1)b) \(x^2-y^2+14x+49\)

\(=x^2+14x+49-y^2\)

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

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

\(=\left(x+7-y\right)\left(x+7+y\right)\)

Khôi phục các đa thức sau:

GỌI ƯỚC CHUNG LỚN NHẤT LÀ d

3n+2 chia hết d --->4 (3n+2) chia hết d--->12n+8 chia hết d 

4n+3 chia hết d --->3 (4n+3) chia hết d--->12n+9 chia hết d 

ta có: (12n+9)-(12n+8) chia hết d

---> 1 chia hết d ---> d bằng 1

--->3n+2 và 4n+3 là 2 số nguyên tố cùng nhau

><  ​TICK MÌNH NHA

a) \( A=5x-x^2\)

\(\Leftrightarrow A=-x^2+2.x\dfrac{5}{2}-\left(\dfrac{5}{2}\right)^2+\left(\dfrac{5}{2}\right)^2\)

\(\Leftrightarrow A=-\left[x^2-2.x\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2\right]+\left(\dfrac{5}{2}\right)^2\)

\(\Leftrightarrow A=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\)

Vậy GTLN của \(A=\dfrac{25}{4}\) khi \(x=\dfrac{5}{2}\)

b) \(B=x-x^2\)

\(\Leftrightarrow B=-x^2+2.x\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^2\)

\(\Leftrightarrow B=-\left[x^2-2.x\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]+\left(\dfrac{1}{2}\right)^2\)

\(\Leftrightarrow B=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\)

Vậy GTLN của \(B=\dfrac{1}{4}\) khi \(x=\dfrac{1}{2}\)

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

\(\Leftrightarrow C=-x^2+4x-4+7\)

\(\Leftrightarrow C=-\left(x^2-2.x.2+2^2\right)+7\)

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

Vậy GTLN của \(C=7\) khi \(x=2\)

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

\(\Leftrightarrow\left(x^2-2.x.2+2^2\right)-\left(x^2-3^2\right)-6=0\)

\(\Leftrightarrow x^2-4x+4-x^2+9-6=0\)

\(\Leftrightarrow-4x+7=0\)

\(\Leftrightarrow-4x=-7\)

\(\Leftrightarrow x=\dfrac{-7}{-4}\)

\(\Leftrightarrow x=1,75\)

Vậy x=1,75

a) \(\left(x^2-9\right)^2+12x\left(x-3\right)^2\)

\(=\left[\left(x-3\right)\left(x+3\right)\right]^2+12x\left(x-3\right)^2\)

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

\(=\left(x-3\right)^2\left[\left(x+3\right)^2+12x\right]\)

\(=\left(x-3\right)^2\left(x^2+6x+3^2+12x\right)\)

\(=\left(x-3\right)^2\left(x^2+18x+9\right)\)

\(x^3\left(x^2+1\right)^2-49x\)

\(=x\left[x^2\left(x^2+1\right)^2-49\right]\)

\(=x\left[\left(x^3+x\right)^2-7^2\right]\)

\(=x\left(x^3+x-7\right)\left(x^3+x+7\right)\)

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

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

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

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

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

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

\(=\left(x+2\right)\left[x\left(x-1\right)+2\left(x-1\right)\right]\)

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

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

\(A=4x^2-8x+1\)

\(\Leftrightarrow A=4x^2-8x+4-3\)

\(\Leftrightarrow A=\left(2x-2\right)^2-3\)

Vậy GTNN của \(A=-3\) khi \(2x-2=0\Leftrightarrow x=1\)

a,Để 4n-5 chia hết 13 thì 4n-5 sẽ có dạng 13k

=>4n-5=13k

<=>4n=13k+5

=>\(n=\frac{13k+5}{4}\)