ĐKXĐ: \(sinx+cosx\ne0\)

=>\(\sqrt{2}\cdot sin\left(x+\dfrac{\Omega}{4}\right)\ne0\)

=>\(sin\left(x+\dfrac{\Omega}{4}\right)\ne0\)

=>\(x+\dfrac{\Omega}{4}\ne k\Omega\)

=>\(x=k\Omega-\dfrac{\Omega}{4}\)

Vậy: TXĐ là \(D=R\backslash\left\{-\dfrac{\Omega}{4}+k\Omega\right\}\)

a: \(\left|\overrightarrow{a}\right|=\sqrt{1^2+7^2}=\sqrt{50}=5\sqrt{2}\)

b: \(\left|\overrightarrow{b}\right|=\sqrt{7^2+\left(-1\right)^2}=\sqrt{50}=5\sqrt{2}\)

c: \(\overrightarrow{a}\cdot\overrightarrow{b}=1\cdot7+7\cdot\left(-1\right)=7-7=0\)

d: \(\overrightarrow{a}\cdot\overrightarrow{b}=1\cdot7+7\cdot\left(-1\right)=7-7=0\)

=>\(\overrightarrow{a}\perp\overrightarrow{b}\)

e: \(\overrightarrow{a}\perp\overrightarrow{c}\)

=>\(1\cdot4+7\cdot\left(m-4\right)=0\)

=>7(m-4)+4=0

=>7m-28+4=0

=>7m-24=0

=>7m=24

=>\(m=\dfrac{24}{7}\)

f: \(\overrightarrow{b};\overrightarrow{c}\) cùng phương khi \(\dfrac{7}{4}=\dfrac{-1}{m-4}\)

=>\(m-4=-\dfrac{4}{7}\)

=>\(m=4-\dfrac{4}{7}=\dfrac{24}{7}\)

g: \(\left|\overrightarrow{c}\right|=5\)

=>\(\sqrt{4^2+\left(m-4\right)^2}=5\)

=>\(\left(m-4\right)^2+16=25\)

=>\(\left(m-4\right)^2=9\)

=>\(\left[{}\begin{matrix}m-4=3\\m-4=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=7\\m=1\end{matrix}\right.\)

a: \(\left|\overrightarrow{a}\right|=\sqrt{3^2+\left(-6\right)^2}=\sqrt{9+36}=\sqrt{45}=3\sqrt{5}\)

b: \(\left|\overrightarrow{b}\right|=\sqrt{4^2+2^2}=\sqrt{16+4}=\sqrt{20}=2\sqrt{5}\)

c: \(\overrightarrow{a}\cdot\overrightarrow{b}=3\cdot4+\left(-6\right)\cdot2=12-12=0\)

d: \(\overrightarrow{a}\cdot\overrightarrow{b}=0\)

=>\(\overrightarrow{a}\perp\overrightarrow{b}\)

e: \(\overrightarrow{a}\perp\overrightarrow{c}\)

=>\(3\cdot8+\left(-6\right)\left(m+1\right)=0\)

=>6(m+1)=24

=>m+1=4

=>m=3

f: Để \(\overrightarrow{b};\overrightarrow{c}\) cùng phương thì \(\dfrac{4}{8}=\dfrac{2}{m+1}\)

=>\(m+1=2\cdot\dfrac{8}{4}=2\cdot2=4\)

=>m=4-1=3

g: \(\left|\overrightarrow{c}\right|=5\)

=>\(\sqrt{8^2+\left(m+1\right)^2}=5\)

=>\(\left(m+1\right)^2+64=25\)

=>\(\left(m+1\right)^2=-39< 0\left(loại\right)\)

=>\(m\in\varnothing\)

\(y=cot2x\) là hàm số lẻ vì \(cot\left(-2x\right)=-cot2x\)

D đúng, \(cot2x\) là hàm lẻ

ĐKXĐ: \(2x-\dfrac{\Omega}{2}\ne\dfrac{\Omega}{2}+k\Omega\)

=>\(2x\ne\Omega+k\Omega\)

=>\(x\ne\dfrac{\Omega}{2}+\dfrac{k\Omega}{2}\)

Vậy: TXĐ: \(D=R\backslash\left\{\dfrac{\Omega}{2}+\dfrac{k\Omega}{2}\right\}\)

ĐKXĐ: \(cos\left(2x-\dfrac{\pi}{2}\right)\ne0\)

\(\Rightarrow2x-\dfrac{\pi}{2}\ne\dfrac{\pi}{2}+k\pi\)

\(\Rightarrow2x\ne k\pi\)

\(\Rightarrow x\ne\dfrac{k\pi}{2}\)

a.

