Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)
\(\begin{array}{l}\frac{2}{9}:x + \frac{5}{6} = 0,5\\\frac{2}{9}:x = \frac{1}{2} - \frac{5}{6}\\\frac{2}{9}:x = \frac{3}{6} - \frac{5}{6}\\\frac{2}{9}:x = \frac{{ - 2}}{6}\\x = \frac{2}{9}:\frac{{ - 2}}{6}\\x = \frac{2}{9}.\frac{{ - 6}}{2}\\x = \frac{{ - 2}}{3}\end{array}\)
Vậy \(x = \frac{{ - 2}}{3}\).
b)
\(\begin{array}{l}\frac{3}{4} - \left( {x - \frac{2}{3}} \right) = 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - \frac{4}{3}\\x - \frac{2}{3} = \frac{9}{{12}} - \frac{{16}}{{12}}\\x - \frac{2}{3} = \frac{{ - 7}}{{12}}\\x = \frac{{ - 7}}{{12}} + \frac{2}{3}\\x = \frac{{ - 7}}{{12}} + \frac{8}{{12}}\\x = \frac{1}{12}\end{array}\)
Vậy\(x = \frac{1}{12}\).
c)
\(\begin{array}{l}1\frac{1}{4}:\left( {x - \frac{2}{3}} \right) = 0,75\\\frac{5}{4}:\left( {x - \frac{2}{3}} \right) = \frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}:\frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}.\frac{4}{3}\\x - \frac{2}{3} = \frac{5}{3}\\x = \frac{5}{3} + \frac{2}{3}\\x = \frac{7}{3}\end{array}\)
Vậy \(x = \frac{7}{3}\).
d)
\(\begin{array}{l}\left( { - \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}\\ - \frac{5}{6}x + \frac{5}{4} = \frac{4}{3}.\frac{3}{2}\\ - \frac{5}{6}x + \frac{5}{4} = 2\\ - \frac{5}{6}x = 2 - \frac{5}{4}\\ - \frac{5}{6}x = \frac{8}{4} - \frac{5}{4}\\ - \frac{5}{6}x = \frac{3}{4}\\x = \frac{3}{4}:\left( { - \frac{5}{6}} \right)\\x = \frac{3}{4}.\frac{{ - 6}}{5}\\x = \frac{{ - 9}}{{10}}\end{array}\)
Vậy \(x = \frac{{ - 9}}{{10}}\).
làm hộ mình cái để mai nộp thầy,ai nhanh và đúng thì mình k cho nha
\(a)-3\frac{1}{2}+\frac{1}{3}.\left(x-1\right)=-1\frac{1}{3}:2\frac{1}{3}\)
\(-\frac{7}{2}+\frac{1}{3}.\left(x-1\right)=-\frac{4}{3}:\frac{7}{3}\)
\(-\frac{7}{2}+\frac{1}{3}.\left(x-1\right)=-\frac{4}{7}\)
\(\frac{1}{3}.\left(x-1\right)=-\frac{4}{7}-\frac{-7}{2}\)
\(\frac{1}{3}.\left(x-1\right)=\frac{41}{14}\)
\(\Rightarrow x-1=\frac{41}{14}:\frac{1}{3}\)
\(\Rightarrow x-1=\frac{123}{14}\)
\(\Rightarrow x=\frac{123}{14}+1\)
\(\Rightarrow x=\frac{137}{14}\)
a ) \(\left(\frac{2}{5}-x\right):1\frac{1}{3}+\frac{1}{2}=-4\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}+\frac{1}{2}=-4\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}=-4-\frac{1}{2}\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}=-\frac{9}{2}\)
\(\frac{2}{5}-x=-\frac{9}{2}.\frac{4}{3}\)
\(\frac{2}{5}-x=-3\)
\(x=\frac{2}{5}-\left(-3\right)\)
\(x=\frac{2}{5}+3\)
\(x=\frac{3}{5}-\frac{15}{5}\)
\(x=-\frac{12}{5}\)
Vay \(x=-\frac{12}{5}\)
b ) \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(1+\frac{2}{5}+\frac{2}{3}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15}{15}+\frac{6}{15}+\frac{10}{15}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15+6+10}{15}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\frac{31}{15}=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{5}{4}.\frac{31}{15}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{1}{4}.\frac{31}{3}\)
