( 9x + 1 ) ( 5 + 2x ) = 0
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a) -3x(2x-5)-2x(2-3x)=7
=> -6x2 + 15 - 4x + 6x2 = 7
=> -6x2 + 6x2 + 15 -4x =7
=> 15 - 4x =7
=> 4x = 15-7 =8
=> x= 8:4 = 2
b) \(\left(9x-12x+4\right)\left(2-5x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}9x-12x+4=0\\2-5x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\left(9-12\right)=-4\\5x=2\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}-3x=-4\\x=\frac{2}{5}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{2}{5}\end{cases}}\)
Vay...
c) (4-3x) = (5+2x)
=> 4-3x=5+2x
=> -3x - 2x = 5-4
=> x(-3-2) = 1
=> -5x = 1
=> x= \(\frac{-1}{5}\)
d) (2x-1)-3(2x-1)=0
=> 2x-1 - 6x + 3 =0
=> 2x - 6x = 1 -3
=> x(2-6)=-2
=> -4x= -2
=> x = \(\frac{1}{2}\)
d)(2x-1)-3(2x-1)
=>1(2x-1)-3(2x-1)=0
=>(1-3).(2x-1)=0
=>-2(2x-1)=0
=>2x-1=0
=>2x=-1
=>x=-0,5
vay x =-0,5
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(2x-5\right)^2-x^2=0\)
\(\left(2x-5-x\right)\left(2x-5+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{5}{3}\end{matrix}\right.\)
b) \(\left(x^2-2x+1\right)-9x^2=0\)
\(\left(x-1\right)^2-\left(3x\right)^2=0\)
\(\left(x-1-3x\right)\left(x-1+3x\right)=0\)
\(\left[{}\begin{matrix}-2x=1\\4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{4}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow\left(2x-5-x\right)\left(2x-5+x\right)=0\)
=>(x-5)(3x-5)=0
=>x=5 hoặc x=5/3
b: \(\Leftrightarrow\left(x-1-3x\right)\left(x-1+3x\right)=0\)
=>(2x+1)(4x-1)=0
=>x=-1/2 hoặc x=1/4
![](https://rs.olm.vn/images/avt/0.png?1311)
c) x2 + 9x = 10
x2 + 9x - 10 = 0
=> x2 - x + 10x - 10 = 0
=> x(x - 1) + 10(x - 1) = 0
=> (x + 10)(x - 1) = 0
=> \(\orbr{\begin{cases}x=-10\\x=1\end{cases}}\)
d) 2x2 + 9x = 35
=> 2x2 + 9x - 35 = 0
=> 2x2 + 14x - 5x - 35 = 0
=> 2x(x + 7) - 5(x + 7) = 0
=> (x + 7)(2x - 5) = 0
=> \(\orbr{\begin{cases}x=-7\\x=\frac{5}{3}\end{cases}}\)
(x2 - 2x - 1)2 - 5(x2 - 2x - 1) - 14 = 0
=> (x2 - 2x - 1)2 + 2(x2 - 2x - 1) - 7(x2 - 2x - 1) - 14 = 0
=> (x2 - 2x - 1)(x2 - 2x + 1) - 7(x2 - 2x + 1) = 0
=> (x2 - 2x + 1)(x2 - 2x - 8) = 0
=> (x - 1)2 (x - 4)(x + 2) = 0
=> x = 1 hoặc x = 4 hoặc x = -2
e) (2k2 + 5k + 1)2 - 12(2k2 + 5k + 1) + 32 = 0
=> (2k2 + 5x + 1)2 - 4(2k2 + 5k + 1) - 8(2k2 + 5k + 1) + 32 = 0
=> (2k2 + 5k + 1)(2k2 + 5k - 3) - 8(2k2 + 5k - 3) = 0
=> (2k2 + 5k - 3)(2k2 + 5k - 7) = 0
