A)2013x3,64+7,26x2013–2013x10,89

=2013x(3,64+7,26–10,89)

=2013x0,01

=20,13

B) \(3\frac{1}{4}\) giờ —1 giờ 15 phút

=\(3\frac{1}{4}\)giờ —\(1\frac{1}{4}\)giờ

= 2 giờ

a) 2013 x 3,64 + 7,26 x 2013 - 2013 x 10,89

= ( 3,64 + 7,26 -10,89 ) x 2013

= 0,01 x 2013

= 20,13.

b) \(3\frac{1}{4}\)giờ - 1 giờ 25 phút

=  3 giờ 15 phút - 1 giờ 25 phút

= 1 giờ 50 phút.

\(a)\) Đặt \(A=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2013}\) ta có : 

\(A=\frac{2014-1}{2014}+\frac{2015-1}{2015}+\frac{2013+2}{2013}\)

\(A=\frac{2014}{2014}-\frac{1}{2014}+\frac{2015}{2015}-\frac{1}{2015}+\frac{2013}{2013}+\frac{2}{2013}\)

\(A=1-\frac{1}{2014}+1-\frac{1}{2015}+1+\frac{2}{2013}\)

\(A=\left(1+1+1\right)-\left(\frac{1}{2014}+\frac{1}{2015}-\frac{2}{2013}\right)\)

\(A=3-\left[\frac{1}{2014}+\frac{1}{2015}-\left(\frac{1}{2013}+\frac{1}{2013}\right)\right]\)

\(A=3-\left[\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2013}\right]\)

\(A=3-\left[\left(\frac{1}{2014}-\frac{1}{2013}\right)+\left(\frac{1}{2015}-\frac{1}{2013}\right)\right]\)

Mà : 

\(\frac{1}{2014}< \frac{1}{2013}\)\(\Rightarrow\)\(\frac{1}{2014}-\frac{1}{2013}< 0\)

\(\frac{1}{2015}< \frac{1}{2013}\)\(\Rightarrow\)\(\frac{1}{2015}-\frac{1}{2013}< 0\)

Từ (1) và (2) suy ra : \(\left(\frac{1}{2014}-\frac{1}{2013}\right)+\left(\frac{1}{2015}-\frac{1}{2013}\right)< 0\) ( cộng theo vế ) 

\(\Rightarrow\)\(-\left[\left(\frac{1}{2014}-\frac{1}{2013}\right)+\left(\frac{1}{2015}-\frac{1}{2013}\right)\right]>0\)

\(\Rightarrow\)\(A=3-\left[\left(\frac{1}{2014}-\frac{1}{2013}\right)+\left(\frac{1}{2015}-\frac{1}{2013}\right)\right]>3\) ( cộng hai vế cho 3 ) 

\(\Rightarrow\)\(A>3\) ( điều phải chứng minh ) 

Vậy \(A>3\)

Chúc đệ học tốt ~ 

c, 

\(C=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{9999}{10000}\)

vì \(\frac{1}{2}< \frac{2}{3}\)

\(\frac{3}{4}< \frac{4}{5}\)

\(\frac{5}{6}< \frac{6}{7}\)

.............................

\(\frac{9999}{10000}< \frac{10000}{10001}\)

nên \(C^2< \frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{10000}{10001}\)

\(\Rightarrow C^2< \frac{1}{10001}< \frac{1}{10000}\)

\(\Rightarrow C< \frac{1}{100}\)

bt lm mỗi một câu :v

ae nhiệt tình lên nào

,mình sửa lại đề:

\(\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2013}< 3\)

xóa các chữ số ở tử và mẫu: 2014 và 2014,2015 và 2015

=\(\frac{2013}{2013}\)

=\(1\)

vì \(1>3\) nên \(\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2013}>3\)

1. A = 75(4²⁰⁰⁴ + 4²⁰⁰³ +...+ 4² + 4 + 1) + 25

    A = 25 . [3 . (4²⁰⁰⁴ + 4²⁰⁰³ +...+ 4² + 4 + 1) + 1]

    A = 25 . (3 . 4²⁰⁰⁴ + 3 . 4²⁰⁰³ +...+ 3 . 4² + 3 . 4 + 3 + 1)

    A = 25 . (3 . 4²⁰⁰⁴ + 3 . 4²⁰⁰³ +...+ 3 . 4² + 3 . 4 + 4)

    A = 25 . 4 . (3 . 4²⁰⁰³ + 3 . 4²⁰⁰² +...+ 3 . 4 + 3 + 1)

    A =100 . (3 . 4²⁰⁰³ + 3 . 4²⁰⁰² +...+ 3 . 4 + 3 + 1) \(⋮\) 100

3a) |x| = 1/2 

=> x = 1/2 hoặc x = -1/2

với x = 1/2:

A = \(3.\left(\frac{1}{2}\right)^2-2.\frac{1}{2}+1\)

\(A=\frac{3}{4}-1+1=\frac{3}{4}\)

với x = -1/2

A = \(3.\left(-\frac{1}{2}\right)^2-2\left(-\frac{1}{2}\right)+1\)

\(A=\frac{3}{4}+1+1=\frac{3}{4}+2=\frac{11}{4}\)

2.

A=\(\frac{\left(1+2+3+.....+99+100\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+......+99-100}\)\

A=0

\(A=2\frac{1}{2}\)

a) (1-1/2)(1-1/3)...(1-1/100)=lx-1 99/100l

=> (1-1/2)(1-1/3)...(1-1/100)=1/2.2/3.3/4...99/100

=> (1-1/2)(1-1/3)...(1-1/100)=1.2.3.4....99/2.3.4....100

=>(1-1/2)(1-1/3)...(1-1/100)=1/100      (1)

từ (1)=>1/100= l x-1 99/100 l

TH₁:x-1 99/100 =1/100                 TH₂ : x-1 99/100= -1/100

=>x- 199/100 =1/100                           =>x- 199/100= -1/100

=>x=1/100+199/100                            =>x=-1/100+199/100

=>x=200/100                                       =>x=198/100

=>x=2                                                  =>x=99/50

Vậy x=2 hoặc x=99/50

a) 100.75

b)\(\frac{-7}{6}\)

n=ghi lộn nhé !!

a)\(10.\sqrt{0,01.\sqrt{ }\frac{16}{9}}+3\sqrt{49-\frac{1}{6}}\sqrt{4}\)

a, \(10.\sqrt{100}-\sqrt{\frac{1}{16}}+\left(\frac{1}{3}\right)^0\)

= 10 . 10 - \(\frac{1}{4}\) + 1

= 100 - \(\frac{1}{4}+1\)

= 99,75 + 1 = 100,75

b, \(\left(\frac{1}{3}\right)^{50}.\left(-9\right)^{25}-\frac{2}{3}:4\)

= \(\left(\frac{1}{9}\right)^{25}.\left(-9\right)^{25}-\frac{1}{6}\)

= \(\left(\frac{1}{9}.-9\right)^{25}-\frac{1}{6}\)

\(\left(-1\right)^{25}-\frac{1}{6}\)

= \(-1-\frac{1}{6}=\frac{-7}{6}\)

1+1=3 :)))