a) \(mx^2-4mx+4m-nx^2+4nx-4n\)

\(=\left(mx^2-nx^2\right)-\left(4mx-4nx\right)+\left(4m-4n\right)\)

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

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

\(=\left(m-n\right)\left(x-2\right)^2\)

b) \(3x^2+48+24x-12y^2\)

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

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

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

a) \(mx^2-4mx+4m-nx^2+4nx-4n\)

\(=\left(mx^2-nx^2\right)-\left(4mx+4nx\right)+\left(4m-4n\right)\)

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

\(=\left(m-n\right).\left(x^2-4x+4\right)\)

\(=\left(m-n\right).\left(x-2\right)^2\)

b) \(3x^2+48+24x-12y^2\)

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

\(=3\left[\left(x^2+8x+16\right)-\left(2y\right)^2\right]\)

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

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

2. Tìm x:

( x - 3 )² - x + 3 = 0

=> x² - 6x + 9 - x + 3 = 0

=> x² - 7x + 12 = 0

=> ( x² - 3x ) + ( 4x - 12 ) = 0

=> x.(x - 3) + 4.(x - 3) = 0

=> ( x - 3 ).( x + 4 ) = 0

=> x - 3 = 0 => x = 3

     x + 4 = 0 => x = -4

Trl:

1.

a. \(75^2+150\text{.}25+25^2\)

\(=75^2+2\text{.}75\text{.}25+25^2\)

\(=\left(75+25\right)^2\)

\(=100^2\)

\(=10000\)

b. \(2019^2-2019.19-19^2-19.1981\)

(Đề bài có sai ko vậy???)~ hoặc lak do mk ngu quá k bt lm

2. \(\left(\text{x}-3\right)^2-\text{x}+3=0\)

\(\text{x}^2-6\text{x}+9-\text{x}+3=0\)

\(\text{x}^2-7\text{x}+12=0\)

\(\text{x}^2-3\text{x}-4\text{x}+12=0\)

\(\text{x}\left(\text{x}-3\right)-4\left(\text{x}-3\right)=0\)

\(\left(\text{x}-3\right)\left(\text{x}-4\right)=0\)

\(\orbr{\begin{cases}\text{x}-3=0\\\text{x}-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}\text{x}=3\\\text{x}=4\end{cases}}}\)

Vậy ....

#HuyềnAnh#

\(P=\frac{8x^3-12x^2+6x-1}{4x^2-4x+1}\)

a) ĐKXĐ: x \(\ne\pm\frac{1}{2}\)

b) Theo đề bài ta có:

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

\(\Rightarrow x\left(2x+1\right)=0\)

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

Thay x = 0 (thỏa mãn điều kiện) vào P ta có:

\(P=\frac{0-0+0-1}{0-0+1}=\frac{-1}{1}=-1\)

Vậy khi x = 0 thì P = -1

c) \(P=\frac{8x^3-12x^2+6x-1}{4x^2-4x+1}=\frac{\left(2x-1\right)^3}{\left(2x-1\right)^2}=2x-1\)

Để P \(\inℤ\Leftrightarrow2x-1\inℤ\)

Mà -1\(\inℤ;x\inℤ\Rightarrow-1⋮2x\)

\(\Rightarrow2x\inƯ\left(-1\right)=\left\{1;-1\right\}\)

Ta có bảng giá trị:

Vậy không có x thỏa mãn P \(\inℤ\)

d) Với x \(\ne\pm\frac{1}{2};P=2\)

\(\Leftrightarrow2x-1=2\)

\(\Leftrightarrow2x=3\)

\(\Leftrightarrow x=\frac{3}{2}\)

Vậy \(x=\frac{3}{2}\)thì \(P=2\)

\(3x+3y-x^2-xy\)

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

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

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