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a) \(x^3-\dfrac{1}{9}x=0\)
\(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\)
\(\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\\x+\dfrac{1}{3}=0\Leftrightarrow x=-\dfrac{1}{3}\end{matrix}\right.\)
b) \(x\left(x-3\right)+x-3=0\)
\(\Rightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\Rightarrow x=3\\x+1=0\Rightarrow x=-1\end{matrix}\right.\)
c) \(2x-2y-x^2+2xy-y^2=0\) (thêm đề)
\(\Rightarrow2\left(x-y\right)-\left(x-y\right)^2=0\)
\(\Rightarrow\left(x-y\right)\left(2-x+y\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-y=0\Rightarrow x=y\\2-x+y=0\Rightarrow x-y=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=y\left(1\right)\\\left(1\right)\Rightarrow x-x=2\left(loại\right)\end{matrix}\right.\)
d) \(x^2\left(x-3\right)+27-9x=0\)
\(\Rightarrow x^2\left(x-3\right)+\left(x-3\right).9=0\)
\(\Rightarrow\left(x-3\right)\left(x^2+9\right)=0\)
\(\Rightarrow x-3=0\Rightarrow x=3.\)
a, Vì x2 ≥ 0 , 2y2 ≥ 0 với mọi x,y
=>x2+2y2+ 1 ≥ 1
=>Phân thức trên luôn có nghĩa
\(A=3x^2-12x+10\\ A=3x^2-12x+12-2\\ A=\left(3x^2-12x+12\right)-2\\ A=3\left(x^2-4x+4\right)-2\\ A=3\left(x^2-2\cdot x\cdot2+2^2\right)-2\\ A=3\left(x-2\right)^2-2\\ Do\left(x-2\right)^2\ge0\forall x\\ \Rightarrow3\left(x-2\right)^2\ge0\forall x\\ \Rightarrow A=3\left(x-2\right)^2-2\ge-2\forall x\\ \text{Dấu “=” xảy ra khi : }\\ \left(x-2\right)^2=0\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\\ \text{ Vậy }A_{\left(Min\right)}=-2\text{ khi }x=2\)
A=3x2 - 12x + 10
A= (3x2- 2.3x.2+22)-22+10
A= (3x-2)2+6 \(\ge\) +6
Vậy min A = 6 . Dấu = xảy ra khi 3x -2 = 0
3x= 2
x= \(\dfrac{2}{3}\)
a) \(7x^2-28=0\Leftrightarrow7\left(x^2-4\right)=0\Leftrightarrow x^2-4=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\) vậy \(x=2;x=-2\)
b) \(\left(2x+1\right)+x\left(2x+1\right)=0\Leftrightarrow\left(x+1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\2x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\2x=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=\dfrac{-1}{2}\end{matrix}\right.\) vậy \(x=-1;x=\dfrac{-1}{2}\)
c) \(2x^3-50x=0\Leftrightarrow2x\left(x^2-25\right)=0\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=0\\x-5=0\\x+5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\) vậy \(x=0;x=5;x=-5\)
d) \(9\left(3x-2\right)=x\left(2-3x\right)\Leftrightarrow9\left(3x-2\right)=-x\left(3x-2\right)\)
\(\Leftrightarrow9\left(3x-2\right)+x\left(3x-2\right)=0\Leftrightarrow\left(9+x\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}9+x=0\\3x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-9\\3x=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-9\\x=\dfrac{2}{3}\end{matrix}\right.\) vậy \(x=-9;x=\dfrac{2}{3}\)
e) \(5x\left(x-3\right)-2x+6=0\Leftrightarrow5x\left(x-3\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(5x-2\right)\left(x-3\right)=0\) \(\Leftrightarrow\left\{{}\begin{matrix}5x-2=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}5x=2\\x=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\x=3\end{matrix}\right.\) vậy \(x=\dfrac{2}{5};x=3\)
x11+x4+1
= x11+x10+x9-x10-x9-x8+x8+x7+x6-x7-x6-x5+x5+x4+x3-x3-x2-x+x2+x+1
= x9(x2+x+1)-x8(x2+x+1)+x6(x2+x+1)-x5(x2+x+1)+x3(x2+x+1)-x(x2+x+1)+(x2+x+1)
= (x2+x+1)(x9-x8+x6-x5+x3-x+1)
Đặt tính \(2n^2-n+2\) : \(2n+1\) sẽ bằng n - 1 dư 3
Để chia hết thì 3 phải chia hết cho 2n + 1 hay 2n + 1 là ước của 3
Ư(3) = {\(\pm\) 3; \(\pm\) 1}
\(2n+1=1\Leftrightarrow2n=0\Leftrightarrow n=0\)
\(2n+1=-1\Leftrightarrow2n=-2\Leftrightarrow n=-1\)
\(2n+1=3\Leftrightarrow2n=2\Leftrightarrow n=1\)
\(2n+1=-3\Leftrightarrow2n=-4\Leftrightarrow n=-2\)
Vậy \(n=\left\{0;-2;\pm1\right\}\)
a: \(=\dfrac{2a^2-6a+3a+9-3a^2-3}{\left(a-3\right)\left(a+3\right)}\cdot\dfrac{a-3}{a+1}\)
\(=\dfrac{-a^2-3a+6}{\left(a+3\right)}\cdot\dfrac{1}{â+1}=\dfrac{-a^2-3a+6}{\left(a+3\right)\left(a+1\right)}\)
b: |a|=2
=>a=2 hoặc a=-2
Khi a=2 thì \(A=\dfrac{-2^2-3\cdot2+6}{\left(2+3\right)\left(2+1\right)}=\dfrac{-4}{15}\)
Khi a=-2 thì \(A=\dfrac{-\left(-2\right)^2-3\cdot\left(-2\right)+6}{\left(-2+3\right)\left(-2+1\right)}=-8\)
em c.ơn nhiều ạ