:v lớp 10

D

a) Vì tia Ot nằm giữa 2 tia Ox và Oy

\(\Rightarrow\) \(\widehat{xOt}+\widehat{yOt}=\widehat{xOy}\)

\(\Rightarrow\) 30^o + 70^o = \(\widehat{xOy}\)

\(\Rightarrow\) \(\widehat{xOy}\) = 100^o

Vậy \(\widehat{xOy}\) = 100^o

b) Vì tia Ot nằm giữa 2 tia Ox và Oy

\(\Rightarrow\) \(\widehat{xOt}+\widehat{yOt}=\widehat{xOy}\)

\(\Rightarrow\) \(\dfrac{1}{3}\widehat{yOt}+\widehat{yOt}=108^o\)

\(\Rightarrow\) \(\widehat{yOt}\left(\dfrac{1}{3}+1\right)\) = 108^o

\(\Rightarrow\) \(\widehat{yOt}\dfrac{1}{4}\) = 108^o

\(\Rightarrow\) \(\widehat{yOt}\)= 108^o : \(\dfrac{4}{3}\) = 81^o

\(\Rightarrow\) \(\widehat{xOt}\)= 81^o : 3 = 27^o

Vậy \(\widehat{yOt}\) = 81^o và \(\widehat{xOt}\) = 27^o

c) Vì tia Ot nằm giữa 2 tia Ox và Oy

\(\Rightarrow\) \(\widehat{yOt}+\widehat{xOt}=\widehat{xOy}\)

\(\Rightarrow\) \(\widehat{yOt}+\widehat{xOt}=80^o\)(1)

Theo bài ra, ta có: \(\widehat{yOt}-\widehat{xOt}=20^o\) (2)

Từ (1) và (2) suy ra:

\(\widehat{xOt}\) = (80^o - 20^o) : 2 = 30^o

\(\Rightarrow\) \(\widehat{yOt}\) = 80^o - 30^o = 50^o

Vậy \(\widehat{xOt}\) = 30^o và \(\widehat{yOt}\) = 50^o

c) Vì tia Ot nằm giưa 2 tia Ox và Oy

\(\Rightarrow\) \(\widehat{xOt}+\widehat{yOt}=\widehat{xOy}\)

\(\Rightarrow\) 50^o + \(\widehat{yOt}\) = 100^o

\(\Rightarrow\) \(\widehat{yOt}\) = 100^o - 50^o = 50^o

Vậy \(\widehat{yOt}\) = 50^o

d) Vì tia Ot nằm giữa 2 tia Ox và Oy

\(\Rightarrow\) \(\widehat{xOt}+\widehat{yOt}=\widehat{xOy}\)

\(\Rightarrow\) a^o + b^o = \(\widehat{xOy}\)

Vậy \(\widehat{xOy}\)= a^o + b^o (với 0 \(\le\) a,b \(\le\) 180)

oh

a) 4126

b) 615

c) 927

b) 615

c) 927

B5

a)\(A=\left(1-\dfrac{1}{2010}\right)\left(1-\dfrac{2}{2010}\right)\left(1-\dfrac{3}{2010}\right)\cdot...\cdot\left(1-\dfrac{2010}{2010}\right)\left(1-\dfrac{2011}{2010}\right)\\ =\left(1-\dfrac{1}{2010}\right)\left(1-\dfrac{2}{2010}\right)\left(1-\dfrac{3}{2010}\right)\cdot...\cdot\left(1-1\right)\left(1-\dfrac{2011}{2010}\right)\\ =\left(1-\dfrac{1}{2010}\right)\left(1-\dfrac{2}{2010}\right)\left(1-\dfrac{3}{2010}\right)\cdot...\cdot0\cdot\left(1-\dfrac{2011}{2010}\right)\\ =0\)

b)

\(A=\dfrac{1946}{1986}=\dfrac{1986-40}{1986}=\dfrac{1986}{1986}-\dfrac{40}{1986}=1-\dfrac{40}{1986}\\ B=\dfrac{1968}{2008}=\dfrac{2008-40}{2008}=\dfrac{2008}{2008}-\dfrac{40}{2008}=1-\dfrac{40}{2008}\)

Vì \(\dfrac{40}{1986}>\dfrac{40}{2008}\) nên \(1-\dfrac{40}{1986}< 1-\dfrac{40}{2008}\) hay \(A< B\)

B6

a) Đề sai

Sửa lại:

\(B=\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{28\cdot31}\\ =\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{28}-\dfrac{1}{31}\\ =1-\dfrac{1}{31}\\ =\dfrac{30}{31}\)

b)

\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+\dfrac{1}{8^2}\)

Ta thấy:

\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}=\dfrac{1}{1}-\dfrac{1}{2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)

\(\dfrac{1}{4^2}< \dfrac{1}{3\cdot4}=\dfrac{1}{3}-\dfrac{1}{4}\)

...

