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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 ...
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\(\left(6x-3^3\right).5^3=3.5^4\)
\(\Rightarrow\left(6x-27\right)=3\left(5^4\div5^3\right)\)
\(\Rightarrow\left(6x-27\right)=3.5\)
\(\Rightarrow6x-27=15\)
\(\Rightarrow6x=42\)
\(\Rightarrow x=7\)
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Bài 2:
Ta có: \(16x+40=10\cdot3^2+5\left(1+2+3\right)\)
\(\Leftrightarrow16x+40=90+30\)
\(\Leftrightarrow16x=80\)
hay x=5
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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; \(1^3+2^3+...+9^3=2025\)
\(\Leftrightarrow2^3.\left(1^3+2^3+....+9^3\right)=2^3.2025\)
\(\Leftrightarrow2^3+4^5+...+18^3=16200\)
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2x - 138 = 23 . 32
2x - 138 = 72
2x = 72 + 138
2x = 210
x = 105
10 + 2x = 45 : 43
10 + 2x = 16
2x = 16 - 10
2x = 6
x= 3
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\(2^x< 27^3\)
\(\Rightarrow2^x< \left(3^3\right)^3\)
\(\Rightarrow2^x< 3^9\)
=> x = 9
Vậy x = 9
1 + 2 + 3 + .... + x = 630
Ta có :
1 + 2 + 3 + ... + x
= ( 1 + x ) . x : 2 = 630
=> x ( x + 1 ) = 1260
Do x và x + 1 là 2 số tự nhiên liên tiếp
Ta có : 1260 = \(2^2.3^2.5.7\)= 36 . 35 ( là 2 số tự nhiên liên tiếp )
Mà x < x + 1 => x = 35
Vậy x = 35
\(\left(x+2\right)+1^3=6\\ =>\left(x+2\right)+1=6\\ =>x+2=6-1\\ =>x+2=5\\ =>x=5-2\\ =>x=3\)
(x+2)+13=6
x=6-1-2
x=3