I\(\dfrac{3}{4}\)x-\(\dfrac{1}{2}\)I -1=\(\dfrac{1}{4}\)
1. Tính :
a, \(A=\dfrac{\dfrac{1}{3}-\dfrac{5}{2}}{\dfrac{3}{4}-\dfrac{1}{2}}.\dfrac{\dfrac{5}{6}+\dfrac{7}{3}}{1-\dfrac{5}{6}}.\dfrac{\dfrac{-2}{5}+1}{\dfrac{2}{5}-1}\).
b, \(B=\dfrac{\dfrac{1}{3}-\dfrac{4}{5}}{\dfrac{1}{3}+\dfrac{4}{5}}.\dfrac{\dfrac{3}{4}-\dfrac{5}{3}}{\dfrac{3}{4}+\dfrac{5}{3}}:\dfrac{\dfrac{4}{5}-1}{1-\dfrac{2}{3}}\).
1. Tìm x \(\in\) Z: \(4\dfrac{1}{2}.\left(\dfrac{1}{6}-\dfrac{1}{2}\right)\le x\le\dfrac{2}{3}\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)\)
\(\dfrac{9}{2}\left(\dfrac{2}{12}-\dfrac{6}{12}\right)\le x\le\dfrac{2}{3}\left(\dfrac{4}{12}-\dfrac{6}{12}-\dfrac{9}{12}\right)\)
\(\Leftrightarrow\dfrac{9}{2}.\dfrac{-4}{12}\le x\le\dfrac{2}{3}.\dfrac{-11}{12}\)\(\Leftrightarrow\dfrac{-3}{2}\le x\le\dfrac{-11}{18}\)
\(\Rightarrow x=-1\)
a)\([x.\dfrac{1}{2}]^{3}=\dfrac{1}{27}\)
b)\([x+\dfrac{1}{2} ]^{2}=\dfrac{4}{5} \)
c) I 3x-4/5 I = 11/5
d) I 2x - 2I = 0
\(a,\left(x.\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x.\dfrac{1}{2}=\dfrac{1}{3}\\ \Rightarrow x=\dfrac{1}{3}:\dfrac{1}{2}=\dfrac{2}{3}\\ ---\\ b,\left(x+\dfrac{1}{2}\right)^2=\dfrac{4}{5}=\left(\dfrac{2}{\sqrt{5}}\right)^2=\left(-\dfrac{2}{\sqrt{5}}\right)^2 \\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{2}{\sqrt{5}}\\x+\dfrac{1}{2}=-\dfrac{2}{\sqrt{5}}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{\sqrt{5}}-\dfrac{1}{2}\\x=-\dfrac{2}{\sqrt{5}}-\dfrac{1}{2}\end{matrix}\right.\\ Vậy:x=\pm\dfrac{2}{\sqrt{5}}-\dfrac{1}{2}\)
\(c,\left|3x-\dfrac{4}{5}\right|=\dfrac{11}{5}\\ \Rightarrow\left[{}\begin{matrix}3x-\dfrac{4}{5}=\dfrac{11}{5}\\3x-\dfrac{4}{5}=-\dfrac{11}{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}3x=\dfrac{11}{5}+\dfrac{4}{5}=3\\3x=-\dfrac{11}{5}+\dfrac{4}{5}=-\dfrac{7}{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{3}=1\\x=-\dfrac{7}{5}:3=-\dfrac{7}{15}\end{matrix}\right.\\ ---\\ d,\left|2x-2\right|=0\\ \Leftrightarrow2x-2=0\\ \Leftrightarrow2x=2\\ \Leftrightarrow x=1\)
a: (x*1/2)^3=1/27
=>x*1/2=1/3
=>x=1/3:1/2=2/3
b: \(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{2}{\sqrt{5}}\\x+\dfrac{1}{2}=-\dfrac{2}{\sqrt{5}}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2\sqrt{5}}{5}-\dfrac{1}{2}=\dfrac{4\sqrt{5}-5}{10}\\x=\dfrac{-4\sqrt{5}-5}{10}\end{matrix}\right.\)
c: =>3x-4/5=11/5 hoặc 3x-4/5=-11/5
=>3x=3 hoặc 3x=-7/5
=>x=-7/15 hoặc x=1
d: =>2x-2=0
=>2x=2
=>x=1
Bài 1: Tính
\(\dfrac{1}{2}.\dfrac{1}{-3}+\dfrac{1}{-3}.\dfrac{1}{4}+\dfrac{1}{4}.\dfrac{1}{-5}+\dfrac{1}{-5}.\dfrac{1}{6}\)
Bài 2: Tìm x ∈ Q:
\(\left(x-\dfrac{4}{7}\right).\left(x+\dfrac{1}{2}\right)>0\)
Câu 2:
(x-4/7)(x+1/2)>0
=>x-4/7>0 hoặc x+1/2<0
=>x>4/7 hoặc x<-1/2
Bài 1:Thực hiện phép tính.
