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a, \(x^2-6x+9=4< =>\left(x-3\right)^2=4< =>\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\)
\(< =>\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
b,\(x^2\left(x-3\right)-4\left(x-3\right)=0< =>\left(x-2\right)\left(x+2\right)\left(x-3\right)=0\)
\(< =>\orbr{\begin{cases}x=2\\x=-2\end{cases}orx=3}\)
c nhường mấy bn khácccc
a) x^2-6x+9=4.
x=1, x=5
b) x^2(x-3)-(4X-12)=0
x=-2, x=2, x=3
c) (2x+3)^2-4(x+2)^2=12
x=-19/4
Mk sửa đề nhá : (x + 2)2 - (x - 4)(x + 4) + 16
Ta có : (x + 2)2 - (x - 4)(x + 4) + 16 = 12
<=> x2 + 4x + 4 - (x2 - 42) + 16 = 12
<=> x2 + 4x + 4 - x2 + 16 + 16 = 12
<=> 4x + 36 = 12
<=> 4x = -24
=> x = -6
x3 - 8 - (x - 2).(x - 12) = 0
<=> x3 - 23 - (x - 2).(x - 12) = 0
<=> (x - 2).(x2 + 2x + 4) - (x - 2).(x - 12) = 0
<=> (x - 2).(x2 + 2x + 4 - x + 12) = 0
<=> (x - 2).(x2 + x + 16) = 0
<=> x - 2 = 0
<=> x = 2
Vậy: x = 2
x3 - 8 - ( x - 2 )( x - 12 ) = 0
⇔ ( x - 2 )( x2 + 2x + 4 ) - ( x - 2 )( x - 12 ) = 0
⇔ ( x - 2 )( x2 + 2x + 4 - x + 12 ) = 0
⇔ ( x - 2 )( x2 + x + 16 ) = 0
⇔ x - 2 = 0 hoặc x2 + x + 16 = 0
⇔ x = 2 < do x2 + x + 16 = ( x2 + x + 1/4 ) + 63/4 = ( x + 1/2 )2 + 63/4 ≥ 63/4 > 0 ∀ x >
a)Ta có: 132=(12+1)2
=122+2.12.1+1
=122+25
=>x=25
b)
Ta có: 132=(12+1)2
=122+2.12.1+1
=122+25
=122-(-25)
=>x=-25
\(\left(x-3\right)\left(x+x^2\right)+\left(x-5\right)\left(x+1\right)-x^3=12\)
\(\Leftrightarrow x^2+x^3-3x-3x^2+x^2+x-5x-5-x^3=12\)
\(\Leftrightarrow-x^2-7x-5=12\Leftrightarrow-x^2-7x-17=0\)
Ta có : \(\left(-7\right)^2-4\left(-17\right)\left(-1\right)< 0\)Vậy phương trình vô nghiệm
a) x^2 + 14x + 49 - x^2 + 3x = 12
<=> 17x = -37
<=> x = -37/17
b) x^2 + 2x +1 - x^2 + 4 = 0
<=> 2x = -5
<=> x = -5/2
a) \(\left(x+7\right)^2-x\left(x-3\right)=12\)
\(\Leftrightarrow x^2+14x+49-x^2+3x=12\)
\(\Leftrightarrow17x=-37\)
\(\Leftrightarrow x=\frac{-37}{17}\)
Vậy x = -37/17
b) \(\left(x+1\right)^2-\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow x^2+2x+1-\left(x^2-4\right)=0\)
\(\Leftrightarrow x^2+2x+1-x^2+4=0\)
\(\Leftrightarrow2x=-5\)
\(\Leftrightarrow x=\frac{-5}{2}\)
Vậy x = -5/2
A xác định \(\Leftrightarrow x^2-1\ne0\Leftrightarrow x\ne\left\{1;-1\right\}\)
B xác định \(\Leftrightarrow x^3-8\ne0\Leftrightarrow x^3\ne8\Leftrightarrow x\ne2\)
\(x^2+x=12\)
\(\Rightarrow x^2+x-12=0\)
\(\Rightarrow x\left(x-3\right)+4\left(x-3\right)=0\Rightarrow\left(x-3\right)\left(x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\)