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1

SAI

a, Ta có :

A = \(\dfrac{a}{a+b}\) + \(\dfrac{b}{b+c}\) + \(\dfrac{c}{c+a}\) (a; b; c thuộc N*)

Ta có :

\(\dfrac{a}{a+b}\) < 1 => \(\dfrac{a}{a+b}\) < \(\dfrac{a+c}{a+b+c}\)

Tương tự :

\(\dfrac{b}{b+c}\) < \(\dfrac{b+a}{b+c+a}\)

\(\dfrac{c}{c+a}\) < \(\dfrac{c+b}{c+a+b}\)

=> A < \(\dfrac{2\left(a+b+c\right)}{a+b+c}\)= 2 (1)

Mặt khác :

\(\dfrac{a}{a+b+c}\) < \(\dfrac{a}{a+b}\)

\(\dfrac{b}{a+b+c}\) < \(\dfrac{b}{b+c}\)

\(\dfrac{c}{a+b+c}\) < \(\dfrac{c}{c+a}\)

=> \(\dfrac{a+b+c}{a+b+c}\) < A

1 < A (2)

Từ (1) và (2) => 1 < A < 2

=> A ko thể là 1 số nguyên ( do 1 và 2 là 2 số nguyên liên tiếp)

Câu b tương tự nha bn!!

Chúc bn học tốt!!

b) Ta có: \(\left\{{}\begin{matrix}\dfrac{a}{a+b+c}< \dfrac{a+d}{a+b+c+d}\\\dfrac{b}{b+c+d}< \dfrac{b+a}{a+b+c+d}\\\dfrac{c}{c+d+a}< \dfrac{c+b}{a+b+c+d}\\\dfrac{d}{d+a+b}< \dfrac{b+c}{a+b+c+d}\end{matrix}\right.\)

\(\Rightarrow B=\dfrac{a}{a+b+c}+\dfrac{b}{b+c+d}+\dfrac{c}{c+d+a}+\dfrac{d}{d+a+b}< \dfrac{a+d}{a+b+c+d}+\dfrac{b+a}{a+b+c+d}+\dfrac{c+b}{a+b+c+d}+\dfrac{b+c}{a+b+c+d}\)

\(=\dfrac{2a+2b+2c+2d}{a+b+c+d}=\dfrac{2\left(a+b+c+d\right)}{a+b+c+d}=2\)

\(\Rightarrow B< 2\) (1)

\(\left\{{}\begin{matrix}\dfrac{a}{a+b+c}>\dfrac{a}{a+b+c+d}\\\dfrac{b}{b+c+d}>\dfrac{b}{a+b+c+d}\\\dfrac{c}{c+d+a}>\dfrac{c}{a+b+c+d}\\\dfrac{d}{d+a+b}>\dfrac{d}{a+b+c+d}\end{matrix}\right.\)

\(\Rightarrow B=\dfrac{a}{a+b+c}+\dfrac{b}{b+c+d}+\dfrac{c}{c+d+a}+\dfrac{d}{d+a+b}>\dfrac{a+b+c+d}{a+b+c+d}=1\)

\(\Rightarrow B>1\) (2)

Từ (1) và (2) \(\Rightarrow1< B< 2\)

\(\Rightarrow B\notin Z\left(đpcm\right)\)

Vậy...

II.

1. height

2. beautiful

3. friends

4. teaches

5. jogging

6. activities

7. oldest

8. dangerous

9. bigger

10. friendly

III.

1. watches

2. don't read

3. plays

4. Do your students play

5. Is Nam working

SGK đó nha

TAO LA AI: lq j đến cái ảnh đâu bn?

\(C=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2013.2015}\)

\(C=\frac{1}{2}\left(1-\frac{1}{3}\right)+\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}\left(\frac{1}{5}-\frac{1}{7}\right)+...+\frac{1}{2}\left(\frac{1}{2013}-\frac{1}{2015}\right)\)

\(C=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)

\(C=\frac{1}{2}\left(1-\frac{1}{2015}\right)\)

\(C=\frac{1}{2}.\frac{2014}{2015}=\frac{1007}{2015}\)

mấy bài còn lại dễ nên bạn tự làm nha

c) \(\left|2x-\dfrac{3}{7}\right|-\dfrac{1}{2}=\dfrac{3}{4}\)

\(\left|2x-\dfrac{3}{7}\right|=\dfrac{3}{4}+\dfrac{1}{2}\)

\(\left|2x-\dfrac{3}{7}\right|=\dfrac{5}{4}\)

\(\Rightarrow\left\{{}\begin{matrix}2x-\dfrac{3}{7}=\dfrac{5}{4}\\2x-\dfrac{3}{7}=\dfrac{-5}{4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=\dfrac{47}{28}\\2x=\dfrac{-23}{28}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{47}{56}\\x=\dfrac{-23}{56}\end{matrix}\right.\)

Vậy \(x=\dfrac{47}{56}\) hoặc \(x=\dfrac{-23}{56}\)

d) \(\dfrac{2x+1}{3}=\dfrac{x-5}{2}\)

\(\Rightarrow\left(2x+1\right).2=\left(x-5\right).3\)

\(2x.2+1.2=x.3-5.3\)

\(4x-3x=\left(-15\right)-2\)

\(x=-17\)

Vậy \(x=-17\)

Người ra đề chắc là fan Sơn Tùng

Ban tổ chức trong đề cũng vậy

Cau 5 (1,5 d )

Đặt: 10²⁰¹²+10²⁰¹¹+10²⁰¹⁰+10²⁰⁰⁹+8 = A

A = 10²⁰¹²+10²⁰¹¹+10²⁰¹⁰+10²⁰⁰⁹+8 = 111100...008(2009 c/s 0)

\(\Rightarrow\) S(A)=1+1+1+1+8=12

\(\Rightarrow\) A \(⋮\) 3 (1)

Mà 3 CSTC Của A Là 008

\(\Rightarrow\) A \(⋮\) 8 (2)

Tu (1) va (2)

\(\Rightarrow\)A \(⋮\) BCNN(3; 8)

\(\Rightarrow\)A \(⋮\) 24

a)0,5-|x-3,5|

         Vì |x-3,5|\(\ge0\)

                   Do đó 0,5-|x-3,5|\(\ge0,5\)

 Dấu = xảy ra khi x-3,5=0

                            x=3,5

Vậy Max A=0,5 khi x=3,5

Mỏi cổ quá khi đọc đề bài của bn nên mk làm câu a thôi

     Vậy 
 

c) \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2015}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2014}{2015}\)

\(=\frac{1.2.3.4...2014}{2.3.4.5...2015}=\frac{\left(1.2.3.4...2014\right)}{\left(2.3.4.5...2014\right).2015}=\frac{1}{2015}\)