2x +1/6=5/12
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
bạn làm sai rồi !
\(\Leftrightarrow x\left(3x+2\right)+\left(x+1\right)^2-\left(2x-5\right)\left(2x+5\right)=-12\)
\(\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=-12\)
\(\Leftrightarrow4x+26=-12\)
\(\Leftrightarrow4x=-38\)
\(\Leftrightarrow x=-\frac{19}{2}\)
Vậy tập nghiệm của phương trình là \(S=\left\{-\frac{19}{2}\right\}\)
1: Ta có: \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)
\(\Leftrightarrow5x+20+12x-28=7x+2\)
\(\Leftrightarrow17x-7x=2+8=10\)
hay x=1
2: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)
\(\Leftrightarrow\dfrac{6x}{36}+\dfrac{4\left(1-3x\right)}{36}=\dfrac{3\left(-x+1\right)}{36}\)
\(\Leftrightarrow6x+4-12x=-3x+3\)
\(\Leftrightarrow-6x+3x=3-4\)
hay \(x=\dfrac{1}{3}\)
3: Ta có: \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)
\(\Leftrightarrow4x-12-x-2=6x-3\)
\(\Leftrightarrow3x-14-6x+3=0\)
\(\Leftrightarrow-3x=11\)
hay \(x=-\dfrac{11}{3}\)
4: Ta có: \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)
\(\Leftrightarrow3x-6-8x-12=x+6\)
\(\Leftrightarrow-5x-x=6+18\)
hay x=-4
5: Ta có: \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)
\(\Leftrightarrow6x-3+2x-6=-1\)
\(\Leftrightarrow8x=8\)
hay x=1
5^x + 5^ ( x + 2 ) = 650
5x + 5x . 52 = 650
5x .( 1 + 25 ) = 650
5x . 26 = 650
5x = 650 : 26
5x = 25
5x = 52
=> x = 2
Vậy x = 2
a) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}\\ \dfrac{x}{5}+\dfrac{1}{2}=\dfrac{3}{5}\\ \dfrac{x}{5}=\dfrac{3}{5}-\dfrac{1}{2}\\ \dfrac{x}{5}=\dfrac{6}{10}-\dfrac{5}{10}\\ \dfrac{x}{5}=\dfrac{1}{10}\\ \dfrac{2x}{10}=\dfrac{1}{10}\\ \Rightarrow2x=1\\ x=1:2\\ x=0,5=\dfrac{1}{2}\)
b) \(x+\dfrac{3}{15}=\dfrac{1}{3}\\ x=\dfrac{1}{3}-\dfrac{3}{15}\\ x=\dfrac{5}{15}-\dfrac{3}{15}\\ x=\dfrac{2}{15}\)
c) \(x-\dfrac{12}{4}=\dfrac{1}{2}\\ x-3=\dfrac{1}{2}\\ x=\dfrac{1}{2}+3\\ x=\dfrac{1}{2}+\dfrac{6}{2}\\ x=\dfrac{7}{2}\)
d) \(\dfrac{1}{2}x+\dfrac{1}{2}=\dfrac{5}{2}\\ \dfrac{1}{2}x=\dfrac{5}{2}-\dfrac{1}{2}\\ \dfrac{1}{2}x=2\\ x=2:\dfrac{1}{2}\\ x=4\)
a. \(\dfrac{2x+5}{10}=\dfrac{6}{10}\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\)
b. \(\dfrac{15x+3}{15}=\dfrac{5}{15}\Leftrightarrow15x=2\Leftrightarrow x=\dfrac{2}{15}\)
c. \(\dfrac{4x-12}{4}=\dfrac{2}{4}\Leftrightarrow4x=14\Leftrightarrow x=\dfrac{7}{2}\)
d. \(\dfrac{1+x}{2x}=\dfrac{5x}{2x}\Leftrightarrow-4x=-1\Leftrightarrow x=\dfrac{1}{4}\)
e. \(\dfrac{-4\left(2x-5\right)}{6\left(2x-5\right)}-\dfrac{2}{6\left(2x-5\right)}=\dfrac{9\left(2x-5\right)}{6\left(2x-5\right)}\)
\(\Leftrightarrow-8x+20-2=18x-45\)
\(\Leftrightarrow-26x=-63\Leftrightarrow x=\dfrac{63}{26}\)
1. \(3-|2x+1|=-5\)
\(\Rightarrow|2x+1|=8\)
