Bài 1:

a) \(x^2-7x=x\left(x-7\right)\)

b) \(5x-10y=5\left(x-2y\right)\)

c) \(5x^2-15xy=5x\left(x-3y\right)\)

d) \(3\left(x-y\right)-y+x=3\left(x-y\right)+x-y=4\left(x-y\right)\)

e) \(5\left(x-y\right)-x^2+xy=5\left(x-y\right)-x\left(x-y\right)=\left(x-y\right)\left(5-x\right)\)

Bài 2 : 

a, \(75.20,9+5^2.20,9=20,9\left(75+25\right)=20,9.100=2090\)

b, \(85.15+150.1,4=85.15+15.14=15.99=1485\)

c, \(93.32+14.16=186.16+14.16=16.200=3200\)

d, \(98,6.199-990.9,86=98,6.199-99.98,6=98,6.100=9860\)

Bài 2: 

a) Ta có: \(75\cdot20,9+5^2\cdot20,9\)

\(=20,9\left(75+25\right)\)

\(=20,9\cdot100=2090\)

b) Ta có: \(85\cdot15+150\cdot1,4\)

\(=85\cdot15+14\cdot15\)

\(=15\left(85+14\right)\)

\(=15\cdot99=1500-15=1485\)

c) Ta có: \(93\cdot32+14\cdot16\)

\(=93\cdot2\cdot16+14\cdot16\)

\(=16\left(186+14\right)\)

\(=16\cdot200=3200\)

a) Ta có: \(x^2-2x=0\)

\(\Leftrightarrow x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

b) Ta có: \(\left(x+1\right)^2-\left(2x+3\right)^2=0\)

\(\Leftrightarrow\left(x+1-2x-3\right)\left(x+1+2x+3\right)=0\)

\(\Leftrightarrow\left(-x-2\right)\left(3x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x-2=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=2\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{-4}{3}\end{matrix}\right.\)

c) Ta có: \(27x^3-27x+9x-1=0\)

\(\Leftrightarrow\left(3x-1\right)^3=0\)

\(\Leftrightarrow3x-1=0\)

hay \(x=\dfrac{1}{3}\)

b: Ta có: \(x\left(x+1\right)-\left(2x+3\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-2x-3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)

d: Ta có: \(\left(x-1\right)^2-4\left(x+2\right)^2=0\)

\(\Leftrightarrow\left(x-1-2x-4\right)\left(x-1+2x+4\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(3x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-1\end{matrix}\right.\)

BÀI 3:

bài 4:

Là mấy bài này em làm được bài nào chưa?

3.

Do \(sin\left(x+k2\pi\right)=sinx\Rightarrow sin\left(x+2020\pi\right)=sinx\)

\(sin\left(\dfrac{\pi}{2}+x\right)=cos\left(\dfrac{\pi}{2}-\dfrac{\pi}{2}-x\right)=cos\left(-x\right)=cosx\)

\(A=\dfrac{sinx+sin3x+sin5x}{cosx+cos3x+cos5x}=\dfrac{sinx+sin5x+sin3x}{cosx+cos5x+cos3x}\)

\(=\dfrac{2sin3x.cosx+sin3x}{2cos3x.cosx+cos3x}=\dfrac{sin3x\left(2cosx+1\right)}{cos3x\left(2cosx+1\right)}\)

\(=\dfrac{sin3x}{cos3x}=tan3x\)

4.

a.

\(\overrightarrow{CB}=\left(2;-2\right)=2\left(1;-1\right)\)

Do đường thẳng d vuông góc BC nên nhận \(\left(1;-1\right)\) là 1 vtpt

Phương trình đường thẳng d đi qua \(A\left(-1;2\right)\) và có 1 vtpt là \(\left(1;-1\right)\) là:

\(1\left(x+1\right)-1\left(y-2\right)=0\Leftrightarrow x-y+3=0\)

b.

Gọi \(I\left(a;b\right)\) là tâm đường tròn, ta có \(\left\{{}\begin{matrix}\overrightarrow{AI}=\left(a+1;b-2\right)\\\overrightarrow{BI}=\left(a-3;b-2\right)\\\overrightarrow{CI}=\left(a-1;b-4\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}AI^2=\left(a+1\right)^2+\left(b-2\right)^2\\BI^2=\left(a-3\right)^2+\left(b-2\right)^2\\CI^2=\left(a-1\right)^2+\left(b-4\right)^2\end{matrix}\right.\)

Do I là tâm đường tròn qua 3 điểm nên: \(\left\{{}\begin{matrix}AI=BI\\AI=CI\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}AI^2=BI^2\\AI^2=CI^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(a+1\right)^2+\left(b-2\right)^2=\left(a-3\right)^2+\left(b-2\right)^2\\\left(a+1\right)^2+\left(b-2\right)^2=\left(a-1\right)^2+\left(b-4\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}8a=8\\4a+4b=12\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=1\\b=2\end{matrix}\right.\) \(\Rightarrow I\left(1;2\right)\)

\(\overrightarrow{AI}=\left(2;0\right)\Rightarrow R=AI=\sqrt{2^2+0^2}=2\)

Pt đường tròn có dạng:

\(\left(x-1\right)^2+\left(y-2\right)^2=4\) 

\(\Rightarrow\left[\left(-37\right)+129\right]-\left[529-117\right]\)
\(\Rightarrow92-412\)
\(\Rightarrow-320\)

a) Thay m=3 vào hệ pt, ta được:

\(\left\{{}\begin{matrix}x+3y=3\\3x+4y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+9y=9\\3x+4y=6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5y=3\\x+3y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{3}{5}\\x=3-3y=3-3\cdot\dfrac{3}{5}=\dfrac{6}{5}\end{matrix}\right.\)

Vậy: Khi m=3 thì hệ phương trình có nghiệm duy nhất là \(\left(x,y\right)=\left(\dfrac{6}{5};\dfrac{3}{5}\right)\)