\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=0\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=0\)

\(\Leftrightarrow x-\dfrac{\pi}{3}=k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{3}+k\pi\)

b.

\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=0\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=0\)

\(\Leftrightarrow x+\dfrac{\pi}{4}=k\pi\)

\(\Leftrightarrow x=-\dfrac{\pi}{4}+k\pi\)

a: \(sinx-\sqrt{3}\cdot cosx=0\)

=>\(\sqrt{3}\cdot cosx=sinx\)

=>\(\dfrac{sinx}{cosx}=\sqrt{3}\)

=>\(tanx=\sqrt{3}\)

=>\(x=\dfrac{\Omega}{3}+k\Omega\)

b: \(sinx+cosx=0\)

=>\(\sqrt{2}\cdot cos\left(x-\dfrac{\Omega}{4}\right)=0\)

=>\(cos\left(x-\dfrac{\Omega}{4}\right)=0\)

=>\(x-\dfrac{\Omega}{4}=\dfrac{\Omega}{2}+k\Omega\)

=>\(x=\dfrac{3}{4}\Omega+k\Omega\)

Trong mp (SBC), nối IK và BC kéo dài cắt nhau tại E

Trong mp (SCD), nối JK và DC kéo dài cắt nhau tại F

\(\Rightarrow EF\in\left(IJK\right)\)

Trong mp (ABCD), nối EF cắt AC kéo dài tại G \(\Rightarrow G\in\left(SAC\right)\cap\left(IJK\right)\)

Trong mp (SAC), nối GK kéo dài cắt SA tại H \(\Rightarrow H\in\left(IJK\right)\)

\(\Rightarrow HK=\left(SAC\right)\cap\left(IJK\right)\)

\(HI=\left(SAB\right)\cap\left(IJK\right)\)

\(HJ=\left(SAD\right)\cap\left(IJK\right)\)

Tứ giác IHJK là thiết diện của (IJK) với (ABCD)

Bài 1: \(sin^2x+cos^2x=1\)

=>\(sin^2x=1-\left(-\dfrac{1}{5}\right)=\dfrac{24}{25}\)

mà sin x>0(90⁰<x<180⁰)

nên \(sinx=\dfrac{2\sqrt{6}}{5}\)

a: \(sin\left(x+\dfrac{\Omega}{6}\right)=sinx\cdot cos\left(\dfrac{\Omega}{6}\right)+cosx\cdot sin\left(\dfrac{\Omega}{6}\right)\)

\(=\dfrac{2\sqrt{6}}{5}\cdot\dfrac{\sqrt{3}}{2}+\dfrac{-1}{5}\cdot\dfrac{1}{2}=\dfrac{6\sqrt{2}-1}{10}\)

b: \(cos2x=2\cdot cos^2x-1=2\cdot\left(-\dfrac{1}{5}\right)^2-1\)

\(=\dfrac{2}{25}-1=-\dfrac{23}{25}\)

c:\(tanx=\dfrac{sinx}{cosx}=\dfrac{2\sqrt{6}}{5}:\dfrac{-1}{5}=-2\sqrt{6}\)

\(tan\left(x-\dfrac{\Omega}{3}\right)=\dfrac{tanx-tan\left(\dfrac{\Omega}{3}\right)}{1+tanx\cdot tan\left(\dfrac{\Omega}{3}\right)}\)

\(=\dfrac{-2\sqrt{6}-\sqrt{3}}{1+\left(-2\sqrt{6}\right)\cdot\sqrt{3}}=\dfrac{-2\sqrt{6}-\sqrt{3}}{1-2\sqrt{18}}=\dfrac{8\sqrt{6}+25\sqrt{3}}{71}\)

d: \(cos\left(2x+\dfrac{\Omega}{4}\right)=cos2x\cdot cos\left(\dfrac{\Omega}{4}\right)-sin2x\cdot sin\left(\dfrac{\Omega}{4}\right)\)

