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DT
1
20 tháng 3 2023
\(\Leftrightarrow\dfrac{3}{x\left(x+3\right)}+\dfrac{3}{\left(x+3\right)\left(x+6\right)}+...+\dfrac{3}{\left(x+9\right)\left(x+12\right)}=\dfrac{3}{16}\)
=>\(\dfrac{1}{x}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+6}+...+\dfrac{1}{x+9}-\dfrac{1}{x+12}=\dfrac{3}{16}\)=>\(\dfrac{1}{x}-\dfrac{1}{x+12}=\dfrac{3}{16}\)
=>\(\dfrac{x+12-x}{x\left(x+12\right)}=\dfrac{3}{16}\)
=>12/x(x+12)=3/16
=>4/x(x+12)=1/16
=>x(x+12)=64
=>x^2+12x-64=0
=>x^2+16x-4x-64=0
=>(x+16)(x-4)=0
=>x=4 hoặc x=-16
6 tháng 8 2018
\(a,\left(3x+x\right)\left(x^2-9\right)-\left(x-3\right)\left(x^2+3x+9\right)\)
\(=4x\left(x^2-9\right)-x^3+27\)
\(=4x^3-36x-x^3+27\)
\(=3x^3-36x+27\)
6 tháng 8 2018
\(\left(x+6\right)^2-2x.\left(x+6\right)+\left(x-6\right).\left(x+6\right)\)
\(=\left(x+6\right).\left(x+6-2x+x-6\right)\)
\(=\left(x+6\right).0\)
\(=0\)
28 tháng 7 2021
a) Ta có: \(\dfrac{x}{x-3}-\dfrac{6}{x}-\dfrac{9}{x^2-3x}\)
\(=\dfrac{x^2}{x\left(x-3\right)}-\dfrac{6\left(x-3\right)}{x\left(x-3\right)}-\dfrac{9}{x\left(x-3\right)}\)
\(=\dfrac{x^2-6x+18-9}{x\left(x-3\right)}\)
\(=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)
b) Ta có: \(\dfrac{7}{x}-\dfrac{x}{x+6}+\dfrac{36}{x^2+6x}\)
\(=\dfrac{7\left(x+6\right)-x^2+36}{x\left(x+6\right)}\)
\(=\dfrac{7x+42-x^2+36}{x\left(x+6\right)}\)
\(=\dfrac{-\left(x^2-7x-78\right)}{x\left(x+6\right)}\)
\(=\dfrac{-\left(x^2-13x+6x-78\right)}{x\left(x+6\right)}\)
\(=\dfrac{-\left[x\left(x-13\right)+6\left(x-13\right)\right]}{x\left(x+6\right)}\)
\(=\dfrac{13-x}{x}\)
c) Ta có: \(\dfrac{6}{x-3}-\dfrac{2x-6}{x^2-9}-\dfrac{4}{x+3}\)
\(=\dfrac{6\left(x+3\right)-2x+6-4\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{6x+18-2x+6-4x+12}{\left(x-3\right)\left(x+3\right)}=\dfrac{36}{\left(x-3\right)\left(x+3\right)}\)
NL
0
3
1
Y
23 tháng 3 2023
\(a,\left(x+6\right)\left(x-3\right)=2\left(x-3\right)\)
\(\Leftrightarrow\left(x+6\right)\left(x-3\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+6-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\)
\(b,x^2+5x+6=0\)
\(\Leftrightarrow x^2+2x+3x+6=0\)
\(\Leftrightarrow x\left(x+2\right)+3\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)
\(c,\left(x-1\right)^2=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
\(d,\dfrac{x+1}{x-3}-\dfrac{41}{x+3}+\dfrac{x^2-3}{9-x^2}=0\left(dkxd:x\ne\pm3\right)\)
\(\Leftrightarrow\dfrac{x+1}{x-3}-\dfrac{41}{x+3}-\dfrac{x^2-3}{x^2-9}\)\(=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+3\right)-41\left(x-3\right)-x^2+3=0\)
\(\Leftrightarrow x^2+4x+3-41x+123-x^2+3=0\)
\(\Leftrightarrow-37x=-129\)
\(\Leftrightarrow x=\dfrac{129}{37}\left(n\right)\)