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26 tháng 12 2021
a) \(\Rightarrow\dfrac{1}{3}x\left(x-2\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
b) \(\Rightarrow\left(x+5\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)
c) \(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
e) \(\Rightarrow\left(x+2\right)\left(x+2-x+2\right)=0\Rightarrow\left(x+2\right).4=0\Rightarrow x=-2\)
f) \(\Rightarrow x\left(2x-3\right)+2\left(2x-3\right)=0\Rightarrow\left(2x-3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)
g) \(\Rightarrow2\left(3x-2\right)^2-\left(3x-2\right)\left(3x+2\right)=0\Rightarrow\left(3x-2\right)\left(3x-6\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)
h) \(\Rightarrow x\left(x+1\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)
i) \(\Rightarrow4x\left(x+1\right)+5\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(4x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{5}{4}\end{matrix}\right.\)
25 tháng 10 2021
\(b,P=\left[\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-1\right]:\dfrac{9-x^2+\left(x-3\right)\left(x+3\right)-\left(x-2\right)^2}{\left(x-2\right)\left(x+3\right)}\left(x\ne\pm3;x\ne2\right)\\ P=\left(\dfrac{x}{x+3}-1\right)\cdot\dfrac{\left(x-2\right)\left(x+3\right)}{9-x^2+x^2-9-\left(x-2\right)^2}\\ P=\dfrac{x-x-3}{x+3}\cdot\dfrac{\left(x-2\right)\left(x+3\right)}{-\left(x-2\right)^2}\\ P=\dfrac{-3}{-\left(x-2\right)}=\dfrac{3}{x-2}\)
Với \(x^3-4x=0\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\left(ktm\right)\\x=-2\end{matrix}\right.\)
Với \(x=0\Leftrightarrow P=\dfrac{3}{0-2}=-\dfrac{3}{2}\)
Với \(x=-2\Leftrightarrow P=\dfrac{3}{-2-2}=-\dfrac{3}{4}\)
29 tháng 12 2021
Bài 1:
\(a,=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+2y^2}{2\left(x-y\right)\left(x+y\right)}=\dfrac{2y\left(x+y\right)}{2\left(x-y\right)\left(x+y\right)}=\dfrac{y}{x-y}\\ b,Sửa:\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\\ =\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}:\dfrac{3x-9-x^2}{3x\left(x+3\right)}=\dfrac{x^2+3x+9}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{-3x\left(x+3\right)}{x^2-3x+9}\\ =\dfrac{-3}{x-3}\)
Bài 2:
\(a,\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\\ b,\Leftrightarrow x^3+x^2+x+a=\left(x+1\right)\cdot a\left(x\right)\\ \text{Thay }x=-1\Leftrightarrow-1+1-1+a=0\Leftrightarrow a=1\)
HP
9 tháng 5 2022
a. \(x-\dfrac{x+2}{3}< 3x+\dfrac{x}{2}+5\)
\(\Leftrightarrow\dfrac{6x}{6}-\dfrac{2\left(x+2\right)}{6}< \dfrac{18x}{6}+\dfrac{3x}{6}+\dfrac{30}{6}\)
\(\Rightarrow6x-2x-4-18x-3x-30< 0\)
\(\Leftrightarrow-17x< 34\)
\(\Leftrightarrow x>-2\)
b. \(\dfrac{x}{2}+\dfrac{1-x}{3}>0\)
\(\Leftrightarrow3x+2-2x>0\)
\(\Leftrightarrow x>-2\)
c. \(\left(x-9\right)^2-x\left(x+9\right)< 0\)
\(\Leftrightarrow x^2-18x+81-x^2-9x< 0\)
\(\Leftrightarrow-27x< -81\)
\(\Leftrightarrow x>3\)
DH
18 tháng 9 2017
a)(x+1)(x2+2x)=(x+1)x(x+2)=0
\(=>\left\{{}\begin{matrix}x+1=0=>x=-1\\x=0\\x+2=0=>x=-2\end{matrix}\right.\)
b)x(3x-2)-5(2-3x)=x(3x-2)+5(3x-2)=(3x-2)(x+5)=0
\(=>\left\{{}\begin{matrix}3x-2=0=>x=\dfrac{2}{3}\\x+5=0=>x=-5\end{matrix}\right.\)
c)\(\dfrac{4}{9}-25x^2=\left(\dfrac{2}{3}\right)^2-\left(5x\right)^2=\left(\dfrac{2}{3}-5x\right)\left(\dfrac{2}{3}+5x\right)\)
=0
\(=>\left\{{}\begin{matrix}\dfrac{2}{3}-5x=0=>x=\dfrac{2}{15}\\\dfrac{2}{3}+5x=0=>x=\dfrac{-2}{15}\end{matrix}\right.\)
d)\(x^2-x+\dfrac{1}{4}=x^2-2.\dfrac{1}{2}.x+\left(\dfrac{1}{2}\right)^2=\left(x-\dfrac{1}{2}\right)^2=0\)
\(=>x-\dfrac{1}{2}=0=>x=\dfrac{1}{2}\)
KK
5 tháng 9 2021
a. (x - 22) - 1 = 0
<=> x - 4 - 1 = 0
<=> x = 5
b. 4 - (x - 2)2 = 0
<=> 22 - (x - 2)2 = 0
<=> (2 - x + 2)(2 + x - 2) = 0
<=> x(4 - x) = 0
<=> \(\left[{}\begin{matrix}x=0\\4-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
KK
5 tháng 9 2021
d. (3x - 2)2 - (2x + 3)2 = 5(x + 4)(x - 4)
<=> (3x - 2 - 2x - 3)(3x - 2 + 2x + 3) = 5(x2 - 16)
<=> (x - 5)(5x + 1) = 5x2 - 80
<=> 5x2 + x - 25x - 5 = 5x2 - 80
<=> 5x2 - 5x2 + x - 25x = -80 + 5
<=> -24x = -75
<=> x = \(\dfrac{25}{8}\)
29 tháng 4 2021
ĐKXĐ: \(x\notin\left\{3;-3\right\}\)
Ta có: \(\dfrac{x+9}{x^2-9}-\dfrac{3x}{x+3}=0\)
\(\Leftrightarrow\dfrac{x+9}{\left(x+3\right)\left(x-3\right)}-\dfrac{3x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=0\)
Suy ra: \(x+9-3x\left(x-3\right)=0\)
\(\Leftrightarrow x+9-3x^2+9x=0\)
\(\Leftrightarrow-3x^2+10x+9=0\)
\(\Delta=10^2-4\cdot\left(-3\right)\cdot9=100+108=208\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-10-4\sqrt{13}}{-6}=\dfrac{5+2\sqrt{13}}{3}\left(nhận\right)\\x_2=\dfrac{-10+4\sqrt{13}}{-6}=\dfrac{5-2\sqrt{13}}{3}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{5+2\sqrt{13}}{3};\dfrac{5-2\sqrt{13}}{3}\right\}\)
em c.ơn trước
21 tháng 2 2021
a)\(x\in R\)
b)\(x\ne1\)
c) \(x\notin\left\{1;2\right\}\)
d) \(x\notin\left\{3;-3\right\}\)
e) \(x\ne1\)
f) \(x\notin\left\{2;3\right\}\)
\(ĐK:x\ne\pm3\\ PT\Leftrightarrow3x-x^2=0\\ \Leftrightarrow x\left(3-x\right)=0\\ \Leftrightarrow x=0\left(x\ne3\right)\)