Question

[nguyễn minh anh]

cứu tui với !!!!!!!!!!!!!!!!!!!!!!!

Answer 1

[Yen Nhi]

Answer:

Bài 1:

\(6x^2-3xy=3x.2x-3x.y=3x.\left(2x-y\right)\)

\(x^2-y^2-6x+9=\left(x^2-6x+9\right)-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right).\left(x-3+y\right)\)

\(x^2+5x=x.\left(x+5\right)\)

Bài 2:

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

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

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

\(=6x+7\)

\(\left(x^2-2x^2+5x-10\right):\left(x-2\right)\)

\(=[x^2.\left(x-2\right)+5.\left(x-2\right)]:\left(x-2\right)\)

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

\(=x^2+5\)

Bài 3:

a) Ta có: 

\(x^2-3x=0\Rightarrow x.\left(x-3\right)=0\)

Mà đề ra: ĐKXĐ: \(x\ne0\)

\(\Rightarrow x-3=0\)

\(\Rightarrow x=3\)

Ta thay vào biểu thức A được

\(A=\frac{3-5}{3-4}=\frac{-2}{-1}=2\)

b) \(B=\frac{x+5}{2x}-\frac{x-6}{5-x}-\frac{2x^2-2x-50}{2x^2-10x}\)

\(=\frac{x+5}{2x}+\frac{x-6}{x-5}-\frac{2x^2-2x-50}{2x.\left(x-5\right)}\)

\(=\frac{\left(x+5\right).\left(x-5\right)}{2x.\left(x-5\right)}+\frac{\left(x-6\right).2x}{2x.\left(x-5\right)}-\frac{2x^2-2x-50}{2x.\left(x-5\right)}\)

\(=\frac{x^2-25+2x^2-12x-2x^2+2x+50}{2x.\left(x-5\right)}\)

\(=\frac{x^2-25+2x^2-12x-2x^2+2x+50}{2x.\left(x-5\right)}\)

\(=\frac{x^2-10x+25}{2x.\left(x-5\right)}\)

\(=\frac{x^2-2x.5+5^2}{2x.\left(x-5\right)}\)

\(=\frac{x-5}{2x}\)

c) Ta xét: \(P=A:B=\frac{x-5}{x-4}:\frac{x-5}{2x}=\frac{x-5}{x-4}.\frac{2x}{x-5}=2+\frac{8}{x-4}\)

Mà để \(P\inℤ\Rightarrow8⋮x-4\)

\(\Rightarrow x-4\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

\(\Rightarrow x\in\left\{-4;0;2;3;5;6;8;12\right\}\)

Mà điều kiện đề bài ra: \(x\ne0;x\ne4;x\ne5\)

\(\Rightarrow x\in\left\{-4;2;3;6;8;12\right\}\)