a.A= \(x^2+10x+27\)

\(=x^2+2.x.5+25+2\)

\(\left(x+5\right)^2+2\ge2\forall x\)

Dấu " = " xảy ra <=> x + 5 = 0

=> x = -5

Vậy Min A = 2 <=> x = -5

b.B = \(x^2-12x+37\)

\(=x^2-2.x.6+36+1\)

\(=\left(x-6\right)^2+1\ge1\forall x\)

Dấu " = " xảy ra <=> x - 6 = 0

=> x = 6

Vậy Min B = 1 <=> x = 6

c. \(x^2+x+7\)

\(=x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{27}{4}\)

\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{27}{4}\ge\dfrac{27}{4}\forall x\)

Dấu " =" xảy ra <=> \(x+\dfrac{1}{2}=0\)

\(x=\dfrac{-1}{2}\)

Vậy Min C = \(\dfrac{27}{4}\Leftrightarrow x=\dfrac{-1}{2}\)

Hà An bn ơi lm giúp mk mấy cau còn lại vs mai mk đi hok rồi

a, x² + 10x + 27

Đặt A = x² + 2. x. 5 + 5² + 2

= ( x + 5 )² + 2

Vì ( x + 5 )² \(\ge\)0 với mọi x

=> ( x + 5 )² + 2 \(\ge\)2 với mọi x

Hay A \(\ge\)2

Dấu " = " xảy ra khi:

( x + 5 )² = 0

x + 5 = 0

x = - 5

Vậy Min A = 2 khi x = - 5

b, x² + x + 7

Đặt B = x2 + x + 7

\(=x^2+x+\frac{1}{4}+\frac{27}{4}\)

\(=\left[x^2+2\cdot x\cdot\frac{1}{2}+\left(\frac{1}{2}\right)^2\right]+\frac{27}{4}\)

\(=\left(x+\frac{1}{2}\right)^2+\frac{27}{4}\)

Vì \(\left(x+\frac{1}{2}\right)^2\ge0\)với mọi x

\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{27}{4}\ge\frac{27}{4}\)với mọi x

Hay B \(\ge\frac{27}{4}\)

Dấu " = " xảy ra khi:

\(\left(x+\frac{1}{2}\right)^2=0\)

\(x+\frac{1}{2}=0\)

\(x=-\frac{1}{2}\)

Vậy Min B = \(\frac{27}{4}\)khi x = \(-\frac{1}{2}\)

a) x² + 10 x + 27 =( x² + 2. 5 . x + 5² ) + 2 = ( x + 5 ) ² + 2 

Vì ( x + 5 ) ² \(\ge\) 0 với mọi x nên ( x + 5 ) ² + 2 \(\ge\) 2 với mọi x

Dấu bằng xảy ra \(\Leftrightarrow\)x + 5 = 0 \(\Leftrightarrow\) x = -5

b) x² + x + 7 = 0 \(\Leftrightarrow\) x² + 2. x . \(\frac{1}{2}\)+  \(\left(\frac{1}{2}\right)^2\) + \(\frac{27}{4}\) = 0 \(\Leftrightarrow\)( x + 1/2) ² + 27/4  = 0

Vì  ( x + 1/2 )² \(\ge\) 0 với mọi x nên ( x + 1/2) ² + 27/4 \(\ge\)27/4 với mọi x

Dấu bằng xảy ra \(\Leftrightarrow\)x+ 1/2 = 0 \(\Leftrightarrow\) x = ---\(\frac{1}{2}\) 

c + d ) Tương tự a, b

e) x² + 14 x + y² - 2y +7 = 0 \(\Leftrightarrow\) ( x² + 2. x. 7 + 7² ) + ( y² -- 2y + 1 ) -43 = 0 \(\Leftrightarrow\) ( x + 7 ) ² + ( y -- 1 ) ²  --43 = 0 ( 1 ) 

Vì ( x + 7 )² \(\ge\)  0 và ( y -- 1 )² \(\ge\) 0 với mọi x, y nên  ( 1 ) \(\ge\) --43 với mọi x, y

Dấu bằng xảy ra \(\Leftrightarrow\) \(\hept{\begin{cases}x+7=0\\y-1=0\end{cases}}\) \(\Leftrightarrow\) \(\hept{\begin{cases}x=-7\\y=1\end{cases}}\)

a: \(=\dfrac{3x\left(x-y\right)^2\cdot\left(x-1\right)}{3x\left(x-1\right)\cdot\left(x-y\right)^2\cdot2\cdot\left(x-y\right)}=\dfrac{1}{2\left(x-y\right)}\)

b: =(x+1)^2/(x+1)=x+1

c: \(=\dfrac{a\left(a^2-4a+4\right)}{\left(a-2\right)\left(a+2\right)}=\dfrac{a\left(a-2\right)^2}{\left(a-2\right)\left(a+2\right)}=\dfrac{a\left(a-2\right)}{a+2}\)

d: \(=\dfrac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\dfrac{7\left(x+1\right)}{3x}\)

a. 2x

b.\({3x}\over x^2-1\)

1) x² - 4 = 0

=> x² = 4

=> x = \(\pm\)2

2) 2x² - 8 = 0

=> 2x² = 8

=> x² = 4

=> x = \(\pm2\)

3) (x + 3)² = 4 => (x + 3)² = 2²

=> \(\orbr{\begin{cases}x+3=2\\x+3=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=-5\end{cases}}\)

4) (x - 7)² = 36

=> (x - 7)² = 6²

=> \(\orbr{\begin{cases}x-7=6\\x-7=-6\end{cases}}\Rightarrow\orbr{\begin{cases}x=13\\x=1\end{cases}}\)

5) x² - 14x = -49

=> x² - 14x + 49 = 0

=> x² - 7x - 7x + 49 = 0

=> x(x - 7) - 7(x - 7) = 0

=> (x - 7)² = 0

=> x = 7

6) x² + 6x + 5 = 0

=> x² + x + 5x + 5 = 0

=> x(x + 1) + 5(x + 1) = 0

=> (x + 1)(x + 5) = 0

=> \(\orbr{\begin{cases}x=-1\\x=-5\end{cases}}\)

7) x² - 14x + 13 = 0

=> x² - x - 13x + 13 = 0

=> x(x - 1) - 13(x - 1) = 0

=> (x - 1)(x - 13) = 0

=> \(\orbr{\begin{cases}x=1\\x=13\end{cases}}\)

8) x² + 10x +16 = 0

=> x² + 2x + 8x + 16 = 0

=> x(x + 2) + 8(x + 2) = 0

=> (x + 2)(x + 8) = 0

=> \(\orbr{\begin{cases}x=-2\\x=-8\end{cases}}\)

a) Ta có: \(8x^2+30x+7\)

\(=8x^2+28x+2x+7\)

\(=4x\left(2x+7\right)+\left(2x+7\right)\)

\(=\left(2x+7\right)\left(4x+1\right)\)

b) Ta có: \(x^2+14x+48\)

\(=x^2+8x+6x+48\)

\(=x\left(x+8\right)+6\left(x+8\right)\)

\(=\left(x+8\right)\left(x+6\right)\)

c) Ta có: \(x^8+x+1\)

\(=x^8+x^7-x^7+x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x+1\)

\(=\left(x^8+x^7+x^6\right)-\left(x^7+x^6+x^5\right)+\left(x^5+x^4+x^3\right)-\left(x^4+x^3+x^2\right)+\left(x^2+x+1\right)\)

\(=x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)