CTV mới học lớp 7 thui ak                                    

\(2^{x-2}.3^{y-3}.5^{z-1}=144=2^4.3^2.5^0\)

\(\Rightarrow\hept{\begin{cases}x-2=4\Rightarrow x=6\\y-3=2\Rightarrow y=5\\z-1=0\Rightarrow z=1\end{cases}}\)

\(2^{x-2}.3^{y-3}.5^{z-1}=144\)

mà 144 =  2⁴.3²

=> \(2^{x-2}.3^{y-3}.5^{z-1}=2^4.3^2.1=2^4.3^2.5^0\)

=> \(\hept{\begin{cases}x-2=4\\y-3=2\\z-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=6\\y=5\\z=1\end{cases}}}\)

Vậy...

Các bạn ơi! Dấu chấm là dấu nhân nha!

Ta có: \(144=2^4.3^2.5^0\)

Suy ra: \(2^{x-2}.3^{y-3}.5^{z-1}=2^4.3^2.5^0\)

Suy ra: \(2^{x-2}=2^4;3^{y-3}=3^2;5^{z-1}=5^0\)

Suy ra: \(x-2=4;y-3=2\) và \(z-1=0\)

Hay \(x=6;y=5\) và \(z=1\)

Dễ thấy: \(144=2^4\cdot3^2=2^4\cdot3^2\cdot5^0\)

\(pt\Leftrightarrow2^{x-2}\cdot3^{y-3}\cdot5^{z-1}=2^4\cdot3^2\cdot5^0\)

\(\Rightarrow\left\{{}\begin{matrix}x-2=4\\y-3=2\\z-1=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=6\\y=5\\z=1\end{matrix}\right.\)

trong câu hỏi tt cx có '-'

Giải:

\(x-5\sqrt{x}\) = 0 (\(x\) ≥ 0)

\(\sqrt{x}\) .(\(\sqrt{x}\) - 5) = 0

\(\left[\begin{array}{l}\sqrt{x}=0\\ \sqrt{x}-5=0\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ \sqrt{x}=5\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ x=25\end{array}\right.\)

Vậy \(x\in\) {0; 25}

\(x^5\) = 2\(x^7\)

\(x^5\) - 2\(x^7\) = 0

\(x^5\).(1 - 2\(x^2\)) = 0

\(\left[\begin{array}{l}x^5=0\\ 1-2x^2=0\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ 2x^2=1\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ x^2=\frac12\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ x=\pm\sqrt{\frac12}\end{array}\right.\)

Vậy \(x\) ∈ {- \(\sqrt{\frac12}\); 0; \(\sqrt{\frac12}\)}

Câu 3:

2\(^{x-2}\).3\(^{y-3}\).5\(^{z-1}\) = 144

2\(^{x-2}\).3\(^{y-3}\).5\(^{z-1}\) = 2\(^4\).3\(^2\).5\(^0\)

\(\left[\begin{array}{l}x-2=4\\ y-3=2\\ z-1=0\end{array}\right.\)

\(\left[\begin{array}{l}x=4+2\\ y=2+3\\ x=1\end{array}\right.\)

\(\left[\begin{array}{l}x=6\\ y=5\\ z=1\end{array}\right.\)

Vậy \(\left(x;y;z\right)\) = (6; 5; 1)

Giải:

\(x-5\sqrt{x}\) = 0 (\(x\) ≥ 0)

\(\sqrt{x}\) .(\(\sqrt{x}\) - 5) = 0

\(\left[\begin{array}{l}\sqrt{x}=0\\ \sqrt{x}-5=0\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ \sqrt{x}=5\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ x=25\end{array}\right.\)

Vậy \(x\in\) {0; 25}

\(2^{x-2}.3^{y-3}.5^{z-1}=144=>2^{x-2}.3^{y-3}.5^{z-1}=2^4.3^2.5^0\)

\(\hept{\begin{cases}2^{x-2}=2^4\\3^{y-3}=3^2\\5^{z-1}=5^0\end{cases}}=>\hept{\begin{cases}x-2=4\\y-3=2\\z-1=0\end{cases}}=>\hept{\begin{cases}x=4+2\\y=2+3\\z=0+1\end{cases}}=>\hept{\begin{cases}x=6\\y=5\\z=1\end{cases}}\)

vậy \(\hept{\begin{cases}x=6\\y=5\\z=1\end{cases}}\)

Tách số 144 ra ta có : 

\(144=2^4.3^2.1=2^4.3^2.5^0\)

Theo đề bài 

\(\Rightarrow\hept{\begin{cases}x-2=4\\y-3=2\\z-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=6\\y=5\\z=1\end{cases}}}\)