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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
\(C=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2013.2015}\)
\(C=\frac{1}{2}\left(1-\frac{1}{3}\right)+\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}\left(\frac{1}{5}-\frac{1}{7}\right)+...+\frac{1}{2}\left(\frac{1}{2013}-\frac{1}{2015}\right)\)
\(C=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(C=\frac{1}{2}\left(1-\frac{1}{2015}\right)\)
\(C=\frac{1}{2}.\frac{2014}{2015}=\frac{1007}{2015}\)
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 \(x=-17\)
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) \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2015}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2014}{2015}\)
\(=\frac{1.2.3.4...2014}{2.3.4.5...2015}=\frac{\left(1.2.3.4...2014\right)}{\left(2.3.4.5...2014\right).2015}=\frac{1}{2015}\)
Hình E đúng nhất
Các hình bên trái có chấm đỏ gần chấm đen k được nối