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\(3\frac{1}{2}+4\frac{2}{5}=\left(3+4\right)+\left(\frac{1}{2}+\frac{2}{5}\right)=7+\frac{9}{10}=7\frac{9}{10}\)
nha....................................................
12 . ( x - 1 ) : 3 = 43 + 23
12 . ( x - 1 ) : 3 = 64 + 8
12 . ( x - 1 ) : 3 = 72
12 . ( x - 1 ) = 72 . 3
12 . ( x - 1 ) = 216
x - 1 = 216 : 12
x - 1 = 18
x = 18 + 1
x = 19
=> 5 - [ 4 - ( 1 + 2x ) ] = -6
=> 4 - 1 - 2x = 11
=> 2x = 3 - 11 = -8
=> x = -4
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{x\cdot\left(x+2\right)}=\frac{20}{41}\)
\(\frac{1}{2}\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{x\cdot\left(x+2\right)}\right)=\frac{20}{41}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{20}{41}:\frac{1}{2}\)
\(1-\frac{1}{x+2}=\frac{40}{41}\)
\(\frac{1}{x+2}=1-\frac{40}{41}\)
\(\frac{1}{x+2}=\frac{1}{41}\)
\(\Rightarrow x+2=41\Rightarrow x=39\)
\(2.x+2^2.x-10=5^2+3^2\)
\(\Leftrightarrow2.x+4.x-10=25+9\)
\(\Leftrightarrow2.x+4.x-10=34\)
\(\Leftrightarrow x.\left(4+2\right)-10=34\)
\(\Rightarrow x.\left(4+2\right)=34+10=44\)
\(\Rightarrow x=44:6=7\)(dư 2)