\(\dfrac{4}{x}=\dfrac{y}{21}=\dfrac{28}{49}=\dfrac{28:7}{49:7}=\dfrac{4}{9}\\ Vậy:x=\dfrac{4.9}{4}=9\\ y=\dfrac{4.21}{9}=\dfrac{28}{3}\)

\(\dfrac{x}{2}=\dfrac{3}{y}\\ \Leftrightarrow x.y=2.3=6\\ Vậy:\left[{}\begin{matrix}\left(x;y\right)=\left(1;6\right)=\left(6;1\right)\\\left(x;y\right)=\left(2;3\right)=\left(3;2\right)\end{matrix}\right.\)

10^x:5^y=20^y

=>2^x.5^x:5^y=4^y.5^y

=>2^x.5^x-y=2^2y.5^y

=>x=2y:x-y=y

=>x=2y

vậy x=2y  ;   y=x:2

bạn ko biết câu a àk?

\(\left|x+3\right|+\left(y-x\right)^{10}+\left(z+x\right)^{20}=0\)

Ta có: \(\hept{\begin{cases}\left|x+3\right|\ge0\forall x\\\left(y-x\right)^{10}\ge0\forall x;y\\\left(z+x\right)^{20}\ge0\forall x;z\end{cases}}\)

Mà \(\left|x+3\right|+\left(y-x\right)^{10}+\left(z+x\right)^{20}=0\)

\(\Rightarrow\hept{\begin{cases}\left|x+3\right|=0\\\left(y-x\right)^{10}=0\\\left(z+x\right)^{20}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+3=0\\y-x=0\\z+x=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-3\\y=x\\z=-x\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-3\\y=-3\\z=3\end{cases}}}\)

D

a) (x - 3)(y - 3) = 9 = 1.9 = 3.3

Lập bảng:

Vậy ...

b) A = \(\frac{10^{19}+1}{10^{20}+1}\) => 10A = \(\frac{10^{20}+10}{10^{20}+1}=1+\frac{9}{10^{20}+1}\)

B = \(\frac{10^{20}+1}{10^{21}+1}\) => 10B = \(\frac{10^{21}+10}{10^{21}+1}=1+\frac{9}{10^{21}+1}\)

Do \(10^{20}+1< 10^{21}+1\) => \(\frac{9}{10^{20}+1}>\frac{9}{10^{21}+1}\) => 10A > 10B => A > B

\((x-6)(3x-9)>0\)
TH1:
\(\orbr{\begin{cases}x-6< 0\\3x-9< 0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x< 6\\x< 3\end{cases}}\)\(\Rightarrow x< 3\)
TH2:
\(\orbr{\begin{cases}x-6>0\\3x-9>0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x>6\\x>3\end{cases}}\)\(\Rightarrow x>6\)
Vậy \(x< 3\) hoặc \(x>6\)thì \((x-6)(3x-9)>0\)
Học tốt!

20.
\((2x-1)(6-x)>0\)
TH1:
\(\orbr{\begin{cases}2x-1>0\\6-x>0\end{cases}\Rightarrow\orbr{\begin{cases}x< \frac{1}{2}\\x< 6\end{cases}}\Rightarrow x< 6}\)
TH2
\(\orbr{\begin{cases}2x-1< 0\\6-x< 0\end{cases}\Rightarrow\orbr{\begin{cases}x>\frac{1}{2}\\x>6\end{cases}}\Rightarrow x>\frac{1}{2}}\)
Vậy \(x< 6\)hoặc \(x>\frac{1}{2}\)thì \((2x-1)(6-x)>0\)
 

a) 10 - 3x + 3 = -5

=> 13 - 3x = -5

=> 3x = 13 + 5

=> 3x = 18

=> x = 18 : 3 = 6

b) -6|x + 3| = 15 + (-3)

=> -6|x + 3| = 12

=> |x + 3| = 12 : (-6)

=> |x + 3| = -2

=> ko có giá trị x tm

c) 17 - x = 7 - 6x

=> 17 - 7 = -6x + x

=> -5x = 10

=> x = 10 : (-5) = -2

d) Ta có: x + y = 10 

x = y => y + y = 10  

   => 2y = 10 => y = 5 

 => x = 10 - 5 = 5

a , x = 6 

b , ko có giá trị x thỏa mãn 

c , -2 

d , 5 

k và kb nếu có thể