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anh em cần thuê người IQ 33443499999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999934444444444444444444444444444344444444444444444444444444444444444444444444444449999999999999999999999999999999999
Silver Bullet - Silver Bullet đã thêm một ảnh mới.
thánh tìm cái này hả ?
\(z=x+yi\Rightarrow w=\dfrac{z}{2+i}=\dfrac{x+yi}{2+i}=\dfrac{\left(x+yi\right)\left(2-i\right)}{4-i^2}=\dfrac{2x+y}{5}+\dfrac{2y-x}{5}i\)
\(\left(1+3i\right)w+1+7i=\left(1+3i\right)\left(\dfrac{2x+y}{5}+\dfrac{2y-x}{5}i\right)+1+7i\)
\(=x-y+1+\left(x+y+7\right)i\)
\(\Rightarrow\left(x-y+1\right)^2+\left(x+y+7\right)^2=50\)
\(\Leftrightarrow x^2+y^2+8x+6y=0\)
Tập hợp z là đường tròn tâm \(I\left(-4;-3\right)\) bán kính \(R=5\)
\(I=\int\dfrac{2}{2+5sinxcosx}dx=\int\dfrac{2sec^2x}{2sec^2x+5tanx}dx\\ =\int\dfrac{2sec^2x}{2tan^2x+5tanx+2}dx\)
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,
\(I=\dfrac{2}{3}ln\left(\left|\dfrac{t-\dfrac{3}{4}}{t+\dfrac{3}{4}}\right|\right)+C=\dfrac{2}{3}ln\left(\left|\dfrac{tanx+\dfrac{1}{2}}{tanx+2}\right|\right)+C\)
hình này nè:
bai lao