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b) \(\frac{x}{2}+\frac{3}{y}=\frac{5}{4}\)

\(\Rightarrow\frac{x+3}{2+y}=\frac{5}{4}\)

\(\Rightarrow\frac{x}{y}=\frac{5-3}{4-2}\)

\(\Rightarrow\hept{\begin{cases}x=2\\y=2\end{cases}}\)

Vì góc CBA' kề bù với góc ABC.

\(\Rightarrow\)CBA' + ABC = 180^o

Thay ABC = 56^o

\(\Rightarrow\)CBA' + 56^o = 180^o

\(\Rightarrow\)CBA' = 180^o - 56^o

\(\Rightarrow\)CBA' = 124^o

\(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}\Rightarrow\frac{a}{2}=\frac{2b}{6}=\frac{c}{4}=\frac{a+2b+c}{12}\)

\(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=\frac{a+b-c}{2+3-4}=a+b-c\)

\(\Rightarrow\frac{a+2b+c}{12}=a+b-c\Rightarrow H=\frac{a+2b+c}{a+b-c}=12\)

\(5x-2\left(x+3\right)=4\left(x-1\right)\)

\(\Rightarrow5x-2x-6=4x-4\)

\(\Rightarrow3x-6=4x-4\)

\(\Rightarrow4x-3x=-6+4\)

\(\Rightarrow x=-2\)

∫(5x+1)⋅dx(x+3)(x+2)(x−4)

=12⋅ln(x−4)+32⋅ln(x+2)−2ln(x+3)+C

Explanation:

I decomposed integrand into basic fractions,

5x+1(x+3)(x+2)(x−4)

=Ax+3+Bx+2+Cx−4

After expanding denominator,

A(x+2)(x−4)+B(x+3)(x−4)+C(x+3)(x+2)=5x+1

After setting x=−3, 7A=−14, so A=−2

After setting x=−2, −6B=−9, so B=32

After setting x=4, 42C=21, so C=12

Thus,

∫(5x+1)⋅dx(x+3)(x+2)(x−4)

=−2⋅∫dxx+3+32⋅∫dxx+2+12⋅∫dxx−4

=12⋅ln(x−4)+32⋅ln(x+2)−2ln(x+3)+C