\(A=9-\frac{3}{5}+\frac{2}{3}-7-\frac{7}{5}+\frac{3}{2}-3+\frac{9}{5}-\frac{5}{2}\)

\(=\left(9-7-3\right)+\left(\frac{9}{5}-\frac{7}{5}-\frac{3}{5}\right)+\left(\frac{3}{2}-\frac{5}{2}\right)\)

\(=-2-\frac{1}{5}=-\frac{11}{5}\)

A = 27/12

Hok tốt

cả lời giải nữa nha

A=\(\frac{1}{2}\)+\(\frac{1}{2^2}\)+\(\frac{1}{2^3}\)+...+\(\frac{1}{2^{2017}}\)

Ax2=1+\(\frac{1}{2}\)+\(\frac{1}{2^2}\)+...+\(\frac{1}{2^{2016}}\)

Ax2-A=1-\(\frac{1}{2^{2016}}\)

Vậy A=1-\(\frac{1}{2^{2016}}\)

Vì 1-\(\frac{1}{2^{2016}}\)<1(Vì1-\(\frac{1}{2^{2016}}\)>0)

A<1

bai 1

\(\frac{7}{4}\)+ \(\frac{5}{6}\):5 - 0,375.2.\(^{\left(-2\right)^2}\)= \(\frac{7}{4}\)+ \(\frac{5}{6}\)x\(\frac{1}{5}\)- \(\frac{15}{4}\). 2.4=\(\frac{7}{4}\)+\(\frac{1}{6}\)-\(\frac{15}{4}\).8=\(\frac{42}{24}\)+\(\frac{4}{24}\)-30=\(\frac{11}{6}\)-30=-169/6

\(\frac{1}{4}\)+\(\frac{3}{4}\). \(\left(\frac{-1}{2}+\frac{2}{3}\right)\)=\(\frac{1}{4}\)+ \(\frac{3}{4}\).\(\left(\frac{-3}{6}+\frac{4}{6}\right)\)= \(\frac{1}{4}+\frac{3}{4}.\frac{1}{6}=\frac{1}{4}+\frac{3}{8}\)= \(\frac{5}{8}\)

Bài 1 : 

Ta có :

\(A=\frac{10^{17}+1}{10^{18}+1}=\frac{\left(10^{17}+1\right).10}{\left(10^{18}+1\right).10}=\frac{10^{18}+10}{10^{19}+10}\)

Mà : \(\frac{10^{18}+10}{10^{19}+10}>\frac{10^{18}+1}{10^{19}+1}\)

Mà \(A=\frac{10^{18}+10}{10^{19}+10}\)nên \(A>B\)

Vậy \(A>B\)

Bài 2 :

Ta có :

\(S=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2013}\)

\(\Rightarrow S=\frac{2014-1}{2014}+\frac{2015-1}{2015}+\frac{2016-1}{2016}+\frac{2013+3}{2013}\)

\(\Rightarrow S=1-\frac{1}{2014}+1-\frac{1}{2015}+1-\frac{1}{2016}+1+\frac{3}{2013}\)

\(\Rightarrow S=4+\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)\)

Vì \(\frac{1}{2013}>\frac{1}{2014}>\frac{1}{2015}>\frac{1}{2016}\)nên  \(\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)>0\)

Nên : \(M>4\)

Vậy \(M>4\)

Bài 3 : 

Ta có :

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.......+\frac{1}{100^2}\)

Suy ra : \(A< \frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+....+\frac{1}{99.101}\)

\(\Rightarrow A< \frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{2.4}+......+\frac{2}{99.101}\right)\)

\(\Rightarrow A< \frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-......-\frac{1}{101}\right)\)

\(\Rightarrow A< \frac{1}{2}.\left[\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{99}\right)-\left(\frac{1}{3}+\frac{1}{4}+......+\frac{1}{101}\right)\right]\)

\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{100}-\frac{1}{101}\right)\)

\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}\right)\)

\(\Rightarrow A< \frac{3}{4}\)

Vậy \(A< \frac{3}{4}\)

Bài 4 :

\(a)A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2015.2017}\)

\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{1}{2015.2017}\right)\)

\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{2015}-\frac{1}{2017}\right)\)

\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)

\(\Rightarrow A=\frac{1}{2}.\frac{2016}{2017}\)

\(\Rightarrow A=\frac{1008}{2017}\)

Vậy \(A=\frac{1008}{2017}\)

\(b)\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+......+\frac{1}{x\left(x+2\right)}=\frac{1008}{2017}\)

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{x.\left(x+2\right)}=\frac{2016}{2017}\)

\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{x+2}=\frac{2016}{2017}\)

\(1-\frac{1}{x+2}=\frac{2016}{2017}\)

\(\Rightarrow\frac{1}{x+2}=1-\frac{2016}{2017}\)

\(\Rightarrow\frac{1}{x+2}=\frac{1}{2017}\)

\(\Rightarrow x+2=2017\)