\(-3+\frac{3}{x}-\frac{1}{3}=-\frac{31}{12}\)
\(-3+\frac{3}{x}=-\frac{31}{12}+\frac{1}{2}\)
\(-3+\frac{3}{x}=-\frac{31}{12}+\frac{6}{12}\)
\(-3+\frac{3}{x}=\frac{-25}{12}\)
\(\frac{3}{x}=\frac{-25}{12}+3\)
\(\frac{3}{x}=\frac{-25}{12}+\frac{36}{12}\)
\(\frac{3}{x}=\frac{5}{6}\)
\(\frac{18}{6x}=\frac{5x}{6x}\)
Đèn dây , bạn tự làm tiếp nhé , de rồi chứ
\(\frac{2}{3}\left(\frac{3}{5}x+\frac{1}{2}\right)=\frac{4}{5}\left(\frac{5}{6}x-\frac{4}{3}\right)+\frac{1}{2}x-\frac{4}{5}\)
\(\frac{2}{5}x+\frac{1}{3}=\frac{2}{3}x-\frac{16}{15}+\frac{1}{2}x-\frac{4}{5}\)
\(\frac{2}{5}x-\frac{2}{3}x-\frac{1}{2}x=-\frac{16}{15}-\frac{4}{5}-\frac{1}{3}\)
\(\left(\frac{2}{5}-\frac{2}{3}-\frac{1}{2}\right)x=-\frac{16}{15}-\frac{12}{15}-\frac{5}{15}\)
\(\left(\frac{12}{30}-\frac{20}{30}-\frac{15}{30}\right)x=-\frac{33}{15}\)
\(\frac{-23}{30}x=-\frac{33}{15}\)
\(x=\frac{-33}{15}:-\frac{23}{30}=\frac{-33}{15}\cdot-\frac{30}{23}=-\frac{66}{23}\)
mk k chắc nữa, tính nhẩm
a) \(\Leftrightarrow x+\frac{3}{4}x=\frac{1}{3}+\frac{5}{4}\)
\(\Leftrightarrow\frac{7}{4}x=\frac{19}{12}\Leftrightarrow x=\frac{19}{12}:\frac{7}{4}=\frac{19}{21}\)
b) \(\Leftrightarrow\frac{2}{3}x-\frac{1}{2}x=\frac{1}{4}+\frac{1}{5}\Leftrightarrow\frac{1}{6}x=\frac{9}{20}\Leftrightarrow x=\frac{9}{20}:\frac{1}{6}=\frac{27}{10}\)
\(a,\frac{-1}{2}+\left(x-3\right):\frac{-1}{2}=-1\frac{2}{3}.\)
\(\Rightarrow\left(x-3\right):\frac{-1}{2}=-1\frac{2}{3}-\frac{-1}{2}=\frac{-7}{6}\)
\(\Rightarrow x-3=\frac{-7}{6}\cdot\frac{-1}{2}=\frac{7}{12}\)
\(\Rightarrow x=\frac{7}{12}+3=3\frac{7}{12}\)
\(b.2,25+\frac{3}{2}:\left(x-5\right)=2\frac{1}{2}\)
\(\Rightarrow\frac{3}{2}:\left(x-5\right)=2\frac{1}{2}-2,25=\frac{1}{4}\)
\(\Rightarrow x-5=\frac{3}{2}:\frac{1}{4}=6\)
\(\Rightarrow x=6+5=11\)
\(c,\left(\frac{1}{3}-x\right)^2=\frac{1}{4}=\left(\frac{1}{2}\right)^2=\left(-\frac{1}{2}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}\frac{1}{3}-x=\frac{1}{2}\\\frac{1}{3}-x=-\frac{1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}-\frac{1}{2}=-\frac{1}{6}\\x=\frac{1}{3}-\frac{-1}{2}=\frac{5}{6}\end{cases}}\)
\(d,\frac{3}{2}+\frac{x-1}{3}=1\)
\(\Rightarrow\frac{x-1}{3}=1-\frac{3}{2}=-\frac{1}{2}\)
\(\Rightarrow x-1=-\frac{1}{2}\cdot3=-\frac{3}{2}\)
\(\Rightarrow x=-\frac{3}{2}+1=\frac{1}{2}\)
\(e,-\frac{6}{8}+\frac{x}{12}=\frac{5}{6}\)
\(\Rightarrow\frac{x}{12}=\frac{5}{6}-\frac{-6}{8}=\frac{19}{12}\)
\(\Rightarrow x=19\)
\(g,\frac{1}{2}-\frac{1}{3}\left(x-2\right)=-\frac{2}{3}\)
\(\Rightarrow-\frac{1}{3}\left(x-2\right)=-\frac{2}{3}-\frac{1}{2}=-\frac{7}{6}\)
\(\Rightarrow x-2=\frac{-7}{6}:\frac{-1}{3}=\frac{7}{2}\)
\(\Rightarrow x=\frac{7}{2}+2=2\frac{7}{2}\)
\(h,\frac{5}{2}\left(x+1\right)-\frac{1}{2}=3\frac{1}{2}\)
\(\Rightarrow\frac{5}{2}\left(x+1\right)=3\frac{1}{2}-\frac{1}{2}=3\)
\(\Rightarrow x+1=3:\frac{5}{2}=\frac{6}{5}\)
