a: \(\Leftrightarrow4x^2+9x-4x-9=0\)

=>(4x+9)(x-1)=0

=>x=1 hoặc x=-9/4

b: \(\Leftrightarrow x^2-x-4x+4=0\)

=>(x-1)(x-4)=0

=>x=1 hoặc x=4

c: \(\Leftrightarrow5x^2-5x-12x+12=0\)

=>(x-1)(5x-12)=0

=>x=12/5 hoặc x=1

d: \(\Leftrightarrow x^2-4x+x-4=0\)

=>(x-4)(x+1)=0

=>x=4 hoặc x=-1

a, Ta có a + b + c = 4 + 5 - 9 = 0

vậy pt có 2 nghiệm x = 1 ; x = -9/4 

b, Ta có a + b + c = 1 - 5 + 4 = 0 

vậy pt có 2 nghiệm x = 1 ; x = 4 

c, Ta có a + b + c = 5 - 17 + 12 = 0 

vậy pt có 2 nghiệm x = 1 ; x = 12/5 

d, Ta có a - b + c = 1 + 3 - 4 = 0 

vậy pt có 2 nghiệm x = -1 ; x = 4 

a) Ta có: \(3x^2+2x-1=0\)

\(\Leftrightarrow3x^2+3x-x-1=0\)

\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{-1;\dfrac{1}{3}\right\}\)

b) Ta có: \(x^2-5x+6=0\)

\(\Leftrightarrow x^2-2x-3x+6=0\)

\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

Vậy: S={2;3}

c) Ta có: \(x^2-3x+2=0\)

\(\Leftrightarrow x^2-x-2x+2=0\)

\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

Vậy: S={1;2}

d) Ta có: \(2x^2-6x+1=0\)

\(\Leftrightarrow2\left(x^2-3x+\dfrac{1}{3}\right)=0\)

mà \(2\ne0\)

nên \(x^2-3x+\dfrac{1}{3}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{23}{12}=0\)

\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=\dfrac{23}{12}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=\dfrac{\sqrt{69}}{6}\\x-\dfrac{3}{2}=\dfrac{-\sqrt{69}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9+\sqrt{69}}{6}\\x=\dfrac{9-\sqrt{69}}{6}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{9+\sqrt{69}}{6};\dfrac{9-\sqrt{69}}{6}\right\}\)

e) Ta có: \(4x^2-12x+5=0\)

\(\Leftrightarrow4x^2-10x-2x+5=0\)

\(\Leftrightarrow2x\left(2x-5\right)-\left(2x-5\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{5}{2};\dfrac{1}{2}\right\}\)

cho vào máy tính là ra hết

Answer:

\(3x^2-4x=0\)

\(\Rightarrow x\left(3x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{4}{3}\end{cases}}\)

\(\left(x^2-5x\right)+x-5=0\)

\(\Rightarrow x\left(x-5\right)+\left(x-5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)

\(x^2-5x+6=0\)

\(\Rightarrow x^2-2x-3x+6=0\)

\(\Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\)

\(\Rightarrow x\left(x-2\right)-3\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)

\(5x\left(x-3\right)-x+3=0\)

\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)

\(\Rightarrow\left(5x-1\right)\left(x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}5x-1=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=3\end{cases}}\)

\(x^2-2x+5=0\)

\(\Rightarrow\left(x^2-2x+1\right)+4=0\)

\(\Rightarrow\left(x-1\right)^2=-4\) (Vô lý)

Vậy không có giá trị \(x\) thoả mãn

\(x^2+x-6=0\)

\(\Rightarrow x^2+3x-2x-6=0\)

\(\Rightarrow x.\left(x+3\right)-2\left(x+3\right)=0\)

\(\Rightarrow\left(x-2\right)\left(x+3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}}\)

a, \(4x\left(x+1\right)-5\left(x+1\right)=0\)

\(\left(x+1\right)\left(4x-5\right)\)=0

\(\left\{{}\begin{matrix}x+1=0\\4x-5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\left(-1\right)\\4x=5\Rightarrow x=\frac{5}{4}\end{matrix}\right.\)

b, \(5x\left(x-20\right)+5x-100=0\)

\(5x\left(x-20\right)+\left(5x-100\right)=0\)

