Bài 3

\(\dfrac{55}{23}+\dfrac{-22}{23}\le x\le\dfrac{1}{5}-\dfrac{-1}{6}+\dfrac{79}{30}\)

\(=\dfrac{33}{23}\)\(\le x\le\dfrac{90}{30}\)

\(=\dfrac{33}{23}\le x\le3\)

Mà \(x\in Z\) \(\Rightarrow\)\(x=2\)

Có 1 giá trị thỏa mãn 

Chọn A

Bài 4

\(\dfrac{-11}{12}< \dfrac{5}{x}< \dfrac{-11}{15}\)

Chọn D

Bài 5

\(M=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)

\(M=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\)

\(M=1-\dfrac{1}{100}\)

\(M=\dfrac{100}{100}-\dfrac{1}{100}\)

\(M=\dfrac{99}{100}\)

CHọn C

A
D
C

\(5,\dfrac{4}{x-2}+\dfrac{x}{x+1}-\dfrac{x^2-2}{\left(x-2\right)\left(x+1\right)}=0\left(dkxd:x\ne2;-1\right)\)

\(\Rightarrow4\left(x+1\right)+x\left(x-2\right)-x^2-2=0\)

\(\Rightarrow4x+4+x^2-2x-x^2-2=0\)

\(\Rightarrow2x+2=0\)

\(\Rightarrow x=-1\left(loai\right)\)

Vậy \(S=\varnothing\)

\(4,\dfrac{x}{x-3}-\dfrac{1}{x+2}=0\left(dkxd:x\ne3;-2\right)\)

\(\Rightarrow x\left(x+2\right)-\left(x-3\right)=0\)

\(\Rightarrow x^2+3x-x+3=0\)

\(\Rightarrow x^2+2x+3=0\)

\(\Rightarrow S=\varnothing\)

giúp em tl những câu tl trên vs

a, \(\dfrac{x-2}{5}=\dfrac{x}{3}\)

\(\Leftrightarrow3\left(x-2\right)=5x\)

\(\Leftrightarrow3x-6=5x\)

\(\Leftrightarrow5x-3x=6\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=3\)

b, \(\dfrac{x+23}{x+40}=\dfrac{3}{4}\)

\(\Leftrightarrow4\left(x+23\right)=3\left(x+40\right)\)

\(\Leftrightarrow4x+92=2x+80\)

\(\Leftrightarrow4x-2x=80-92\)

\(\Leftrightarrow2x=-12\)

\(\Leftrightarrow x=-6\)

c, \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...........+\dfrac{1}{2^{2017}}\)

\(\Leftrightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...........+\dfrac{1}{2^{2016}}\)

\(\Leftrightarrow2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+........+\dfrac{1}{2^{2016}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+........+\dfrac{1}{2^{2017}}\right)\)

\(\Leftrightarrow A=1-\dfrac{1}{2^{2017}}\)

d, \(B=1+2+2^2+........+2^{2017}\)

\(\Leftrightarrow2B=2+2^2+2^3+......+2^{2018}\)

\(\Leftrightarrow2B-B=\left(2+2^2+.....+2^{2018}\right)-\left(1+2+....+2^{2017}\right)\)

\(\Leftrightarrow B=2^{2018}-1\)

\(\dfrac{x-2}{5}=\dfrac{x}{3}=>3\left(x-2\right)=5x\)

\(< =>3x-6=5x=>x=-3\)

\(\dfrac{x+23}{x+40}=\dfrac{3}{4}=>4\left(x+23\right)=3\left(x+40\right)\)

\(4x+92=3x+120=>x=28\)

a) Ta có: \(\dfrac{9-3x}{x^2+3x+4}-\dfrac{3x-23}{\left(1-x\right)\left(x+4\right)}\)

\(=\dfrac{9-3x}{x^2+3x+4}+\dfrac{3x-23}{x^2+3x-4}\)

\(=\dfrac{\left(9-3x\right)\left(x^2+3x-4\right)}{\left(x^2+3x+4\right)\left(x^2+3x-4\right)}+\dfrac{\left(3x-23\right)\left(x^2+3x+4\right)}{\left(x^2+3x-4\right)\left(x^2+3x+4\right)}\)

\(=\dfrac{9x^2+27x-36-3x^3-9x^2+12x+3x^3+9x^2+12x-23x^2-69x-92}{\left(x^2+3x-4\right)\left(x^2+3x+4\right)}\)

