (\text {đồng vị})\)
`+`\(\widehat {A_3}=\widehat {B_1}=72^0 (\text {sole trong})\)
`+`\(\widehat {A_3}=\widehat {B_3}=72^0 (\text {đồng vị})\)
A = 1 + 2 + 22 + 23 + ...+ 2100
A\(\times\)2 = 2 + 22 + 23 +...+ 2100 + 2101
A \(\times\)2 - A = 2101 - 1
A = 2101 - 1
Đoạn 2x-1/3 là $\frac{2x-1}{3}$ hay $2x-\frac{1}{3}$ vậy bạn?
`@` `\text {Ans}`
`\downarrow`
\(\dfrac{8^{14}}{4^4\cdot64^5}\)
`=`\(\dfrac{2^{42}}{2^8\cdot2^{30}}\)
`=`\(\dfrac{2^{42}}{2^{38}}=2^4=16\)
\(\dfrac{9^{10}\cdot27^7}{81^7\cdot3^{15}}\)
`=`\(\dfrac{3^{20}\cdot3^{21}}{3^{28}\cdot3^{15}}\)
`=`\(\dfrac{3^{41}}{3^{43}}\)
`=`\(\dfrac{1}{3^2}=\dfrac{1}{9}\)
\(\left(\dfrac{3}{10}\right)^4\cdot\left(0,3\right)^5\cdot\left(\dfrac{10}{3}\right)^{10}\)
`=`\(\left(\dfrac{3}{10}\right)^4\cdot\left(\dfrac{3}{10}\right)^5\cdot\left(\dfrac{10}{3}\right)^{10}\)
`=`\(\left(\dfrac{3}{10}\right)^9\cdot\left(\dfrac{10}{3}\right)^{10}=\dfrac{10}{3}\)
\(\dfrac{\left(4^3\right)^2\cdot9^4}{6^7\cdot8^2}\)
`=`\(\dfrac{\left(2^6\right)^2\cdot3^8}{2^7\cdot3^7\cdot2^9}\)
`=`\(\dfrac{2^{12}\cdot3^8}{2^{16}\cdot3^7}\)
`=`\(\dfrac{3}{2^4}=\dfrac{3}{16}\)
\(\dfrac{4^8\cdot9^4}{6^6\cdot8^3}\)
`=`\(\dfrac{2^{16}\cdot3^8}{2^6\cdot3^6\cdot2^9}\)
`=`\(\dfrac{2^{16}\cdot3^8}{2^{15}\cdot3^6}=2\cdot3^2=18\)
\(3^6\cdot\left(\dfrac{1}{3}\right)^6\cdot81^2\cdot\dfrac{1}{27^2}\)
`=`\(\left(3\cdot\dfrac{1}{3}\right)^6\cdot\dfrac{81^2}{27^2}\)
`=`\(1^6\cdot\dfrac{3^8}{3^6}=1\cdot3^2=9\)
`\text {#KaizuulvG}`
`a)` Xét tử số phân số M :
\(2012-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{2012}{2020}\\ =\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...+\left(1-\dfrac{2012}{2020}\right)\\ =\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{2020}\\ =24\left(\dfrac{1}{27}+\dfrac{1}{30}+\dfrac{1}{33}+...+\dfrac{1}{6060}\right)\)
Ta được : \(M=\dfrac{24\left(\dfrac{1}{27}+\dfrac{1}{30}+\dfrac{1}{33}+...+\dfrac{1}{6060}\right)}{\dfrac{1}{27}+\dfrac{1}{30}+\dfrac{1}{33}+...+\dfrac{1}{6060}}=24\)
`b)` Xét tử số phân số N :
\(\dfrac{1}{1.300}+\dfrac{1}{2.301}+\dfrac{1}{3.302}+...+\dfrac{1}{101.400}\\ =\dfrac{1}{299}.\left(\dfrac{299}{1.300}+\dfrac{299}{2.301}+\dfrac{299}{3.302}+...+\dfrac{299}{101.400}\right)\\ =\dfrac{1}{299}.\left(1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+\dfrac{1}{3}-\dfrac{1}{302}+...+\dfrac{1}{101}-\dfrac{1}{400}\right)\)
Xét mẫu số phân số N :
\(\dfrac{1}{1.102}+\dfrac{1}{2.103}+\dfrac{1}{3.104}+...+\dfrac{1}{299.400}\\ =\dfrac{1}{101}.\left(\dfrac{101}{1.102}+\dfrac{101}{2.103}+\dfrac{101}{3.104}+...+\dfrac{101}{299.400}\right)\\ =\dfrac{1}{101}.\left(1-\dfrac{1}{102}+\dfrac{1}{2}-\dfrac{1}{103}+\dfrac{1}{3}-\dfrac{1}{104}+...+\dfrac{1}{299}-\dfrac{1}{400}\right)\)
\(=\dfrac{1}{101}.\left(1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+\dfrac{1}{3}-\dfrac{1}{302}+...+\dfrac{1}{101}-\dfrac{1}{400}\right)\)
Ta được: \(N=\dfrac{\dfrac{1}{299}\left(1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+\dfrac{1}{3}-\dfrac{1}{302}+...+\dfrac{1}{101}-\dfrac{1}{400}\right)}{\dfrac{1}{101}\left(1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+\dfrac{1}{3}-\dfrac{1}{302}+...+\dfrac{1}{101}-\dfrac{1}{400}\right)}\\ =\dfrac{\dfrac{1}{299}}{\dfrac{1}{101}}=\dfrac{101}{299}\)