a) \(A=x^2-10x+5\)

\(A=x^2-10x+25-20\)

\(A=\left(x-5\right)^2-20\ge-20\)

Min A = -20 \(\Leftrightarrow x=5\)

b) \(B=3x^2-6x+11\)

\(B=3\left(x^2-2x+1\right)+8\)

\(B=3\left(x-1\right)^2+8\ge8\)

Min B = 8\(\Leftrightarrow x=1\)

a) \(A=x^2-10x+5=\left(x^2-10x+25\right)-20\)

\(=\left(x-5\right)^2-20\ge-20\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left(x-5\right)^2=0\Rightarrow x=5\)

Vậy \(Min_A=-20\Leftrightarrow x=5\)

b) \(B=3x^2-6x+11=3\left(x^2-2x+1\right)+8\)

\(=3\left(x-1\right)^2+8\ge8\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left(x-1\right)^2=0\Rightarrow x=1\)

Vậy \(Min_B=8\Leftrightarrow x=1\)

c) \(C=8x^2+10x-30=8\left(x^2-\frac{5}{4}x+\frac{25}{64}\right)-\frac{265}{8}\)

\(=8\left(x-\frac{5}{8}\right)^2-\frac{265}{8}\ge-\frac{265}{8}\)

Dấu "=" xảy ra khi: \(\left(x-\frac{5}{8}\right)^2=0\Rightarrow x=\frac{5}{8}\)

Vậy \(Min_C=-\frac{265}{8}\Leftrightarrow x=\frac{5}{8}\)

A = x² - 10x + 5

A = ( x² - 10x + 25 ) - 20

A = ( x - 5 )² - 20

( x - 5 )² ≥ 0 ∀ x => ( x - 5 )² - 20 ≥ -20

Đẳng thức xảy ra <=> x - 5 = 0 => x = 5

=> MinA = -20 <=> x = 5

b) B = 3x² - 6x + 11

B = 3( x² - 2x + 1 ) + 8

B = 3( x - 1 )² + 8

3( x - 1 )² ≥ 0 ∀ x => 3( x - 1 )² + 8 ≥ 8

Đẳng thức xảy ra <=> x - 1 = 0 => x = 1

=> MinB = 8 <=> x = 1

C = 8x² + 10x - 30

C = 8( x² + 5/4x + 25/64 ) - 265/8

C = 8( x + 5/8 )² - 265/8

8( x + 5/8 )² ≥ 0 ∀ x => 8( x + 5/8 )² - 265/8 ≥ -265/8

Đẳng thức xảy ra <=> x + 5/8 = 0 => x = -5/8

=> MinC = -265/8 <=> x = -5/8

\(F=\left(x-2\right)\left(x-5\right)\left(x^2-7x-10\right)\)

\(=\left(x^2-7x+10\right)\left(x^2-7x-10\right)\)

Đặt \(x^2-7x=a\)

\(F=\left(a-10\right)\left(a+10\right)=a^2-100\ge-100\)

\(\Rightarrow F_{min}=-100\Leftrightarrow x^2-7x=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=7\end{cases}}\)

a) \(A=x^2-10+30=\left(x-5\right)^2+5\)

Mà \(\left(x+5\right)^2\ge0\)

=> GTNN của A = 5

b) \(B=x^2-12\text{x}+8=\left(x-6\right)^2-4\)

Mà \(\left(x-6\right)^2\ge0\)

=> GTLN = -4

a,\(x^2-6x-17=x^2-2\cdot3x+9-26=\left(x-3\right)^2-26\ge-26\)

b, \(x^2-10x=x^2-2\cdot5x+25-25=\left(x-5\right)^2-25\ge-25\)

c,\(3x^2-12x+5=3x^2-2\cdot\sqrt{3}x\cdot2\sqrt{3}+12-7=\left(\sqrt{3}x-2\sqrt{3}\right)^2-7\ge-7\)

d,\(2x^2-x-1=2x^2-2\cdot\sqrt{2}x\cdot\dfrac{1}{2\sqrt{2}}+\dfrac{1}{8}-\dfrac{9}{8}=\left(\sqrt{2}x-\dfrac{1}{2\sqrt{2}}\right)^2-\dfrac{9}{8}\ge-\dfrac{9}{8}\)

e,\(x^2+y^2-8x+4y+27=x^2-2\cdot4x+16+y^2+2\cdot2y+4+7=\left(x-4\right)^2+\left(y+2\right)^2+7\ge7\)

f,\(x\left(x-6\right)=x^2-6x=x^2-2\cdot3x+9-9=\left(x-3\right)^2-9\ge-9\)

h,\(\left(x-2\right)\cdot\left(x-5\right)\cdot\left(x^2-7x-10\right)=\left(x^2-7x+10\right)\left(x^2-7x-10\right)=\left(x^2-7x\right)^2-100\ge-100\)

Mình giúp tính biểu thức thôi

còn lại bạn tự làm nhé

Toàn bộ đều tìm Max :)

D = -x² + 30x - 10

D = -( x² - 30x + 225 ) + 215

D = -( x - 15 )² + 215

-( x - 15 )² ≤ 0 ∀ x => -( x - 15 )² + 215 ≤ 215

Đẳng thức xảy ra <=> x - 15 = 0 => x = 15

=> MaxD = 215 <=> x = 15

E = -2x² + 9x + 30

E = -2( x² - 9/2x + 81/16 ) + 321/8

E = -2( x - 9/4 )² + 321/8

-2( x - 9/4 )² ≤ 0 ∀ x => -2( x - 9/4 )² + 321/8 ≤ 321/8

Đẳng thức xảy ra <=> x - 9/4 = 0 => x = 9/4

=> MaxE = 321/8 <=> x = 9/4

F = -5x² - 20x - 4

F = -5( x² + 4x + 4 ) + 16

F = -5( x + 2 )² + 16

-5( x + 2 )² ≤ 0 ∀ x => -5( x + 2 )² + 16 ≤ 16

Đẳng thức xảy ra <=> x + 2 = 0 => x = -2

=> MaxF = 16 <=> x = -2

d) \(D=-x^2+30x-10\)

\(D=-\left(x^2-30x+10\right)\)

\(D=\left(x^2-30x+225-215\right)\)

\(D=-\left(x-15\right)^2+215\le215\)

Max D = 215 \(\Leftrightarrow x=15\)

e) \(E=-2x^2+9x+30\)

\(E=-2\left(x^2-\frac{9}{2}x-15\right)\)

\(E=-2\left(x-\frac{9}{4}\right)^2+\frac{321}{8}\le\frac{321}{8}\)

Max \(E=\frac{321}{8}\Leftrightarrow x=\frac{9}{4}\)

f) \(F=-5x^2-20x-4\)

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

\(F=-5\left(x^2+4x+4+\frac{16}{5}\right)\)

\(F=-5\left(x+2\right)^2-16\le-16\)

Max F = -16 \(\Leftrightarrow x=-2\)

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^

A = 2.(x^2-4x+4) - 18 = 2.(x-2)^2 - 18 >= -18

Dấu "=" xảy ra <=> x-2 = 0 <=> x=2 

Vậy Min A = -18 <=> x=2