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\(\)a//b
\(c\perp a\)
\(\Rightarrow c\perp b\)
\(\widehat{A1}=\widehat{A3}\) (đối đỉnh)
\(\widehat{A3}=\widehat{B3}\) (đồng vị)
\(\widehat{B3}+\widehat{B2}=180^o\) (kề bù)
\(\Rightarrow\widehat{B2}=180^o-115^o=65^o\)
y : 0,25 + y : 0,125 - y : 0,5 = 2,63
y : ( 0,25 + 0,125 - 0,5 ) = 2,63
y : ( 0,375 - 0,5 ) = 2,63
y : ( -0,125) = 2,63
y = 2,63 x ( -0,125 )
y = -0,32875
y:0,25+y:0,125-y:0,5=2,63
y:1/4+y:1/8-y:1/2=2,63
y x 4 +y x 8 - y x 2=2,63
y x (4+8-2)=2,63
y x 10=2,63
y=2,63 : 10
y=0,263
b: \(=\dfrac{2^{10}\cdot2^2\cdot13+2^{12}\cdot5\cdot13}{2^{11}\cdot2^2\cdot13}+\dfrac{3^{10}\cdot11+3^{10}\cdot5}{3^9\cdot2^4}\)
\(=\dfrac{2^{12}\cdot13\left(1+5\right)}{2^{13}\cdot13}+\dfrac{3^{10}\left(11+5\right)}{3^9\cdot2^4}\)
\(=\dfrac{2^{13}\cdot3}{2^{13}}+\dfrac{3^{10}\cdot16}{3^9\cdot2^4}=3+3=6\)
2: \(C=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2023^2}\)
=>\(C< \dfrac{1}{2^2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\)
=>\(C< \dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{2023}=\dfrac{6065}{8092}< \dfrac{6069}{8092}=\dfrac{3}{4}\)