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19 tháng 3 2020
a, (3x-1)2 - (x+3)2 = 0
<=> [(3x-1)-(x+3)][(3x-1)+(x+3)] = 0
<=> (3x-1-x-3)(3x-1+x+3) = 0
<=> (2x-4)(4x+2) = 0
=> 2x-4=0 hoặc 4x+2=0
=> 2x =4 hoặc 4x = -2
=> x = 2 hoặc x = \(\frac{-1}{2}\)
19 tháng 3 2020
\(\begin{array}{l} a){\left( {3x - 1} \right)^2} - {\left( {x + 3} \right)^2} = 0\\ \Leftrightarrow \left( {3x - 1 + x + 3} \right)\left[ {3x - 1 - x - 3} \right] = 0\\ \Leftrightarrow \left( {4x + 2} \right)\left( {2x - 4} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} 4x + 2 = 0\\ 2x - 4 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = - \dfrac{1}{2}\\ x = 2 \end{array} \right.\\ b){x^3} - \dfrac{x}{{49}} = 0\\ \Leftrightarrow 49{x^3} - x = 0\\ \Leftrightarrow x\left( {49{x^2} - 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = 0\\ 49{x^2} - 1 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = \pm \dfrac{1}{7} \end{array} \right.\\ c){x^2} - 7x + 12 = 0\\ \Leftrightarrow {x^2} - 3x - 4x + 12 = 0\\ \Leftrightarrow x\left( {x - 3} \right) - 4\left( {x - 3} \right) = 0\\ \Leftrightarrow \left( {x - 3} \right)\left( {x - 4} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x - 3 = 0\\ x - 4 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 3\\ x = 4 \end{array} \right.\\ d)4{x^2} - 3x - 1 = 0\\ \Leftrightarrow 4{x^2} + x - 4x - 1 = 0\\ \Leftrightarrow x\left( {4x + 1} \right) - \left( {4x + 1} \right) = 0\\ \Leftrightarrow \left( {4x + 1} \right)\left( {x - 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} 4x + 1 = 0\\ x - 1 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = - \dfrac{1}{4}\\ x = 1 \end{array} \right.\\ e){x^3} - 2x - 4 = 0\\ \Leftrightarrow {x^3} - 4x + 2x - 4 = 0\\ \Leftrightarrow x\left( {{x^2} - 4} \right) + 2\left( {x - 2} \right) = 0\\ \Leftrightarrow x\left( {x - 2} \right)\left( {x + 2} \right) + 2\left( {x - 2} \right) = 0\\ \Leftrightarrow \left( {x - 2} \right)\left[ {x\left( {x + 2} \right) + 2} \right] = 0\\ \Leftrightarrow \left( {x - 2} \right)\left( {{x^2} + 2x + 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x - 2 = 0\\ {x^2} + 2x + 2 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 2\\ {x^2} + 2x + 2x = 0\left( {VN} \right) \end{array} \right.\\ f){x^3} + 8{x^2} + 17x + 10 = 0\\ \Leftrightarrow \left( {x + 1} \right)\left( {{x^2} + 7x + 10} \right) = 0\\ \Leftrightarrow \left( {x + 1} \right)\left( {{x^2} + 5x + 2x + 10} \right) = 0\\ \Leftrightarrow \left( {x + 1} \right)\left[ {x\left( {x + 5} \right) + 2\left( {x + 5} \right)} \right] = 0\\ \Leftrightarrow \left( {x + 1} \right)\left( {x + 5} \right)\left( {x + 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x + 1 = 0\\ x + 5 = 0\\ x + 2 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = - 1\\ x = - 5\\ x = - 2 \end{array} \right. \end{array}\)
