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) \(A=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)

\(A=\left[\left(x-1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\)

\(A=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

\(A=\left(x^2+5x\right)^2-6^2\)

\(A=\left(x^2+5x\right)^2-36\)

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

\(\Rightarrow Cmin=-14\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)

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

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

D = (x-1).(x+2).(x+3).(x+6)

= (x² + 5x - 6).(x² + 5x + 6)

= (x² + 5x)² + 6x.(x²+5x)-6(x² + 5x) - 36

= (x² + 5x)² - 36 \(\ge\) -36 với mọi x

Vậy D có GTNN = - 36 khi x² + 5x = 0

hay x = 0; x = 5

A = x² - 2x + y² + 4y + 8

= (x² - 2x + 1) + (y² + 2.2y + 4) + 3

= (x-1)² + (y+2)² + 3 \(\ge\) 3 với mọi x,y

Vậy A có GTNN = 3

C = x² - 4x + y² - 8y + 6

= (x² - 4x + 4) + (y² - 8y + 16) - 12

= (x-2)² + (y-4)² - 12 \(\ge\) -12 với mọi x;y

Vậy C có GTNN = -12

B = 2x² - 4x + 10

= x² + (x² - 4x + 4) + 6

= x² + (x-2)² + 6 \(\ge\) 6 với mọi x

Vậy B có GTNN = 6