\(\Rightarrow x=2017-2=2015\)

Vậy \(x=2015\)

(\(\frac{1}{4.9}+\frac{1}{9.14}+...+\frac{1}{44.49}\)).\(\frac{1-3-5-...-49}{89}\)

= \(\frac{1}{5}.\left(\frac{5}{4.9}+\frac{5}{9.14}+...+\frac{5}{45.49}\right).\frac{1-3-5-...-49}{89}\)

\(=\frac{1}{5}.\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\right).\frac{1-\frac{24.\left(49+3\right)}{2}}{89}\)

\(=\frac{1}{5}.\left(\frac{1}{4}-\frac{1}{49}\right).\left(-7\right)\)

\(=-\frac{9}{28}\)

Có chỗ ghi nhầm 44 thành 45. Tự sửa nhé

Bài 2/ a/

|2x + 3| = x + 2

Điều kiện \(x\ge-2\)

Với x < - 1,5 thì ta có

- 2x - 3 = x + 2

<=> 3x = - 5

<=> \(x=-\frac{5}{3}\)

Với \(x\ge-1,5\)thì ta có

2x + 3 = x + 2

<=> x = - 1

b/ A = |x - 2006| + |2007 - x| \(\ge\)|x - 2006 + 2007 - x| = 1

\(\frac{B}{\sqrt{2}}=\frac{\frac{2+\sqrt{3}}{2}}{\sqrt{2}+\sqrt{\frac{4+2\sqrt{3}}{2}}}+\frac{\frac{2-\sqrt{3}}{2}}{\sqrt{2}-\sqrt{\frac{4-2\sqrt{3}}{2}}}\)

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

\(=\frac{\frac{2+\sqrt{3}}{2}}{\frac{2}{\sqrt{2}}+\frac{\sqrt{3}+1}{\sqrt{2}}}+\frac{\frac{2-\sqrt{3}}{2}}{\frac{2}{\sqrt{2}}-\frac{\sqrt{3}-1}{\sqrt{2}}}=\frac{\frac{2+\sqrt{3}}{2}}{\frac{\sqrt{3}+3}{\sqrt{2}}}+\frac{\frac{2-\sqrt{3}}{2}}{\frac{3-\sqrt{3}}{\sqrt{2}}}\)

\(=\frac{\left(2+\sqrt{3}\right).\sqrt{2}}{2\cdot\left(3+\sqrt{3}\right)}+\frac{\left(2-\sqrt{3}\right).\sqrt{2}}{2.\left(3-\sqrt{3}\right)}\)

=> \(B=\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}=\frac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}+\frac{\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)}\)

\(B=\frac{3+\sqrt{3}}{6}+\frac{3-\sqrt{3}}{6}=1\)

----

Vài chỗ mình làm vắn tắt không hiểu cứ hỏi nhé, còn kết quả mình ấn máy tính ra chính xác rùi :)

=10/17+-5/13+7/17+-8/13+-11/25

=1+-1+-11/25

=11/25

b , 

= 1+-2+3+-4+5+-6+......2011+-2012

=1+1+1+1+1.........+-2012

=1.2011+-2012

=2011+-2012

=-1

2.

2/3-x = 5/4

=>x=2/3-5/4

=>x=-7/12

b,

[124-(20-4x)]:30+7=11

=>124-(20-4x)] :30 =11-7

=>[124-(20-4x)]:30=4

=>124-(20-4x)=4x30

=>124-(20-4x)=120

=>20-4x=120-124

=>20-4x=4

=>4x=20-4

=>4x=16

=>x=16:4

=>x=4

bài 1 

a, ghép cặp phân số 10/17 + 7/17 và 5/13 + -8/13 

b, 1-2+3-4+5-6+.....+2011-2012

= ( 1-2) + ( 3-4) + ( 5-6) +....+(2011-2012)

= (-1) + (-1) + (-1) + ....+ (-1)

< từ 1 đến 2012 có 2012 số số hạng, suy ra có 1006 cặp mà mỗi cặp có giá trị = (-1) Suy ra tổng trên = (-1) * 1006=-1006

Bài 2:

a,2/3 - x =5/4

          x=2/3 - 5/4

           x= -7/12

A) \(\frac{10}{12}\)+\(2\)- /\(\frac{-2}{3}\)/ -\(\frac{3}{4}\)= \(\frac{10}{12}\)+2-\(\frac{2}{3}\)-\(\frac{3}{4}\)= \(\frac{10}{12}\)+\(\frac{24}{12}\)-\(\frac{8}{12}\)-\(\frac{9}{12}\)=\(\frac{17}{12}\)

tương tự bài B= \(\frac{59}{40}\)

mk hk bk ghi dáu GTTĐ nên mk ghi như thế 

bạn tính kết quả trong dấu GT tuyệt đối rồi bạn mở dấu GTTĐ bằng cách cho số đó trở thành số dương là được

chúc bn may mắn