Em nhập GT biểu thức vào ha
=10(3+x)/(x+2)(3-x)(3+x)-12(x+2)/(3-x)(3+x)(x+2)-1(3-x)/(x+3)(3-x)(x+2)
=30+10x-12x-24-3+x/(x+2)(3-x)(3+x)
=1/(x+2)(3+x)
\(P=\dfrac{10}{\left(x+2\right)\left(3-x\right)}-\dfrac{12}{\left(3-x\right)\left(3+x\right)}-\dfrac{1}{\left(x+3\right)\left(x+2\right)}\)
Thay \(x=-0,75\) vào biểu thức P ta có,
\(P=\dfrac{10}{\left(-0,75+2\right)\left(3+0,75\right)}-\dfrac{12}{\left(3+0,75\right)\left(3-0,75\right)}-\dfrac{1}{\left(-0,75+3\right)\left(-0,75+2\right)}\)
\(=\dfrac{32}{15}-\dfrac{64}{45}-\dfrac{16}{45}\)
\(=\dfrac{96}{45}-\dfrac{64}{45}-\dfrac{16}{45}\)
\(=\dfrac{16}{45}\)