\(=\dfrac{-14x^2-18x-128}{\left(x^2+3x-4\right)\left(x^2+3x+4\right)}\)

b) Ta có: \(\dfrac{4-x}{x^3+2x}-\dfrac{x+5}{x^3-x^2+2x-2}\)

\(=\dfrac{4-x}{x\left(x^2+2\right)}-\dfrac{x+5}{x^2\left(x-1\right)+2\left(x-1\right)}\)

\(=\dfrac{4-x}{x\left(x^2+2\right)}-\dfrac{x+5}{\left(x-1\right)\left(x^2+2\right)}\)

\(=\dfrac{\left(4-x\right)\left(x-1\right)}{x\left(x-1\right)\left(x^2+2\right)}-\dfrac{x\left(x+5\right)}{x\left(x-1\right)\left(x^2+2\right)}\)

\(=\dfrac{4x-4-x^2+x-x^2-5x}{x\left(x-1\right)\left(x^2+2\right)}\)

\(=\dfrac{-2x^2-4}{x\left(x-1\right)\left(x^2+2\right)}\)

\(=\dfrac{-2\left(x^2+2\right)}{x\left(x-1\right)\left(x^2+2\right)}\)

\(=\dfrac{-2}{x\left(x-1\right)}\)

a:

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

\(\Leftrightarrow\dfrac{15\left(2x+1\right)-100-2\left(3x+2\right)}{20}=\dfrac{8\left(3x-1\right)}{20}\)

\(\Leftrightarrow15\left(2x+1\right)-100-2\left(3x+2\right)=8\left(3x-1\right)\)

\(\Leftrightarrow30x+15-100-6x+4=24x-8\)\(\Leftrightarrow30x-6x-24x=100-4-8\)

\(\Leftrightarrow0x=88\)

Vậy pt vô nghiệm

b:

\(\dfrac{x-15}{23}+\dfrac{x-23}{15}-2=0\)

\(\Leftrightarrow\dfrac{x-15}{23}+\dfrac{x-23}{15}=2\)

\(\Leftrightarrow\dfrac{x-15}{23}-1+\dfrac{x-23}{15}-1=2-2\)

\(\Leftrightarrow\dfrac{x-15-23}{23}+\dfrac{x-23-15}{15}=0\)

\(\Leftrightarrow\dfrac{x-38}{23}+\dfrac{x-23}{15}=0\)

\(\Leftrightarrow\left(x+38\right)\left(\dfrac{1}{23}+\dfrac{1}{15}\right)=0\)

Vì \(\dfrac{1}{23}+\dfrac{1}{15}\ne0\) nên x + 38 =0 \(\Leftrightarrow x=-38\)

Vậy tập nghiện của pt S= {-38}

c:

\(\dfrac{3\left(2x+1\right)}{4}-\dfrac{5x+3}{6}+\dfrac{x+1}{3}=x+\dfrac{7}{12}\)

\(\Leftrightarrow\dfrac{9\left(2x+1\right)-2\left(5x+3\right)+4\left(x+1\right)}{12}=\dfrac{12x+7}{12}\)

\(\Leftrightarrow9\left(2x+1\right)-2\left(5x+3\right)+4\left(x+1\right)=12x+7\)

\(\Leftrightarrow18x+9-10x-6+4x+4=12x+7\)

\(\Leftrightarrow18x-10x+4x-12x=7-9+6-4\)

\(\Leftrightarrow0x=0\)

Vậy pt vô số nghiệm

đấy nhá

b) Ta có: \(\left|x\right|-\dfrac{3}{4}=\dfrac{5}{3}\)

\(\Leftrightarrow\left|x\right|=\dfrac{5}{3}+\dfrac{3}{4}=\dfrac{20}{12}+\dfrac{9}{12}=\dfrac{29}{12}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{12}\\x=-\dfrac{29}{12}\end{matrix}\right.\)

c) Ta có: \(\left|2x-\dfrac{1}{3}\right|+\dfrac{5}{6}=1\)

\(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{1}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{1}{3}=\dfrac{1}{6}\\2x-\dfrac{1}{3}=\dfrac{-1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{1}{6}+\dfrac{1}{3}=\dfrac{1}{2}\\2x=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{1}{12}\end{matrix}\right.\)

h.