\(5x\left(x-20\right)+5\left(x-20\right)=0\)

\(\left(x-20\right)\left(5x+5\right)\)= 0

\(\left\{{}\begin{matrix}x-20=0\\5x+5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=20\\5x=-5\Rightarrow x=-1\end{matrix}\right.\)

c, \(2\left(x-2\right)+\left(x-2\right)^2=0\)

Rút gọn thừa số chung

Giải phương trình

Giải phương trình

Biệt thức

Biệt thức

Nghiệm

Lời giải thu được

Vậy x= 0 và x = 2

d, \(\left(x-3\right)^2-5x-x^2=12\)

\(\left(x^2-2.x.3+3^2\right)-5x-x^2=12\)

\(x^2-6x+9-5x-x^2=12\)

\(-11x+9=12\)

\(-11x=3\)

=> \(x=-\frac{3}{11}\)

a/ \(\left(5x-1\right)^2-5x\left(5x-1\right)=0\)

\(\Leftrightarrow\left(5x-1\right)\left(5x-1-5x\right)=0\Leftrightarrow1-5x=0\Leftrightarrow x=\dfrac{1}{5}\)

Vaayj........

b/ \(x\left(x+1\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\Leftrightarrow x=-1\\x+2=0\Leftrightarrow x=-2\end{matrix}\right.\)

Vay......

c/ \(\left(3x+2\right)x-3\left(3x+2\right)=0\)

\(\Leftrightarrow\left(3x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=3\end{matrix}\right.\)

Vậy.....

\(a)\) \(\left(5x-1\right)^2-5x\left(5x-1\right)=0\)

\(\Leftrightarrow\)\(\left(5x-1\right)\left(5x-1-5x\right)=0\)

\(\Leftrightarrow\)\(\left(5x-1\right).\left(-1\right)=0\)

\(\Leftrightarrow\)\(5x-1=0\)

\(\Leftrightarrow\)\(5x=1\)

\(\Leftrightarrow\)\(x=\frac{1}{5}\)

Vậy \(x=\frac{1}{5}\)

\(b)\) \(x\left(x+1\right)\left(x+2\right)=0\)

Suy ra \(x=0\) hoặc \(x+1=0\) hoặc \(x+2=0\)

\(\Leftrightarrow\)\(x=0\) hoặc \(x=-1\) hoặc \(x=-2\)

Vậy \(x=0\) hoặc \(x=-1\) hoặc \(x=-2\)

\(c)\) \(\left(3x+2\right)x-3\left(3x+2\right)=0\)

\(\Leftrightarrow\)\(\left(3x+2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}3x+2=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=0-2\\x=0+3\end{cases}}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}3x=-2\\x=3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-2}{3}\\x=3\end{cases}}}\)

Vậy \(x=\frac{-2}{3}\) hoặc \(x=3\)

Chúc bạn học tốt ~ 

a/ \(\left(5x-1\right)^2-5x\left(5x-1\right)=0\)

<=> \(\left(5x-1\right)\left(5x-1-5x\right)=0\)

<=> \(-1\left(5x-1\right)=0\)

<=> \(5x-1=0\)

<=> \(5x=1\)

<=> \(x=\frac{1}{5}\)

b/ \(x\left(x+1\right)\left(x+2\right)=0\)

<=> \(x=0\) hoặc \(\orbr{\begin{cases}x+1=0\\x+2=0\end{cases}}\)

<=> \(x=0\)hoặc \(\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)

c/ \(\left(3x+2\right)x-3\left(3x+2\right)=0\)

<=> \(\left(3x+2\right)\left(x-3\right)=0\)

<=> \(\orbr{\begin{cases}3x+2=0\\x-3=0\end{cases}}\)