\(\Rightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=7\\2x=-9\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{9}{2}\end{cases}}\)
Vậy \(x\in\left\{\frac{7}{2};-\frac{9}{2}\right\}\)
2.\(12+|3-x|=9\)
\(\Rightarrow|3-x|=-3\)
Mà \(|3-x|\ge0\forall x\)
\(\Rightarrow\)Vô lí
Vậy không có x
3.\(|x+9|=12+\left(-9\right)+2\)
\(\Rightarrow|x+9|=5\)
\(\Rightarrow\orbr{\begin{cases}x+9=5\\x+9=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-4\\x=-14\end{cases}}\)
Vậy \(x\in\left\{-4;-14\right\}\)
4.\(5x-16=40+x\)
\(\Rightarrow5x-x=40+16\)
\(\Rightarrow4x=56\)
\(\Rightarrow x=14\)
Vậy \(x=14\)
5.\(5x-7=-21-2x\)
\(\Rightarrow5x+2x=-21+7\)
\(\Rightarrow7x=-14\)
\(\Rightarrow x=-2\)
Vậy \(x=-2\)
6.\(\left(2x-1\right)\left(y-2\right)=12\)
Vì \(x,y\inℤ\)nên \(2x-1;y-2\inℤ\)
\(\Rightarrow2x-1;y-2\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
Ta có bảng : (em tự xét bảng nhé)
a, \(\frac{\left(5-2x\right)}{3}=\frac{\left(4x-1\right)}{-5}\)
\(\Leftrightarrow-5(5-2x)=3\left(4x-1\right)\)
\(\Leftrightarrow10x-25=12x-3\)
\(\Leftrightarrow10x-12x=25-3\)
\(\Leftrightarrow-2x=22\)
\(\Leftrightarrow x=-11\)
b, \(\frac{\left(12-3x\right)}{32}=\frac{6}{\left(4-x\right)}\)
\(\Leftrightarrow\frac{3\left(4-x\right)}{32}=\frac{6}{\left(4-x\right)}\)
\(\Leftrightarrow3(4-x)\left(4-x\right)=32.6\)
\(\Leftrightarrow(4-x)\left(4-x\right)=32.2\)
\(\Leftrightarrow(4-x)^2=64\)
\(\Leftrightarrow\orbr{\begin{cases}4-x=8\\4-x=-8\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-4\\x=12\end{cases}}\)
c, \(\frac{\left(10-2x\right)}{6}=\frac{27}{\left(5-x\right)}\)
\(\Leftrightarrow\frac{2\left(5-x\right)}{6}=\frac{27}{\left(5-x\right)}\)
\(\Leftrightarrow2(5-x)\left(5-x\right)=27.6\)
\(\Leftrightarrow(5-x)\left(5-x\right)=27.3\)
\(\Leftrightarrow(5-x)^2=81\)
\(\Leftrightarrow\orbr{\begin{cases}5-x=9\\5-x=-9\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-4\\x=14\end{cases}}\)
a, \(\frac{5-2x}{3}=\frac{4x-1}{-5}\Leftrightarrow-25+10x=12x-3\Leftrightarrow-22-2x=0\Leftrightarrow x=-11\)
b, \(\frac{12-3x}{32}=\frac{6}{4-x}\Leftrightarrow\frac{12-3x}{32}=\frac{18}{12-3x}\)
\(\Leftrightarrow\left(12-3x\right)^2=576\Leftrightarrow12-3x=\pm2\)\(\Leftrightarrow\orbr{\begin{cases}x=\frac{10}{3}\\x=\frac{14}{3}\end{cases}}\)
c, \(\frac{10-2x}{6}=\frac{27}{5-x}\Leftrightarrow\frac{10-2x}{6}=\frac{54}{10-2x}\)
\(\Leftrightarrow\left(10-2x\right)^2=324\Leftrightarrow10-2x=\pm18\)\(\Leftrightarrow\orbr{\begin{cases}x=14\\x=-4\end{cases}}\)
\(2x+\dfrac{1}{6}=\dfrac{5}{12}\\ \Rightarrow2x=\dfrac{5}{12}-\dfrac{1}{6}\\ \Rightarrow2x=\dfrac{5}{12}-\dfrac{2}{12}\\ \Rightarrow2x=\dfrac{1}{4}\\ \Rightarrow x=\dfrac{1}{4}:2\\ \Rightarrow x=\dfrac{1}{8}\)
Vậy...