1) \(\dfrac{2\cdot6^9-2^5\cdot18^4}{2^2\cdot6^8}\) ; 2)\(\dfrac{15^3+5\cdot15^2-5^3}{18^2+6\cdot18^2-6^3}\)
Bài 2:Tìm x, biết:
1)\(^{\left(5x+1\right)^2}\)=\(\dfrac{36}{49}\) ; 2) \(2^{x-1}=16\) ; 3)\(\left(-\dfrac{3}{4}\right)^{3x-1}=-\dfrac{27}{64}\) ; 4)\(\dfrac{-1}{2}+\dfrac{5}{x+3}=\dfrac{-5}{6}\)
5)\(\dfrac{37-x}{x+13}=\dfrac{3}{7}\) ; 6)\(\dfrac{1}{12}:\dfrac{4}{21}=3\dfrac{1}{2}:\left(3x-2\right)\) ; 7) \(^{2^{2x+1}+4^{x+3}=264}\) ; 8) \(\dfrac{x+7}{-20}=\dfrac{-5}{x+7}\)
9) \(\dfrac{x}{8}=\dfrac{x}{x^3}\) ; 10) \(\dfrac{x-1}{x+5}=\dfrac{6}{7}\)
Mọi người giúp mình với,làm ơn !!!!!
Bài 2:
1: =>5x+1=6/7 hoặc 5x+1=-6/7
=>5x=-1/7 hoặc 5x=-13/7
=>x=-1/35 hoặc x=-13/35
2: =>x-1=4
=>x=5
3: =>3x-1=3
=>3x=4
=>x=4/3
4: \(\Leftrightarrow\dfrac{5}{x+3}=\dfrac{-5}{6}+\dfrac{1}{2}=\dfrac{-5+3}{6}=\dfrac{-2}{6}=\dfrac{-1}{3}\)
=>x+3=-15
=>x=-18
7: \(\Leftrightarrow2^{2x+1}+2^{2x+6}=264\)
=>2^2x+1*(1+2^5)=264
=>2^2x+1=8
=>2x+1=3
=>x=1
9: =>x^4=8x
=>x^4-8x=0
=>x=2
GIÚP MÌNH VỚI
a, x - \(\dfrac{5}{7}\)=\(\dfrac{19}{21}\)
b,\(\dfrac{5}{3}\)-I x - \(\dfrac{1}{5}\)I = \(\dfrac{1}{3}\)
c, (x - \(\dfrac{2}{5}\)) = \(\dfrac{1}{4}\)
d, 5\(\sqrt{x}\) - 30 = 15
\(a,x-\dfrac{5}{7}=\dfrac{19}{21}\\ x=\dfrac{34}{21}\\ b,\dfrac{5}{3}-\left|x-\dfrac{1}{5}\right|=\dfrac{1}{3}\\ \left|x-\dfrac{1}{5}\right|=\dfrac{4}{3}\\ TH1:x-\dfrac{1}{5}=\dfrac{4}{3}\\ x=\dfrac{23}{15}\\ TH2:x-\dfrac{1}{5}=-\dfrac{4}{3}\\ x=-\dfrac{17}{15}\\ c,x-\dfrac{2}{5}=\dfrac{1}{4}\\ x=\dfrac{13}{20}\\ d,5\sqrt{x}-30=15\\ 5\sqrt{x}=45\\ \sqrt{x}=9\\ x=9^2=81\)
Bài 1 : Tìm các số nguyên x , biết :
a) \(\dfrac{2}{3}\left(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{1}{3}\right)\le\dfrac{x}{18}\le\dfrac{7}{13}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\)
b) \(\left(\dfrac{31}{20}-\dfrac{26}{45}\right).-\dfrac{36}{35}< x< \left(\dfrac{51}{56}+\dfrac{8}{21}+\dfrac{1}{3}\right).\dfrac{8}{13}\)
Bài 2 :
C = \(\dfrac{-1}{3}.\dfrac{141}{17}-\dfrac{39}{9}.\dfrac{-1}{7}\)
Bài 1:
a: \(\Leftrightarrow\dfrac{2}{3}\cdot\dfrac{6+9-4}{12}< =\dfrac{x}{18}< =\dfrac{7}{13}\cdot\dfrac{3-1}{6}\)
\(\Leftrightarrow\dfrac{22}{36}< =\dfrac{x}{18}< =\dfrac{14}{78}=\dfrac{7}{39}\)
