\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{8}{125}\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{2}{5}\right)^3\Rightarrow x-\dfrac{1}{2}=\dfrac{2}{5}\)

\(\Rightarrow x=\dfrac{2}{5}+\dfrac{1}{2}\Rightarrow x=\dfrac{9}{10}\)

\(2^x+2^{x+3}=144\Rightarrow2^x+2^x.2^3=144\Rightarrow2^x\left(1+2^3\right)=144\Rightarrow9.2^x=144\Rightarrow2^x=144:9=16\Rightarrow2^x=2^4\Rightarrow x=4\)

thank you!

Thật ra câu này mk làm rồi nhưng chưa chắc chắn cho lắm!

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 \(z=1\)

a: \(\Leftrightarrow2^{2x}\cdot8+2^{2x}\cdot2+2^{2x}=176\)

\(\Leftrightarrow2^{2x}=16\)

=>2x=4

=>x=2

b: \(\Leftrightarrow2^{2x}\left(2^3+2-1\right)=144\)

\(\Leftrightarrow2^{2x}=16\)

=>2x=4

=>x=2

\(2^x+2^{x+3}=144\)

\(\Leftrightarrow2^x+2^x.2^3=144\)

\(\Leftrightarrow2^x+2^x.8=144\)

\(\Leftrightarrow2^x.\left(1+8\right)=144\)

\(\Leftrightarrow2^x.9=144\)

\(\Leftrightarrow2^x=16\)\(\Leftrightarrow2^x=2^4\)

\(\Leftrightarrow x=4\)

Vậy \(x=4\)

\(144=2^4\cdot3^2\cdot1=2^4\cdot3^2\cdot5^0=2^{x-2}\cdot3^{y-3}\cdot5^{z-1}\\\Rightarrow \left\{{}\begin{matrix}x-2=4\\y-3=2\\z-1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=6\\y=5\\z=1\end{matrix}\right.\)

a)\(\frac{10^3+2.5^3+5^3}{55}\)=\(\frac{5^3.2^3+2.5^3+5^3}{5.11}\)=\(\frac{5^3.\left(2^3+2+1\right)}{5.11}\)=\(5^2\)=\(25\)

b) \(2^x+2^{x+3}=144\)

\(\Rightarrow2^x+2^x.2^3=144\)

\(\Rightarrow2^x.\left(1+2^3\right)=144\)

\(\Rightarrow2^x=16=2^4\)

\(\Rightarrow x=4\)

c) \(2\left(x-5\right)+3\left(x-7\right)=10\)

\(\Rightarrow2x-10+3x-21=10\)

\(\Rightarrow5x-31=10\)

\(\Rightarrow5x=41\)

\(\Rightarrow x=\frac{41}{5}=8,2\)

a) \(81^{-2x}.27^x=9^5\)

\(\Rightarrow\left(3^4\right)^{-2x}.\left(3^3\right)^x=3^{10}\)

\(\Rightarrow3^{-8x}.3^{3x}=3^{10}\)

\(\Rightarrow3^{-5}=3^{10}\)

\(\Rightarrow-5x=10\)

\(\Rightarrow x=-2\)

Vậy \(x=-2\)

b) \(2^x+2^{x+3}=144\)

\(\Rightarrow\left(1+2^3\right).2^x=144\)

\(\Rightarrow\left(1+8\right).2^x=144\)

\(\Rightarrow9.2^x=144\)

\(\Rightarrow2^x=16\)

\(\Rightarrow x=4\)

Vậy \(x=4\)

c) \(2^{x-1}+5.2^{x-2}=7.32\)

\(\Rightarrow\left(2+5\right).2^{x-2}=244\)

\(\Rightarrow7.2^{x-2}=244\)

\(\Rightarrow2^{x-2}=32\)

\(\Rightarrow x-2=5\)

\(\Rightarrow x=7\)

Vậy \(x=7\)

a)Có: (x⁴)³=\(\dfrac{x^{15}}{x^5}\)

<=> x¹²=x¹⁰

<=> x¹²-x¹⁰=0

<=> x¹⁰(x²-1)=0

<=> \(\left[{}\begin{matrix}x^{10}=0\\x^2-1=0\end{matrix}\right.\)<=>\(\left[{}\begin{matrix}x=0\\x^2=1\end{matrix}\right.< =>\left[{}\begin{matrix}x=0\\x\in\left\{1;-1\right\}\end{matrix}\right.\)

Vậy x\(\in\left\{1;-1;0\right\}\)

b)Có 2^x+2^x+3=144

<=> 2^x(1+2³⁾=144

<=> 2^x=16=2⁴

=> x=4

c) Có \(1^2+2^2+3^2+...+10^2=385\)

<=> \(100^2\left(1^2+2^2+3^2+...+10^2\right)=385.100^2\)

<=> \(100^2.1^2+100^2.2^2+...+100^2.10^2=3850000\)

<=> \(100^2+200^2+...+1000^2=3850000\)