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a)(x-3)2=49
(x-3)2=72
x-3=7
x=7+3
x=10
b)x13=27.x10
x13:x10=33
x13-10=33
x3=33
x=3
\(\left(x-3\right)^2=16\)
\(\Rightarrow\left(x-3\right)^2=4^2\)
\(\Rightarrow x-3=4\)
\(\Rightarrow x=4+3\)
\(\Rightarrow x=7\)
\(\Leftrightarrow\left(x+2\right)\left(\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}\right)=0\)
Vì \(\left(\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}\right)\ne0\)
=> x+2 =0 => x =-2
Ta có
\(\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}=\frac{x+2}{12^{12}}+\frac{x+2}{13^{13}}\)
<=>\(\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}-\frac{x+2}{12^{12}}-\frac{x+2}{13^{13}}=0\)
<=>\(\left(x+2\right)\left(\frac{1}{10^{10}}+\frac{1}{11^{11}}+\frac{1}{12^{12}}+\frac{1}{13^{13}}\right)=0\)
Vì \(\frac{1}{10^{10}}+\frac{1}{11^{11}}+\frac{1}{12^{12}}+\frac{1}{13^{13}}\ne0\)
=>\(x+2=0\)
<=>\(x=-2\)
Tick nha quachtxuanhong23
\(\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}=\frac{x +2}{12^{12}}+\frac{x+2}{13^{13}}\)
\(\Leftrightarrow\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}-\left(\frac{x+2}{12^{12}}+\frac{x+2}{13^{13}}\right)=0\)
\(\Leftrightarrow\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}-\frac{x+2}{12^{12}}-\frac{x+2}{13^{13}}=0\)
\(\Leftrightarrow\left(x+2\right).\left(\frac{1}{10^{10}}+\frac{1}{11^{11}}+\frac{1}{12^{12}}+\frac{1}{13^{13}}\right)=0\)
Vì \(\left(\frac{1}{10^{10}}+\frac{1}{11^{11}}+\frac{1}{12^{12}}+\frac{1}{13^{13}}\right)\ne0\)nên \(x+2=0\Rightarrow x=-2\)
<=>\(\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}-\frac{x+2}{12^{12}}-\frac{x+2}{13^{13}}=0\)
<=>\(\left(x+2\right)\left(\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}\right)=0\)
Vì \(\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}>0\)
=> \(x+2=0\)
<=>\(x=-2\)
\(x^{13}=x^{10}\)
\(\Rightarrow x^{13}-x^{10}=0\)
\(x^{10}.\left(x^3-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^{10}=0\\x^3-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
Vậy \(x=0\)hoặc \(x=1\)
Tham khảo nhé~
\(x^{13}=x^{10}\)
\(\Leftrightarrow x^{13}-x^{10}=0\)
\(\Leftrightarrow x^{10}.x^3-x^{10}=0\)
\(\Leftrightarrow x^{10}.\left(x^3-1\right)=0\Leftrightarrow\hept{\begin{cases}x^{10}=0\\x^3-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=1\end{cases}}}\)
\(x\in\left\{0;1\right\}\)
\(\left(4x-1\right)^4=\left(4x-1\right)^2\)
\(\Leftrightarrow\left(4x-1\right)^4-\left(4x-1\right)^2=0\)
\(\Leftrightarrow\left(4x-1\right)^2.\left[\left(4x-1\right)^2-\left(4x-1\right)\right]=0\)
\(\Leftrightarrow\hept{\begin{cases}(4x-1)^2=0\\\left[\left(4x-1\right)^2-\left(4x-1\right)\right]=0\end{cases}\Leftrightarrow\hept{\begin{cases}4x=1\Leftrightarrow x=\frac{1}{4}=0,25\\\left(4x-1\right)^2=0+\left(4x-1\right)\Leftrightarrow x=\frac{1}{2}=0,5\end{cases}}}\)
\(x\in\left\{0,25;0,5\right\}\)
x^13 = 27x^10
<=>x^3 . x^10 = 27 . x^10
<=>x^3=27
<=>x^3=3^3
<=>x=3
Vậy x=3
chúc hok tốt
\(x^{13}=27x^{10}\)
<=> \(x^{10}\left(x^3-27\right)=0\)
<=> \(\orbr{\begin{cases}x^{10}=0\\x^3-27=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^3=27\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)