(x+2)+1 mũ3=6
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\(1.\left(x^3-1\right)\left(x^2+1\right)=0\)
\(< =>\left\{{}\begin{matrix}x^3-1=0\\x^2+1=0\end{matrix}\right.\)
\(< =>\left\{{}\begin{matrix}x^3=1\\x^2=-1\left(kxđ\right)\end{matrix}\right.\)
<=>x=1
vậy ...
\(2.\left(2x+6\right)\left(3x^2-12\right)=0\)
\(< =>\left\{{}\begin{matrix}2x+6=0\\3x^2-12=0\end{matrix}\right.\)
\(< =>\left\{{}\begin{matrix}2x=-6\\3x^2=12\end{matrix}\right.\)
\(< =>\left\{{}\begin{matrix}x=-3\\x^2=4\end{matrix}\right.\)
\(< =>\left\{{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)
vậy ...
a) \(\left(x-\frac{1}{2}\right)^3=27\)
=> \(\left(x-\frac{1}{2}\right)^3=3^3\)
=> \(x-\frac{1}{2}=3\)
=> \(x=3+\frac{1}{2}\)
=> \(x=\frac{7}{2}\)
Vậy \(x=\frac{7}{2}.\)
b) \(\left(2x-1\right)^3=-27\)
=> \(\left(2x-1\right)^3=\left(-3\right)^3\)
=> \(2x-1=-3\)
=> \(2x=\left(-3\right)+1\)
=> \(2x=-2\)
=> \(x=\left(-2\right):2\)
=> \(x=-1\)
Vậy \(x=-1.\)
Chúc bạn học tốt!
\(\left(6x-3^3\right).5^3=3.5^4\)
\(\Rightarrow\left(6x-27\right)=3\left(5^4\div5^3\right)\)
\(\Rightarrow\left(6x-27\right)=3.5\)
\(\Rightarrow6x-27=15\)
\(\Rightarrow6x=42\)
\(\Rightarrow x=7\)
\(\frac{-3.7^4+7^3}{7^5.6-7^3.2}=\frac{7^3\left(-3.7+1\right)}{7^3\left(7^2.6-2\right)}=\frac{-21+1}{49.6-2=4}=\frac{-20}{290}\)
Bài 2:
Ta có: \(16x+40=10\cdot3^2+5\left(1+2+3\right)\)
\(\Leftrightarrow16x+40=90+30\)
\(\Leftrightarrow16x=80\)
hay x=5
a) \(\left(3^4.57-9^2.21\right):3^5\)
\(=\left(3^4.57-3^4.21\right):3^5\)
\(=\left[3^4\left(57-21\right)\right]:3^5\)
\(=3^4.36:3^5\)
\(=3^4.2^2.3^2:3^5\)
\(=3.4\)
\(=12\)
b) Ta có; \(1^3+2^3+...+9^3=2025\)
\(\Leftrightarrow2^3.\left(1^3+2^3+....+9^3\right)=2^3.2025\)
\(\Leftrightarrow2^3+4^5+...+18^3=16200\)
\(\left(x+2\right)+1^3=6\\ =>\left(x+2\right)+1=6\\ =>x+2=6-1\\ =>x+2=5\\ =>x=5-2\\ =>x=3\)
(x+2)+13=6
x=6-1-2
x=3