x-y = 5 => x= 5+y 

\(\frac{x-4}{y-3}=\frac{3}{4}\Rightarrow4x-16=3y-9\Leftrightarrow4x-3y=7\)

Thay x= 5+y vào biểu thức ở đề bài: \(4\left(5+y\right)-3y=7\Leftrightarrow20+4y-3y=7\Leftrightarrow y=-13\)

=> x= 5+y = -8

\(\frac{x}{5}=\frac{y}{3}\)và x²-y²=4(x,y>0)

\(\Rightarrow\frac{x}{5}=\frac{y}{3}=\frac{x^2}{5^2}=\frac{y^2}{3^2}=\frac{x^2-y^2}{25-9}=\frac{4}{16}=\frac{1}{4}\)

Áp dụng tính chất của dãy tỉ số bằng nhau , ta có :

\(\Rightarrow\frac{x^2}{25}=\frac{1}{4}\Rightarrow x^2=\frac{25}{4}\Rightarrow x=\frac{5}{2}\)

\(\Rightarrow\frac{y^2}{9}=\frac{1}{4}\Rightarrow y^2=\frac{9}{4}\Rightarrow y=\frac{3}{2}\)

Vậy x =\(\frac{5}{2}\)và y =\(\frac{3}{2}\)

Ta có:

\(\frac{x}{3}=\frac{y}{5}\Rightarrow\frac{x^2}{3}=\frac{y^2}{5}\)

Áp dụng dãy tỉ số bằng nhau, ta có:

\(\frac{x^2}{3^2}=\frac{y^2}{5^2}=\frac{x^2-y^2}{3^2-5^2}=\frac{-4}{-16}=\frac{1}{4}\)

\(\Rightarrow\frac{x^2}{3^2}=\frac{1}{4}\Rightarrow x=\sqrt{3^2.\frac{1}{4}}=\frac{3}{2}\)

\(\frac{y^2}{5^2}=\frac{1}{4}\Rightarrow y=\sqrt{5^2.\frac{1}{4}}=\frac{5}{2}\)

X= 1- 1/2=1/2

X = 8.KB NHA!!!!!

X x 6 = 48

X = 48 : 6

X = 8

X x 6 = 48

     X = 48 : 6

     X = 8

Vậy X=8

1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 = 362880

= 362880

362880

\(x+x:0,5+x:0,25+x:0,125=2022\)

\(x\text{×}1+x\text{×}2+x\text{×}4+x\text{×8}=2022\)

\(x\text{×}\left(1+2+4+8\right)=2022\)

\(x\text{×}15=2022\)

\(x=2022:15\)

\(x=134,8\)

thấy sai sai bạn ạ

\(x^2+2x-10=0\)

\(\Leftrightarrow x^2+2x+1-9=0\)

\(\Leftrightarrow\left(x+1\right)^2-9=0\\\)

\(\Leftrightarrow\left(x+1\right)^2=9\)

\(\Leftrightarrow\left(x+1\right)^2=\pm\sqrt{9}\)

\(\Leftrightarrow\left(x+1\right)^2=\left(\pm3\right)^2\)

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

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

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

Vậy S={2;-4}

\(x^2\) + 2\(x\) - 10 = 0

\(x^2\) + 2\(x\) + 1 - 11 = 0

(\(x^2\) + 2\(x\) + 1) - 11 = 0

(\(x\) + 1)² - 11 = 0

(\(x\) + 1)² = 11

\(\left[{}\begin{matrix}x+1=\sqrt{11}\\x+1=-\sqrt{11}\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\sqrt{11}-1\\x=-\sqrt{11}-1\end{matrix}\right.\)

Em tách ra thành:

x(1+3+5+...+2021)-x(2+4+...+2020)=2022.

Sau đó giải bình thường.

Chúc em học tốt!

\(\Leftrightarrow9x\left(x+2\right)+9y\left(y-\dfrac{2}{3}\right)=10\\ \Leftrightarrow9x^2+18x+9y^2-6y-10=0\\ \Leftrightarrow\left(9x^2+18x+9\right)+\left(9y^2-6y+1\right)=0\\ \Leftrightarrow9\left(x+1\right)^2+\left(3y-1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=\dfrac{1}{3}\end{matrix}\right.\)