Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{-x}{2}+\frac{2x}{3}+\frac{x+1}{4}+\frac{2x+1}{6}=\frac{8}{3}\)
\(\frac{2.-x}{4}+\frac{4x}{6}+\frac{x+1}{4}+\frac{2x+1}{6}=\frac{8}{3}\)
\(\frac{-2x}{4}+\frac{4x}{6}+\frac{x+1}{4}+\frac{2x+1}{6}=\frac{8}{3}\)
\(\frac{-2x+x+1}{4}+\frac{4x+2x+1}{6}=\frac{8}{3}\)
\(\frac{-1x+1}{4}+\frac{6x+1}{6}=\frac{8}{3}\)
\(\frac{3\left(-1x+1\right)}{3.4}+\frac{2\left(6x+1\right)}{2.6}=\frac{4.8}{4.3}\)
\(\frac{-3x+3}{12}+\frac{12x+2}{12}=\frac{24}{12}\)
\(\frac{-3x+3+12x+2}{12}=\frac{24}{12}\)
\(\frac{9x+5}{12}=\frac{24}{12}\)
=> 9x+5=24
=> 9x= 24-5
9x= 19
x= 19:9
x= \(\frac{19}{9}\)
vậy x= \(\frac{19}{9}\)
d,
\(|x-\frac{1}{3}|=\frac{5}{6}\Rightarrow \left[\begin{matrix} x-\frac{1}{3}=\frac{5}{6}\\ x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{7}{6}\\ x=\frac{-1}{2}\end{matrix}\right.\)
e,
\(\frac{3}{4}-2|2x-\frac{2}{3}|=2\)
\(\Leftrightarrow 2|2x-\frac{2}{3}|=\frac{3}{4}-2=\frac{-5}{4}\)
\(\Leftrightarrow |2x-\frac{2}{3}|=-\frac{5}{8}<0\) (vô lý vì trị tuyệt đối của 1 số luôn không âm)
Vậy không tồn tại $x$ thỏa mãn đề bài.
f,
\(\frac{2x-1}{2}=\frac{5+3x}{3}\Leftrightarrow 3(2x-1)=2(5+3x)\)
\(\Leftrightarrow 6x-3=10+6x\)
\(\Leftrightarrow 13=0\) (vô lý)
Vậy không tồn tại $x$ thỏa mãn đề bài.
a,
$0-|x+1|=5$
$|x+1|=0-5=-5<0$ (vô lý do trị tuyệt đối của một số luôn không âm)
Do đó không tồn tại $x$ thỏa mãn điều kiện đề.
b,
\(2-|\frac{3}{4}-x|=\frac{7}{12}\)
\(|\frac{3}{4}-x|=2-\frac{7}{12}=\frac{17}{12}\)
\(\Rightarrow \left[\begin{matrix} \frac{3}{4}-x=\frac{17}{12}\\ \frac{3}{4}-x=\frac{-17}{12}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-2}{3}\\ x=\frac{13}{6}\end{matrix}\right.\)
c,
\(2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}\)
\(2|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{4}\)
\(|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{8}\)
\(\Rightarrow \left[\begin{matrix} \frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\ \frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{29}{12}\\ x=\frac{-13}{12}\end{matrix}\right.\)
a)\(\frac{1}{3}x+\frac{2}{5}\left(x-1\right)=0\)
\(\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0\)
\(\frac{11}{15}x-\frac{2}{5}=0\)
\(\frac{11}{15}x=\frac{2}{5}\)
\(x=\frac{6}{11}\)
b)(2x-3)(6-2x)=0
=>2x-3=0 hoặc 6-2x=0
=>x=3/2 hoặc x=3
c)\(x:\frac{3}{4}+\frac{1}{4}=-\frac{2}{3}\)
\(x:\frac{3}{4}=-\frac{11}{12}\)
\(x=-\frac{11}{16}\)