\(\Leftrightarrow\dfrac{11}{9}< =\dfrac{x}{9}< =\dfrac{7}{13}\)
=>143<=x<=63
hay \(x\in\varnothing\)
b: \(\Leftrightarrow\dfrac{31\cdot9-26\cdot4}{180}\cdot\dfrac{-36}{35}< x< \dfrac{153+64+56}{168}\cdot\dfrac{8}{13}\)
\(\Leftrightarrow-1< x< 1\)
=>x=0
Tính giá trị biểu thức:
\(D=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}+\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}+\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}+\dfrac{2}{2004}}{\dfrac{2}{2002}+\dfrac{3}{2003}+\dfrac{3}{2004}}\)
\(H=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2012}}{\dfrac{2011}{1}+\dfrac{2010}{2}+...+\dfrac{1}{2011}}\)
\(I=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2012}}{\dfrac{2012}{2}+\dfrac{2012}{3}+...+\dfrac{2012}{2011}}\)
Help me!
Chữa lại đề.Bạn xem lại đề xem đúng chưa nhé!
\(D=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}+\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}+\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}+\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}+\dfrac{3}{2004}}\)
\(D=\dfrac{1.\left(\dfrac{1}{2003}+\dfrac{1}{2004}+\dfrac{1}{2005}\right)}{5.\left(\dfrac{1}{2003}+\dfrac{1}{2004}+\dfrac{1}{2005}\right)}-\dfrac{2.\left(\dfrac{1}{2002}+\dfrac{1}{2003}+\dfrac{1}{2004}\right)}{3\left(\dfrac{1}{2002}+\dfrac{1}{2003}+\dfrac{1}{2004}\right)}\)
\(D=\dfrac{1}{5}-\dfrac{2}{3}\)
\(D=-\dfrac{7}{15}\)
Cái này học lâu rồi.Bạn xem lại xem mình làm đúng chưa nhé!
1)Tính giá trị biểu thức:
\(\dfrac{\left(13\dfrac{1}{4}-2\dfrac{5}{27}-10\dfrac{5}{6}\right).230\dfrac{1}{25}+46\dfrac{3}{4}}{\left(1\dfrac{3}{7}+\dfrac{10}{3}\right):\left(12\dfrac{1}{3}-14\dfrac{2}{7}\right)}\)
2)Tìm x biết:
a) \(42+\dfrac{3}{7}.\left|3x-1\right|=12\)
b) \(\left|2x-1\right|=\left|x+2\right|\)
c) \(2.3^x.3^2=18\) với \(x\in N\)
\(\dfrac{\left(13\dfrac{1}{4}-1\dfrac{5}{27}-10\dfrac{5}{6}\right).230\dfrac{1}{25}+46\dfrac{3}{4}}{\left(1\dfrac{3}{7}+\dfrac{10}{3}\right):\left(12\dfrac{1}{3}-14\dfrac{2}{7}\right)}\)
\(=\dfrac{1\dfrac{25}{108}.230\dfrac{1}{25}+46\dfrac{3}{4}}{4\dfrac{16}{21}:\left(-1\dfrac{20}{21}\right)}=\dfrac{330\dfrac{1}{25}}{-2\dfrac{18}{41}}=-135,3164\)