XO
1 tháng 10 2020
a) (2x - 3)2 = (x + 5)2
=> 4x2 - 12x + 9 = x2 + 10x + 25
=> 4x2 - 12x + 9 - (x2 + 10x + 25) = 0
=> 3x2 - 22x - 16 = 0
=> 3x2 - 24x + 2x - 16 = 0
=> 3x(x - 8) + 2(x - 8) = 0
=> (3x + 2)(x - 8) = 0
=> \(\orbr{\begin{cases}3x+2=0\\x-8=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{2}{3}\\x=8\end{cases}}\)
b) x2(x - 1) - 4x2 + 8x - 4 = 0
=> x2(x - 1) - (2x - 2)2 = 0
=> x2(x - 1) - [2(x- 1)]2 = 0
=> x2(x - 1) - 4(x - 1)2 = 0
=> (x - 1)(x2 - 4(x - 1) = 0
=> (x - 1)(x2 - 4x + 4) = 0
=> (x - 1)(x - 2)2 = 0
=> \(\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)
c) x2 + 7x + 12 = 0
=> x2 + 3x + 4x + 12 = 0
=> x(x + 3) + 4(x + 3) = 0
=> (x + 4)(x + 3) = 0
=> \(\orbr{\begin{cases}x+4=0\\x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-4\\x=-3\end{cases}}\)
d) x2 + 3x - 18 = 0
=> x2 + 6x - 3x - 18 = 0
=> x(x + 6) - 3(x + 6) = 0
=> (x - 3)(x + 6) = 0
=> \(\orbr{\begin{cases}x-3=0\\x+6=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-6\end{cases}}\)
e) x(x + 6) - 7x - 42 = 0
=> x(x + 6) - 7(x + 6) = 0
=> (x - 7)(x + 6) = 0
=> \(\orbr{\begin{cases}x-7=0\\x+6=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=7\\x=-6\end{cases}}\)
1 tháng 10 2020
1. ( 2x - 3 )2 = ( x + 5 )2
<=> ( 2x - 3 )2 - ( x + 5 )2 = 0
<=> [ ( 2x - 3 ) - ( x + 5 ) ][ ( 2x - 3 ) + ( x + 5 ) ] = 0
<=> ( 2x - 3 - x - 5 )( 2x - 3 + x + 5 ) = 0
<=> ( x - 8 )( 3x + 2 ) = 0
<=> \(\orbr{\begin{cases}x-8=0\\3x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=-\frac{2}{3}\end{cases}}\)
2. x2( x - 1 ) - 4x2 + 8x - 4 = 0
<=> x2( x - 1 ) - ( 4x2 - 8x + 4 ) = 0
<=> x2( x - 1 ) - 4( x2 - 2x + 1 ) = 0
<=> x2( x - 1 ) - 4( x - 1 )2 = 0
<=> ( x - 1 )[ x2 - 4( x - 1 ) ] = 0
<=> ( x - 1 )( x2 - 4x + 4 ) = 0
<=> ( x - 1 )( x - 2 )2 = 0
<=> \(\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)
3. x2 + 7x + 12 = 0
<=> x2 + 3x + 4x + 12 = 0
<=> x( x + 3 ) + 4( x + 3 ) = 0
<=> ( x + 3 )( x + 4 ) = 0
<=> \(\orbr{\begin{cases}x+3=0\\x+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=-4\end{cases}}\)
4. x2 + 3x - 18 = 0
<=> x2 - 3x + 6x - 18 = 0
<=> x( x - 3 ) + 6( x - 3 ) = 0
<=> ( x - 3 )( x + 6 ) = 0
<=> \(\orbr{\begin{cases}x-3=0\\x+6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-6\end{cases}}\)
5. x( x + 6 ) - 7x - 42 = 0
<=> x( x + 6 ) - 7( x + 6 ) = 0
<=> ( x + 6 )( x - 7 ) = 0
<=> \(\orbr{\begin{cases}x+6=0\\x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-6\\x=7\end{cases}}\)
4 tháng 3 2020
(x2 + x + 1)(6 - 2x) = 0
