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\(\text{a) }\left(\dfrac{1}{2}a^2x^4+\dfrac{4}{3}\:ax^3-\dfrac{2}{3}ax^2\right):\left(-\dfrac{2}{3}\:ax^2\right)\\ =-3ax^2-2x+1\)
\(\text{b) }4\left(\dfrac{3}{4}x-1\right)+\left(12x^2-3x\right):\left(-3x\right)-\left(2x+1\right)\\ =3x-4-4x+1-2x-1\\ =-3x-4\)
kết quả cuối cùng là: a. -\(\dfrac{3}{4}ax^2-2x+1\)
b. \(\)-\(3x-4\)
\(a,\left(x+1\right)^2-\left(x-1\right)^2-3\left(x+1\right)\left(x-1\right)\)
\(=x^2+2x+1-\left(x^2-2x+1\right)-3\left(x^2-1\right)\)
\(=x^2+2x+1-x^2+2x-1-3x^2+2=-3x^2+4x+2\)\(b,5\left(x+2\right)\left(x-2\right)-\left(2x-3\right)^2-x^2+17\)
\(=5\left(x^2-4\right)-\left(4x^2-12x+9\right)-x^2+17\)
\(=5x^2-20-4x^2+12x-9-x^2+17=12x-12\)
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) \(7x^2-28=0\Leftrightarrow7\left(x^2-4\right)=0\Leftrightarrow x^2-4=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\) 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 \(x=\dfrac{2}{5};x=3\)
Bất đẳng thức à
ủa nhưng mà thỏa mãn cái gì mới c.m mấy cái kia chứ