\(\left(-0,6x-\frac{1}{2}\right).\frac{3}{4}-\left(-1\right)=\frac{1}{3}\)

\(\left(-0,6x-\frac{1}{2}\right).\frac{3}{4}+1=\frac{1}{3}\)

\(\left(-0,6x-\frac{1}{2}\right).\frac{3}{4}=\frac{1}{3}-1\)

\(\left(-0,6x-\frac{1}{2}\right).\frac{3}{4}=\frac{-2}{3}\)

\(-0,6x-\frac{1}{2}=-\frac{2}{3}:\frac{3}{4}\)

\(-0,6x=-\frac{8}{9}\)

\(x=\frac{-8}{9}:\left(-0,6\right)\)

\(x=\frac{40}{27}\)

@SKT_NTT cảm ơn bn

cảm ơn jma cụt ngủn Nguyễn Lan Anb

c) \(\left(2x-3\right).\left(6-2x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=3\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{3}{2};3\right\}\)

e) \(2\left|\frac{1}{2}x-\frac{1}{3}\right|-\frac{3}{2}=\frac{1}{4}\)

\(\Leftrightarrow2\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{1}{4}+\frac{3}{2}=\frac{7}{4}\)

\(\Leftrightarrow\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{4}:2=\frac{7}{4}.\frac{1}{2}=\frac{7}{8}\)

\(\Rightarrow\left[{}\begin{matrix}\frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\\frac{1}{2}x-\frac{1}{3}=\left(-\frac{7}{8}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{29}{12}\\x=\frac{-13}{12}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{29}{12};\frac{-13}{12}\right\}\)

Mấy bài này ko quá khó, tải MathPhoto trong đt về nó tự lm

cái đề gì v trời...gi câu tl quá

a=12b=18c=15

Chi tiết giúp mk vs

a) (2x - 3). (6 - 2x) = 0

==>2x - 3 = 0 hoặc 6 - 2x = 0

      2x = 3                  2x = 6

      x = 3/2                  x = 3

vậy x thuộc { 3/2 ; 3}

b) x không tồn tại

c) x = 35/54

mk ko biết làm 

xin lỗi bn nhae

xin lỗi vì đã ko giúp được bn

chcus bn học gioi!

nhae@@@

mình không biết làm

tk nhé@@@@@@@@@@@@@@@@@@@@

LOL

hihi

a) ... \(=\frac{3\left(2x-1\right)+2x\left(3x+3\right)+2x^2+1}{2x\left(2x-1\right)}=\frac{6x-3+6x^2+6x+2x^2+1}{2x\left(2x-1\right)}\)

\(=\frac{8x^2+12x-2}{2x\left(2x-1\right)}=\frac{4x^2+6x-1}{x\left(2x-1\right)}\)(hình như hết đơn giản được rồi, kết quả tạm vậy bạn nhé!)

b) ... \(=\frac{x^3+2x+2x\left(x+1\right)+x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{x^3+2x+2x^2+2x+x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\frac{x^3+3x^2+3x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{\left(x+1\right)^3}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{x^2+2x+1}{x^2-x+1}\)

c) ... \(=\frac{4\left(x-2\right)+2\left(x+2\right)-5x+6}{\left(x-2\right)\left(x+2\right)}=\frac{4x-8+2x+4-5x+6}{\left(x-2\right)\left(x+2\right)}=\frac{x+2}{\left(x-2\right)\left(x+2\right)}=\frac{1}{x-2}\)

1/1+2 + 1/1+2+3 +1/1+2+3+4 +...+1/1+2+3+...+50

Ta có 2/2(1+2)+2/2(1+2+3)+...+2/2(1+2+...+50)

=2/6+2/12+2/20+...+2/2550

=2/2.3+2/3.4+...+2/50.51

=2(1/2.3+1/3.4+...+1/50.51)

=2(1/1-1/2+1/2-...+1/50-1/51)

=2.(1-1/51)

=2.50/51=100/51

\(3.\left(x-\frac{1}{5}\right)-7.\left(\frac{5}{14}-3\right)=20\)

\(3.\left(x-\frac{1}{5}\right)-7.\frac{-37}{14}=20\)

\(3.\left(x-\frac{1}{5}\right)-\frac{-37}{2}=20\)

\(3.\left(x-\frac{1}{5}\right)=20+\frac{-37}{2}\)

\(3.\left(x-\frac{1}{5}\right)=\frac{3}{2}\)

\(x-\frac{1}{5}=\frac{3}{2}:3\)

\(x-\frac{1}{5}=\frac{1}{2}\)

\(x=\frac{1}{2}+\frac{1}{5}\)

\(x=\frac{7}{10}\)

Giải tiếp đi bạn

\(\frac{2}{3}-\frac{1}{5}.\left(x-\frac{1}{4}\right)=\frac{3}{2}-\frac{1}{4}\)

\(\frac{2}{3}-\frac{1}{5}.\left(x-\frac{1}{4}\right)=\frac{5}{4}\)

\(\frac{1}{5}.\left(x-\frac{1}{4}\right)=\frac{2}{5}-\frac{5}{4}\)

\(\frac{1}{5}.\left(x-\frac{1}{4}\right)=\frac{-17}{20}\)

\(x-\frac{1}{4}=\frac{-17}{20}:\frac{1}{5}\)

\(x-\frac{1}{4}=\frac{-17}{4}\)

\(x=\frac{-17}{4}+\frac{1}{4}\)

\(x=-4\)