<=> \(\orbr{\begin{cases}3x=-2\\x=3\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-\frac{2}{3}\\x=3\end{cases}}\)

a) (5x+1)² - (5x-3).(5x+3) = 0

25x² + 10x + 1 - 25x² + 9 = 0

10x + 10 = 0

10.(x+1) = 0

=> x + 1 = 0 => x = - 1

b) (x+3).(x² - 3x + 9) - x.(x-2).(x+2) = 0

x³ + 27 - x.(x² - 4) = 0

x³ + 27 - x³ + 4x = 0

27 + 4x = 0

4x = - 27

x = -27/4

c) 3x.(x-2) - x + 2= 0

3x.(x-2) - (x-2) = 0

(x-2).(3x-1) = 0

=> x - 2 =0 => x = 2

3x-1 = 0 => 3x = 1 => x = 1/3

d) x.(2x-3) - 2.(3-2x) = 0

x.(2x-3) + 2.(2x-3) = 0

(2x-3).(x+2) = 0

=> 2x - 3 = 0 => 2x =  3 => x = 3/2

x+ 2 = 0 => x = -2

KL:...\

a) 5x( x - 1 ) = x - 1

<=> 5x² - 5x = x - 1

<=> 5x² - 5x - x + 1 = 0

<=> 5x² - 6x + 1 = 0

<=> 5x² - 5x - x + 1 = 0

<=> 5x( x - 1 ) - 1( x - 1 ) = 0

<=> ( x - 1 )( 5x - 1 ) = 0

<=> \(\orbr{\begin{cases}x-1=0\\5x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)

b) 2( x + 5 ) - x² - 5x = 0

<=> 2x + 10 - x² - 5x = 0

<=> -x² - 3x + 10 = 0

<=> -x² - 5x + 2x + 10 = 0

<=> -x( x + 5 ) + 2( x + 5 ) = 0

<=> ( x + 5 )( 2 - x ) = 0

<=> \(\orbr{\begin{cases}x+5=0\\2-x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)

c) x² - 2x - 3 = 0

<=> x² + x - 3x - 3 = 0

<=> x( x + 1 ) - 3( x + 1 ) = 0

<=> ( x + 1 )( x - 3 ) = 0

<=> \(\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)

d) 2x² + 5x - 3 = 0

<=> 2x² - x + 6x - 3 = 0

,<=> x( 2x - 1 ) + 3( 2x - 1 ) = 0

<=> ( 2x - 1 )( x + 3 ) = 0

<=> \(\orbr{\begin{cases}2x-1=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-3\end{cases}}\)

a) 5x ( x - 1 ) = x - 1 <=> 5x² - 5x - x + 1 = 0

<=> 5x² - 6x + 1 = 0 <=> 5x² - x - ( 5x - 1 ) = 0 

<=> x ( 5x - 1 ) - ( 5x - 1 ) = 0 <=> ( x - 1 )( 5x - 1 ) = 0

<=> x = 1 hoặc x = 1/5

b) 2 ( x + 5 ) - x² - 5x = 0 <=> 2 ( x + 5 ) - x ( x + 5 ) = 0

<=> ( 2 - x ) ( x + 5 ) = 0 <=> x = 2 hoặc x = -5

c) x² - 2x - 3 = 0 <=> x² + x - 3x - 3 = 0 

<=> x ( x + 1 ) - 3 ( x + 1 ) = 0 <=> ( x - 3 ) ( x + 1 ) = 0 

<=> x = 3 hoặc x = -1

d) 2x²  + 5x - 3 = 0

Ta có : delta = 5² - 4.2.3 = 25 - 24 = 1

Khi đó : x = -1 hoặc x = 3/2  

a) 5x(x-1)=x-1

\(\Leftrightarrow\)5x(x-1)-(x-1)=0

\(\Leftrightarrow\)(x-1)(5x-1)=0

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-1=0\\5x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)

vậy x=1 hoặc x=1/5

b) 2(x+5)-x²-5x=0

\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(2-x\right)\)=0

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\2-x=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)

Vậy x=2 hoặc x=-5

c) x²-2x-3=0

\(\Leftrightarrow x^2-3x+x-3=0\)\(\Leftrightarrow x\left(x-3\right)+\left(x-3\right)=0\)\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)

vậy...

a/ (do (x-3)^2 + 1 ≠0 vs mọi x

a) (x-3)³-3+x=0

=> (x-3)³+(x-3)=0

=> (x-3)(x²-6x+10)

=> \(\left[{}\begin{matrix}x-3=0\\x^2-6x+10=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=3\\\left(x-3\right)^2=1\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=3\\x=4\\x=2\end{matrix}\right.\)