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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{2}\end{matrix}\right.\)

b: Ta có: \(x^2\left(x-6\right)-x^2+36=0\)

\(\Leftrightarrow\left(x-6\right)\left(x^2-x-6\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x-3\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=3\\x=-2\end{matrix}\right.\)

a)=\(3x^3-15x^2+21x\)

b)\(=-2x^4y-10x^2y+2xy\)

c)\(=-x^3+6x^2+5x-4x^2+24x+20=-x^3+2x^2+29x+20\)

d)\(=2x^4-3x^3+4x^2-2x^2+3x-4=2x^4-3x^32x^2+3x-4\)

e)\(=x^2-4y^2\)

f)\(=-2x^2y^3+y-3\)

g)\(=3xy^4-\dfrac{1}{2}y^2+2x^2y\)

h)\(=9x^2-6x+1-7x^2-14=2x^2-6x-13\)

i)\(=x^2-x-3\)

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

Tại sao ý b có dấu - trước ngoặc đâu mà đổi dấu mong bn giải đáp

1: \(cos^2\left(x-\frac{\pi}{5}\right)=\sin^2\left(2x+\frac45\pi\right)\)

=>\(\left[\begin{array}{l}cos\left(x-\frac{\pi}{5}\right)=\sin\left(2x+\frac45\pi\right)=cos\left(\frac{\pi}{2}-2x-\frac45\pi\right)=cos\left(-2x-\frac{3}{10}\pi\right)\\ cos\left(x-\frac{\pi}{5}\right)=-\sin\left(2x+\frac45\pi\right)=\sin\left(-2x-\frac45\pi\right)=cos\left(\frac{\pi}{2}+2x+\frac45\pi\right)=cos\left(2x+\frac{13}{10}\pi\right)\end{array}\right.\)

TH1: \(cos\left(x-\frac{\pi}{5}\right)=cos\left(-2x-\frac{3}{10}\pi\right)\)

=>\(\left[\begin{array}{l}x-\frac{\pi}{5}=-2x-\frac{3}{10}\pi+k2\pi\\ x-\frac{\pi}{5}=2x+\frac{3}{10}\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}3x=-\frac{3}{10}\pi+\frac{\pi}{5}+k2\pi=-\frac{1}{10}\pi+k2\pi\\ -x=\frac{3}{10}\pi+\frac{\pi}{5}+k2\pi=-\frac12\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac{1}{30}\pi+\frac{k2\pi}{3}\\ x=\frac12\pi-k2\pi\end{array}\right.\)

TH2: \(cos\left(x-\frac{\pi}{5}\right)=cos\left(2x+\frac{13}{10}\pi\right)\)

=>\(\left[\begin{array}{l}2x+\frac{13}{10}\pi=x-\frac{\pi}{5}+k2\pi\\ 2x+\frac{13}{10}\pi=-x+\frac{\pi}{5}+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}2x-x=-\frac{\pi}{5}-\frac{13}{10}\pi+k2\pi\\ 2x+x=\frac{\pi}{5}-\frac{13}{10}\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}x=-\frac{15}{10}\pi+k2\pi=-\frac32\pi+k2\pi\\ 3x=-\frac{11}{10}\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac32\pi+k2\pi\\ x=-\frac{11}{30}\pi+\frac{k2\pi}{3}\end{array}\right.\)

2: \(\sin3x=\sqrt2\cdot cos\left(x-\frac{\pi}{5}\right)+cos3x\)

=>\(\sin3x-cos3x=\sqrt2\cdot cos\left(x-\frac{\pi}{5}\right)\)

=>\(\sqrt2\cdot\sin\left(3x-\frac{\pi}{4}\right)=\sqrt2\cdot cos\left(x-\frac{\pi}{5}\right)\)

=>\(\sin\left(3x-\frac{\pi}{4}\right)=cos\left(x-\frac{\pi}{5}\right)=\sin\left(\frac{\pi}{2}-x+\frac{\pi}{5}\right)=\sin\left(-x+\frac{7}{10}\pi\right)\)

=>\(\left[\begin{array}{l}3x-\frac{\pi}{4}=-x+\frac{7}{10}\pi+k2\pi\\ 3x-\frac{\pi}{4}=\pi+x-\frac{7}{10}\pi+k2\pi=x+\frac{3}{10}\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}4x=\frac{7}{10}\pi+\frac{\pi}{4}+k2\pi=\frac{19}{20}\pi+k2\pi\\ 2x=\frac{3}{10}\pi+\frac{\pi}{4}+k2\pi=\frac{11}{20}\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{19}{80}\pi+\frac{k\pi}{2}\\ x=\frac{11}{40}\pi+k\pi\end{array}\right.\)

Câu 2: 

\(\left(A\cup B\right)\cap C=A\cap C=[1;+\infty)\cap\left(0;4\right)=[1;4)\)

Tập này có 3 phần tử nguyên

Ta có: \(A=x^2+2y^2+2xy+2x-4y+2022\)

\(=x^2+2xy+y^2+2x+2y+y^2-6y+2022\)

\(=\left(x+y\right)^2+2\left(x+y\right)+1+y^2-6y+9+2012\)

\(=\left(x+y+1\right)^2+\left(y-3\right)^2+2012\ge2012\forall x,y\)

Dấu '=' xảy ra khi y-3=0 và x+y+1=0

=>y=3 và x=-y-1=-3-1=-4

Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ bên trái khung soạn thảo) để được hỗ trợ tốt hơn.