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31 tháng 5 2018
3) \(x^2-7x+6=0\)
\(\Leftrightarrow x^2-6x-x+6=0\)
\(\Leftrightarrow x\left(x-6\right)-\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)
S=\(\left\{6;1\right\}\)
\(\)
NN
Tìm x
a) 9(3x-2)=x(2-3x)
b) 25x2-2=0
c) x2-25=6x-9
d) (x+2)2-(x-2)(x+2)=0
e) x3-8=(x-2)3
f) x3+5x2-4x-20=0
1
9 tháng 6 2017
a) 9(3x - 2) = x(2 - 3x)
\(\Leftrightarrow\)-9(2 - 3x) = x(2 - 3x)
\(\Leftrightarrow\)-9(2 - 3x) - x(2 - 3x) = 0
\(\Leftrightarrow\)(2 - 3x)(- 9 - x) = 0
\(\Leftrightarrow\)2 - 3x = 0 hay - 9 - x = 0
\(\Leftrightarrow\) 3x = 2 \(\Leftrightarrow\) x = - 9
\(\Leftrightarrow\) x = 2/3
b) 25x2 - 2 = 0
\(\Leftrightarrow\)(5x)2 - (\(\sqrt{2}\))2 = 0
\(\Leftrightarrow\)(5x - \(\sqrt{2}\))(5x + \(\sqrt{2}\)) = 0
\(\Leftrightarrow\)5x - \(\sqrt{2}\)= 0 hay 5x + \(\sqrt{2}\)= 0
\(\Leftrightarrow\)5x = \(\sqrt{2}\) \(\Leftrightarrow\)5x = -\(\sqrt{2}\)
\(\Leftrightarrow\) x = \(\sqrt{2}\)/5 \(\Leftrightarrow\) x = -\(\sqrt{2}\)/5
c) x2 - 25 = 6x - 9
\(\Leftrightarrow\)(x2 - 6x + 9) - 25 = 0
\(\Leftrightarrow\)(x - 3)2 - 52 = 0
\(\Leftrightarrow\)(x - 3 - 5)(x - 3 + 5) = 0
\(\Leftrightarrow\)(x - 7)(x + 2) = 0
\(\Leftrightarrow\)x - 7 = 0 hay x + 2 = 0
\(\Leftrightarrow\)x = 7 \(\Leftrightarrow\)x = -2
d) (x + 2)2 - (x - 2)(x + 2) = 0
\(\Leftrightarrow\)(x + 2)(x + 2) - (x - 2)(x + 2) = 0
\(\Leftrightarrow\)(x + 2)(x + 2 - x + 2) = 0
\(\Leftrightarrow\)(x + 2)4 = 0 (hay 4(x + 2) = 0)
\(\Leftrightarrow\)x + 2 = 0 (vì 4 \(\ne\)0)
\(\Leftrightarrow\)x = -2
e) x3 - 8 = (x - 2)3
\(\Leftrightarrow\)x3 - 23 = (x - 2)3
\(\Leftrightarrow\)(x - 2)(x2 + 2x + 4) = (x - 2)3
\(\Leftrightarrow\)(x - 2)(x2 + 2x + 4) - (x - 2)3 = 0
\(\Leftrightarrow\)(x - 2)(x2 + 2x + 4) - (x - 2)(x - 2)2 = 0
\(\Leftrightarrow\)(x - 2)[x2 + 2x + 4 - (x - 2)2] = 0
\(\Leftrightarrow\)(x - 2)[x2 + 2x + 4 - (x2 - 4x + 4)] = 0
\(\Leftrightarrow\)(x - 2)(x2 + 2x + 4 - x2 + 4x - 4) = 0
\(\Leftrightarrow\)(x - 2)6x = 0 (hay 6x(x - 2) = 0)
\(\Leftrightarrow\)x - 2 = 0 hay x = 0 (vì 6\(\ne\)0)
\(\Leftrightarrow\)x = 2
f) x3 + 5x2 - 4x - 20 = 0
\(\Leftrightarrow\)x2(x + 5) - 4(x + 5) = 0
\(\Leftrightarrow\)(x + 5)(x2 - 4) = 0
\(\Leftrightarrow\)(x + 5)(x - 2)(x + 2) = 0
\(\Leftrightarrow\)x + 5 = 0 hay x - 2 = 0 hay x + 2 = 0
\(\Leftrightarrow\)x = -5 \(\Leftrightarrow\)x = 2 \(\Leftrightarrow\)x = -2
14 tháng 10 2020
9x2 - 4 - ( 3x - 2 )( x + 5 ) = 0
<=> ( 3x - 2 )( 3x + 2 ) - ( 3x - 2 )( x + 5 ) = 0
<=> ( 3x - 2 )( 3x + 2 - x - 5 ) = 0
<=> ( 3x - 2 )( 2x - 3 ) = 0
<=> \(\orbr{\begin{cases}3x-2=0\\2x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{3}{2}\end{cases}}\)
x3 + 64 + ( x + 4 )( 2x - 3 ) = 0
<=> ( x + 4 )( x2 - 4x + 16 ) + ( x + 4 )( 2x - 3 ) = 0
<=> ( x + 4 )( x2 - 4x + 16 + 2x - 3 ) = 0
