a) \(\left|5x+4\right|+\left|5x-1\right|=\frac{10}{\left|y-7\right|^2+2}\)(1)

Ta có: 

\(\left|5x+4\right|+\left|5x-1\right|=\left|5x+4\right|+\left|1-5x\right|\ge\left|5x+4+1-5x\right|=5\)

Dấu \(=\)khi \(\left(5x+4\right)\left(1-5x\right)\ge0\Leftrightarrow\frac{-4}{5}\le x\le\frac{1}{5}\).

\(\frac{10}{\left|y-7\right|^2+2}\le\frac{10}{2}=5\)

Dấu \(=\)khi \(y-7=0\Leftrightarrow y=7\).

Do đó (1) xảy ra khi \(\left|5x+4\right|+\left|5x-1\right|=\frac{10}{\left|y-7\right|^2+2}=5\)

\(\Leftrightarrow\hept{\begin{cases}-\frac{4}{5}\le x\le\frac{1}{5}\\y=7\end{cases}}\)

mà \(x\)nguyên nên \(x=0\).

Nghiệm nguyên của phương trình đã cho là \(\left(0,7\right)\).

b) Tương tự a).

\(B=\left(x-3\right)^2+\left(x-11\right)^2\)

\(=x^2-6x+9+x^2-22x+121\)

\(=2\left(x^2-14x+49\right)+32\)

\(=2\left(x-7\right)^2+32\)

Ta có: \(\left(x-7\right)^2\ge0\Leftrightarrow2\left(x-7\right)^2+32\ge32\)

Vậy \(MinB=32\Leftrightarrow x=7\)

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

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

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

Đặt \(c=x^2-5x\)lúc này \(C\)thành: \(C=\left(c-6\right)\left(c+6\right)=c^2-36\)

Mà: \(c^2\ge0\forall c\Leftrightarrow c^2-36\ge-36\Leftrightarrow C\ge-36\)

Dấu '' = '' xảy ra: \(c=0\)

\(\Leftrightarrow x^2-5x=0\)

\(\Leftrightarrow x\left(x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)

Vậy \(MinC=-36\Leftrightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)

1, \(\left(x-2\right)^3-\left(x-5\right)^3=\left(x-2-x+5\right)\left[\left(x-2\right)^2+\left(x-2\right)\left(x-5\right)+\left(x-5\right)^2\right]\)

\(=3\left(x^2-4x+4+x^2-7x+10+x^2-10x+25\right)\)

\(=3\left(3x^2-21x-11\right)=9x^2-63x-33\)

2, \(\left(x+2\right)\left(x^2-2x+4\right)-\left(x^3-8\right)=x^3+8-x^3+8=16\)

3, \(\left(x-3\right)\left(x^2+3x+9\right)-\left(x^2-1\right)\left(x+27\right)\)

\(=x^3-27-x^3-27x^2+x+27=-27x^2+x\)

bạn đăng tách ra cho mn cùng giúp nhé 

9, \(x\left(x-14\right)-10\left(x-1\right)^2=x^2-14x-10\left(x^2-2x+1\right)\)

\(=x^2-14x-10x^2+20x-10=-9x^2+6x-10\)

10, \(\left(x+3\right)\left(x-4\right)-\left(x+4\right)^2=x^2-x-12-x^2-8x-16=-9x-28\)

11, \(2x\left(x+2\right)-\left(x+2\right)\left(x-2\right)=2x^2+4x-x^2+4=x^2+4x+4\)

12, \(\left(2x-3\right)^2-\left(2x+1\right)\left(2x-1\right)+3\left(2x-3\right)\)

\(=4x^2-12x+9-4x^2+1+6x-9=-6x+1\)

1, \(2x^2-8xy-5x+20y=2x\left(x-4y\right)-5\left(x-4y\right)=\left(2x-5\right)\left(x-4y\right)\)

2, \(x^3-x^2y-xy+y^2=x^2\left(x-y\right)-y\left(x-y\right)=\left(x^2-y\right)\left(x-y\right)\)

Ta có A = 25x² + 3y² - 10x + 11

= (25x² - 10x + 1) + 3y² + 10

= (5x - 1)² + 3y² + 10 \(\ge\)10

=> Min A = 10

Dấu "=" xảy ra <=> \(\hept{\begin{cases}5x-1=0\\y=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{5}\\y=0\end{cases}}\)

Vậy Min A = 10 <=> x = 1/5 ; y = 0 

Trả lời:

\(\sqrt{4x^2}=\sqrt{\left(2x\right)^2}=\left|2x\right|\)

+) TH1:  Nếu \(x\ge0\Rightarrow\left|2x\right|=2x\)

+) TH2: Nếu \(x< 0\Rightarrow\left|2x\right|=-2x\)