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\(1,2x+3x-4x=\left(-2\right)^3\)
<=>\(x=-8\)
\(2,x-2x=4^2+4^0\)
<=>\(-x=16+1\)
<=>\(-x=17\)
<=>\(x=-17\)
\(3,2^3x-3^2x=|12-21|\)
<=>\(-x=9\)
<=>\(x=-9\)
\(4,x-45=2x+54\)
<=>\(x-2x=54+45\)
<=>\(-x=99\)
<=>\(x=-99\)
\(5,5x-12+23=6^7:6^5\)
<=>\(5x+11=6^2\)
<=>\(5x+11=36\)
<=>\(5x=25\)
<=>\(x=5\)
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(2x - 6)5 = (2x - 6)2
=> (2x - 6)5 - (2x - 6)2 = 0
=> (2x - 6)2.[(2x - 6)3 - 1] = 0
=> \(\orbr{\begin{cases}\left(2x-6\right)^2=0\\\left(2x-6\right)^3-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}2x-6=0\\\left(2x-6\right)^3=1\end{cases}}\)
=> \(\orbr{\begin{cases}2x=6\\2x-6=1\end{cases}}\)
=> \(\orbr{\begin{cases}x=3\\2x=7\end{cases}}\)
=> \(\orbr{\begin{cases}x=3\\x=\frac{7}{2}\left(ktm\right)\end{cases}}\)
33x - 4 - x0 = 8
=> 33x - 4 - 1 = 8
=> 33x - 4 = 8 +1
=> 33x - 4 = 9
=> 33x - 4= 32
=> 3x - 4 = 2
=> 3x = 2 + 4
=> 3x = 6
=> x = 6 : 3 = 2
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\(2x-8x^2=0\Rightarrow2x\left(1-4x\right)=0\Rightarrow\orbr{\begin{cases}2x=0\\1-4x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{4}\end{cases}}}\)
\(x-x^2=0\Rightarrow x\left(1-x\right)=0\Rightarrow\orbr{\begin{cases}x=0\\1-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
Cn lại lm tương tự nha e!
=.= hok tốt!!
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1.
a) ( 57 + 59 ) . ( 68 + 610 ) . ( 24 - 42 )
= ( 57 + 59 ) . ( 68 + 610 ) . 0
= 0
b) 9 < 3x < 27
32 < 3x < 33
2 < x < 3
Vậy 2 < x < 3
2.
a) xy - 2x = 0
x ( y - 2 ) = 0
\(\Rightarrow\orbr{\begin{cases}x=0\\y-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\y=2\end{cases}}}\)
b) ( x- 4 ) . ( x - 3 ) = 0
\(\Rightarrow\orbr{\begin{cases}x-4=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=3\end{cases}}\)
c) Ta có : 3n+2 + 3n = 3n . 32 + 3n = 3n . ( 32 + 1 ) = 3n . 10 \(⋮\)10
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Ta có : (2x + 1)2 + (x + 3)2 - 5(x + 6)(x - 6) = 0
<=> 4x2 + 4x + 1 + x2 + 6x + 9 - 5(x2 - 36) = 0
<=> 4x2 + 4x + 1 + x2 + 6x + 9 - 5x2 + 180 = 0
<=> 4x2 + x2 - 5x2 + 4x + 6x + 1 + 9 + 180 = 0
<=> 10x + 190 = 0
<=> 10x = -190
=> x = -19
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\(2^x=2\Rightarrow x=1\)
\(2^{2x+2}=8^2\Rightarrow2^{2x+2}=2^6\Rightarrow2x+2=6\)\(2x=6-2=3\Rightarrow x=3:2=\frac{3}{2}\)
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:v Không ai giải à? Thế thì mình giải =)
\(\left(2x-1\right)^5-27\left(2x-1\right)^2=0\)
\(\Leftrightarrow\left[\left(2x-1\right)^3\right]^2-\left(54x-27\right)^2=0\)
\(\Leftrightarrow\left(2x-1\right)^3-\left(54x-27\right)=0\) (1)
\(\Rightarrow\left(2x-1\right)^3=\left(54x-27\right)\) (2)
Từ (2) kết hợp với (1) suy ra: \(\left(2x-1\right)^3=\left(54x-27\right)=0\Leftrightarrow\hept{\begin{cases}\left(2x-1\right)^3=0\\54x-27=0\end{cases}\Leftrightarrow\hept{\begin{cases}2x=1\\54x=27\end{cases}}\Leftrightarrow}x=\frac{1}{2}\)
Chết,ban nãy giải nhầm! Thế mà vẫn ra kết quả đúng. Thế mới lạ! Giải lại thôi kẻo mấy đứa nói mình giải xàm thì phiền =)
\(\left(2x-1\right)^5-27\left(2x-1\right)^2=0\)
\(\Leftrightarrow1\left(2x-1\right)^2\left(2x-1\right)^3-27\left(2x-1\right)^2=0\)
\(\Leftrightarrow\left(1-27\right)\left(2x-1\right)^2\left(2x-1\right)^3=0\)
\(\Leftrightarrow-26\left(2x-1\right)^5=0\Leftrightarrow\left(2x-1\right)^5=0\)
\(\Leftrightarrow2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)
\(\left(2x-6\right)^5+27\left(2x-6\right)^2=0\)
\(\Leftrightarrow\left(2x-6\right)^2\left[\left(2x-6\right)^3+27\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(2x-6\right)^2=0\\\left(2x-6\right)^3+27=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-6=0\\\left(2x-6\right)^3=-27\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\2x-6=-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{3}{2}\end{cases}}\)
Vậy \(x\in\left\{3;\frac{3}{2}\right\}\)