=> (2k2 + 6k - k - 3)(2k2 - 2x + 7k - 7) = 0
=> (k + 3)(2k - 1)(k - 1)(2k + 7) = 0
=> k = -3 hoặc k = 1/2 hoặc k = 1 hoặc k = -7/2
1.x2 + 6x = 0 < như này nhỉ ? >
⇔ x( x + 6 ) = 0
⇔ x = 0 hoặc x + 6 = 0
⇔ x = 0 hoặc x = -6
2. x2 - 25x + 250 = 0
⇔ ( x2 - 25x + 625/4 ) + 375/4 = 0
⇔ ( x - 25/2 )2 = -375/4 ( vô lí )
=> Phương trình vô nghiệm
3. x2 + 9x = 10
⇔ x2 + 9x - 10 = 0
⇔ x2 - x + 10x - 10 = 0
⇔ x( x - 1 ) + 10( x - 1 ) = 0
⇔ ( x - 1 )( x + 10 ) = 0
⇔ x - 1 = 0 hoặc x + 10 = 0
⇔ x = 1 hoặc x = -10
4. 2x2 + 9x = 35
⇔ 2x2 + 9x - 35 = 0
⇔ 2x2 + 14x - 5x - 35 = 0
⇔ 2x( x + 7 ) - 5( x + 7 ) = 0
⇔ ( x + 7 )( 2x - 5 ) = 0
⇔ x + 7 = 0 hoặc 2x - 5 = 0
⇔ x = -7 hoặc x = 5/2
5. ( x2 - 2x - 1 )2 - 5( x2 - 2x - 1 ) - 14 = 0
Đặt t = x2 - 2x - 1
bthuc ⇔ t2 - 5t - 14 = 0
⇔ t2 - 7t + 2t - 14 = 0
⇔ t( t - 7 ) + 2( t - 7 ) = 0
⇔ ( t - 7 )( t + 2 ) = 0
⇔ ( x2 - 2x - 1 - 7 )( x2 - 2x - 1 + 2 ) = 0
⇔ ( x2 - 4x + 2x - 8 )( x - 1 )2 = 0
⇔ ( x - 4 )( x + 2 )( x - 1 )2 = 0
⇔ x - 4 = 0 hoặc x + 2 = 0 hoặc x - 1 = 0
⇔ x = 4 hoặc x = -2 hoặc x = 1
6. ( 2k2 + 5k + 1 )2 - 12( 2k2 + 5k + 1 ) + 32 = 0
Đặt t = 2k2 + 5k + 1
bthuc ⇔ t2 - 12t + 32 = 0
⇔ t2 - 8t - 4t + 32 = 0
⇔ t( t - 8 ) - 4( t - 8 ) = 0
⇔ ( t - 8 )( t - 4 ) = 0
⇔ ( 2k2 + 5k + 1 - 8 )( 2k2 + 5k + 1 - 4 ) = 0
⇔ ( 2k2 - 2k + 7k - 7 )( 2k2 - k + 6k - 3 ) = 0
⇔ ( k - 1 )( 2k + 7 )( 2k - 1 )( k + 3 ) = 0
⇔ k = 1 hoặc k = -7/2 hoặc k = 1/2 hoặc k = -3
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\sqrt[]{x^2-4x+4}=x+3\)
\(\Leftrightarrow\sqrt[]{\left(x-2\right)^2}=x+3\)
\(\Leftrightarrow\left|x-2\right|=x+3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=x+3\\x-2=-\left(x+3\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}0x=5\left(loại\right)\\x-2=-x-3\end{matrix}\right.\)
\(\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)
b) \(2x^2-\sqrt[]{9x^2-6x+1}=5\)
\(\Leftrightarrow2x^2-\sqrt[]{\left(3x-1\right)^2}=5\)
\(\Leftrightarrow2x^2-\left|3x-1\right|=5\)
\(\Leftrightarrow\left|3x-1\right|=2x^2-5\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=2x^2-5\\3x-1=-2x^2+5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x-4=0\left(1\right)\\2x^2+3x-6=0\left(2\right)\end{matrix}\right.\)
Giải pt (1)
\(\Delta=9+32=41>0\)
Pt \(\left(1\right)\) \(\Leftrightarrow x=\dfrac{3\pm\sqrt[]{41}}{4}\)
Giải pt (2)
\(\Delta=9+48=57>0\)