\(\Rightarrow x=\frac{6}{5}-1=\frac{1}{5}\)
\(k,\frac{x}{3}-\frac{1}{2}=-2\left(x+1\right)+3\)
\(\Rightarrow x\cdot\frac{1}{3}-\frac{1}{2}=-2x-2+3\)
\(\Rightarrow\frac{1}{3}x+2x=-2+3+\frac{1}{2}\)
\(\Rightarrow\frac{7}{3}x=\frac{3}{2}\Rightarrow x=\frac{3}{2}:\frac{7}{2}=\frac{3}{7}\)
1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)
=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)
b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c) TT
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)
=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)
=> \(\left|50x-140\right|=\left|25x+24\right|\)
=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)
Bài 2 : a. |2x - 5| = x + 1
TH1 : 2x - 5 = x + 1
=> 2x - 5 - x = 1
=> 2x - x - 5 = 1
=> 2x - x = 6
=> x = 6
TH2 : -2x + 5 = x + 1
=> -2x + 5 - x = 1
=> -2x - x + 5 = 1
=> -3x = -4
=> x = 4/3
Ba bài còn lại tương tự
1. a) \(\frac{3}{4}-\frac{-1}{2}+\frac{1}{3}=\frac{3}{4}+\frac{1}{2}+\frac{1}{3}=\frac{9}{12}+\frac{6}{12}+\frac{4}{12}=\frac{19}{12}\)
b) \(5\frac{5}{27}+\frac{7}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}\)
\(=\frac{140}{27}-\frac{5}{27}+\frac{7}{23}+\frac{16}{23}+\frac{1}{2}\)
\(=\frac{135}{27}+\frac{23}{23}+\frac{1}{2}\)
\(=5+1+0,5=6,5\)
2) a) 1/2 + 2/3x = 1/4
=> 2/3x = 1/4 - 1/2
=> 2/3x = -1/4
=> x = -1/4 : 2/3
=> x = -3/8
b) 3/5 + 2/5 : x = 3 1/2
=> 3/5 + 2/5 : x = 7/2
=> 2/5 : x = 7/2 - 3/5
=> 2/5 : x = 29/10
=> x = 2/5 : 29/10
=> x = 4/29
c) x+4/2004 + x+3/2005 = x+2/2006 + x+1/2007
=> x+4/2004 + 1 + x+3/2005 + 1 = x+2/2006 + 1 + x+1/2007 + 1
=> x+2008/2004 + x+2008/2005 = x+2008/2006 + x+2008/2007
=> x+2008/2004 + x+2008/2005 - x+2008/2006 - x+2008/2007 = 0
=> (x+2008). (1/2004 + 1/2005 - 1/2006 - 1/2007) = 0
Vì 1/2004 + 1/2005 - 1/2006 - 1/2007 khác 0
Nên x + 2008 = 0 <=> x = -2008
Vậy x = -2008
1,a,\(\frac{3}{4}-\frac{-1}{2}+\frac{1}{3}=\frac{3}{4}+\frac{2}{4}+\frac{1}{3}=\frac{5}{4}+\frac{1}{3}=\frac{15}{12}+\frac{4}{12}=\frac{19}{12}\)
b, \(5\frac{5}{27}+\frac{7}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}=\frac{140}{27}-\frac{5}{27}+\frac{7}{23}+\frac{16}{23}+\frac{1}{2}=\frac{135}{27}+\frac{23}{23}+\frac{1}{2}=5+1+\frac{1}{2}=\frac{13}{2}\)2,a,\(\frac{1}{2}+\frac{2}{3}.x=\frac{1}{4}\)
<=>\(\frac{2}{3}.x=-\frac{1}{2}\)
<=>\(x=-\frac{3}{4}\)
b,\(\frac{3}{5}+\frac{2}{5}\div x=3\frac{1}{2}\)
<=>\(\frac{2}{5x}=\frac{29}{10}\)
<=>\(x=\frac{29}{4}\)
c,\(\frac{x+4}{2004}+\frac{x+3}{2005}=\frac{x+2}{2006}+\frac{x+1}{2007}\)
<=> \(\frac{x+4}{2004}+1+\frac{x+3}{2005}+1=\frac{x+2}{2006}+1+\frac{x+1}{2007}+1\)
<=>\(\frac{x+2008}{2004}+\frac{x+2008}{2005}=\frac{x+2008}{2006}+\frac{x+2008}{2007}\)
<=>\(\left(x+2008\right)\left(\frac{1}{2004}+\frac{1}{2005}-\frac{1}{2006}-\frac{1}{2007}\right)\)=0
<=>x+2008=0 vì cái ngoặc còn lại\(\ne0\)
<=>x=-2008
Vậy x=-2008
Bạn nhớ tk cho mình vì mình đã chăm chỉ làm hết bài bạn hỏi nha!