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a, \(4x+6y-x^2-y^2+2\)
\(=-\left(x^2+y^2-4x-6y-2\right)\)
\(=-\left(x^2-2x-2x+4+y^2-3y-3y+9-15\right)\)
\(=-\left[\left(x^2-2x\right)-\left(2x-4\right)+\left(y^2-3y\right)-\left(3y-9\right)-15\right]\)
\(=-\left[\left(x-2\right)^2+\left(y-3\right)^2-15\right]\)
Với mọi giá trị của \(x;y\in R\) ta có:
\(\left(x-2\right)^2\ge0;\left(y-3\right)^2\ge0\)
\(\Rightarrow\left(x-2\right)^2+\left(y-3\right)^2\ge0\)
\(\Rightarrow\left(x-2\right)^2+\left(y-3\right)^2-15\ge-15\)
\(\Rightarrow-\left[\left(x-2\right)^2+\left(y-3\right)^2-15\right]\le15\)
Để \(-\left[\left(x-2\right)^2+\left(y-3\right)^2-15\right]=15\) thì \(\left(x-2\right)^2+\left(y-3\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x-2\right)^2=0\\\left(y-3\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
Vậy GTLN của biểu thức là 15 đạt được khi và chỉ khi \(x=2;y=3\)
Câu b làm tương tự! Chúc bạn học tốt!!!
Thui đang chán không có bài :) làm lun:
b, \(-x^2-4y^2-z^2+2x+12y-4z-10\)
\(=-\left(x^2+4y^2+z^2-2x-12y+4z+10\right)\)
\(=-\left(x^2-x-x+1+4y^2-6y-6y+9+z^2+2z+2z+4-4\right)\)
\(=-\left[\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2-4\right]\)
Với mọi giá trị của \(x;y;z\in R\) ta có:
\(\left(x-1\right)^2\ge0;\left(2y-3\right)^2\ge0;\left(z+2\right)^2\ge0\)
\(\Rightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2-4\ge-4\)
\(\Rightarrow-\left[\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2-4\right]\le4\)
với mọi giá trị của \(x;y;z\in R\).
Để \(-\left[\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2-4\right]=4\) thì
\(\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(2y-3\right)^2=0\\\left(z+2\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)
Vậy .....
Chúc bạn học tốt!!!
\(\dfrac{x^4+x^2y^2-x^3y-xy^3}{x^2+y^2}\)
\(=\dfrac{x^2\left(x^2+y^2\right)-xy\left(x^2+y^2\right)}{x^2+y^2}=x^2-xy\)
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)
c)(x2+x)2-2(x2+x)-15
đặt x2+x=a ta có
a2-2a-15
=a2+3a-5a-15
=(a2+3a)-(5a+15)
=a(a+3)-5(a+3)
=(a+3)(a-5)
thay a=x2+x
(x2+x+3)(x2+x-5)
\(\text{a) }\left(\dfrac{1}{2}a^2x^4+\dfrac{4}{3}\:ax^3-\dfrac{2}{3}ax^2\right):\left(-\dfrac{2}{3}\:ax^2\right)\\ =-3ax^2-2x+1\)
\(\text{b) }4\left(\dfrac{3}{4}x-1\right)+\left(12x^2-3x\right):\left(-3x\right)-\left(2x+1\right)\\ =3x-4-4x+1-2x-1\\ =-3x-4\)
kết quả cuối cùng là: a. -\(\dfrac{3}{4}ax^2-2x+1\)
b. \(\)-\(3x-4\)
a) \(7x^2-28=0\Leftrightarrow7\left(x^2-4\right)=0\Leftrightarrow x^2-4=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\) 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 \(x=\dfrac{2}{5};x=3\)
\(x^2-x+\dfrac{1}{2}=x^2-2\cdot\dfrac{1}{2}x+\dfrac{1}{4}-\dfrac{1}{4}+\dfrac{1}{2}\\ =\left(x^2-2\cdot\dfrac{1}{2}x+\dfrac{1}{4}\right)-\dfrac{1}{4}+\dfrac{1}{2}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\)
ta có: \(\left(x-\dfrac{1}{2}^{ }\right)^2\ge0\forall x\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}>0\forall x\left(vì\dfrac{1}{4}>0\right)\)
hay \(x^2-x+\dfrac{1}{2}>0\forall x\)