\(a,\Leftrightarrow\left\{{}\begin{matrix}10x-5-4x+4y=0\\2x-2y-3x-y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+4y=5\\x+3y=-3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}18x+12y=15\\4x+12y=-12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}14x=27\\x+3y=-3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{27}{14}\\y=-\dfrac{23}{14}\end{matrix}\right.\)

\(f,ĐK:x\ge-4;y\ge1\)

Đặt \(\sqrt{x+4}=a;\sqrt{y-1}=b\left(a,b\ge0\right)\), hpt trở thành:

\(\left\{{}\begin{matrix}2a-b=1\\a+2b=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a-b=1\\2a+4b=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a-b=1\\5b=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{6}{5}\\b=\dfrac{7}{5}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+4=\dfrac{36}{25}\\y-1=\dfrac{49}{25}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{64}{25}\left(tm\right)\\y=\dfrac{74}{25}\left(tm\right)\end{matrix}\right.\)

Vậy pt có nghiệm \(\left(x;y\right)=\left(-\dfrac{64}{25};\dfrac{74}{25}\right)\)

\(k,\) Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\y-2=b\ne0\end{matrix}\right.\), hpt trở thành:

\(\left\{{}\begin{matrix}3a-\dfrac{4}{b}=1\\a+\dfrac{2}{b}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a-\dfrac{4}{b}=1\\2a+\dfrac{4}{b}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5a=5\\a+\dfrac{2}{b}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=2\end{matrix}\right.\)

Từ đó thế vào tìm đc x,y

\(m,\) Đặt \(\left\{{}\begin{matrix}\left|x-1\right|=a\ge0\\y-2=b\ne0\end{matrix}\right.\), hpt trở thành:

\(\left\{{}\begin{matrix}2a+\dfrac{3}{b}=3\\3a-\dfrac{1}{b}=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a+\dfrac{3}{b}=3\\9a-\dfrac{3}{b}=30\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}11a=33\\2a+\dfrac{3}{b}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=3\\b=-1\end{matrix}\right.\)

Từ đó thế vào tìm x,y

\(ac=-\dfrac{1}{2}< 0\Rightarrow\) pt luôn có 2 nghiệm phân biệt trái dấu

Do \(x_1< x_2\Rightarrow\left\{{}\begin{matrix}x_1< 0\\x_2>0\\\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left|x_1\right|=-x_1\\\left|x_2\right|=x_2\end{matrix}\right.\)

Đồng thời theo Viet: \(x_1+x_2=m\)

Ta có:

\(\left|x_2\right|-\left|x_1\right|=2021\)

\(\Leftrightarrow x_2-\left(-x_1\right)=2021\)

\(\Leftrightarrow x_1+x_2=2021\)

\(\Leftrightarrow m=2021\)

Bài này trong sách là trên mạng có ấy c lên ytb shears cho nhanh 

Polluted

\(\dfrac{1}{x^2-4}+\dfrac{2x}{x+2}=\dfrac{1}{\left(x-2\right)\left(x+2\right)}+\dfrac{2x}{x+2}=\dfrac{1+2x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{1+2x^2-4x}{\left(x+2\right)\left(x-2\right)}\)

trên bài mink đã ẩn đi bước quy đồng!!

\(\dfrac{18}{\left(x-3\right)\left(x^2-9\right)}-\dfrac{3}{x^2-6x+9}-\dfrac{x}{x^2-9}=\dfrac{18}{\left(x-3\right)\left(x+3\right)\left(x-3\right)}-\dfrac{3}{\left(x-3\right)^2}-\dfrac{x}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{18}{\left(x-3\right)^2\left(x+3\right)}-\dfrac{3}{\left(x-3\right)^2}-\dfrac{x}{\left(x-3\right)\left(x+3\right)}=\dfrac{18-3\left(x+3\right)-x\left(x-3\right)}{\left(x-3\right)^2\left(x+3\right)}\)

\(=\dfrac{18-3x-9-x^2+3x}{\left(x-3\right)^2\left(x+3\right)}=\dfrac{9-x^2}{\left(x-3\right)^2\left(x+3\right)}=\dfrac{-\left(x-3\right)\left(x+3\right)}{\left(x-3\right)^2\left(x+3\right)}=\dfrac{-1}{x-3}\)

Bạn ơi làm ý nào đéy ??

2 ý trên à ?????

\(\dfrac{x}{12}+\dfrac{1}{4}=\dfrac{x}{10}\)

\(\leftrightarrow\)\(\dfrac{5x}{60}+\dfrac{15}{60}=\dfrac{6x}{60}\)

\(\leftrightarrow\)\(5x+15=6x\)

\(\leftrightarrow\)\(15=6x-5x\)

\(\leftrightarrow\)\(15=x\)

5:

a: \(-120x^5y^4=20x^5y^2\cdot\left(-6y^2\right)\)

b: \(60x^6y^2=20x^5y^2\cdot3x\)

c: \(-5x^{15}y^3=20x^5y^2\cdot\left(-\dfrac{1}{4}x^{10}y\right)\)

d: \(2x^{12}y^{10}=20x^5y^2\cdot\left(\dfrac{1}{10}x^7y^8\right)\)