7³ x 5⁸ = 32¹¹

\(7^3x5^8=>=343x687,5=235812,5\)

CHúc bạn THÀNH Công !! 

!!

phân số là 2592/390625

số của phân số là 0,006663552

=-1x1x-1x1x-1x1x-1x1x-1x1=-1

\(=\left(-1\right).1.\left(-1\right).1.\left(-1\right).1.\left(-1\right).1.\left(-1\right).1=-1\)

Chọn B :>

B OK HOK TTO

Ta có \(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)

\(=\frac{y+z+1+x+z+2+x+y-3}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)

=> \(\frac{1}{x+y+z}=2\Rightarrow x+y+z=\frac{1}{2}\)

Khi đó \(\hept{\begin{cases}\frac{y+z+1}{x}=2\\\frac{x+z+2}{y}=2\\x+y+z=\frac{1}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y+z+1=3x\\x+y+z+2=3y\\x+y+z=\frac{1}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{1}{2}+1=3x\\\frac{1}{2}+2=3y\\x+y+z=\frac{1}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{5}{6}\\z=-\frac{5}{6}\end{cases}}\)

Theo tính chất dãy tỉ số bằng nhau 

\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{y+z+1+x+z+2+x+y-3}{x+y+z}\)

\(=\frac{2\left(x+y+z\right)}{x+y+z}=2\)

\(\Rightarrow\frac{1}{x+y+z}=2\Rightarrow x+y+z=\frac{1}{2}\)

\(\Rightarrow x=\frac{1}{2}-y-z;y=\frac{1}{2}-x-z;z=\frac{1}{2}-y-x\)

*, \(\frac{y+z+1}{x}=2\Rightarrow2x=y+z+1\)

\(\Leftrightarrow2x=\frac{1}{2}-x-z+z+1\Leftrightarrow2x=\frac{1}{2}-x+1\Leftrightarrow3x=\frac{3}{2}\Leftrightarrow x=\frac{1}{2}\)

*, \(\frac{x+z+2}{y}=2\Rightarrow x+z+2=2y\)

\(\Leftrightarrow2y=\frac{1}{2}-y-z+z+2\Leftrightarrow2y=\frac{1}{2}-y+2\Leftrightarrow3y=\frac{5}{2}\Leftrightarrow y=\frac{5}{6}\)

*, \(\frac{1}{x+y+z}=2\Rightarrow x+y+z=\frac{1}{2}\Leftrightarrow\frac{1}{2}+\frac{5}{6}+z=\frac{1}{2}\Leftrightarrow z=-\frac{5}{6}\)

a).  x^4  =  x^7

=>  x   =  \(\orbr{\begin{cases}0\\1\end{cases}}\)          

b)  (2y+1)^6 = (2y+1)^8

=>    \(\orbr{\begin{cases}2y+1=0\\2y+1=1\end{cases}}\)       

=>    \(\orbr{\begin{cases}y=-\frac{1}{2}\\y=0\end{cases}}\)           

`a,`

`x^4=x^7`

`->x^4-x^7=0`

`->x^4 (1 - x^3)=0`

TH1 : `x^4=0 ->x=0`

TH2 : `1-x^3=0 ->x^3=1^3 ->x=1`

Vậy `x=0,x=1`

`b,`

`(2y+1)^6 = (2y+1)^8`

`-> (2y+1)^6 - (2y+1)^8=0`

`-> (2y+1)^6 [1-(2y+1)^2]=0`

TH1 : `(2y+1)^6 =0 ->2y+1=0 ->y=(-1)/2`

TH2 : `1 - (2y+1)^2=0`

`-> (2y+1)^2=1`

`-> 2y+1=1` hoặc `2y+1=-1`

`->y=0` hoặc `y=-1`

Vậy `y=0,y=-1,y=(-1)/2`