Đề bài không cho biết phần tô đậm trong hình vẽ chỗ nào!!!

\(\left\{{}\begin{matrix}\dfrac{3}{x-1}+\dfrac{4}{y}=13\\\dfrac{2}{x-1}-\dfrac{5}{y}=1\end{matrix}\right.\)(1)

ĐK: \(\left\{{}\begin{matrix}x-1\ne0\\y\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\y\ne0\end{matrix}\right.\)

Đặt \(u=\dfrac{1}{x-1};v=\dfrac{1}{y}\)

\(\left(1\right)\Rightarrow\left\{{}\begin{matrix}3u+4v=13\\2u-5v=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6u+8v=26\\6u-15v=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}23v=23\\2u-5v=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}v=1\\2u=1-5v=1+5.1=6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}v=1\\u=\dfrac{6}{2}=3\end{matrix}\right.\)

- Khi u= 3, ta có \(\dfrac{1}{x-1}=3\Leftrightarrow1=3\left(x-1\right)\Leftrightarrow1=3x-3\)

\(\Leftrightarrow3x=4\Leftrightarrow x=\dfrac{4}{3}\)(thỏa mãn)

- Khi v= 1, ta có: \(\dfrac{1}{y}=1\Leftrightarrow y=1\)(thỏa mãn)

Vậy nghiệm của hệ phương trình là: \(\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=1\end{matrix}\right.\)

1) y = x^2/5
Tọa độ A có tung độ bằng 5 -> y = 5
-> x^2 = 25
-> x = 5 hoặc x = -5
2)Do phương trình có 2 nghiệm phân biệt ta có
 x1x2 = c/a = m^2
 x1+x2 = -b/a = 2(m+3)
 (x1+x2)^2 = x1^2 + 2x1x2 +x2^2
 = 15 + x1x2 = 15 + m^2
 -> (2(m+3) )^2 = 15 + m^2
 -> 4(m^2 +6m + 9) = 15 + m^2
 -> 3m^2 +24m + 21 = 0
 delta = 24*24 -4*3*21 =324, sqrt(delta) = 18
 
 m1 = (-24+18)/6 = -1
 m2 = (-24-18)/6 = -7
 Vậy m = -1 hoặc m = -7

\(\Leftrightarrow40+2xy=x\left(x\ne0\right)\)

\(\Leftrightarrow x\left(1-2y\right)=40\Leftrightarrow x=\dfrac{40}{1-2y}\)

Do 2y chẵn => 1-2y lẻ

Để x nguyên thì 1-2y là ước của 40

\(\Rightarrow1-2y=\left\{-5;-1;1;5\right\}\Rightarrow y=\left\{3;1;0;-2\right\}\)

\(\Rightarrow x=\left\{-8;-40;40;8\right\}\)

\(\Leftrightarrow x\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}\right)=\dfrac{1}{21}\)

\(\Leftrightarrow x\left(\dfrac{5-2}{2.5}+\dfrac{8-5}{5.8}+\dfrac{11-8}{8.11}+\dfrac{14-11}{11.14}\right)=\dfrac{1}{21}\)

\(\Leftrightarrow x\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{11}-\dfrac{1}{14}\right)=\dfrac{1}{21}\)

\(\Leftrightarrow x\left(\dfrac{1}{2}-\dfrac{1}{14}\right)=\dfrac{1}{21}\Leftrightarrow x.\dfrac{3}{7}=\dfrac{1}{21}\Leftrightarrow x=\dfrac{1}{9}\)

x=0.1111111111

1, vt : \(\left(1-\dfrac{5+\sqrt{2}}{\sqrt{2}+1}\right).\sqrt{3+2\sqrt{2}}\)

=\(\dfrac{\sqrt{2}+1-5-\sqrt{2}}{\sqrt{2}+1}.\sqrt{\left(\sqrt{2}\right)^2+2\sqrt{2}+1}\)

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

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

=-4

2, A=\(\left(\dfrac{\sqrt{x}}{x+\sqrt{x}}-\dfrac{1}{\sqrt{x}-1}\right)\div\dfrac{2}{x+\sqrt{x}-2}\)

=\(\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{2}\)

=\(\left(\dfrac{x-\sqrt{x}-x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2}\)

=\(\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{2}\)

=\(\dfrac{-\sqrt{x}-2}{\sqrt{x}+1}\)

5/14a=3

=.>a=3:5/14

=>a=8.4

Gọi số cần tìm là abcdef. theo đề bài

bcdefa=3xabcdef

10xbcdef+a=300000xa+3xbcdef

7xbcdef=299999xa

=> bcdef=299999xa:7=42857xa

Do bcdef là số có 5 chữ số nên a=1 hoặc a=2

+ Với a=1 => bcdef=42857 => abcdef=142857

+ Với a=2 => bcdef=42857x2=85714 => abcdef=285714

\(\dfrac{x-3}{10}-\dfrac{x-3}{15}=\dfrac{x-3}{12}-\dfrac{x-3}{18}\)

\(\Rightarrow\dfrac{x-3}{1}-\dfrac{x-3}{15}-\dfrac{x-3}{12}+\dfrac{x-3}{18}=0\)

\(\Rightarrow\left(x-3\right)\left(1-\dfrac{1}{15}-\dfrac{1}{12}+\dfrac{1}{18}\right)=0\)

Mà \(1-\dfrac{1}{15}-\dfrac{1}{12}+\dfrac{1}{18}\ne0\)

\(\Rightarrow x-3=0\Leftrightarrow x=3\)