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Bài làm
\(4x\left(x-1\right)-\left(2x-1\right)^2=4x^2-4x-\left(4x^2-4x+1\right)\)
\(=4x^2-4x-4x^2+4x+1=1\)
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a, \(5x-20y=5\left(x-4y\right)\)
b, \(x^3-9x=x\left(x^2-9\right)=x\left(x-3\right)\left(x+3\right)\)
c, \(x^2-2xy-y^2-16=-\left(x+y\right)^2-16=-\left(x+y-4\right)\left(x+y+4\right)\)
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a) ĐKXĐ : \(x\ne-3;x\ne2\)
b) \(A=\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x-2\right)}=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}\)
\(=\frac{x^2-9}{\left(x+3\right)\left(x-2\right)}=\frac{\left(x-3\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x-3}{x-2}\)
Để \(A\inℤ\Rightarrow x-3⋮x-2\)
=> \(x-2-1⋮x-2\)
Vì \(x-2⋮x-2\)
=> \(1⋮x-2\)
=> \(x-2\inƯ\left(1\right)\)
=> \(x-2\in\left\{1;-1\right\}\)
=> \(x\in\left\{3;1\right\}\)
Vậy \(x\in\left\{3;1\right\}\)là giá trị cần tìm
a + b , ĐK \(x\ne2;-3\)
\(A=\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x-2\right)}\)
\(=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}=\frac{x^2-4-5}{\left(x+3\right)\left(x-2\right)}\)
\(=\frac{\left(x-3\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x-3}{x-2}\)
\(\left(3x-2\right)-5\left(2-3x\right)=0\)
\(\Leftrightarrow\left(3x-2\right)+5\left(3x-2\right)=0\)
\(\Leftrightarrow6\left(3x-2\right)=0\Leftrightarrow x=\frac{2}{3}\)
\(2x^2-2x\left(x-3\right)=30\)
\(\Leftrightarrow2x^2-2x^2+6x=30\Leftrightarrow6x=30\Leftrightarrow x=5\)