d)\(-\frac{2}{3}-\frac{1}{3}\left(2x-5\right)=\frac{3}{2}\)
\(-\frac{2}{3}-\frac{2}{3}x+\frac{5}{3}=\frac{3}{2}\)
\(-\frac{2}{3}x+1=\frac{3}{2}\)
\(-\frac{2}{3}x=\frac{1}{2}\)
\(x=-\frac{3}{4}\)
\(60\%x+\frac{2}{3}x=\frac{1}{3}.6\frac{1}{3}\)
\(\frac{3}{5}x+\frac{2}{3}x=\frac{19}{9}\)
\(\frac{19}{15}x=\frac{19}{9}\)
\(x=\frac{5}{3}\)
mk làm câu c cho nó dễ
c)1/1.2+1/2.3+...+1/x.(x+1)=2009/2010
=1-1/2+1/2-1/3+...+1/x-1/x+1=2009/2010
=1-1/x+1=2009/2010
=1/x+1=1-2009/2010
=1/x+1=1/2010
=) x+1=2010
x =2010-1
x =2009
a) \(\left(x+1\right)-\frac{x+1}{3}=\frac{5\left(x+1\right)-1}{6}\)
\(\Leftrightarrow6\left(x+1\right)-2\left(x+1\right)=5\left(x+1\right)-1\)
\(\Leftrightarrow6x+6-2x-2=5x+5-1\)
\(\Leftrightarrow6x-2x-5x=5-1-6+2\)
\(\Leftrightarrow-x=0\)
\(\Leftrightarrow x=0\)
b) \(\left(1-x\right)^2+\left(x+2\right)^2=2x\left(x-3\right)-7\)
\(\Leftrightarrow1-2x+x^2+x^2+4x+4=2x^2-6x-7\)
\(\Leftrightarrow2x^2+2x+5=2x^2-6x-7\)
\(\Leftrightarrow2x+6x=-7-5\)
\(\Leftrightarrow8x=-12\)
\(\Leftrightarrow x=-\frac{3}{2}\)
c) \(2+\frac{x-2}{2}-\frac{2x-4}{3}-\frac{5}{6}\left(2-x\right)=0\)
\(\Leftrightarrow2+\frac{x}{2}-1-\frac{2}{3}x+\frac{4}{3}-\frac{5}{3}+\frac{5}{6}x=0\)
\(\Leftrightarrow\frac{x}{2}-\frac{2}{3}x+\frac{5}{6}x=-2+1-\frac{4}{3}+\frac{5}{3}\)
\(\Leftrightarrow\frac{2}{3}x=-\frac{2}{3}\)
\(\Leftrightarrow x=-1\)
a) ta có: \(\frac{x-3}{6}=\frac{3}{2x-6}\)
\(\Rightarrow\left(x-3\right).\left(2x-6\right)=6.3\)
\(\Rightarrow x.\left(2x-6\right)-3.\left(2x-6\right)=18\)
\(2x^2-6x-6x+18=18\)
\(2x^2-12x+18=18\)
\(2x^2-12x=0\)
\(2x.\left(x-6\right)=0\)
\(\Rightarrow2x=0\Rightarrow x=0\)
\(x-6=0\Rightarrow x=6\)
KL: x =0 hoặc x = 6
b) ta có: \(\frac{x+6}{x+2}=\frac{2x+3}{2x-1}\)
\(\Rightarrow\left(x+6\right).\left(2x-1\right)=\left(x+2\right).\left(2x+3\right)\)
\(\Rightarrow x.\left(2x-1\right)+6.\left(2x-1\right)=x.\left(2x+3\right)+2.\left(2x+3\right)\)
\(2x^2-x+12x-6=2x^2+3x+4x+6\)
\(2x^2+11x-6=2x^2+7x+6\)
\(\Rightarrow2x^2+11x-2x^2-7x=6+6\)
\(3x=12\)
\(x=12:3\)
\(x=4\)
\(\frac{x-3}{6}=\frac{3}{2x-6}\Leftrightarrow\left(x-3\right).\left(2x-6\right)=18\Leftrightarrow2x^2-12x+18=18\Leftrightarrow2x^2-12x=0\)
\(\Leftrightarrow x^2-6x=0\Leftrightarrow x.\left(x-6\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}\)
\(\frac{x+6}{x+2}=\frac{2x+3}{2x-1}\Leftrightarrow\left(x+6\right).\left(2x-1\right)=\left(2x+3\right).\left(x+2\right)\)
\(\Leftrightarrow2x^2+11x-6=2x^2+7x+6\Leftrightarrow4x-12=0\Leftrightarrow x=3\)