Sử dụng phép chia đa thức \(2n^2+5n-1\)cho n-1. Ta có được

\(2n^2+5n-1=\left(n-1\right)\left(2n+7\right)+6\)

Để \(2n^2+5n-1\)chia hết cho n-1 thì 6 phải chia hết cho n-1 => n-1 là ước của 6 ,

\(n-1\in U\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)và n-1 khác 0.

Bạn tự làm tiếp nhé!

Để 2n² + 5n - 1 chia hết cho n - 1

=> 2n² - 2n + 7n - 7 + 6 chia hết cho n - 1

2n.(n-1) + 7.(n-1) + 6 chia hết cho n - 1

(n-1).(2n+7) + 6 chia hết cho n - 1

mà (n-1).(2n+1) chia hết cho n - 1

=> 6 chia hết cho n - 1

=>  n - 1 thuộc Ư(6)={1;-1;2;-2;3;-3;6;-6}

nếu n - 1 = 1 => n = 2 (TM)

...

bn tự xét tiếp nha!

\(A=x^2\left(x+y\right)+y^2\left(x+y\right)+2x^2y+2xy^2\)

\(A=x^2\left(x+y\right)+y^2\left(x+y\right)+2xy\left(x+y\right)\)

\(A=\left(x+y\right)\left(x^2+2xy+y^2\right)\)

\(A=\left(x+y\right)\left(x^2+2xy+y^2\right)\)

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

\(A=\left(x+y\right)^3\)

\(x\left(3x+2\right)+\left(x+1\right)^2-\left(2x-5\right)\left(2x+5\right)=-12\)

\(3x^2+2x+x^2+2x+1-4x^2+25=-12\)

\(4x+26=-12\)

\(4x=-38\)

\(x=\frac{-19}{2}\)

x(3x+2) + (x+1)² - (2x-5)(2x+5)= -12

(3x²+2x) + (x²+2x+1) - (4x² - 25) = -12

3x² + 2x + x² + 2x + 1 - 4x² +25 = -12

(3x² + x² - 4x²) + ( 2x+2x) + (1+25) = -12

0 + 4x + 26 = -12

4x = -12 - 26

4x = -38

x = -9.5

a, \(x^2-y^2-2x-2y\)

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

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

\(18m^2-36mn+18n^2-72p^2\)

\(=18\left(m^2-2mn+n^2-4p^2\right)\)

\(=18.\left[\left(m-n\right)^2-\left(2p\right)^2\right]\)

\(=18.\left(m-n-2p\right)\left(m-n+2p\right)\)

a) \(\left(2x+1\right)^2-4\left(x+2\right)^2=9\)

\(\left(2x+1\right)^2-\left[2\left(x+2\right)\right]^2=9\)

\(\left[2x+1-2\left(x+2\right)\right]\left[2x+1+2\left(x+2\right)\right]=9\)

\(\left(2x+1-2x-4\right)\left(2x+1+2x+4\right)=9\)

\(-3\left(4x+5\right)=9\)

\(4x+5=-3\)

\(4x=-8\)

\(x=-2\)

b) \(x^2-2x-15=0\)

\(x^2-5x+3x-15=0\)

\(x\left(x-5\right)+3\left(x-5\right)=0\)

\(\left(x-5\right)\left(x+3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-3\end{cases}}}\)

c) \(2x^2+3x-5=0\)

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

\(2x\left(x-1\right)+5\left(x-1\right)=0\)

\(\left(x-1\right)\left(2x+5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\2x+5=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{-5}{2}\end{cases}}}\)

\(x^2-7x+9\)

\(=x^2-2.x.\frac{7}{2}+\left(\frac{7}{2}\right)^2-\frac{13}{4}\)

\(=\left(x-\frac{7}{2}\right)^2-\left(\frac{\sqrt{13}}{2}\right)^2\)

\(=\left(x-\frac{7}{2}-\frac{\sqrt{13}}{2}\right)\left(x-\frac{7}{2}+\frac{\sqrt{13}}{2}\right)\)

\(=\left(x-\frac{7+\sqrt{13}}{2}\right)\left(x-\frac{7-\sqrt{13}}{2}\right)\)

Với a, b dương:

\(8^2=\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}\right)^2\ge\frac{4}{\sqrt{ab}}\)

\(\Rightarrow\frac{1}{\sqrt{ab}}\le\frac{64}{4}=16\)

max A=16 khi a=b=1/4