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a/ \(25x^2-9=0\)
<=> \(\left(5x-3\right)\left(5x+3\right)=0\)
<=> \(\orbr{\begin{cases}5x-3=0\\5x+3=0\end{cases}}\)
<=> \(\orbr{\begin{cases}5x=3\\5x=-3\end{cases}}\)
<=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=-\frac{3}{5}\end{cases}}\)
b/ \(\left(x+4\right)^2-\left(x+9\right)\left(x-1\right)=16\)
<=> \(x^2+8x+16-x^2+8x-9=16\)
<=> \(16x+7=16\)
<=> \(16x=9\)
<=> \(x=\frac{9}{16}\)
a) \(25x^2-9=0\)
\(\Leftrightarrow\left(5x-3\right)\left(5x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x-3=0\\5x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}5x=3\\5x=-3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{3}{5}\\x=-\frac{3}{5}\end{cases}}}\)
Vậy S = {3/5 ; -3/5}
b) \(\left(x+4\right)^2-\left(x+9\right)\left(x-1\right)=16\)
\(\Leftrightarrow\left(x+4\right)^2-4^2-\left(x+9\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x+4-4\right)\left(x+4+4\right)-\left(x+9\right)\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x+8\right)-\left(x+9\right)\left(x-1\right)=0\)
\(\Leftrightarrow x^2+8x-x^2-8x+9=0\)
\(\Leftrightarrow9=0\left(vl\right)\)
Vậy S = \(\varnothing\)
a, \(\left(2x+1\right)\left(1-2x\right)+\left(1-2x\right)^2=0\)
\(\Leftrightarrow\left(1-2x\right)\left(2x+1+1-2x\right)=0\Leftrightarrow x=\frac{1}{2}\)
b, \(2\left(x+1\right)^2-\left(x-3\right)\left(x+3\right)-\left(x-4\right)^2=0\)
\(\Leftrightarrow2\left(x^2+2x+1\right)-\left(x^2-9\right)-\left(x^2-8x+16\right)=0\)
\(\Leftrightarrow2x^2+4x+2-x^2+9-x^2+8x-16=0\Leftrightarrow12x-5=0\Leftrightarrow x=\frac{5}{12}\)
c, \(\left(x-5\right)^2-x\left(x-4\right)=9\Leftrightarrow x^2-10x+25-x^2+4x=9\)
\(\Leftrightarrow-6x+16=0\Leftrightarrow x=\frac{8}{3}\)
d, \(\left(x-5\right)^2+\left(x-4\right)\left(1-x\right)=0\)
\(\Leftrightarrow x^2-10x+25+x-x^2-4+4x=0\)
\(\Leftrightarrow-5x+21=0\Leftrightarrow x=\frac{21}{5}\)
a) 4x2 - 9=0
(2x)2 - 32 = 0
=》(2x - 3)(2x+3) =0
=》 2x - 3 = 0 hoặc 2x +3 = 0
=》x = 1,5 hoặc x = - 1,5
b) (x + 1)2 - 16 = 0
=》( x + 1)2 - 42 = 0
=》( x - 3 )( x + 5 ) =0
=》 x - 3 = 0 hoặc x + 5 = 0
=》 x = 3 hoặc x = -5
c) ( x + 1)2 - (2x + 3)2 = 0
=》 ( x + 1 - 2x - 3)(x+1 +2x +3 ) =0
=》 ( -x - 2 )( 3x + 4 ) = 0
=》 -x -2 =0 hoặc 3x + 4 = 0
=》 x = -2 hoặc x = -4/3
d) 4(3x +2)2 - 9( x + 1 )2 =0
=》 [ 2(3x +2) ]2 - [3 (x + 1)] 2 = 0
=> ( 6x +4 )2 - ( 3x + 3)2 = 0
=》 ( 6x +4 -3x -3 )( 6x + 4 + 3x + 3 )=0
=》 (3x +1)(9x + 7 ) =0
=》 3x + 1 =0 hoặc 9x + 7 =0
=》 x = -1/3 hoặc x = -7/9
\(x\left(2x-7\right)-4x+14=0\Leftrightarrow\left(x-2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{7}{2}\end{matrix}\right.\)
\(x^2\left(x-1\right)-4\left(x-1\right)=\left(x^2-4\right)\left(x-1\right)=\left(x-2\right)\left(x+2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=1\end{matrix}\right.\)
\(x^4-x^3-x^2+x=x\left(x^3+1\right)-x^2\left(x+1\right)=x\left(x+1\right)\left(x^2-x+1-x^2\right)=x\left(x+1\right)\left(1-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\\1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)
a) \(x\left(2x-7\right)-4x+14-0\Leftrightarrow2x^2-11x+14=0\Leftrightarrow2x^2-4x-7x+14=0\Leftrightarrow2x\left(x-2\right)-7\left(x-2\right)=0\Leftrightarrow\left(2x-7\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3,5\\x=2\end{matrix}\right.\)
b) \(x^2\left(x-1\right)-4x+4=0\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-2\end{matrix}\right.\)
c) \(x+x^2-x^3-x^4=0\Leftrightarrow x\left(x^3+x^2-x-1\right)=0\Leftrightarrow x\left[x\left(x^2-1\right)+\left(x^2-1\right)\right]=0\Leftrightarrow x\left(x+1\right)\left(x^2-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
