Question

Tìm x,y,z ,biết:

\(a,5^{3x+1}=25^{x+2}\)

\(b,\left(3x-1\right)^{200}=\left(1-3x\right)^{197}\)

\(c,\left(x-\frac{1}{2}\right)^{100}+\left(y-4\right)^{102}\)

\(đ,\left(\frac{1}{2}x+1\right)^2+\left(\frac{2}{3}y-1\right)^2+|x-y-z|\le0\)

Tìm a,b,c, biết\(ab=2,bc=3,ca=54\)

 

 

Answer 1

Làm đầy đủ hộ mình, mai nộp rùi

Answer 2

a) \(5^{3x+1}=25^{x+2}\)

\(\Leftrightarrow5^{3x+1}=\left(5^2\right)^{x+2}\)

\(\Leftrightarrow5^{3x+1}=5^{2x+4}\)

\(\Leftrightarrow3x+1=2x+4\)

\(\Leftrightarrow3x-2x=4-1\)

\(\Leftrightarrow x=3\)

Answer 3

b) \(\left(3x-1\right)^{200}=\left(1-3x\right)^{197}\)

\(\Leftrightarrow\left(1-3x\right)^{200}=\left(1-3x\right)^{197}\)

\(\Leftrightarrow\left(1-3x\right)^{200}-\left(1-3x\right)^{197}=0\)

\(\Leftrightarrow\left(1-3x\right)^{197}\left[\left(1-3x\right)^3-1\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=0\end{cases}}\)

Answer 4

\(a,\text{ }5^{3x+1}=25^{x+2}\)

\(5^{3x+1}=\left(5^2\right)^{x+2}\)

\(5^{3x+1}=5^{2x+4}\)

\(\Rightarrow\text{ }3x+1=2x+4\)

\(3x-2x=4-1\)

\(x=3\)

Answer 5

\(b,\text{ }\) Ta thấy : 

\(\left(3x-1\right)\text{ và }\left(1-3x\right)\) là hai số đối nhau. Mà :

\(\left(3x-1\right)^{200}=\left(1-3x\right)^{197}\)

\(\Rightarrow\hept{\begin{cases}3x-1\in\left\{0\text{ ; }1\right\}\\1-3x\in\left\{0\text{ ; }1\right\}\end{cases}}\)

\(\text{*}\) Với 3x - 1 = 0 thì 3x = 0 + 1 \(\Rightarrow\) 3x = 1        \(\Rightarrow\) x = \(\frac{1}{3}\)

\(\text{*}\) Với 3x - 1 = 1 thì 3x = 1 + 1 \(\Rightarrow\) 3x = 2        \(\Rightarrow\) x = \(\frac{2}{3}\)

\(\text{ }\text{*}\) Với 1 - 3x = 0 thì 3x = 1 - 0  \(\Rightarrow\) 3x = 1        \(\Rightarrow\) x = \(\frac{1}{3}\)

\(\text{*}\) Với 1 - 3x = 1 thì 3x = 1 - 1 \(\Rightarrow\) 3x = 0        \(\Rightarrow\) x = 0

             Vậy \(x\in\left\{\frac{1}{3}\text{ ; }\frac{2}{3}\text{ ; }0\right\}\)

Answer 6

BloomKo bt làm thì đg ns nhé, nếu \(x=\frac{2}{3}\)thì :

\(\left(3x-1\right)^{200}=1\)và \(\left(1-3x\right)^{197}=-1\)

Không có bằng nhau nhé