\(\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}\)

\(\Leftrightarrow\dfrac{2-x}{2002}+1-2=\dfrac{1-x}{2003}+1+1-\dfrac{x}{2004}-2\)

\(\Leftrightarrow\dfrac{2004-x}{2002}=\dfrac{2004-x}{2003}+\dfrac{2004-x}{2004}\)

\(\Leftrightarrow\dfrac{2004-x}{2002}-\dfrac{2004-x}{2003}-\dfrac{2004-x}{2004}=0\)

\(\Leftrightarrow\left(2004-x\right)\left(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\right)=0\)

Vì: \(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\ne0\)

Suy ra: 2004 - x = 0

Vậy x = 2004

\(a,\dfrac{x-23}{24}+\dfrac{x-23}{25}=\dfrac{x-23}{26}+\dfrac{x-23}{27}\)

\(\Leftrightarrow\dfrac{x-23}{24}+\dfrac{x-23}{25}-\dfrac{x-23}{26}-\dfrac{x-23}{27}=0\)

\(\Leftrightarrow\left(x-23\right)\left(\dfrac{1}{24}+\dfrac{1}{25}-\dfrac{1}{26}-\dfrac{1}{27}\right)=0\)

\(\Leftrightarrow x-23=0\) ( vì \(\dfrac{1}{24}+\dfrac{1}{25}-\dfrac{1}{26}-\dfrac{1}{27}\ne0\) )

\(\Leftrightarrow x=23\)

Vậy pt có tập nghiệm S = { 23 }

\(b,\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)=\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+5}{95}+1\right)\)

\(\Leftrightarrow\dfrac{x+2+98}{98}+\dfrac{x+3+97}{97}-\dfrac{x+4+96}{96}-\dfrac{x+5+95}{95}=0\)

\(\Leftrightarrow\dfrac{x+100}{98}+\dfrac{x+100}{97}-\dfrac{x+100}{96}-\dfrac{x+100}{95}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\right)=0\)

\(\Leftrightarrow x+100=0\)

\(\Leftrightarrow x=-100\)

Vậy pt có tập nghiệm S = { 100 }

\(c,\dfrac{x+1}{2004}+\dfrac{x+2}{2003}=\dfrac{x+3}{2002}+\dfrac{x+4}{2001}\)

\(\Leftrightarrow\dfrac{x+1}{2004}+1+\dfrac{x+2}{2003}+1=\dfrac{x+3}{2002}+1+\dfrac{x+4}{2001}+1\)

\(\Leftrightarrow\dfrac{x+1+2004}{2004}+\dfrac{x+2+2003}{2003}-\dfrac{x+3+2002}{2002}-\dfrac{x+4+2001}{2001}=0\)

\(\Leftrightarrow\dfrac{x+2005}{2004}+\dfrac{x+2005}{2003}-\dfrac{x+2005}{2002}-\dfrac{x+2005}{2001}=0\)

\(\Leftrightarrow\left(x+2005\right)\left(\dfrac{1}{2004}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2001}\right)=0\)

\(\Leftrightarrow x+2005=0\)

\(\Leftrightarrow x=-2005\)

Vậy pt có tập nghiệm S = { 2005 }

\(d,\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0\)

\(\Leftrightarrow\dfrac{201-x}{99}+1+\dfrac{203-x}{97}+1+\dfrac{205-x}{95}+1=0\)

\(\Leftrightarrow\dfrac{201-x+99}{99}+\dfrac{203-x+97}{97}+\dfrac{205-x+95}{95}=0\)

\(\Leftrightarrow\dfrac{300-x}{99}+\dfrac{300-x}{97}+\dfrac{300-x}{95}=0\)

\(\Leftrightarrow\left(300-x\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\right)=0\)

\(\Leftrightarrow300-x=0\)

\(\Leftrightarrow x=300\)

Vậy pt có tập nghiệm S = { 300 }

\(e,\dfrac{x-45}{55}+\dfrac{x-47}{53}=\dfrac{x-55}{45}+\dfrac{x-53}{47}\)

\(\Leftrightarrow\dfrac{x-45}{55}-1+\dfrac{x-47}{53}-1=\dfrac{x-55}{45}-1+\dfrac{x-53}{47}-1\)

\(\Leftrightarrow\dfrac{x-45-55}{55}+\dfrac{x-47-53}{53}-\dfrac{x-55-45}{45}-\dfrac{x-53-47}{47}=0\)

