:v lớp 10

D

There are 12 cubes

There are ... count is out

đầy đủ câu trả lời mới đc nhé các bạn!

1.b

2.d

3.c

4.a

5.a

6.a

7.b

8.c

9.a

10.c

\(\Rightarrow\dfrac{5}{4}-\dfrac{1}{4}x=\dfrac{3}{10}x-\dfrac{2}{5}\)

\(\Rightarrow\dfrac{5}{4}+\dfrac{2}{5}=\dfrac{3}{10}x-\dfrac{1}{4}x\)

\(\Rightarrow\dfrac{33}{20}=\dfrac{11}{20}x\)

\(\Rightarrow x=\dfrac{33}{20}\div\dfrac{11}{20}\)

\(\Rightarrow x=3\)

\(1\dfrac{1}{4}-x\dfrac{1}{4}=x\cdot30\%\cdot\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{5}{4}-x\dfrac{1}{4}=x\cdot\dfrac{3}{10}-\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{5}{4}-\dfrac{1}{4}x=\dfrac{3}{10}x-\dfrac{2}{5}\)

\(\Leftrightarrow25-5x=6x-8\)

\(\Leftrightarrow-5x-6x=-8-25\)

\(\Leftrightarrow-11x=-33\)

\(\Leftrightarrow x=3\)

Vậy x = 3

a) 4126

b) 615

c) 927

b) 615

c) 927

sao thi muộn vậy bn ! mk thi xong lâu rùi

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

Từ đề bài ta có:

\(T=\dfrac{1+2}{2}.\dfrac{1+3}{3}.\dfrac{1+4}{4}...\dfrac{1+98}{98}.\dfrac{1+99}{99}\)

\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{99}{98}.\dfrac{100}{99}\)

\(=\dfrac{100}{2}\)

\(=50\).

\(T=\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{98}+1\right)\left(\dfrac{1}{99}+1\right)\)
\(T=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}....\dfrac{99}{98}.\dfrac{100}{99}\)
\(T=\dfrac{3.4.5......99}{3.4.5......99}.\dfrac{100}{2}\)
\(T=50\)

\(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\\ =\dfrac{200-\left(2+1+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{4}\right)+...+\left(1-\dfrac{99}{100}\right)}\\ =\dfrac{200-2-1-\dfrac{2}{3}-\dfrac{2}{4}-\dfrac{2}{5}-...-\dfrac{2}{100}}{\left(1+1+1+...+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\\ =\dfrac{198-\left(\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\\ =\dfrac{2\cdot99-2\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}{99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\\ =\dfrac{2\cdot\left[99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\right]}{99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\\ =2\)

Đề nhỏ quá!! mà t 4 mắt. cẩn thận

Đặt :

\(A=\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+.............+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+....................+\dfrac{99}{100}}\)

\(A=\dfrac{200-2-\left(\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}+..............+\dfrac{2}{100}\right)}{1-\dfrac{1}{2}+1-\dfrac{1}{3}+.................+1-\dfrac{1}{100}}\)

\(A=\dfrac{198-\left(\dfrac{2}{2}+\dfrac{2}{3}+..................+\dfrac{2}{100}\right)}{\left(1+1+.....+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{3}+...........+\dfrac{1}{100}\right)}\)

\(A=\dfrac{2\left[99-\left(\dfrac{1}{2}+\dfrac{1}{3}+.............+\dfrac{1}{100}\right)\right]}{99-\left(\dfrac{1}{2}+\dfrac{1}{3}+..............+\dfrac{1}{100}\right)}\)

\(A=2\)

Vậy \(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+............+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...............+\dfrac{99}{100}}=2\rightarrowđpcm\)

shinichi

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