C NHA BN CÂU 45 KO LÀM ĐC

\(I=\int\dfrac{2}{2+5sinxcosx}dx=\int\dfrac{2sec^2x}{2sec^2x+5tanx}dx\\ =\int\dfrac{2sec^2x}{2tan^2x+5tanx+2}dx\)

We substitute :

\(u=tanx,du=sec^2xdx\\ I=\int\dfrac{2}{2u^2+5u+2}du\\ =\int\dfrac{2}{2\left(u+\dfrac{5}{4}\right)^2-\dfrac{9}{8}}du\\ =\int\dfrac{1}{\left(u+\dfrac{5}{4}\right)^2-\dfrac{9}{16}}du\\ \)

Then, 

\(t=u+\dfrac{5}{4}\\I=\int\dfrac{1}{t^2-\dfrac{9}{16}}dt\\ =\int\dfrac{\dfrac{2}{3}}{t-\dfrac{3}{4}}-\dfrac{\dfrac{2}{3}}{t+\dfrac{3}{4}}dt\)

Finally,

\(I=\dfrac{2}{3}ln\left(\left|\dfrac{t-\dfrac{3}{4}}{t+\dfrac{3}{4}}\right|\right)+C=\dfrac{2}{3}ln\left(\left|\dfrac{tanx+\dfrac{1}{2}}{tanx+2}\right|\right)+C\)

31B

32C

33B

34 underline "they"?

\(\dfrac{x}{y}=\dfrac{5}{2}\) ⇒\(\dfrac{x}{5}=\dfrac{y}{2}\)

\(\dfrac{y}{z}=\dfrac{1}{3}\) ⇒\(\dfrac{y}{1}=\dfrac{z}{3}\) ⇒\(\dfrac{y}{2}=\dfrac{z}{6}\)

⇒\(\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{z}{6}\) ⇒\(\dfrac{x^2}{25}=\dfrac{y^2}{4}=\dfrac{2z^2}{72}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{x^2}{25}=\dfrac{y^2}{4}=\dfrac{2z^2}{72}=\dfrac{x^2-y^2+2z^2}{25-4+72}=\dfrac{372}{93}=4\)

⇒\(\left\{{}\begin{matrix}x=4.5=20\\y=4.2=8\\z=4.6=24\end{matrix}\right.\)

R₁ và R₂ mắc như nào vậy

Đề bài thiếu thì phải

32, A

33,B

34,C