Đặt \(\hept{\begin{cases}a-23=m^2\\a+22=n^2\end{cases}}\left(m,n\inℕ\right)\)

Ta có : \(a+22>a-23\Rightarrow n^2>m^2\)

\(\Rightarrow n^2-m^2=a+22-\left(a-23\right)\)

\(\Rightarrow n^2-m^2=a+22-a+23\)

\(\Rightarrow\left(n-m\right)\left(n+m\right)=45\)

Từ đây ta lập bảng các ước dương của 45

Vì m, n ∈ N => \(\hept{\begin{cases}n\in\left\{23;9;7\right\}\\m\in\left\{22;6;2\right\}\end{cases}}\)=> \(\hept{\begin{cases}n^2\in\left\{529;81;49\right\}\\m^2\in\left\{484;36;4\right\}\end{cases}}\)

=> \(\hept{\begin{cases}a-23\in\left\{484;36;4\right\}\\a+22\in\left\{529;84;49\right\}\end{cases}}\Rightarrow a\in\left\{507;59;27\right\}\)

Chắc là có sai sót ;-;

Đề:............

<=> - (1 - 2018x) + 2019x.(1 - 2018x) = 0

<=> (1 - 2018x).[(-1) + 2019x] = 0

Xét 2 trường hợp, ta có:

TH₁: 1 - 2018x = 0          TH₂: -1 + 2019x = 0

<=> 2018x = 1                 <=> 2019x = 1

<=> x = 1/2018                <=> x = 1/2019

Vậy x = 1/2018; 1/2019

Bài 1 : 

\(49\left(x-2\right)^2-25\left(2x+1\right)^2=0\)

\(\Leftrightarrow\left[7\left(x-2\right)-5\left(2x+1\right)\right]\left[7\left(x-2\right)+5\left(2x+1\right)\right]=0\)

\(\Leftrightarrow\left(7x-14-10x-5\right)\left(7x-14+10x+5\right)=0\)

\(\Leftrightarrow\left(-3x-19\right)\left(17x-9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}-3x=19\\17x=9\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-19}{3}\\x=\frac{9}{17}\end{cases}}}\)

Bài 2 : 

+) \(9x^2-6xy+y^2-21x+7y\)

\(=\left(3x-y\right)^2-7\left(3x-y\right)\)

\(=\left(3x-y\right)\left(3x-y-7\right)\)

+) \(x^2+2x-35\)

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

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

\(=\left(x-5\right)\left(x+7\right)\)

+) \(2x^2+9x-5\)

\(=2x^2-x+10x-5\)

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

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

+) \(6x^2+23x+15\)

\(=6x^2+18x+5x+15\)

\(=6x\left(x+3\right)+5\left(x+3\right)\)

\(=\left(x+3\right)\left(6x+5\right)\)

\(a=x^3-6x^2+12x-8-x^3+4x^2+8\)

\(a=-2x^2+12x\)

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

\(a=x^3-6x^2+12x-8-x^3+4x^2+8\)

\(a=-2x^2+12x\)