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Nguyễn Sao Mai
Giới thiệu về bản thân
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A = \(3^2\) + \(6^2\) + \(9^2\) + ... + \(30^2\)
= \(3^2\) .( \(1^2\) + \(2^2\) + \(3^2\) + .... + \(10^2\) )
= \(3^2\) . 385 = 3465
3 quyển sách mua hết số tiền là:
120000 x 3 = 360000 (đồng)
Số tiền Lan phải trả khi có thẻ thành viên là:
360000 x ( 100% - 10%) = 324000 (đồng)
Số tiền Lan được trả lại là:
350000 - 324000 = 26000(đồng)
Đ/s: 26000 đồng
a) \(\widehat{BAC}\) + \(\widehat{CAx}\) = \(^{180^0}\) (tc 2 góc kề bù)
⇒ \(100^0\) + \(\widehat{CAx}\) = \(180^0\)
⇒ \(\widehat{CAx}\) = \(80^0\)
b) \(\widehat{yAC}\) = \(\widehat{ACB}\) = \(40^0\) ( tc 2 góc so le trong)
⇒ Ay // BC
c) có Ay // BC ⇒ \(\widehat{xAy}\) = \(\widehat{ABC}\) ( 2 góc đồng vị)
⇒ \(\widehat{ABC}\) = \(40^0\)
a) x - \(\dfrac{2}{3}\) = \(\dfrac{1}{6}\)
x = \(\dfrac{1}{6}\) + \(\dfrac{2}{3}\)
x = \(\dfrac{5}{6}\)
b) 2x + \(\dfrac{1}{2}\) = - \(\dfrac{5}{3}\)
2x = \(\dfrac{-5}{3}\) - \(\dfrac{1}{2}\)
2x = \(\dfrac{-13}{6}\)
x = \(\dfrac{-13}{6}\) : 2
x = \(\dfrac{-13}{12}\)
c) 3x + \(\dfrac{3}{2}\) = x - \(\dfrac{5}{3}\)
3x - x = \(\dfrac{-5}{3}\) - \(\dfrac{3}{2}\)
2x = \(\dfrac{-19}{6}\)
x = \(\dfrac{-19}{6}\) : 2
x = \(\dfrac{-19}{12}\)
a) \(\dfrac{11}{24}\) - \(\dfrac{5}{41}\)+ \(\dfrac{13}{24}\) + 0,5 - \(\dfrac{36}{41}\) = (\(\dfrac{11}{24}\) + \(\dfrac{13}{24}\)) - (\(\dfrac{5}{41}\) + \(\dfrac{36}{41}\)) + 0,5
= 1 - 1 + 0,5
= 0,5
b) \(\dfrac{1}{2}\) . \(\dfrac{3}{4}\) + \(\dfrac{1}{2}\) . \(\dfrac{1}{4}\) + \(\dfrac{1}{2}\) = \(\dfrac{1}{2}\) . (\(\dfrac{3}{4}\) + \(\dfrac{1}{4}\) + 1)
= \(\dfrac{1}{2}\). 2
= 1
c) \(^{\left(\dfrac{-3}{4}\right)^2}\) : \(^{\left(\dfrac{-1}{4}\right)^2}\)+ 9 . \(\left(\dfrac{-1}{9}\right)\) + \(\left(\dfrac{-3}{2}\right)\) = \(\left(\dfrac{-3}{4}:\dfrac{-1}{4}\right)^2\) - 1 - \(\dfrac{3}{2}\)
= 9 - 1 - \(\dfrac{3}{2}\)
= \(\dfrac{13}{2}\)
d) \(\sqrt{0,25}\) . \(\left(-3\right)^3\) - \(\sqrt{\dfrac{1}{84}}\) : \(\left(\dfrac{-1}{3^{ }}\right)^3\) = \(\dfrac{1}{2}\) . (-27) - \(\dfrac{1}{9}\) : \(\dfrac{-1}{27}\)
= \(\dfrac{-27}{2}\) - ( -3)
= \(\dfrac{-21}{2}\)