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a: \(A=x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{1}{4}\)
\(=\left(x-\dfrac{5}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
Dấu '=' xảy ra khi x=5/2
b: \(B=x^2-4x+4+y^2-8y+16-14\)
\(=\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\)
Dấu '=' xảy ra khi x=2 và y=4
a: \(A=x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{1}{4}\)
\(=\left(x-\dfrac{5}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
Dấu '=' xảy ra khi x=5/2
b: \(B=x^2-4x+4+y^2-8y+16-14\)
\(=\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\)
Dấu '=' xảy ra khi x=2 và y=4
a, đặt ( x2+x)=y ta có :
y2+4y=12 <=> y2+4y-12=0
<=> y2+4y+4-16 =0
<=>(y2+4y+4)-16+=0
<=> (y+2)2-16=0
<=>(y-2)(y+6)=0
<=>y-2=0 hoặc y+6=0
<=> y=2 hoặc y=-6
<=> x2+x=2 hoặc x2+x=-6
<=> x2+x -2=0 hoặc x2+x+6=0(vô lý)
<=> (x-1)(x+2)=0 <=> x-1=0 hoặc x+2=0
<=> x=1 hoặc x=-2
vậy pt có nghiệm là x=1 và x=-2
b,6x4-5x3-38x2-5x+6=0
<=>6x4-18x3+13x3-39x2+x2-3x-2x+6=0
<=>6x3(x-3)+13x2(x-3)+x(x-3)-2(x-3)=0
<=>(x-3)(6x3+13x2+x-2)=0
<=>(x-3)(6x3+12x2+x2+2x-x-2)=0
<=>(x-3)(6x2(x+2)+x(x+2)-(x+2))=0
<=>(x-3)(x+2)(6x2+x-1)=0
<=>(x-3)(x+2)(3x-1)(2x+1)=0
tới đây tự làm
a) \(A=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(A=\left[\left(x-1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\)
\(A=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(A=\left(x^2+5x\right)^2-6^2\)
\(A=\left(x^2+5x\right)^2-36\)
Vì \(\left(x^2+5x\right)^2\ge0\) với mọi x
\(\Rightarrow\left(x^2+5x\right)^2-36\ge-36\)
\(\Rightarrow Amin=-36\Leftrightarrow x^2+5x=0\)
\(\Rightarrow x\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
b) \(B=x^2-2x+y^2+4y+8\)
\(B=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+3\)
\(B=\left(x-1\right)^2+\left(y+2\right)^2+3\)
Vì \(\left(x-1\right)^2\ge0\) với mọi x
\(\left(y+2\right)^2\ge0\) với mọi y
\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\) với mọi x,y
\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2+3\ge3\)
\(\Rightarrow Bmin=3\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
c) \(C=x^2-4x+y^2-8y+6\)
\(C=\left(x^2-4x+4\right)+\left(y^2-8y+16\right)-14\)
\(C=\left(x-2\right)^2+\left(y-4\right)^2-14\)
Vì \(\left(x-2\right)^2\ge0\) với mọi x
\(\left(y-4\right)^2\ge0\) với mọi y
\(\Rightarrow\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\) với mọi x,y
\(\Rightarrow Cmin=-14\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)
1, \(A=3x^2+5x-1\)
\(=3\left(x^2+\dfrac{5}{3}x-\dfrac{1}{3}\right)\)
\(=3\left(x^2+\dfrac{5}{6}.x.2+\dfrac{25}{36}-\dfrac{37}{36}\right)\)
\(=3\left(x+\dfrac{5}{6}\right)^2-\dfrac{37}{12}\ge\dfrac{-37}{12}\)
Dấu " = " khi \(3\left(x+\dfrac{5}{6}\right)^2=0\Leftrightarrow x=\dfrac{-5}{6}\)
Vậy \(MIN_A=\dfrac{-37}{12}\) khi \(x=\dfrac{-5}{6}\)
2,3 tương tự
4, \(A=2x^2+7x\)
\(=2\left(x^2+\dfrac{7}{4}.x.2+\dfrac{49}{16}-\dfrac{49}{16}\right)\)
\(=2\left(x+\dfrac{7}{4}\right)^2-\dfrac{49}{8}\ge\dfrac{-49}{8}\)
Dấu " = " khi \(2\left(x+\dfrac{7}{4}\right)^2=0\Leftrightarrow x=\dfrac{-7}{4}\)
Vậy \(MIN_A=\dfrac{-49}{8}\) khi \(x=\dfrac{-7}{4}\)
5, 6 tương tự
7, \(A=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(=\left(x^2+5x\right)^2-36\ge-36\)
Dấu " = " khi \(\left(x^2+5x\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Vậy \(MIN_A=-36\) khi x = 0 hoặc x = -5
8, \(A=x^2-4x+y^2-8x+6\)
\(=x^2-4x+4+y^2-8x+16-14\)
\(=\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\)
Dấu " = " khi \(\left\{{}\begin{matrix}\left(x-2\right)^2=0\\\left(y-4\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)
Vậy \(MIN_A=-14\) khi x = 2 và y = 4
1) \(A=36x^2+12x+1=\left(6x+1\right)^2\ge0\)
\(minA=0\Leftrightarrow x=-\dfrac{1}{6}\)
2) \(B=9x^2+6x+1=\left(3x+1\right)^2\ge0\)
\(minB=0\Leftrightarrow x=-\dfrac{1}{3}\)
4) \(D=x^2-4x+y^2-8y+6=\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\)
\(minD=-14\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)
3) \(C=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)=\left(x^2-5x-6\right)\left(x^2-5x+6\right)=\left(x^2-5x\right)^2-36\ge-36\)
\(minC\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
5) \(E=\left(x-8\right)^2+\left(x+7\right)^2=2x^2-2x+113=2\left(x-\dfrac{1}{2}\right)^2+\dfrac{225}{2}\ge\dfrac{225}{2}\)
\(minE=\dfrac{225}{2}\Leftrightarrow x=\dfrac{1}{2}\)
a: =-x^2+6x-4
=-(x^2-6x+4)
=-(x^2-6x+9-5)
=-(x-3)^2+5<=5
Dấu = xảy ra khi x=3
b: =3(x^2-5/3x+7/3)
=3(x^2-2*x*5/6+25/36+59/36)
=3(x-5/6)^2+59/12>=59/12
Dấu = xảy ra khi x=5/6
c: \(=-\left(x-3\right)^2+2\left|x-3\right|\)
\(=-\left[\left(\left|x-3\right|\right)^2-2\left|x-3\right|+1-1\right]\)
\(=-\left(\left|x-3\right|-1\right)^2+1< =1\)
Dấu = xảy ra khi x=4 hoặc x=2