Pt \(\left(2\right)\) \(\Leftrightarrow x=\dfrac{-3\pm\sqrt[]{57}}{4}\)
Vậy nghiệm pt là \(\left[{}\begin{matrix}x=\dfrac{3\pm\sqrt[]{41}}{4}\\x=\dfrac{-3\pm\sqrt[]{57}}{4}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(6x\left(x-5\right)+3x\left(7-2x\right)=18\)
\(\Leftrightarrow6x^2-30x+21x-6x^2=18\)
\(\Leftrightarrow-9x=18\)
hay x=-2
Vậy: S={-2}
b) Ta có: \(2x\left(3x+1\right)+\left(4-2x\right)\cdot3x=7\)
\(\Leftrightarrow6x^2+2x+12x-6x^2=7\)
\(\Leftrightarrow14x=7\)
hay \(x=\dfrac{1}{2}\)
Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)
c) Ta có: \(0.5x\left(0.4-4x\right)+\left(2x+5\right)\cdot x=-6.5\)
\(\Leftrightarrow0.2x-2x^2+2x^2+5x=-6.5\)
\(\Leftrightarrow5.2x=-6.5\)
hay \(x=-\dfrac{5}{4}\)
Vậy: \(S=\left\{-\dfrac{5}{4}\right\}\)
d) Ta có: \(\left(x+3\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)=6\)
\(\Leftrightarrow x^2+5x+6-\left(x^2+3x-10\right)=6\)
\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)
\(\Leftrightarrow2x+16=6\)
\(\Leftrightarrow2x=-10\)
hay x=-5
Vậy: S={-5}
e) Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)
\(\Leftrightarrow14x=0\)
hay x=0
Vậy: S={0}
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(x^2-2x-3=0\)
\(\Leftrightarrow x^2-3x+x-3=0\)
\(\Leftrightarrow x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
Vậy: \(S_1=\left\{3;-1\right\}\)(1)
Ta có: \(\left(x+1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
Vậy: \(S_2=\left\{-3;-1\right\}\)(2)
Từ (1) và (2) suy ra \(S_1\ne S_2\)
hay Hai phương trình \(x^2-2x-3=0\) và \(\left(x+1\right)\left(x+3\right)=0\) không tương đương với nhau
![](https://rs.olm.vn/images/avt/0.png?1311)
6x(3x + 5) - 2x(9x - 2) = 17
⇒ 18x2 + 30x - (18x2 - 4x) = 17
⇒ 18x2 + 30x - 18x2 + 4x = 17
⇒ 30x + 4x = 17
⇒ 34x = 17
⇒ x = 17 : 34
⇒ x = 1/2
2x(3x - 1) - 3x(2x + 11) - 70 = 0
⇒ 6x2 - 2x - (6x2 + 33x) - 70 = 0
⇒ 6x2 - 2x - 6x2 -33x = 70
⇒ -2x - 33x = 70
⇒ -35x = 70
⇒ x = -2
\(\left(9x+1\right)\left(5+2x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}9x+1=0\\5+2x+0\end{cases}\Leftrightarrow\orbr{\begin{cases}9x=-1\\2x=-5\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{-1}{9}\\x=\frac{-5}{2}\end{cases}}}\)
Vậy x = -1/9 hoặc x = -5/2
\(\left(9x+1\right)\left(5+2x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}9x+1=0\\5+2x=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{9}\\x=-\frac{5}{2}\end{cases}}\)