d) \(2x^3+3x^2+2x+3=0\Leftrightarrow x^2\left(2x+3\right)+2x+3=0\Leftrightarrow\left(x^2+1\right)\left(2x+3\right)=0\Leftrightarrow x=-1,5\left(x^2+1>0\forall x\right)\)
e) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\Leftrightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\Leftrightarrow2x-5=0\Leftrightarrow x=2,5\)
g) \(x^3+27+\left(x+3\right)\left(x-9\right)=0\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\Leftrightarrow x\left(x+3\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=2\end{matrix}\right.\)
a) 16x^2 - (4x - 5)^2 = 15
<=> 16x^2 - 16x^2 + 40x - 25 = 15
<=> 40x = 40
<=> x = 1
b) (2x + 3)^2 - 4(x - 1)(x + 1) = 49
<=> 4x^2 + 12x + 9 - 4x^2 - 4x + 4x + 4 = 49
<=> 12x + 13 = 49
<=> 12x = 36
<=> x = 3
c) (2x + 1)(1 - 2x) + (1 - 2x)^2 = 18
<=> 1 - 4x^2 + 1 - 4x + 4x^2 = 18
<=> 2 - 4x = 18
<=> -4x = 16
<=> x = -4
d)2(x + 1)^2 - (x - 3)(x + 3) - (x - 4)^2 = 0
<=> 2x^2 + 4x + 2 - x^2 + 3^2 - x^2 + 8x - 16 = 0
<=> 12x - 5 = 0
<=> 12x = 5
<=> x = 5/12
e) (x - 5)^2 - x(x - 4) = 9
<=> x^2 - 10x + 25 - x^2 + 4x = 9
<=> -6x + 25 = 9
<=> -6x = 9 - 25
<=> -6x = -16
<=> x = -16/-6 = 8/3
f) (x - 5)^2 + (x - 4)(1 - x) = 0
<=> x^2 - 10x + 25 + x - x^2 - x - 4 + 4x = 0
<=> -5x + 21 = 0
<=> -5x = -21
<=> x = 21/5
Bài 1 :
a, \(\left(x-3\right)^2-4=0\Leftrightarrow\left(x-3\right)^2=4\Leftrightarrow\left(x-3\right)^2=\left(\pm2\right)^2\)
TH1 : \(x-3=2\Leftrightarrow x=5\)
TH2 : \(x-3=-2\Leftrightarrow x=1\)
b, \(x^2-2x=24\Leftrightarrow x^2-2x-24=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+4\right)=0\)
TH1 : \(x-6=0\Leftrightarrow x=6\)
TH2 : \(x+4=0\Leftrightarrow x=-4\)
c, \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-4\right)=0\)
\(\Leftrightarrow2x+30=0\Leftrightarrow x=-15\)
d, tương tự
a) \(4x^3-9x=0\)
\(\Leftrightarrow x\left(4x^2-9\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\4x^2=9\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm\frac{3}{2}\end{cases}}\)
b) \(3x\left(x-2\right)-5x+10=0\)
\(\Leftrightarrow\left(3x-5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=2\end{cases}}\)
c) \(4x\left(x+3\right)-x^2+9=0\)
\(\Leftrightarrow4x\left(x+3\right)-\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(3x+3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-3\end{cases}}\)
d) \(\left(2x+5\right)\left(x-4\right)=\left(x-4\right)\left(5-x\right)\)
\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow3x\left(x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)
e) \(16x^2-25=\left(4x-5\right)\left(2x+1\right)\)
\(\Leftrightarrow\left(4x-5\right)\left(4x+5\right)-\left(4x-5\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left(4x-5\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{4}\\x=-2\end{cases}}\)
f) \(\left(x+\frac{1}{5}\right)^2=\frac{64}{9}\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{5}=\frac{8}{3}\\x+\frac{1}{5}=-\frac{8}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{37}{15}\\x=-\frac{43}{15}\end{cases}}\)
g) \(9\left(x+2\right)^2=\left(x+3\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}3x+6=x+3\\3x+6=-x-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=-3\\4x=-9\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=-\frac{9}{4}\end{cases}}\)
a: =>2x^2=4
=>x^2=2
=>\(x=\pm\sqrt{2}\)
b: =>(x+1)^2-4=0
=>(x+1+2)(x+1-2)=0
=>(x+3)(x-1)=0
=>x=1 hoặc x=-3
c: =>(2x-1)^2-3^2=0
=>(2x-1-3)(2x-1+3)=0
=>(2x-4)(2x+2)=0
=>x=2 hoặc x=-1
d: x^2-x=0
=>x(x-1)=0
=>x=0 hoặc x=1