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\(5^{11}và7^7=\left(5.7\right)^4và\left(7.5\right)^2=35^4>35^2\)
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13:
Qua G, kẻ mn//a//b
mn//a
=>góc G1=góc A1(hai góc so le trong)
=>góc G1=42 độ
mn//b
=>góc G2+góc B2=180 độ(trong cùng phía)
=>góc G2=180-138=42 độ
=>góc AGB=42+42=84 độ
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`#040911`
`b)`
\(A=\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{19\times21}\)
`=`\(\dfrac{1}{2}\times\left(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+...+\dfrac{2}{19\times21}\right)\)
`=`\(\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)\)
`=`\(\dfrac{1}{2}\times\left(1-\dfrac{1}{21}\right)\)
\(=\dfrac{1}{2}\times\dfrac{20}{21}\\ =\dfrac{10}{21}\\ \text{ Vậy, A = }\dfrac{10}{21}\)
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2:
a: góc BEC=góc BDC=90 độ
=>BEDC nội tiếp
b: Xét ΔADB vuông tại D và ΔAEC vuông tại E có
góc DAB chung
=>ΔADB đồng dạng với ΔAEC
=>AD/AE=AB/AC
=>AD*AC=AB*AE
c: góc ABF=góc ACF=1/2*sđ cung AF=90 độ
=>BF//CH và CF//BH
=>BFCH là hình bình hành
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= 28/15 . 3/4 - ( 11/20 + 1/4 ) : 7/3
= 28/15 . 3/4 - 4/5 : 7/3
= 7/5 - 12/35
= 37/35
= \(\dfrac{28}{15}\) . \(\dfrac{3}{4}\) - (\(\dfrac{11}{20}\) + \(\dfrac{1}{4}\)) : \(\dfrac{7}{3}\)
= \(\dfrac{7}{5}\) - (\(\dfrac{11}{20}\) + \(\dfrac{5}{20}\)) : \(\dfrac{7}{3}\)
= \(\dfrac{7}{5}\) - \(\dfrac{16}{20}\) : \(\dfrac{7}{3}\)
= \(\dfrac{7}{5}\) - \(\dfrac{16}{20}\) x \(\dfrac{3}{7}\)
= \(\dfrac{7}{5}\) - \(\dfrac{12}{35}\)
= \(\dfrac{49}{35}\) - \(\dfrac{12}{35}\)
= \(\dfrac{37}{35}\)
P(x) = ( x + 5 )( x + 10 )( x + 15 )( x + 20 ) + 2026
= [ ( x + 5 )( x + 20 ) ][ ( x + 10 )( x + 15 ) ] + 2026
= ( x2 + 25x + 100 )( x2 + 25x + 150 ) + 2026
P(x) : ( x2 + 25x + 110 ) = [ ( x2 + 25x + 100 )( x2 + 25x + 150 ) + 2026 ] : ( x2 + 25x + 110 ) (1)
Đặt y = x2 + 25x + 110
(1) trở thành [ ( y - 10 )( y + 40 ) + 2026 ] : y
= ( y2 + 30y - 400 + 2026 ) : y
= [ ( x2 + 25x + 110 )2 + 30( x2 + 25x + 110 ) + 1626 ] : ( x2 + 25x + 110 )
Đến đây dễ thấy P(x) chia ( x2 + 25x + 110 ) dư 1626