GB
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
29 tháng 4 2016
\(\frac{1}{4}x-1-\frac{2}{3}x+1+\frac{4}{5}x-1=\frac{2}{3}\)
\(\left(\frac{1}{4}x-\frac{2}{3}x+\frac{4}{5}x\right)+\left(1-1-1\right)=\frac{2}{3}\)
\(\frac{23}{60}x-1=\frac{2}{3}\)
\(\frac{23}{60}x=\frac{2}{3}+1\)
\(\frac{23}{60}x=\frac{2+3}{3}\)
\(\frac{23}{60}x=\frac{5}{3}\)
\(x=\frac{5}{3}\div\frac{23}{60}\)
\(x=\frac{5}{3}\times\frac{60}{23}\)
\(x=\frac{100}{23}\)
29 tháng 4 2016
\(\left(\frac{1}{4}x-1\right)-\left(\frac{2}{3}x-1\right)+\left(\frac{4}{5}x-1\right)=\frac{2}{3}\)
<=> \(\frac{1}{4}x-1-\frac{2}{3}x+1+\frac{4}{5}x-1=\frac{2}{3}\)
<=> \(\frac{1}{4}x-\frac{2}{3}x+\frac{4}{5}x-1+1-1=\frac{2}{3}\)
<=> \(\frac{23}{60x}=\frac{2}{3}\)=> x=\(\frac{40}{23}\)
Giải cụ thể theo cách lớp 7 đó...còn giải theo cách lớp 8 đơn giản hơn nhiều..nhưng làm theo lớp 8 sợ khó hiểu với lớp 7
>.<
P1
6 tháng 8 2017
Câu b thôi các bạn nhé, câu a mình ko cần nx với cả mình ghi sai dữ liệu câu a r
DP
6 tháng 8 2017
a, \(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot\frac{4}{10}\cdot...\cdot\frac{30}{62}\cdot\frac{31}{64}=2x\)
\(\Leftrightarrow\frac{1\cdot2\cdot3\cdot4\cdot...\cdot30\cdot31}{4\cdot6\cdot8\cdot10\cdot...\cdot62\cdot64}=2x\)
\(\Leftrightarrow\frac{1\cdot2\cdot3\cdot4\cdot...\cdot30\cdot31}{2\cdot2\cdot3\cdot2\cdot4\cdot2\cdot5\cdot2\cdot....\cdot31\cdot2\cdot32\cdot2}=2x\)
\(\Leftrightarrow\frac{1}{2\cdot2\cdot2\cdot2\cdot....\cdot2\cdot2\cdot32}=2x\)
Có : (31 - 1) : 1 + 1 = 31 (thừa số 2)
\(\Rightarrow\frac{1}{2^{31}.32}=2x\)
\(\Rightarrow x=\frac{1}{2^{31}.32}\div2\)
b, \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
\(\Leftrightarrow x+1=x+4\)
\(\Leftrightarrow0=3\text{ (vô lý) }\)
16 tháng 8 2019
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
16 tháng 8 2019
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
AK
12 tháng 7 2018
\(\left|x-\frac{1}{2}\right|+\frac{3}{4}=\left|-1,6+\frac{3}{5}\right|\)
\(\Rightarrow\left|x-\frac{1}{2}\right|+\frac{3}{4}=\left|-1,6+0,6\right|\)
\(\Rightarrow\left|x-\frac{1}{2}\right|+\frac{3}{4}=\left|-1\right|\)
\(\Rightarrow\left|x-\frac{1}{2}\right|+\frac{3}{4}=1\)
\(\Rightarrow\left|x-\frac{1}{2}\right|=1-\frac{3}{4}\)
\(\Rightarrow\left|x-\frac{1}{2}\right|=\frac{1}{4}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{2}=\frac{1}{4}\\x-\frac{1}{2}=-\frac{1}{4}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=\frac{1}{4}\end{cases}}}\)
Vậy ...