\(\dfrac{1}{8^2}< \dfrac{1}{7\cdot8}=\dfrac{1}{7}-\dfrac{1}{8}\)

\(\Rightarrow B< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\\ B< 1-\dfrac{1}{8}\\ B< \dfrac{7}{8}\left(1\right)\)

Mà \(\dfrac{7}{8}< 1\left(2\right)\)

Từ (1) và (2) ta có \(B< 1\)

i) \(5\dfrac{8}{17}:x+\left(-\dfrac{4}{17}\right):x+3\dfrac{1}{7}:17\dfrac{1}{3}=\dfrac{4}{11}\)

\(\Rightarrow\dfrac{93}{17}:x-\dfrac{4}{17}:x+\dfrac{33}{182}=\dfrac{4}{11}\)

\(\Rightarrow\left(\dfrac{93}{17}-\dfrac{4}{17}\right):x=\dfrac{4}{11}-\dfrac{33}{182}\)

\(\Rightarrow\dfrac{89}{17}:x=\dfrac{365}{2002}\)

\(\Rightarrow x=\dfrac{89}{17}:\dfrac{365}{2002}=\dfrac{178178}{6205}\)

j) \(\dfrac{17}{2}-\left|2x-\dfrac{3}{4}\right|=-\dfrac{7}{4}\)

\(\Rightarrow\left|2x-\dfrac{3}{4}\right|=\dfrac{17}{2}-\left(-\dfrac{7}{4}\right)=\dfrac{41}{4}\)

\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=\dfrac{41}{4}\\2x-\dfrac{3}{4}=-\dfrac{41}{4}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}2x=11\Rightarrow x=\dfrac{11}{2}\\2x=-\dfrac{19}{2}\Rightarrow x=-\dfrac{19}{4}\end{matrix}\right.\)

k) \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)

\(\Rightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{26}{25}-\dfrac{17}{25}=\dfrac{9}{25}=\left(\dfrac{3}{5}\right)^2\)\(=\left(-\dfrac{3}{5}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\Rightarrow x=\dfrac{2}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\Rightarrow x=-\dfrac{4}{5}\end{matrix}\right.\)

l) \(-1\dfrac{5}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)

\(\Rightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-32}{27}-\left(-\dfrac{24}{27}\right)=-\dfrac{8}{27}=\left(-\dfrac{2}{3}\right)^3\)

\(\Rightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)

\(\Rightarrow3x=-\dfrac{2}{3}+\dfrac{7}{9}=\dfrac{1}{9}\)

\(\Rightarrow x=\dfrac{1}{27}\)

j, \(\dfrac{17}{2}-\left|2x-\dfrac{3}{4}\right|=\dfrac{-7}{4}\)

\(\Rightarrow-\left|2x-\dfrac{3}{4}\right|=\dfrac{-7}{4}-\dfrac{17}{2}\)

\(\Rightarrow-\left|2x-\dfrac{3}{4}\right|=\dfrac{-41}{4}\)

\(\Rightarrow\left|2x-\dfrac{3}{4}\right|=\dfrac{41}{4}\)

\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=\dfrac{41}{4}\\2x-\dfrac{3}{4}=\dfrac{-41}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=\dfrac{-19}{4}\end{matrix}\right.\)

k, \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)

\(\Rightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)

\(\Rightarrow x+\dfrac{1}{5}=\pm\dfrac{3}{5}\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=\dfrac{-3}{5}\end{matrix}\right.\Rightarrow}\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=\dfrac{-4}{5}\end{matrix}\right.\)

l, \(-1\dfrac{5}{27}-\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-24}{27}\)

\(\Rightarrow-\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-19}{27}\)

\(\Rightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{19}{27}\)

\(\Rightarrow3x-\dfrac{7}{9}=\dfrac{\sqrt[3]{19}}{3}\)

\(\Rightarrow3x=\dfrac{\sqrt[3]{19}}{3}+\dfrac{7}{19}\)

\(\Rightarrow...\)

Đổi: \(1h30'=1,5h\),

Tổng vận tốc của hai xe là: 

\(150\div1,5=100\left(km/h\right)\)

Vận tốc xe tải là \(2\)phần thì vận tốc taxi là \(3\)phần.

Tổng số phần bằng nhau là: 

\(2+3=5\)(phần) 

Vận tốc taxi là: 

\(100\div5\times3=60\left(km/h\right)\)

Vận tốc xe tải là: 

\(100-60=40\left(km/h\right)\)

sao thi muộn vậy bn ! mk thi xong lâu rùi

There are 12 cubes

There are ... count is out

A = 10,11 + 11,12 + 12,13 + . . .+ 98,99 + 99,10
Ta có :
10,11 = 10 + 0,11
11,12 = 11 + 0,12
12,13 = 12 + 0,13
. . . . . . . . . . . . . .
97,98 = 97 + 0,98
98,99 = 98 + 0,99
99,10 = 99 + 0,10

Đặt B = 10 + 11 + 12 + 13 + . .. +98 + 99

và C = 0,11 + 0,12 + 0,13 + . . . .+ 0,98 + 0,99 + 0,10

- - > 100C = 11 + 12 + 13 + . . .+ 98 + 99 + 10

Ta chỉ việc tính B là suy ra C !

B = 10 + 11 + 12 + 13 + . .. +98 + 99

B = (10+99)+(11+98)+(12+97)+. . . +(44+65) + (45 + 64)

Vì từ 10 đến 99 có tất cả 90 số . Ta sẽ có 90/2 = 45 cặp

Mỗi cặp có tổng là 10 + 99 = 11 + 98 = . .= 45 +64 = 109

Vậy ta có B = 45.109 = 4905

Với A = 4905 . Ta thấy 100C = 10 + 11 + 12 +. . + 98 + 99 =B

- - > 100C = 4905 . Hay C = 4905/100 = 49,05

Vậy A = B + C = 4905 + 49,05 = 4954,05

dài vãi