<=> ( x + 4 )( x2 - 2x + 13 ) = 0
<=> \(\orbr{\begin{cases}x+4=0\\x^2-2x+13=0\end{cases}}\Leftrightarrow x=-4\)( vì x2 - 2x + 13 = ( x2 - 2x + 1 ) + 12 = ( x - 1 )2 + 12 ≥ 12 > 0 ∀ x )
( x - 3 )( x2 + 4x + 9 ) + 2( x2 - 9 ) - 10( x - 3 ) = 0
<=> ( x - 3 )( x2 + 4x + 9 ) + 2( x - 3 )( x + 3 ) - 10( x - 3 ) = 0
<=> ( x - 3 )( x2 + 4x + 9 + 2x + 6 - 10 ) = 0
<=> ( x - 3 )( x2 + 6x + 5 ) = 0
<=> ( x - 3 )( x + 1 )( x + 5 ) = 0
<=> x = 3 hoặc x = -1 hoặc x = -5
<=> ( x - 3 )(
27 tháng 12 2015
b)(x+3)\(^2\)-x\(^2\)=9
(x+3-x)(x+3+x)=9
3(2x+3)=9
2x+3=3
2x=0
x=0
c)\(x^2+4x+4+x^2-6x+9-2x^2+2=9\)
-2x+15=9
-2x=-6
x=3
DN
27 tháng 12 2015
a) x(x-2) - 5x + 10 = 0
x(x - 2) - 5(x - 2)=0
(x - 2)(x - 5)=0
=> x=2 ; x=5
28 tháng 7 2018
\(x\left(x-3\right)+x-3=0\)
\(\left(x-3\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}}\)
KL:......................
\(x^3-5x=0\)
\(x\left(x^2-5\right)=0\)
Làm tương tự như câu a
@_@ n...h..i......ề....u q...u.....................á!
HH
1 tháng 7 2018
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/ \(\left(x+4\right)^2-\left(x+9\right)\left(x-1\right)=16\)
<=> \(x^2+8x+16-x^2+8x-9=16\)
<=> \(16x+7=16\)
<=> \(16x=9\)
<=> \(x=\frac{9}{16}\)
1 tháng 7 2018
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 = \(\varnothing\)
23 tháng 8 2020
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
31 tháng 3 2020
\(a,\left(2x-1\right)^2=49\)
\(\left[{}\begin{matrix}2x-1=7\\2x-1=-7\end{matrix}\right.\)
\(\left[{}\begin{matrix}2x=8\\2x=-6\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
\(b,\left(2x+7\right)^2=9\left(x+2\right)^2\)
\(4x^2+28x+49=9x^2+36x+36\)
\(4x^2+28x+49-9x^2-36x-36=0\)
\(-5x^2-8x+13=0\)
\(5x^2+13-5x-13=0\)
\(x\left(5x+13\right)-1\left(5x+13\right)=0\)
\(\left(x-1\right)\left(5x+13\right)=0\)
\(\left[{}\begin{matrix}x=1\\5x=-13\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=1\\x=-\frac{13}{5}\end{matrix}\right.\)
31 tháng 3 2020
\(c,4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
\(\left[2\left(2x+7\right)\right]^2-\left[3\left(x+3\right)\right]^2=0\)
\(\left(4x+14\right)^2-\left(3x+9\right)^2=0\)
\(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
\(x=-5\)
\(d,\left(5x-3\right)^2-\left(4x-7\right)^2=0\)
\(25x^2-30x+9-16x^2+56x-49=0\)
\(9x^2+26x-40=0\)
\(9x^2+36x-10x-40=0\)
\(9x\left(x+4\right)-10\left(x+4\right)=0\)
\(\left(9x-10\right)\left(x+4\right)=0\)
\(\left[{}\begin{matrix}9x-10=0\\x+4=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\frac{10}{9}\\x=-4\end{matrix}\right.\)
6 tháng 9 2020
1. (x + 2)(x2 - 2x + 4) - (x3 + 2x2) = 5
=> x(x2 - 2x + 4) + 2(x2 - 2x + 4) - x3 - 2x2 - 5 = 0
=> x3 - 2x2 + 4x + 2x2 - 4x + 8 - x3 - 2x2 - 5 = 0
=> (x3 - x3) + (-2x2 + 2x2 - 2x2) + (4x - 4x) + (8 - 5) = 0
=> -2x2 + 3 = 0
=> -2x2 = -3
=> x2 = 3/2
=> x = \(\pm\sqrt{\frac{3}{2}}\)
