\(4,7\div0,25+5,3\times4\)

\(=18,8+21,2\)

\(=40\)

\(3\times\left(a-2\right)+150=240\)

\(3\times\left(a-2\right)=90\)

\(a-2=30\)

\(a=32\)

\(\dfrac{1}{9}+a+\dfrac{7}{12}=\dfrac{17}{18}\)

\(\dfrac{1}{9}+a=\dfrac{13}{36}\)

\(a=\dfrac{1}{4}\)

\(\left(\dfrac{1}{2}\times\dfrac{1}{3}+\dfrac{1}{3}\times\dfrac{1}{4}+\dfrac{1}{4}\times\dfrac{1}{5}+\dfrac{1}{5}\times\dfrac{1}{6}+\dfrac{1}{6}\times\dfrac{1}{7}+\dfrac{1}{7}\times\dfrac{1}{8}\right)\times a=\dfrac{9}{16}\)

\(\left(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+\dfrac{1}{7\times8}\right)\times a=\dfrac{9}{16}\)

\(\left(\dfrac{1}{2}-\dfrac{1}{8}\right)\times a=\dfrac{9}{16}\)

\(\dfrac{3}{8}\times a=\dfrac{9}{16}\)

\(a=\dfrac{3}{2}\)

\(1,\left(3x+2\right)\left(5-x^2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\5-x^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\-x^2=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=\pm\sqrt{5}\end{matrix}\right.\)

Vậy \(S=\left\{-\dfrac{2}{3};-\sqrt{5};\sqrt{5}\right\}\)

\(2,-2x-\dfrac{2}{3}\left(\dfrac{3}{4}-\dfrac{1}{8}x\right)=\left(-\dfrac{1}{2}\right)^3\)

\(\Leftrightarrow-2x-\dfrac{1}{2}+\dfrac{1}{12}x=-\dfrac{1}{8}\)

\(\Leftrightarrow-2x+\dfrac{1}{12}x=-\dfrac{1}{8}+\dfrac{1}{2}\)

\(\Leftrightarrow-\dfrac{23}{12}=\dfrac{3}{8}\)

\(\Leftrightarrow x=-\dfrac{9}{46}\)

Vậy \(S=\left\{-\dfrac{9}{46}\right\}\)

\(3,\dfrac{1}{12}:\dfrac{4}{21}=3\dfrac{1}{2}:\left(3x-2\right)\)

\(\Leftrightarrow\dfrac{1}{12}.\dfrac{21}{4}=\dfrac{7}{2}.\dfrac{1}{3x-2}\)

\(\Leftrightarrow\dfrac{7}{16}=\dfrac{7}{6x-4}\)

\(\Leftrightarrow6x-4=7:\dfrac{7}{16}\)

\(\Leftrightarrow6x-4=16\)

\(\Leftrightarrow x=\dfrac{10}{3}\)

Vậy \(S=\left\{\dfrac{10}{3}\right\}\)

\(4,\dfrac{x-1}{x+2}=\dfrac{4}{5}\left(dk:x\ne-2\right)\)

\(\Rightarrow5\left(x-1\right)=4\left(x+2\right)\)

\(\Rightarrow5x-5=4x+8\)

\(\Rightarrow x=13\left(tmdk\right)\)

Vậy \(S=\left\{13\right\}\)

mk c.ơn bn

b)

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2007}{2009}\)

\(=\frac{1}{1.3}+\frac{1}{2.3}+\frac{1}{2.5}+...+\frac{2}{x.\left(x+1\right)}=\frac{2007}{2009}\)

\(=\frac{1}{2}.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2007}{2009}\)

\(=\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2007}{2009}:\frac{1}{2}\)

\(=\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{4018}\)

\(=\frac{1}{x-1}=\frac{1}{2009}\Leftrightarrow x+1=2009\)

\(\Rightarrow x=2009-1=2008\)

Bạn Phúc Trần Tấn bạn có biết làm phần a ko?Giúp mk với ạ!Mai mk cần rùi

\(1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{2009^2}\)

\(=1-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2009^2}\right)\)

\(>1-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2008.2009}\right)\)

\(=1-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2008}-\frac{1}{2009}\right)\)

\(=1-\left(1-\frac{1}{2009}\right)\)

\(=\frac{1}{2009}\)

1/1x2 + 1/2x3+ 1/3x4+....+1/2009x2010

= 1/1-1/2 + 1/2-1/3+ 1/3-1/4+...+1/2009-1/2010

= 1/1-1/2010

= 2009/2010

\(\frac{1}{1\cdot2}+\cdot\cdot\cdot+\frac{1}{2009\cdot2010}\)

\(=1-\frac{1}{2}+\cdot\cdot\cdot+\frac{1}{2009}-\frac{1}{2010}\)

\(=1-\frac{1}{2010}\)

\(=\frac{2009}{2010}\)

1.A= 1.2.3+2.3.4+...+29.30.31+x=15

\(4A=1.2.3.4+2.3.4.\left(5-1\right)+...+29.30.31.\left(32-28\right)+4x=60\)

\(\Rightarrow4A=1.2.3.4+2.3.4.5-1.2.3.4+...+29.30.31.32-28.29.30.31+4x=60\)

Từ đó suy ra nha bạn

2.\(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)

\(=\frac{2}{2\left(2+1\right)}+\frac{2}{3.\left(3+1\right)}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)

\(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)

\(=2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2007}{2009}\\ =1-\frac{2}{\left(x+1\right)}=\frac{2007}{2009}\)

\(\Rightarrow\frac{2}{x+1}=\frac{2}{2009}\Rightarrow x+1=2009\Rightarrow x=2008\)