(2 + x).( 7 – x) = 0
giải giùm nhé mik tich choooo
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(x - 12) - 15 = (20 - 7) - (18 + x)
vậy x = ?
nhanh giùm mik nha, cám ơn, ai nhanh nhất mik tích nhé !
(x - 12) - 15 = (20 - 7) - (18 + x)
x - 12 - 15 = 13- 18 - x
x - 27 = -5 - x
x +x = -5 +27
2x=22
x=11
\(5x^2-3=0\Leftrightarrow x^2=\dfrac{3}{5}\Leftrightarrow x=\pm\sqrt{\dfrac{3}{5}}=\pm\dfrac{\sqrt{15}}{5}\)
\(4x^3+x=0\Leftrightarrow x\left(4x^2+1\right)=0\Leftrightarrow x=0;4x^2+1>0\)
\(5x^2-3=0\\ \Leftrightarrow5x^2=3\\ \Leftrightarrow x^2=\dfrac{3}{5}\\\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{3}{5}}\\x=-\sqrt{\dfrac{3}{5}}\end{matrix}\right. \)
vậy \(x=\sqrt{\dfrac{3}{5}}\) ;\(x=-\sqrt{\dfrac{3}{5}}\)
\(4x^3+x=0\\ \Leftrightarrow x\left(4x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\4x^2+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\4x^2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=\dfrac{-1}{4}\left(vl\right)\end{matrix}\right.\)
vậy x=0
\(=x^3+ax^2+bcx+abc+\left(b+c\right)x^2+a\left(b+c\right)x\)
\(=x^2\left(x+a\right)+bc\left(x+a\right)+\left(b+c\right)x\left(x+a\right)\)
\(=\left(x+a\right)\left(x^2+bc+\left(b+c\right)x\right)\)
\(=\left(x+a\right)\left(x^2+bx+cx+bc\right)\)
\(=\left(x+a\right)\left[x\left(x+b\right)+c\left(x+b\right)\right]\)
\(=\left(x+a\right)\left(x+b\right)\left(x+c\right)\)
a,\(\left(x-4-5\right)\left(x-4+5\right)=0\Leftrightarrow\left(x-9\right)\left(x+1\right)=0\Leftrightarrow x=9;x=-1\)
b, \(\left(x-3-x-1\right)\left(x-3+x+1\right)=0\Leftrightarrow2x-2=0\Leftrightarrow x=1\)
c, \(\left(x^2-4\right)\left(2x-3\right)-\left(x^2-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(2x-3-x+1\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-2\right)=0\Leftrightarrow x=-2;x=2\)
d, \(\left(3x-7\right)^2-\left(2x+2\right)^2=0\Leftrightarrow\left(3x-7-2x-2\right)\left(3x-7+2x+2\right)=0\)
\(\Leftrightarrow\left(x-9\right)\left(5x-5\right)=0\Leftrightarrow x=1;x=9\)
a: \(\Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)
\(\left(2+x\right).\left(7-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2+x=0\\7-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\x=7\end{cases}}}\)
Vậy \(x\in\left\{-2;7\right\}\)
hok tốt!!