4n - 5 chia hết cho n-3

=> 4n - 12 + 7 chia hết cho n - 3

=> 7 chia hết cho n-3

=> n - 3 \(\in\)U(7)

U(7) = {-7;-1;1;7}

n - 3 = -7

=> n = -4

n - 3 = -1

n = 2

n - 3 = 1

n = 4

n - 3 = 7

n = 10

Vậy x \(\in\){-4;2;4;10}

1) \(\left|x-\frac{3}{5}\right|< \frac{1}{3}\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{3}{5}< \frac{1}{3}\\x-\frac{3}{5}< -\frac{1}{3}\end{cases}}\Rightarrow\orbr{\begin{cases}x< \frac{1}{3}+\frac{3}{5}\\x< \frac{-1}{3}+\frac{3}{5}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x< \frac{5}{15}+\frac{9}{15}\\x< \frac{-5}{15}+\frac{9}{15}\end{cases}}\Rightarrow\orbr{\begin{cases}x< \frac{14}{15}\\x< \frac{4}{15}\end{cases}}\)

                vay \(\orbr{\begin{cases}x< \frac{14}{15}\\x< \frac{4}{15}\end{cases}}\)

2) \(\left|x+\frac{11}{2}\right|>\left|-5,5\right|\)

\(\left|x+\frac{11}{2}\right|>5,5\)

\(\Rightarrow\orbr{\begin{cases}x+\frac{11}{2}>\frac{11}{2}\\x+\frac{11}{2}>-\frac{11}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x>\frac{11}{2}-\frac{11}{2}\\x>\frac{-11}{2}-\frac{11}{2}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x>0\\x>-11\end{cases}}\)

vay \(\orbr{\begin{cases}x>0\\x>-11\end{cases}}\)

3) \(\frac{2}{5}< \left|x-\frac{7}{5}\right|< \frac{3}{5}\)

\(\Rightarrow\left|x-\frac{7}{5}\right|>\frac{2}{5}\) va \(\left|x-\frac{7}{5}\right|< \frac{3}{5}\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{7}{5}>\frac{2}{5}\\x-\frac{7}{5}>\frac{-2}{5}\end{cases}}\Rightarrow\orbr{\begin{cases}x>\frac{2}{5}+\frac{7}{5}\\x>\frac{-2}{5}+\frac{7}{5}\end{cases}}\)va \(\orbr{\begin{cases}x-\frac{7}{5}< \frac{3}{5}\\x-\frac{7}{5}< \frac{-3}{5}\end{cases}}\Rightarrow\orbr{\begin{cases}x< \frac{3}{5}+\frac{7}{5}\\x< \frac{-3}{5}+\frac{7}{5}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x>\frac{9}{5}\\x>1\end{cases}}\)va \(\orbr{\begin{cases}x< 2\\x< \frac{4}{5}\end{cases}}\)

vay ....

\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)

\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)

\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\)

Dễ thấy: \(\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\ne0\Rightarrow x+2004=0\Leftrightarrow x=-2014\)

x = -2014

ti-ck nha

.........

áp dụng tính chất của dãy  tỉ số bằng nhau : 

ta có :

\(\frac{x}{3}=\frac{y}{4}=\frac{x+y}{3+4}=\frac{42}{7}=6.\)

\(\Rightarrow\frac{x}{3}=6\Rightarrow x=18\)

\(\frac{y}{4}=6\Rightarrow y=24\)

\(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{2}\) và \(x-3y=20\)

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

\(\dfrac{x}{5}=\dfrac{3y}{9}=\dfrac{z}{2}=\dfrac{x-3y}{5-9}=\dfrac{20}{-4}=-5\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=-5< =>x=-25\\\dfrac{y}{3}=-5< =>y=-15\\\dfrac{z}{2}=-5< =>z=-10\end{matrix}\right.\)

Vậy ....