M=x(x-6)+74

=x²-6x+9+65

=(x-3)²+65

vì =(x-3)² >=0 nên M>=65

bạn làm tiêp nhé

x^2-6x+74

=x^2-6x +9+65

=(x-3)^2+65 > hoac=65

=>M_min=65

Tìm GTNN 

M=x(x-6)+74

x^2-6x+74

=x^2-6x +9+65

=(x-3)^2+65 > =65

=>M_min=65

A = x( 6 - x ) + 74 + x

A = 6x - x² + 74 + x

A = - x² + 7x + 74

A = - ( x² - 7x - 74 )

A = - [ x² - 2 . 7 / 2 + ( 7 / 2 )² - ( 7 / 2 )² - 74 ]

A = - ( x - 7 / 2 )² - 345 / 2 \(\le\)- 345 / 2

Dấu= xảy ra \(\Leftrightarrow\)x - 7 / 2 = 0

                       \(\Rightarrow\)x              = 7 / 2

Vậy : Max A = - 345 / 2 \(\Leftrightarrow\)x = 7 / 2

\(x\left(x-6\right)+74+x\)

\(=x^2-6x+74+x\)

\(=x^2-5x+74\)

\(=\left(x^2-2.x.\frac{5}{2}+\frac{25}{4}\right)+\frac{271}{4}\)

\(=\left(x-\frac{5}{2}\right)^2+\frac{271}{4}\ge\frac{271}{4}\)

Dấu '' = '' xảy ra 

\(\Leftrightarrow x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\)

Vậy..................

P/s : chưa kt lại bài nên sai bỏ qua

e tưởng\(\le\frac{-345}{4}\)

\(A=\frac{x\left(x-6\right)+74}{13}=\frac{x^2-6x+74}{13}=\frac{\left(x^2-6x+9\right)+65}{13}=\frac{\left(x-3\right)^2}{13}+\frac{65}{13}\ge\frac{65}{13}\)

Dấu "=" xảy ra tại \(x=3\)

\(A=\frac{x\left(x-6\right)+74}{13}=\frac{x^2-6x+9+65}{13}=\frac{\left(x-3\right)^2+65}{13}=\frac{\left(x-3\right)^2}{13}+5\)

Vì \(\left(x-3\right)^2\ge0\)\(\Rightarrow\frac{\left(x-3\right)^2}{13}\ge0\)\(\Rightarrow A\ge5\)

Dấu bằng xảy ra khi \(x-3=0\)\(\Leftrightarrow x=3\)

Vậy \(minA=5\Leftrightarrow x=3\)

\(M=x\left(x-6\right)+74=x^2-6x+74=\left(x^2-6x+9\right)+65=\left(x-3\right)^2+65\ge65\)

\(minM=65\Leftrightarrow x=3\)

\(M=x\left(x-6\right)+74\)

\(=x^2-6x+9+65\)

\(=\left(x-3\right)^2+65\ge65\forall x\)

Dấu '=' xảy ra khi x=3

em cảm ơn ah 

M = x(6 - x) + 74 + x

M = 6x - x² + 74 + x

M = 74 + 7x - x²

M = 345/4 - 49/4 + 7/2x + 7/2x - x²

M = 345/4 - 7/2.(7/2 - x) + x.(7/2 - x)

M = 345/4 - (7/2 - x)² \(\le\frac{345}{4}\)

Dấu "=" xảy ra khi (7/2 - x)² = 0

<=> 7/2 - x = 0

<=> x = 7/2

Vậy Max M = 345/4 khi x = 7/2

86,25 tại x=3,5

\(M=\dfrac{x+6\sqrt{x}+9+25}{\sqrt{x}+3}=\dfrac{\left(\sqrt{x}+3\right)^2+25}{\sqrt{x}+3}=\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}\)Áp dụng Cô si có

\(M\ge2\sqrt{\left(\sqrt{x}+3\right).\dfrac{25}{\sqrt{x}+3}}=10\)

Dấu "=" \(\sqrt{x}+3=\dfrac{25}{\sqrt{x}+3}\leftrightarrow x=4\)

Vậy GTNN của M = 10 <=> x = 4

\(M=\dfrac{\left(x+6\sqrt{x}+9\right)+25}{\sqrt{x}+3}=\dfrac{\left(\sqrt{x}+3\right)^2+25}{\sqrt{x}+3}=\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}\)

