Lời giải:

$2x=3y\Rightarrow \frac{x}{3}=\frac{y}{2}$

$\Rightarrow \frac{x}{21}=\frac{y}{14}$

$5y=7z\Rightarrow \frac{y}{7}=\frac{z}{5}\Rightarrow \frac{y}{14}=\frac{z}{10}$

Vậy:

$\frac{x}{21}=\frac{y}{14}=\frac{z}{10}$

$=\frac{3x}{63}=\frac{7y}{98}=\frac{5z}{50}=\frac{3x-7y+5z}{63-98+50}=\frac{30}{15}=2$

$\Rightarrow x=21.2=42; y=14.2=28; z=10.2=20$

ghê zậy 

2=2.1

3=3.1

4=2²

5=5.1

6=2.3

7=7.1

8=2³

9=3²

10=2.5

=> BCNN (2,3,4,5,6,7,8,9,10)=1.2³.3².5.7=2520

\(3^{x+1}+2x+3^x-18x-27=0\)

<=> \(3^x\left(3+2x\right)-9\left(2x+3\right)=0\)

<=> \(\left(2x+3\right)\left(3^x-9\right)=0\)

<=>\(\orbr{\begin{cases}x=-\frac{3}{2}\\x=2\end{cases}}\)

vậy.......

3^{x + 1} + 2x .3^x - 18x - 27 = 0

<=> 3^x ( 3 + 2x ) - 9 ( 2x + 3 ) = 0

<=> ( 3^x - 9 ) ( 2x + 3 ) = 0

<=> \(\orbr{\begin{cases}3^x-9=0\\2x+3=0\end{cases}}\)<=>\(\orbr{\begin{cases}3^x=9\\2x=-3\end{cases}}\)

<=>\(\orbr{\begin{cases}3^x=3^2\\x=-\frac{3}{2}\end{cases}}\)<=>\(\orbr{\begin{cases}x=2\\x=-\frac{3}{2}\end{cases}}\)

4x = 3y => x/3 = y/4 => x/9 = y/12 ( 1 )

5y = 6z => y/6 = z/5 => y/12 = z/10 ( 2 )

Từ ( 1 ) và ( 2 ) => x/9 = y/12 = z/10

=> 2x/18 = y/12 = z/10 

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

2x/18 = y/12 = z/10 = 2x+y-z/18+12-10 = 40/20 = 2

=> x = 18 ; y = 24 ; z = 20

Vậy ...

   \(\left(2x-1\right)^{10}=\left(2x-1\right)^{11}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=1\\2x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)

Vậy ...........

\(\Rightarrow\left(2x-1\right)^{11}-\left(2x-1\right)^{10}=0\)

\(\Rightarrow\left(2x-1\right)^{10}.\left[\left(2x-1\right)-1\right]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(2x-1\right)^{10}=0\\\left(2x-1\right)-1=0\end{cases}\Rightarrow\orbr{\begin{cases}2x-1=0\\2x-1=1\end{cases}}}\)

                                                     \(\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=1\end{cases}}\)

Vậy \(x\in\left\{\frac{1}{2};1\right\}\)

Ta có: \(2x=3y\Rightarrow\frac{x}{3}=\frac{y}{2};5y=7z\Rightarrow\frac{y}{7}=\frac{z}{5}\)

\(\Rightarrow\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\)

Áp dụng t/c dãy tỉ số bằng nhau, ta có:

\(\frac{3x}{63}=\frac{7y}{98}=\frac{5z}{50}=\frac{3x-7y+5z}{63-98+50}=\frac{-30}{15}=-2\)

=> x = (-2).21 = -42

     y = (-2).14 = -28

     z = (-2).10 = -20

Vậy ...

\(2x=3y\)\(\Rightarrow\)\(\frac{x}{3}=\frac{y}{2}\)hay   \(\frac{x}{21}=\frac{y}{14}\)

\(5y=7z\) \(\Rightarrow\)\(\frac{y}{7}=\frac{z}{5}\)hay  \(\frac{y}{14}=\frac{z}{10}\)

suy ra:   \(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\) hay   \(\frac{3x}{63}=\frac{7y}{98}=\frac{5z}{50}\)

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

      \(\frac{3x}{63}=\frac{7y}{98}=\frac{5z}{50}=\frac{3x-7y+5z}{63-98+50}=-2\)

suy ra:   \(\frac{3x}{63}=-2\)\(\Rightarrow\)\(x=-42\)

             \(\frac{7y}{98}=-2\)\(\Rightarrow\)\(y=-28\)

             \(\frac{5z}{50}=-2\) \(\Rightarrow\)\(z=-10\)

\(\left|3x+5y\right|+\left|2x-10\right|=0\)

Vì \(\left|3x+5y\right|\ge0;\left|2x-10\right|\ge10\forall x;y\)

\(\Rightarrow\hept{\begin{cases}3x+5y=0\\2x-10=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}3\cdot5+5y=0\\x=5\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}y=-3\\x=5\end{cases}}\)

Vậy x = 5; y = -3

a) \(x-2xy+x=0\Leftrightarrow2x-2xy=0\)

\(\Leftrightarrow2x\left(1-y\right)=0\Leftrightarrow\hept{\begin{cases}2x=0\\1-y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\y=1\end{cases}}\)

b) \(2xy+x-2y=4\)

\(\Leftrightarrow x\left(2y+1\right)-2y-1=3\)

\(\Leftrightarrow x\left(2y+1\right)-\left(2y+1\right)=3\)

\(\Leftrightarrow\left(2y+1\right)\left(x-1\right)=3\)

Đến đây bí =) Alibaba!

a,\(x-2xy+x=0=>2x-2xy=0=>2x\left(1-y\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\1-y=0\end{cases}\Rightarrow\orbr{\begin{cases}0\\1\end{cases}}}\)