Pi=3,14

Pi còn là tỉ lệ vàng(xem trên tv chỉ ngớ đc bằng đó)

Phương trình tham số:

\(\left\{{}\begin{matrix}y=-2+t\\z=5+4t\end{matrix}\right.\)

a/ \(\left(P\right):3\left(x-0\right)+0\left(y-2\right)+1\left(z+5\right)=0\Rightarrow\left(P\right):3x+z+5=0\)

b/\(\overrightarrow{AB}\left(2;4;-9\right);\overrightarrow{AC}\left(4;0;-7\right)\)

 \(\overrightarrow{n_{\left(P\right)}}=\left[\overrightarrow{AB},\overrightarrow{AC}\right]=\left(4.\left(-7\right)-0.\left(-9\right);\left(-9\right).4-\left(-7\right).2;2.0-4.4\right)=\left(-28;-22;-16\right)\)

\(\Rightarrow\left(P\right):-28\left(x-0\right)-22\left(y-1\right)-16\left(z-7\right)=0\Rightarrow\left(P\right):28x+22y+16z-134=0\)

c/ Truc Oy di qua O(0;0;0) va co vtcp \(\overrightarrow{j}\left(0;1;0\right)\)

\(\overrightarrow{OD}\left(3;-6;2\right)\)

\(\Rightarrow\overrightarrow{n_{\left(P\right)}}=\left[\overrightarrow{j};\overrightarrow{OD}\right]=\left(2;0;-3\right)\)

\(\Rightarrow\left(P\right):2\left(x-0\right)+0\left(y-1\right)-3\left(z-0\right)=0\Rightarrow\left(P\right):2x-3z=0\)

d/ \(\overrightarrow{Oz}\left(0;0;1\right)\)

\(\overrightarrow{DE}\left(5;-2;-7\right)\)

\(\Rightarrow\overrightarrow{n_{\left(P\right)}}=\left[\overrightarrow{Oz};\overrightarrow{DE}\right]=\left(2;5;0\right)\)

\(\Rightarrow\left(P\right):2\left(x-0\right)+5\left(y-0\right)+0\left(z-1\right)=0\Rightarrow\left(P\right):2x+5y=0\)

e/ \(\overrightarrow{n_{Oyz}}=\overrightarrow{n_{\left(P\right)}}=\left(1;0;0\right)\)

\(\Rightarrow\left(P\right):1\left(x-3\right)=0\Rightarrow\left(P\right):x-3=0\)

f/ Cách làm giống câu b

g/ \(\overrightarrow{HI}=\overrightarrow{IK}\Rightarrow\left\{{}\begin{matrix}x_I=\dfrac{3-1}{2}=1\\y_I=\dfrac{-1+5}{2}=2\\z_1=\dfrac{2-4}{2}=-1\end{matrix}\right.\Rightarrow I\left(1;2;-1\right)\)

\(\overrightarrow{n_{\left(P\right)}}=\overrightarrow{HK}\left(-4;6;-6\right)\)

\(\Rightarrow\left(P\right):-4\left(x-1\right)+6\left(y-2\right)-6\left(z+1\right)=0\Rightarrow\left(P\right):-4x+6y-6z+2=0\)

P/s: Bạn tính toán lại kết quả hộ mình nhé !

\(\dfrac{d}{dx}\left(f\left(x\right)\right)\equiv f'\left(x\right)\)

\(\dfrac{1}{sinx}dx=\dfrac{sinx}{sin^2x}dx=\dfrac{sinx}{1-cos^2x}dx=\dfrac{d\left(cosx\right)}{cos^2x-1}\)

\(\int\dfrac{e^xdx}{e^x-e^{-x}}\) 

Một cách làm khác ngoài sử dụng nguyên hàm phụ ạ!

Hệ số bất định:

\(\dfrac{x^2-3}{x\left(x^2+1\right)\left(x^2+2\right)}=\dfrac{a}{x}+\dfrac{bx}{x^2+1}+\dfrac{cx}{x^2+2}\)

30 = XXX

40 = XL

50 = L

60 = LX

70 = LXX

80 = LXXX

90 = XC

100 = C

#Chúc em học tốt

30 là XXX; 40 là XL; 50 = L; 60 = LX; 70 = LXX; 80 = LXXX; 90 = XC; 100 = C

\(G\left(\dfrac{4}{3};-\dfrac{2}{3};\dfrac{1}{3}\right)\) ; \(\overrightarrow{DC}=\left(1;3;1\right)\)

Pt mặt phẳng qua G vuông góc CD và nhận \(\overrightarrow{DC}\) là 1 vtpt có dạng:

\(1\left(x-\dfrac{4}{3}\right)+3\left(y+\dfrac{2}{3}\right)+1\left(z-\dfrac{1}{3}\right)=0\)

\(\Leftrightarrow x+3y+z-\dfrac{1}{3}=0\)

Câu 14 vs 16 

Câu 10 

Đặt \(a=t^{16}\)

\(\Rightarrow A=\frac{\sqrt{t^{16}\sqrt{t^{16}\sqrt{t^{16}\sqrt{t^{16}}}}}}{\left[\left(\sqrt{\sqrt{t^{16}}}\right)^{\sqrt{\sqrt{16^{-2}}}}\right]^{11}}\)

\(=\frac{t^{15}}{t^{11}}=t^4\)

\(\Rightarrow A=\sqrt[4]{a}\)