\(=-\dfrac{23}{25}\cdot\dfrac{\sqrt{2}}{2}-2\cdot sinx\cdot cosx\cdot\dfrac{\sqrt{2}}{2}\)

\(=\dfrac{-23\sqrt{2}}{50}-2\cdot\dfrac{2\sqrt{6}}{5}\cdot\dfrac{-1}{5}\cdot\dfrac{\sqrt{2}}{2}\)

\(=\dfrac{-23\sqrt{2}+8\sqrt{3}}{50}\)

Bài 3:

sinx+sin4x+sin7x

\(=\left(sinx+sin7x\right)+sin4x\)

\(=2\cdot sin\left(\dfrac{7x+x}{2}\right)\cdot cos\left(\dfrac{7x-x}{2}\right)+sin4x\)

\(=2\cdot sin4x\cdot cos3x+sin4x=sin4x\left(2\cdot cos3x+1\right)\)

cosx+cos4x+cos7x

=(cosx+cos7x)+cos4x

\(=2\cdot cos\left(\dfrac{7x+x}{2}\right)\cdot cos\left(\dfrac{7x-x}{2}\right)+cos4x\)

\(=2\cdot cos4x\cdot cos3x+cos4x=cos4x\left(2\cdot cos3x+1\right)\)

\(A=\dfrac{sinx+sin4x+sin7x}{cosx+cos4x+cos7x}\)

\(=\dfrac{sin4x\left(2\cdot cos3x+1\right)}{cos4x\left(2\cdot cos3x+1\right)}=\dfrac{sin4x}{cos4x}=tan4x\)

Hình vẽ coi như người ta đã giải sẵn cho mình rồi.

Trong mp (ACD), nối DK kéo dài cắt AC tại M, trong mp (BCD) nối DJ kéo dài cắt BC tại M

Trong mp (DMN), nối JK và MN kéo dài cắt nhau tại H \(\Rightarrow H\in\left(IJK\right)\)

Trong mp (ABC), nối IH cắt BC tại P \(\Rightarrow P\in\left(IJK\right)\)

\(\Rightarrow IP=\left(IJK\right)\cap\left(ABC\right)\)

Trong mp (BCD), nối PJ kéo dài cắt CD tại Q

\(\Rightarrow PQ=\left(IJK\right)\cap\left(BCD\right)\)

Trong mp (ACD), nối QK kéo dài cắt AD tại T

\(\Rightarrow QT=\left(IJK\right)\cap\left(ACD\right)\)

Đồng thời \(IT=\left(IJK\right)\cap\left(ABD\right)\)

Any reader of classic English literature will (1) have heard of the Brontë sisters: Charlotte (1816-55), Emily (1818-48), and Anne (1820-49). (2) What they may not know, however, is that there were originally (3) two other sisters, Maria and Elizabeth, as well as a brother called Branwell. Sadly, these two sisters became seriously ill when they were away at boarding school and died in 1825 soon after (4) returning home. At the same time, Charlotte and Emily were also removed from the school, despite (5) having attended for only a year. The pair spent the next six years of their childhood at home with Branwell, who, unlike the girls, was tutored by their father, and their younger sister Anne. The four children had a fondness (6) for storytelling, inspired by twelve wooden soldiers given to Branwell by his father on his ninth birthday. The children spent a (7) great deal of time together inventing adventures in imaginary kingdoms for the soldiers, and at an early age, they began writing them down. The influences of these early stories (8) can be seen in the Brontës' later works. Not until two decades later, nevertheless, (9) did the sisters attempt to get any of their writing published. Their first publication in 1845 was a book of poems, which appeared under the pseudonyms Currer (Charlotte), Ellis (Emily), and Acton (Anne) Bell - no respectable woman (10) would have dared to try and publish a book under her own name at the time. Two years later, three novels were published, one by each sister: Jane Eyre by Charlotte, Wuthering Heights by Emily, and Agnes Grey by Anne. Since then, these novels have gained worldwide acclaim, earning the authors a place alongside Shakespeare and Dickens as some of the greatest British authors.

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