int (3 cdot 2^x - 2 cdot 3^x)/(2^x) dx =A. $3x - 2\ln \frac{3}{2} \cdot \left(\f
int dfrac (2cdot {3)^x-5cdot (2)^x}({3)^x}dx-|||-__;;
(4)lim _(xarrow infty )dfrac ({3)^x+4cdot (2)^x}({2)^x-3cdot (5)^x}(4)
([ {(Si{O)_(2))}_(m)cdot n(S{O)_(3)}^2-cdot 2(n-x)(H)^+] }^2x--|||-C. ((Si{O)_(2
已知 (x)cdot (int )_(0)^2f(x)dx=8, 且 (0)=0, (x)geqslant 0, 则 f(x)= __
求下列不定积分:-|||-int dfrac (2cdot {3)^x-5cdot (2)^x}({3)^x}dx =
关于x的方程((dfrac {1)(2))}^2x-5cdot (2)^-x+4=0有()个实数解.A.1 B.2 C.3 D.0
已知二次型((x)_(1),(x)_(2),(x)_(3))=({x)_(1)}^2+a({x)_(2)}^2+({x)_(3)}^2+2b(x)_(1)(x)
1.求极限 lim _(xarrow 0)dfrac ({int )_(0)^x(tan t-sin t)dt}(({e)^(x^2)-1)cdot ln (1
int ((2)^xsin (x)^2)dx= ()-|||-(4) ^xsin (x)^2-|||-circled (8) ln 2cdot (2)^xcdo