<=> 6 - 2x = 0 (do x2 + x + 1 > 0)
<=> 2x = 6
<=> x = 3
Vậy S = {3}
(8x - 4)(x2 + 2x + 2) = 0
<=> 8x - 4 = 0 (vì x2 + 2x + 2 > 0)
<=> 8x = 4
<=> x = 1/2
Vậy S = {1/2}
x3 - 7x + 6 = 0
<=> x3 - x - 6x + 6 = 0
<=> x(x2 - 1) - 6(x - 1) = 0
<=> x(x - 1)(x + 1) - 6(x - 1) = 0
<=> (x2 + x - 6)(x - 1) = 0
<=> (x2 + 3x - 2x - 6)(x - 1) = 0
<=> (x + 3)(x - 2)(x - 1) = 0
<=> x + 3 = 0
hoặc x - 2 = 0
hoặc x - 1 = 0
<=> x = -3
hoặc x = 2
hoặc x = 1
Vậy S = {-3; 1; 2}
x5 - 5x3 + 4x = 0
<=> x(x4 - 5x2 + 4) = 0
<=> x(x4 - x2 - 4x2 + 4) = 0
<=> x[x2(x2 - 1) - 4(x2 - 1)] = 0
<=> x(x - 2)(x + 2)(x - 1)(x + 1) = 0
<=> x = 0 hoặc x - 2 = 0 hoặc x + 2 = 0 hoặc x - 1 = 0 hoặc x + 1 = 0
<=> x = 0 hoặc x = 2 hoặc x = -2 hoặc x = 1 hoặc x = -1
Vậy S = {-2; -1; 0; 1; 2}
4 tháng 3 2020
+ Ta có: \(\left(x^2+x+1\right).\left(6-2x\right)=0\)
- Ta lại có: \(x^2+x+1=\left(x^2+x+\frac{1}{4}\right)+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)
- Vì \(x^2+x+1>0\forall x\)mà \(\left(x^2+x+1\right).\left(6-2x\right)=0\)
\(\Rightarrow6-2x=0\Leftrightarrow-2x=-6\Leftrightarrow x=3\left(TM\right)\)
Vậy \(S=\left\{3\right\}\)
+ Ta có: \(\left(8x-4\right).\left(x^2+2x+2\right)=0\)
- Ta lại có: \(x^2+2x+2=\left(x^2+2x+1\right)+1=\left(x+1\right)^2+1\ge1>0\forall x\)
- Vì \(x^2+2x+2>0\forall x\)mà \(\left(8x-4\right).\left(x^2+2x+2\right)=0\)
\(\Rightarrow8x-4=0\Leftrightarrow8x=4\Leftrightarrow x=\frac{1}{2}\left(TM\right)\)
Vậy \(S=\left\{\frac{1}{2}\right\}\)
+ Ta có: \(x^3-7x+6=0\)
\(\Leftrightarrow\left(x^3-x^2\right)+\left(x^2-x\right)+\left(6x-6\right)=0\)
\(\Leftrightarrow x^2.\left(x-1\right)+x.\left(x-1\right)-6.\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right).\left(x^2+x-6\right)=0\)
\(\Leftrightarrow\left(x-1\right).\left[\left(x^2-2x\right)+\left(3x-6\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right).\left[x.\left(x-2\right)+3.\left(x-2\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right).\left(x-2\right).\left(x+3\right)=0\)
\(\Leftrightarrow x=1\left(TM\right)\)hoặc \(x=2\left(TM\right)\)hoặc \(x=-3\left(TM\right)\)
Vậy \(S=\left\{-3;1;2\right\}\)
+ Ta có: \(x^5-5x^3+4x=0\)
\(\Leftrightarrow x.\left(x^4-5x^2+4\right)=0\)
\(\Leftrightarrow x.\left[\left(x^4-x^2\right)-\left(4x^2-4\right)\right]=0\)
\(\Leftrightarrow x.\left[x^2.\left(x^2-1\right)-4.\left(x^2-1\right)\right]=0\)
\(\Leftrightarrow x.\left(x^2-1\right).\left(x^2-4\right)=0\)
\(\Leftrightarrow x=0\left(TM\right)\)
hoặc \(x^2-1=0\Leftrightarrow x^2=1\Leftrightarrow x=\pm1\left(TM\right)\)
hoặc \(x^2-4=0\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\left(TM\right)\)
Vậy \(S=\left\{-2;-1;0;1;2\right\}\)
!!@@# ^_^ Chúc bạn hok tốt ^_^#@@!!
