Tìm x:
x(x-1)(x+3)-x2(x+3)=-4
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\(\Leftrightarrow x^3+2x^2-3x-x^3-3x^2=-4\)
\(\Leftrightarrow x^2+3x-4=0\)
=>(x+4)(x-1)=0
=>x=-4 hoặc x=1
x * 2 + x * 3 + x * 4 + x = 2130
x * (2 + 3 + 4 + 1) = 2130
x * 10 = 2130
x = 2130 : 10 = 213
\(x=\dfrac{4}{5}\times\dfrac{4}{3}\)
\(x=\dfrac{16}{15}\)
-----------------------
\(x=\dfrac{5}{9}\times\dfrac{3}{8}\)
\(x=\dfrac{5}{24}\)
\(6x\left(1-3x\right)+9x\left(2x-7\right)+171=0\)
\(\Leftrightarrow6x-18x^2+18x^2-63x+171=0\)
\(\Leftrightarrow-57x=-171\)
\(\Leftrightarrow x=3\)
\(\frac{x+1}{2015}+\frac{x+2}{2014}=\frac{x+3}{2013}+\frac{x+4}{2012}\)
\(\Leftrightarrow\left(\frac{x+1}{2015}+1\right)+\left(\frac{x+2}{2014}+1\right)-\left(\frac{x+3}{2013}+1\right)-\left(\frac{x+4}{2012}+1\right)=0\)
\(\Leftrightarrow\)\(\frac{x+2016}{2015}+\frac{x+2016}{2014}-\frac{x+2016}{2013}+\frac{x+2016}{2012}=0\)
\(\Leftrightarrow\left(x+2016\right)\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)
\(\Leftrightarrow x+2016=0\) ( vì \(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\ne0\) )
\(\Leftrightarrow x=-2016\)
\(x\left(x-1\right)\left(x+3\right)-x^2\left(x+3\right)=-4\)
\(\Leftrightarrow\left(x+3\right)\left[x\left(x-1\right)-x^2\right]=-4\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-x-x^2\right)=-4\)
\(\Leftrightarrow-x\left(x+3\right)=-4\)
\(\Leftrightarrow-x^2-3x=-4\)
\(\Leftrightarrow x^2+3x-4=0\)
\(\Leftrightarrow x^2-x+4x-4=0\)
\(\Leftrightarrow x\left(x-1\right)+4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+4=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-4\end{cases}}\)
vậy........
x( x - 1 )( x + 3 ) - x2( x + 3 ) = -4
⇔ ( x + 3 )[ x( x - 1 ) - x2 ] + 4 = 0
⇔ ( x + 3 )( x2 - x - x2 ) + 4 = 0
⇔ ( x + 3 ).(-x) + 4 = 0
⇔ -x2 - 3x + 4 = 0
⇔ -( x2 + 3x - 4 ) = 0
⇔ -( x2 - x + 4x - 4 ) = 0
⇔ -[ x( x - 1 ) + 4( x - 1 ) ] = 0
⇔ -( x - 1 )( x + 4 ) = 0
⇔ x - 1 = 0 hoặc x + 4 = 0
⇔ x = 1 hoặc x = -4