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a) xem lại đề
b) 3x-1=27
=>3x-1=33
=>x-1=3
=>x=3+1
=>x=4
c)3x+1=9
=>3x+1=32
=>x+1=2
=>x=2-1
=>x=1
a) x2=x3
⇒ x2-x3=0
⇒ x2-x2.x=0
⇒ x2.(x+1)=0
⇒\(\left\{{}\begin{matrix}\text{x}^2=0\\\text{x}+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\text{x}=0\\\text{x}=0+1=1\end{matrix}\right.\)
⇒ \(\text{x}\in\left\{0;1\right\}\)
b) 3x-1=27
⇒ 3x-1=33
⇒ x-1=3
⇒ x= 3+1=4
c) 3x+1=9
⇒ 3x+1= 33
⇒ x+1=3
⇒ x=3-1=2
\(\frac{5.2^{30}.3^{18}-4.3^{20}.2^{27}}{5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}}\)
\(=\frac{5.2^{28}.2^2.3^{18}-4.3^{18}.3^2.2^{27}}{5.2^{28}.3^{18}.3-7.2^{27}.2^2.3^{18}}\)
\(=\frac{3^{18}.\left(5.2^{28}.4-4.9.2^{27}\right)}{3^{18}.\left(5.3.2^{28}-7.2^{27}.4\right)}\)
\(=\frac{20.2^{28}-9.2.2.2^{27}}{15.2^{28}-14.2.2^{27}}\)
\(=\frac{2^{28}.20-18.2^{28}}{15.2^{28}-14.2^{28}}\)
\(=\frac{2^{28}.\left(20-18\right)}{2^{28}.\left(15-14\right)}\)
\(=\frac{2^{28}.2}{2^{28}1}\)
\(=2\)
\(\left(x^3+27\right)\left(x^2+4\right)=0\) (đề ntn phải ko ạ?)
\(=>\left[{}\begin{matrix}x^3+27=0\\x^2+4=0\end{matrix}\right.\\ =>\left[{}\begin{matrix}x^3=-27\\x^2=-4\left(voli\right)\end{matrix}\right.\)
\(=>x=-3\)
a) x = 4
b) x = 3
c) x = 2
d) x = 1
e) x = 3
f) x = 2
g) x = 4
h) x = 3
9 x 9 x 9 x 9 = 9^4
10 x 10 x 10 x 10 = 10^4
2 x 2 x 2 x 3 x 3 = 2^3 x 3^2
Tíc
`(x^3 + 27)(x^2 +4)=9` hả cậu ?
uh