Xác định a, b để $\frac{1}{x^{2}-4}$ = $\frac{a}{x-2}$ + $\frac{b}{x+2}$
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1x2−4=ax−2+bx+2⇔1x2−4=a(x+2)+b(x−2)x2−4⇔a(x+2)+b(x−2)=1⇔ax+2a+bx−2b=1⇔a=−bx−2b−1x+21×2−4=ax−2+bx+2⇔1×2−4=a(x+2)+b(x−2)x2−4⇔a(x+2)+b(x−2)=1⇔ax+2a+bx−2b=1⇔a=−bx−2b−1x+a=−bx−2b−1x+2b=−ax+2a−1x−2a=−bx−2b−1x+2b=−ax+2a−1x−2Tương tự ta có :b=−ax+2a−1x−2
Giải thích các bước giải: $$\frac{1}{x^2-4}=\frac{a}{x-2}+\frac{b}{x+2}\\\Leftrightarrow \frac{1}{x^2-4}=\frac{a(x+2)+b(x-2)}{x^2-4}\\\Leftrightarrow a(x+2)+b(x-2)=1\\\Leftrightarrow ax+2a+bx-2b=1\\\Leftrightarrow a=-\frac{bx-2b-1}{x+2}$$
1x2−4=ax−2+bx+2⇔1x2−4=a(x+2)+b(x−2)x2−4⇔a(x+2)+b(x−2)=1⇔ax+2a+bx−2b=1⇔a=−bx−2b−1x+21×2−4=ax−2+bx+2⇔1×2−4=a(x+2)+b(x−2)x2−4⇔a(x+2)+b(x−2)=1⇔ax+2a+bx−2b=1⇔a=−bx−2b−1x+a=−bx−2b−1x+2b=−ax+2a−1x−2a=−bx−2b−1x+2b=−ax+2a−1x−2Tương tự ta có : b=−ax+2a−1x−2
Đáp án:$$a=-\frac{bx-2b-1}{x+2}\\b=-\frac{ax+2a-1}{x-2}$$
Giải thích các bước giải: $$\frac{1}{x^2-4}=\frac{a}{x-2}+\frac{b}{x+2}\\\Leftrightarrow \frac{1}{x^2-4}=\frac{a(x+2)+b(x-2)}{x^2-4}\\\Leftrightarrow a(x+2)+b(x-2)=1\\\Leftrightarrow ax+2a+bx-2b=1\\\Leftrightarrow a=-\frac{bx-2b-1}{x+2}$$
Tương tự ta có : $b=-\frac{ax+2a-1}{x-2}$