2
25 tháng 10 2021
\(a,x^2-5x\)
\(=x\left(x-5\right)\)
\(b,5x\left(x+5\right)+4x+20\)
\(=5x\left(x+5\right)+4\left(x+5\right)\)
\(=\left(5x+4\right)\left(x+5\right)\)
\(c,7x\left(2x-1\right)-4x+2\)
\(=7x\left(2x-1\right)-2\left(2x-1\right)\)
\(=\left(7x-2\right)-\left(2x-1\right)\)
2
25 tháng 10 2021
\(d,x^2-16+2\left(x+4\right)\)
\(=x^2-16+2x+8\)
\(=x\left(x-2\right)-8\) ( Ý này thì k chắc lắm, sai thông cảm :)) )
\(e,x^2-10x+9\)
\(=x^2-x-9x+9\)
\(=x\left(x-1\right)-9\left(x-1\right)\)
\(=\left(x-9\right)\left(x-1\right)\)
\(f,\left(2x-1\right)^2-\left(x-3\right)^2=0\) ( mk đoán bài này là tìm x, sai thì bảo mk để mk sửa nhé )
\(\Rightarrow\left(2x-1\right)^2=\left(x-3\right)^2\)
\(\Leftrightarrow\pm\left(2x-1\right)=\pm\left(x-3\right)\)
\(\Rightarrow\hept{\begin{cases}2x-1=x-3\\-\left(2x-1\right)=-\left(x-3\right)\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}2x-1-x+3=0\\-2x+1-x+3=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x+2=0\\-3x+4=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\left(-2\right)\\x=\frac{4}{3}\end{cases}}\)
Vậy ...
TX
1
26 tháng 7 2017
a)\(x^2-2x-24=0\Leftrightarrow x^2-2x+1-25=0\)
\(\Leftrightarrow\left(x-1\right)^2-5^2=0\Leftrightarrow\left(x-1-5\right)\left(x-1+5\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+4\right)=0\Leftrightarrow\hept{\begin{cases}x=6\\x=-4\end{cases}}\)
b)\(x^2+8x+12=0\Leftrightarrow x^2+8x+16-4=0\)
\(\Leftrightarrow\left(x+4\right)^2-2^2=0\Leftrightarrow\left(x+4-2\right)\left(x+4+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+6\right)=0\Leftrightarrow\hept{\begin{cases}x=-2\\x=-6\end{cases}}\)
c)\(4x^2+4x-63=0\Leftrightarrow4x^2+4x+1-64=0\)
\(\Leftrightarrow\left(2x+1\right)^2-8^2=0\Leftrightarrow\left(2x+1-8\right)\left(2x+1+8=0\right)\)
\(\Leftrightarrow\left(2x-7\right)\left(2x+9\right)=0\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{9}{2}\end{cases}}\)
23 tháng 7 2022
b: Đặt \(x^2-6x-2=a\)
Theo đề, ta có: \(a+\dfrac{14}{a+9}=0\)
=>(a+2)(a+7)=0
\(\Leftrightarrow\left(x^2-6x\right)\left(x^2-6x+5\right)=0\)
=>x(x-6)(x-1)(x-5)=0
hay \(x\in\left\{0;1;6;5\right\}\)
c: \(\Leftrightarrow\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)
\(\Leftrightarrow-32x^2=8x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)\)
\(\Leftrightarrow-32x^2=16x^2+8x-3\left(16x^2-8x+2x-1\right)\)
\(\Leftrightarrow-48x^2=8x-48x^2+18x+3\)
=>26x=-3
hay x=-3/26
\(a,\Rightarrow\left(2x-1\right)\left(2x+1\right)-x\left(2x+1\right)=0\\ \Rightarrow\left(2x+1\right)\left(2x-1-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=1\end{matrix}\right.\\ b,\Rightarrow\left(x-3\right)\left(x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\\ c,\Rightarrow\left(x^2-8x+16\right)-10=0\\ \Rightarrow\left(x-4\right)^2-10=0\\ \Rightarrow\left(x-4-\sqrt{10}\right)\left(x-4+\sqrt{10}\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=4+\sqrt{10}\\x=4-\sqrt{10}\end{matrix}\right.\)