12 tháng 7 2018
\(1)\) Ta có :
\(3x=4y\)\(\Leftrightarrow\)\(\frac{x}{4}=\frac{y}{3}\)\(\Leftrightarrow\)\(\frac{x}{8}=\frac{y}{6}\)
\(5y=6z\)\(\Leftrightarrow\)\(\frac{y}{6}=\frac{z}{5}\)
\(\Rightarrow\)\(\frac{x}{8}=\frac{y}{6}=\frac{z}{5}\)
Đặt \(\frac{x}{8}=\frac{y}{6}=\frac{z}{5}=k\)\(\Rightarrow\)\(\hept{\begin{cases}x=8k\\y=6k\\z=5k\end{cases}}\) \(\left(1\right)\)
Thay \(\left(1\right)\) vào \(xyz=30\) ta được :
\(8k.6k.5k=30\)
\(\Leftrightarrow\)\(240k^3=30\)
\(\Leftrightarrow\)\(k^3=\frac{30}{240}\)
\(\Leftrightarrow\)\(k^3=\frac{1}{8}\)
\(\Leftrightarrow\)\(k^3=\left(\frac{1}{2}\right)^3\)
\(\Leftrightarrow\)\(k=\frac{1}{2}\)
Suy ra :
\(x=8k=8.\frac{1}{2}=\frac{8}{2}=4\)
\(y=6k=6.\frac{1}{2}=\frac{6}{2}=3\)
\(z=5k=5.\frac{1}{2}=\frac{5}{2}\)
Vậy \(x=4\)\(;\)\(y=3\) và \(z=\frac{5}{2}\)
Chúc bạn học tốt ~
26 tháng 10 2016
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 \(x=-\frac{12}{5}\)
26 tháng 10 2016
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
26 tháng 6 2015
\(\frac{2}{3}\left(\frac{3}{5}x+\frac{1}{2}\right)=\frac{4}{5}\left(\frac{5}{6}x-\frac{4}{3}\right)+\frac{1}{2}x-\frac{4}{5}\)
\(\frac{2}{5}x+\frac{1}{3}=\frac{2}{3}x-\frac{16}{15}+\frac{1}{2}x-\frac{4}{5}\)
\(\frac{2}{5}x-\frac{2}{3}x-\frac{1}{2}x=-\frac{16}{15}-\frac{4}{5}-\frac{1}{3}\)
\(\left(\frac{2}{5}-\frac{2}{3}-\frac{1}{2}\right)x=-\frac{16}{15}-\frac{12}{15}-\frac{5}{15}\)
\(\left(\frac{12}{30}-\frac{20}{30}-\frac{15}{30}\right)x=-\frac{33}{15}\)
\(\frac{-23}{30}x=-\frac{33}{15}\)
\(x=\frac{-33}{15}:-\frac{23}{30}=\frac{-33}{15}\cdot-\frac{30}{23}=-\frac{66}{23}\)
mk k chắc nữa, tính nhẩm
KT
25 tháng 11 2017
1) \(\frac{x+4}{2005}\)\(+\)\(\frac{x+3}{2006}\)= \(\frac{x+2}{2007}\)\(+\)\(\frac{x+1}{2008}\)
\(\Leftrightarrow\) \(\frac{x+4}{2005}\)\(+\)1 \(+\)\(\frac{x+3}{2006}\)\(+\)1 = \(\frac{x+2}{2007}\)\(+\)1 \(+\)\(\frac{x+1}{2008}\)\(+\)1
\(\Leftrightarrow\)\(\frac{x+2009}{2005}\)+ \(\frac{x +2009}{2006}\)= \(\frac{x+2009}{2007}\)+\(\frac{x+2009}{2008}\)
\(\Leftrightarrow\)(x + 2009)(1/2005 + 1/2006) = (x + 2009)(1/2007 + 1/2008)
\(\Leftrightarrow\)(x + 2009)(1/2005 + 1/2006 - 1/2007 - 1/2008) = 0
Ta thấy: 1/2005 + 1/2006 - 1/2007 - 1/2008 \(\ne\)0
\(\Leftrightarrow\)x + 2009 = 0
\(\Leftrightarrow\)x = -2009
\(\left(\frac{2}{3}-x\right)^2=\frac{1}{4}\)
\(\left(\frac{2}{3}-x\right)^2=\left(\frac{1}{2}\right)^2\)
\(\Rightarrow\frac{2}{3}-x=\frac{1}{2}\)
\(x=\frac{2}{3}-\frac{1}{2}\)
\(x=\frac{1}{6}\)
\(\left(\frac{2}{3}-x\right)^2=\frac{1}{4}\)
\(\left(\frac{2}{3}-x\right)^2=\left(\pm\frac{1}{2}\right)\)
\(\Rightarrow\hept{\begin{cases}\frac{2}{3}-x=\frac{1}{2}\\\frac{2}{3}-x=-\frac{1}{2}\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{1}{6}\\x=\frac{7}{6}\end{cases}}}\)
Vậy \(x\left\{\frac{1}{6};\frac{7}{6}\right\}\)