\(P=\left(x^2-4x-5\right)\left(x^2-4x-5-14\right)+49\)

Đặt \(x^2-4x-5=a\)

\(P=a\left(a-14\right)+49=a^2-14a+49=\left(a-7\right)^2\ge0\)

\(\Rightarrow P_{min}=0\) khi \(a=7\Rightarrow x^2-4x-5=7\Rightarrow\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)

tim giá trị lớn nhất hoặc nhỏ nhất

A=\(\left(x^2-4x-5\right)\left(x^2-4x-19\right)+49\)

Đặt \(x^2-4x-12=y\)

\(\Rightarrow\)A=\(\left(y+7\right)\left(y-7\right)+49\)

\(\Leftrightarrow\)A= \(y^2-49+49\)

\(\Leftrightarrow\)A=\(y^2\)

Ta có \(y^2\ge0\forall y\)

Hay \(\left(x^2-4x-12\right)^2\ge0\forall x\)

\(\Rightarrow\)Min A=0\(\Leftrightarrow\)\(x^2-4x-12=0\)

\(\Leftrightarrow\)\(\left(x+2\right)\left(x-6\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x+2=0\\x-6=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=-2\\x=6\end{cases}}\)

Đặt \(a=x^2-4x-12\) thay vào N:

\(N=\left(x^2-4x-5\right)\left(x^2-4x-19\right)+49\)

\(=\left(a+7\right)\left(a-7\right)+49\)\(=a^2-49+49\)\(=a^2\)

Ta có: N = \(a^2\ge0\) \(\left(\forall a\right)\)

\(\Rightarrow\)MIN N = 0 \(\Leftrightarrow a^2=0\Leftrightarrow a=0\)

Hay \(x^2-4x-12=0\Leftrightarrow x^2-4x+4-16=0\)\(\Leftrightarrow\left(x-2\right)^2-4^2=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-6\right)=0\)

\(\Leftrightarrow x=-2;x=6\)

Vậy Min A = 0 \(\Leftrightarrow x=-2;x=6\)

Bài 2: 

a: \(A=x^2+8x\)

\(=x^2+8x+16-16\)

\(=\left(x+4\right)^2-16\ge-16\)

Dấu '=' xảy ra khi x=-4

b: \(B=-2x^2+8x-15\)

\(=-2\left(x^2-4x+\dfrac{15}{2}\right)\)

\(=-2\left(x^2-4x+4+\dfrac{7}{2}\right)\)

\(=-2\left(x-2\right)^2-7\le-7\)

Dấu '=' xảy ra khi x=2

c: \(C=x^2-4x+7\)

\(=x^2-4x+4+3\)

\(=\left(x-2\right)^2+3\ge3\)

Dấu '=' xảy ra khi x=2

e: \(E=x^2-6x+y^2-2y+12\)

\(=x^2-6x+9+y^2-2y+1+2\)

\(=\left(x-3\right)^2+\left(y-1\right)^2+2\ge2\)

Dấu '=' xảy ra khi x=3 và y=1

1. \(4x^2-49=0\)

\(\Leftrightarrow\left(2x+7\right)\left(2x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+7=0\Leftrightarrow x=-\dfrac{7}{2}\\2x-7=0\Leftrightarrow x=\dfrac{7}{2}\end{matrix}\right.\)

Vậy: \(x=-\dfrac{7}{2}\) hoặc \(x=\dfrac{7}{2}\)

===========

2. \(x^2+36=12x\)

\(\Leftrightarrow x^2-12x+36=0\)

\(\Leftrightarrow\left(x-6\right)^2=0\)

\(\Leftrightarrow x=6\)

Vậy: \(x=6\)

===========

3. \(10\left(x-5\right)-8x\left(5-x\right)=0\)

\(\Leftrightarrow10\left(x-5\right)+8x\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(10+8x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\Leftrightarrow x=5\\10+8x=0\Leftrightarrow x=-\dfrac{5}{4}\end{matrix}\right.\)

Vậy: \(x=5\) hoặc \(x=-\dfrac{5}{4}\)

1: Ta có: \(4x^2-49=0\)

\(\Leftrightarrow\left(2x-7\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

2: Ta có: \(x^2+36=12x\)

\(\Leftrightarrow x^2-12x+36=0\)

\(\Leftrightarrow\left(x-6\right)^2=0\)

\(\Leftrightarrow x-6=0\)

hay x=6

a. 4x²-25=0

=> (2x)²-5²=0

=> (2x-5)(2x+5)=0

=> 2x-5=0    hoặc 2x+5=0

=> 2x=5        hoặc 2x=-5

=> x=5:2       hoặc x=-5:2

=> x=2,5        hoặc x=-2,5

b. (x-1)(4x²-49)=0

=> (x-1)[(2x)²-7²   ]=0

=> (x-1)(2x-7)(2x+7)=0

=> x-1=0     hoặc 2x-7=0           hoặc 2x+7=0

=> x=1         hoặc 2x=7               hoặc 2x=-7

=> x=1         hoặc x=7:2=3,5       hoặc x=-7:2=-3,5