\(\Leftrightarrow\dfrac{x-100}{55}+\dfrac{x-100}{53}-\dfrac{x-100}{45}-\dfrac{x-100}{47}=0\)

\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{55}+\dfrac{1}{53}-\dfrac{1}{45}-\dfrac{1}{47}\right)=0\)

\(\Leftrightarrow x-100=0\)

\(\Leftrightarrow x=100\)

Vậy pt có tập nghiệm S = { 100 }

\(f,\dfrac{x+1}{9}+\dfrac{x+2}{8}=\dfrac{x+3}{7}+\dfrac{x+4}{6}\)

\(\Leftrightarrow\dfrac{x+1}{9}+1+\dfrac{x+2}{8}+1=\dfrac{x+3}{7}+1+\dfrac{x+4}{6}+1\)

\(\Leftrightarrow\dfrac{x+10}{9}+\dfrac{x+10}{8}-\dfrac{x+10}{7}-\dfrac{x+10}{6}=0\)

\(\Leftrightarrow\left(x+10\right)\left(\dfrac{1}{9}+\dfrac{1}{8}-\dfrac{1}{7}-\dfrac{1}{6}\right)=0\)

\(\Leftrightarrow x+10=0\)

\(\Leftrightarrow x=-10\)

Vậy pt có tập nghiệm S = { 10 }

\(h,\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}\)

\(\Leftrightarrow\dfrac{2-x}{2002}=\dfrac{1-x}{2003}+\dfrac{-x}{2004}+1\)

\(\Leftrightarrow\dfrac{2-x}{2002}+1=\dfrac{1-x}{2003}+1+\dfrac{-x}{2004}+1\)

\(\Leftrightarrow\dfrac{2-x+2002}{2002}-\dfrac{1-x+2003}{2003}-\dfrac{2004-x}{2004}=0\)

\(\Leftrightarrow\dfrac{2004-x}{2002}-\dfrac{2004-x}{2003}-\dfrac{2004-x}{2004}=0\)

\(\Leftrightarrow\left(2004-x\right)\left(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\right)=0\)

\(\Leftrightarrow2004-x=0\)

\(\Leftrightarrow x=2004\)

Vậy pt có tập nghiệm S = { 2004 }

\(g,\dfrac{x+2}{98}+\dfrac{x+4}{96}=\dfrac{x+6}{94}+\dfrac{x+8}{92}\)

\(\Leftrightarrow\dfrac{x+2}{98}+1+\dfrac{x+4}{96}+1=\dfrac{x+6}{94}+1+\dfrac{x+8}{92}+1\)

\(\Leftrightarrow\dfrac{x+100}{98}+\dfrac{x+100}{96}-\dfrac{x+100}{94}-\dfrac{x+100}{92}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{96}-\dfrac{1}{94}-\dfrac{1}{92}\right)=0\)

\(\Leftrightarrow x+100=0\)

\(\Leftrightarrow x=-100\)

Vậy pt có tập nghiệm S = { -100 }

a.

\(\dfrac{x-23}{24}+\dfrac{x-23}{25}=\dfrac{x-23}{26}+\dfrac{x-23}{27}\)

\(\Leftrightarrow\left(x-23\right)\left(\dfrac{1}{24}+\dfrac{1}{25}-\dfrac{1}{26}-\dfrac{1}{27}\right)=0\)

Vì: \(\dfrac{1}{24}+\dfrac{1}{25}-\dfrac{1}{26}-\dfrac{1}{27}\ne0\)

Suy ra x - 23 = 0

\(\Leftrightarrow x=23\)

a/  => \(\dfrac{3}{5}.\dfrac{1}{x}=\dfrac{6}{25}\)

=> \(\dfrac{1}{x}=\dfrac{2}{5}\)

=> x = 5/2

b/ \(\Rightarrow2\left(x-\dfrac{1}{3}\right)=\dfrac{2}{15}\)

=> \(x-\dfrac{1}{3}=\dfrac{1}{15}\)

=> \(x=\dfrac{2}{5}\)

c/ => | x + 1| = 10/21

=> \(\left[{}\begin{matrix}x=-\dfrac{11}{21}\\x=-\dfrac{31}{21}\end{matrix}\right.\)

d/ => \(5x+5=6x-3\)

=> x = 8