BÀI 1
B=\(\frac{5X}{X+2}-\frac{3X-23}{X-2}+\frac{40}{4-X^2}\)
a) tìm điều kiện xác định và rút gọn B
b)tính B khi X2-1=0
BÀI 2 tìm X\(\inℤ\)để các biểu thức sau có giá trị nguyên
A= \(\frac{5}{X-1}\)
B=\(\frac{-7}{X+3}\)
C=\(\frac{X^2}{X-2}\)
D=\(\frac{X-4}{X-3}\)
\(B=\frac{5x}{x+2}-\frac{3x-23}{x-2}+\frac{40}{4-x^2}\)
a) ĐKXĐ : \(x\ne\pm2\)
\(B=\frac{5x}{x+2}-\frac{3x-23}{x-2}+\frac{40}{4-x^2}\)
\(B=\frac{5x}{x+2}-\frac{3x-23}{x-2}-\frac{40}{x^2-4}\)
\(B=\frac{5x}{x+2}-\frac{3x-23}{x-2}-\frac{40}{\left(x+2\right)\left(x-2\right)}\)
\(B=\frac{5x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{\left(3x-23\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{40}{\left(x+2\right)\left(x-2\right)}\)
\(B=\frac{5x^2-10x}{\left(x+2\right)\left(x-2\right)}-\frac{\left(3x^2-17x-46\right)}{\left(x+2\right)\left(x-2\right)}-\frac{40}{\left(x+2\right)\left(x-2\right)}\)
\(B=\frac{5x^2-10x-\left(3x^2-17x-46\right)-40}{\left(x+2\right)\left(x-2\right)}\)
\(B=\frac{5x^2-10x-3x^2+17x+46-40}{\left(x+2\right)\left(x-2\right)}\)
\(B=\frac{2x^2+7x+6}{\left(x+2\right)\left(x-2\right)}=\frac{\left(x+2\right)\left(2x+3\right)}{\left(x+2\right)\left(x-2\right)}=\frac{2x+3}{x-2}\)
b) x2 - 1 = 0 <=> x2 = 1 <=> x = ±1
Với x = 1
\(B=\frac{2\cdot1+3}{1-2}=-5\)
Với x = -1
\(B=\frac{2\cdot\left(-1\right)+3}{\left(-1\right)-2}=-\frac{1}{3}\)