Do \(\sqrt{x}\ge0\Rightarrow\left\{{}\begin{matrix}\sqrt{x}+3>0\\\dfrac{25}{\sqrt{x}+3}>0\end{matrix}\right.\)

Áp dụng bđt cô-si ta có: 

\(\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}\ge2\sqrt{\left(\sqrt{x}+3\right)\cdot\dfrac{25}{\sqrt{x}+3}}=2\sqrt{25}=10\)

hay \(M\ge10\)

Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}+3=\dfrac{25}{\sqrt{x}+3}\Leftrightarrow x=4\)

Vậy GTNN của M = 10 khi x = 4

\(\dfrac{x+6\sqrt{x}+9+25}{\sqrt{x+3}}\)

=\(\dfrac{\sqrt{x}+2.3.\sqrt{x}+3^2+25}{\sqrt{x}+3}\)

=\(\dfrac{\left(\sqrt{x}+3\right)^2+25}{\sqrt{x}+3}\)

=\(\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}\)

áp dụng cosi

M≥\(^2\sqrt{\left(\sqrt{x}+3\right).\dfrac{25}{\sqrt{x}+3}}\)=10

\(\sqrt{x}+3\)=\(\dfrac{25}{\sqrt{x}+3}\)⇔x=4

vậy...

Ta có:

\(\hept{\begin{cases}x+y=m\\x^2+y^2=-m^2+6\end{cases}\Leftrightarrow\hept{\begin{cases}x+y=m\\\left(x+y\right)^2-2xy=-m^2+6\end{cases}\Leftrightarrow}\hept{\begin{cases}x+y=m\\xy=m^2-3\end{cases}}}\)

Suy ra:

\(P=xy+2\left(x+y\right)=m^2-3+2m=\left(m^2+2m+1\right)-4=\left(m+1\right)^2-4\ge-4\)

Vậy GTNN của P là -4 khi m = -1.

vậy GTNN p = -4 khi m=-1

1.

$x(x+2)(x+4)(x+6)+8$

$=x(x+6)(x+2)(x+4)+8=(x^2+6x)(x^2+6x+8)+8$

$=a(a+8)+8$ (đặt $x^2+6x=a$)

$=a^2+8a+8=(a+4)^2-8=(x^2+6x+4)^2-8\geq -8$

Vậy $A_{\min}=-8$ khi $x^2+6x+4=0\Leftrightarrow x=-3\pm \sqrt{5}$

2.

$B=5+(1-x)(x+2)(x+3)(x+6)=5-(x-1)(x+6)(x+2)(x+3)$

$=5-(x^2+5x-6)(x^2+5x+6)$

$=5-[(x^2+5x)^2-6^2]$

$=41-(x^2+5x)^2\leq 41$

Vậy $B_{\max}=41$. Giá trị này đạt tại $x^2+5x=0\Leftrightarrow x=0$ hoặc $x=-5$

3.

Đặt $x+3=a; 7-x=b$ thì $a+b=10$ 

$C=a^4+b^4$

Áp dụng BĐT Bunhiacopxky:

$(a^4+b^4)(1+1)\geq (a^2+b^2)^2$

$\Rightarrow C\geq \frac{(a^2+b^2)^2}{2}$
$(a^2+b^2)(1+1)\geq (a+b)^2=100$

$\Rightarrow a^2+b^2\geq 50$

$\Rightarrow C\geq \frac{50^2}{2}=1250$

Vậy $C_{\min}=1250$

Giá trị này đạt tại $a=b=5\Leftrightarrow x=2$

\(M=\text{(x+1)(x-2)(x-3)(x-6)}\)

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

\(=\left(x^2-5x\right)^2-6^2=\left(x^2-5x\right)^2-36\ge-36\)( Vì \(\left(x^2-5x\right)^2\ge0\))

Vậy \(MinM=-36\Leftrightarrow\left(x^2-5x\right)^2=0\Leftrightarrow x^2-5x=0\Leftrightarrow x\left(x-5\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}}\)

nỏ hiểu giải thích chữgtnn

M=(x+1)(x-2)(x-3)(x-6)

=(x2−5x−6)(x2−5x+6)

=(x2−5x)2−62=(x2−5x)2−36≥−36( Vì (x2−5x)2≥0)

Vậy MinM=−36⇔(x2−5x)2=0⇔x2−5x=0⇔x(x−5)=0⇔[

⇔[