\(...\Rightarrow\left[{}\begin{matrix}288:\left(x-3\right)^2-2=0\\x^2-169=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}288:\left(x-3\right)^2=2\\x^2=169\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-3\right)^2=288:2\\x^2=13^2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left(x-3\right)^2=144=12^2\\x^2=13^2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-3=12\\x-3=-12\\x=13\\x=-13\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=15\\x=-9\\x=13\\x=-13\end{matrix}\right.\) \(\Rightarrow x\in\left\{-9;15;\pm13\right\}\)

a, \(4.3^{2x}-2.9^x-54=0\)

\(\Rightarrow3^{2x}\left(4-2\right)=54\)

\(\Rightarrow3^{2x}=27=3^3\)

Vì \(3\ne\pm1;3\ne0\) nên \(2x=3\Rightarrow x=\dfrac{3}{2}\)

b, \(\dfrac{1}{2}.2^x+4.2^x-288=0\)

\(\Rightarrow2^x\left(\dfrac{1}{2}+4\right)=288\)

\(\Rightarrow2^x=64=2^6\)

Vì \(2\ne\pm1;2\ne0\) nên \(x=6\)

\(3.2^x+2^{x+1}+2^{x+2}=288\\ \Leftrightarrow2^x\left(3+2+2^2\right)=288\\ \Leftrightarrow2^x=32=2^5\\ \Leftrightarrow x=5\)

tik mik nha

Ta có: \(3\cdot2^x+2^{x+1}+2^{x+2}=288\)

\(\Leftrightarrow2^x\cdot\left(3+2+4\right)=288\)

\(\Leftrightarrow2^x=32\)

hay x=5

a: =>3(x+3)²=26+49=75

=>(x+3)²=25

=>x+3=5 hoặc x+3=-5

=>x=2 hoặc x=-8

b: =>288:(x-3)²=50-8x6=2

=>(x-3)²=144

=>x-3=12 hoặc x-3=-12

=>x=15 hoặc x=-9

2^x+2^x+3=288

=>2(2^x)=288-3=285

2^x=142,5(vô lí)

=>ko có x thỏa mãn

\(2^x+2^{x+3}=288\)

\(\Leftrightarrow2^x+2^x.2^3=288\)

\(\Leftrightarrow2^x\left(1+2^3\right)=288\)

\(\Leftrightarrow2^x=32=2^5\)

\(\Leftrightarrow x=5\)

=>2^x.(1+2³)=288

=>2^x.9=288

=>2^x=288:9

=>2^x=32

=>x=5

2^x+2^x+3=288

2^x.1+2^x.2³=288

2^x(1+2³)=288

2^x.9=288

2^x=288:9

2^x=32

2^x=2⁵

=> x=5

Vậy x=5

\(\Rightarrow2^x\left(\frac{1}{2}+4\right)=288\Rightarrow2^x.\frac{9}{2}=288\Rightarrow2^x=64=2^6\Rightarrow x=6\)

\(\)\(\left(\frac{1}{2}+4\right).2^x=288\)

\(\frac{9}{2}.2^x=288\)

\(2^x=64\)

\(2^x=2^6\)

=> x=6

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{6}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=6k\end{matrix}\right.\)

Ta có: \(x^2y^2z^2=288^2\)

\(\Leftrightarrow36k^2=288^2\)

\(\Leftrightarrow k^2=2304\)

Trường hợp 1: k=48

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k=96\\y=3k=144\\z=6k=288\end{matrix}\right.\)

Trường hợp 2: k=-48

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k=-96\\y=3k=-144\\z=6k=-288\end{matrix}\right.\)

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{6}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=6k\end{matrix}\right.\)

\(\Rightarrow4k^29k^236k^2=288^2\)

\(\Rightarrow k^6=64\)

\(\Leftrightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=4,y=6,z=12\\x=-4,y=-6,z=-12\end{matrix}\right.\)