2. \(\left(x+5\right)^2-6=0\)
=> x2 + 10x + 25 - 6 = 0
=> x2 + 10x + 19 = 0
=> x vô nghiệm(do mình không để căn nên ghi vô nghiệm thôi nhá)
3. \(\left(x+3\right)\left(x^2-3x+9\right)-x^3=2x\)
=> x(x2 - 3x + 9) + 3(x2 - 3x + 9) - x3 - 2x = 0
=> x3 - 3x2 + 9x + 3x2 - 9x + 27 - x3 - 2x = 0
=> (x3 - x3) + (-3x2 + 3x2) + (9x - 9x - 2x) + 27 = 0
=> -2x + 27 = 0
=> -2x = -27
=> x = 27/2
4. \(\left(x-2\right)^3-x^3+6x^2=7\)
=> x3 - 6x2 + 12x - 8 - x3 + 6x2 = 7
=> (x3 - x3) + (-6x2 + 6x2) + 12x - 8 = 7
=> 12x - 8 = 7
=> 12x = 15
=> x = 5/4
5. \(3\left(x-2\right)^2+9\left(x-1\right)-3\left(x^2+x-3\right)=12\)
=> 3x2 - 12x + 12 + 9x - 9 - 3x2 - 3x + 9 = 12
=> (3x2 - 3x2) + (-12x + 9x - 3x) + (12 - 9 + 9) = 12
=> -6x + 12 = 12
=> -6x = 0
=> x = 0
6. \(\left(4x+3\right)^2-\left(4x-3\right)^2-5x-2=0\)
=> 48x - 5x - 2 = 0
=> 43x - 2 = 0
=> 43x = 2
=> x = 2/43
Còn bài cuối tự làm :>
6 tháng 9 2020
Anh Sang làm cầu kì quá ;-;
1. ( x + 2 )( x2 - 2x + 4 ) - ( x3 + 2x2 ) = 5
<=> x3 + 8 - x3 - 2x2 = 5
<=> 8 - 2x2 = 5
<=> 2x2 = 3
<=> x2 = 3/2
<=> \(x^2=\left(\pm\sqrt{\frac{3}{2}}\right)^2\)
<=> \(x=\pm\sqrt{\frac{3}{2}}\)
2. ( x + 5 )2 - 6 = 0
<=> ( x + 5 )2 - ( √6 )2 = 0
<=> ( x + 5 - √6 )( x + 5 + √6 ) = 0
<=> \(\orbr{\begin{cases}x+5-\sqrt{6}=0\\x+5+\sqrt{6}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{6}-5\\x=-\sqrt{6}-5\end{cases}}\)
3. ( x + 3 )( x2 - 3x + 9 ) - x3 = 2x
<=> x3 + 27 - x3 = 2x
<=> 27 = 2x
<=> x = 27/2
4. ( x - 2 )3 - x3 + 6x2 = 7
<=> x3 - 6x2 + 12x - 8 - x3 + 6x2 = 7
<=> 12x - 8 = 7
<=> 12x = 15
<=> x = 15/12 = 5/4
5. 3( x - 2 )2 + 9( x - 1 ) - 3( x2 + x - 3 ) = 12
<=> 3( x2 - 4x + 4 ) + 9x - 9 - 3x2 - 3x + 9 = 12
<=> 3x2 - 12x + 12 + 6x - 3x2 = 12
<=> -6x + 12 = 12
<=> -6x = 0
<=> x = 0
6. ( 4x + 3 )2 - ( 4x - 3 )2 - 5x - 2 = 0
<=> 16x2 + 24x + 9 - ( 16x2 - 24x + 9 ) - 5x - 2 = 0
<=> 16x2 + 24x + 9 - 16x2 + 24x - 9 - 5x - 2 = 0
<=> 43x - 2 = 0
<=> 43x = 2
<=> x = 2/43
7, ( 4x + 7 )( 2 - 3x ) - ( 6x + 2 )( 5 - 2x ) = 0
<=> -12x2 - 13x + 14 - ( -12x2 + 26x + 10 ) = 0
<=> -12x2 - 13x + 14 + 12x2 - 26x - 10 = 0
<=> -39x + 4 = 0
<=> -39x = -4
<=> x = 4/39
\(\left(x^2-9\right)^2-9\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(x-3\right)^2\left(x+3\right)^2-9\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(x-3\right)^2\left[\left(x+3\right)^2-9\right]=0\)
\(\Leftrightarrow\left(x-3\right)^2\left[x^2+6x+9-9\right]=0\)
\(\Leftrightarrow\left(x-3\right)^2\left(x^2+6x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x\left(x+6\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\end{matrix}\right.\)
Vậy \(S=\left\{0;3;-6\right\}\)
\(\left(x-3\right)^2\left(x+3\right)^2-9\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(x-3\right)^2\left(x+3-3\right)\left(x+3+3\right)=0\Leftrightarrow x=3;x=0;x=-6\)