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a) \(ĐKXĐ:x\ne\pm3\)
\(A=\frac{5}{x+3}-\frac{2}{3-x}+\frac{3x^2-2x-9}{x^2-9}\)
\(\Leftrightarrow A=\frac{5\left(x-3\right)+2\left(x+3\right)-3x^2+2x+9}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{5x-15+2x+6-3x^2+2x+9}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{-3x^2+9x}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{-3x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow A=\frac{-3x}{x+3}\)
b) Khi \(\left|x-2\right|=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\2-x=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\left(ktm\right)\\x=1\left(tm\right)\end{cases}}\)
Thay x = 1 vào A, ta được :
\(A=\frac{-3}{1+3}=\frac{-3}{4}\)
Vậy khi \(\left|x-2\right|=1\Leftrightarrow A=-\frac{3}{4}\)
c) Để \(A\inℤ\)
\(\Leftrightarrow\frac{-3x}{x+3}\inℤ\)
\(\Leftrightarrow-3x⋮x+3\)
\(\Leftrightarrow-3\left(x+3\right)+9⋮x+3\)
\(\Leftrightarrow9⋮x+3\)
\(\Leftrightarrow x+3\inƯ\left(9\right)=\left\{\pm1;\pm3;\pm9\right\}\)
\(\Leftrightarrow x\in\left\{-2;-4;0;-6;-12;6\right\}\)
Vậy để \(A\inℤ\Leftrightarrow x\in\left\{-2;-4;0;-6;-12;6\right\}\)
\(\left(5x-4\right)\left(2x+3\right)=10x^2+15x-8x-12=10x^2+7x-12\)
\(b,\frac{x-4}{x-2}+\frac{5x-8}{x-2}=\frac{x-4+5x-8}{x-2}=\frac{6\left(x-2\right)}{x-2}=6\)
\(c,\frac{x-9}{x^2-9}-\frac{3}{x^2+3x}=\frac{x-9}{\left(x+3\right)\left(x-3\right)}-\frac{3}{x\left(x+3\right)}\)
\(=\frac{x^2-9x}{x\left(x+3\right)\left(x-3\right)}-\frac{3x-9}{x\left(x+3\right)\left(x-3\right)}=\frac{x^2-9x-3x+9}{x\left(x+3\right)\left(x-3\right)}=\frac{x^2-6x+9}{x\left(x+3\right)\left(x-3\right)}\)
\(=\frac{x-3}{x\left(x+3\right)}\)
CÂU 1 :
a, ( 5x-4 ) ( 2x + 3 )
= 10x + 15x -8x -12
= 17x - 12
b, \(\frac{x-4}{x-2}\)+ \(\frac{5x-8}{x-2}\)
= \(\frac{x-4+5x-8}{x-2}\)
= \(\frac{6x-12}{x-2}\)
= \(\frac{6\left(x-2\right)}{x-2}\)
= 6
c, \(\frac{x-9}{x^2-9}\)- \(\frac{3}{x^2+3x}\)
= \(\frac{x-9}{\left(x-3\right)\left(x+3\right)}\)- \(\frac{3}{x\left(x+3\right)}\)
= \(\frac{\left(x-9\right).x}{x\left(x-3\right).\left(x+3\right)}\)- \(\frac{3.\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}\)
= \(\frac{x^2-9x}{x\left(x-3\right)\left(x+3\right)}\)- \(\frac{3x-9}{x\left(x-3\right)\left(x+3\right)}\)
= \(\frac{x^2-9x-3x+9}{x\left(x-3\right)\left(x+3\right)}\)
= \(\frac{x^2-12x+9}{x\left(x-3\right)\left(x+3\right)}\)
3/
(2x+3)2 - 2(2x+3)(2x+5) + (2x+5)2
= [(2x+3) - (2x+5)]2
= (2x+3-2x-5)2
= (-2)2 = 4
Câu 1:
a: \(=a^2+2ab+b^2-a^2-2ab-b^2=0\)
b: \(=x^3+27-54-x^3=-27\)
Câu 4:
\(\Leftrightarrow3x^3+x^2+9x^2+3x-3x-1-4⋮3x+1\)
\(\Leftrightarrow3x+1\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(x\in\left\{0;1\right\}\)