(3x + 1)² - (3x - 1).(3x + 1) = 1

<=> (3x + 1).[(3x + 1) - (3x - 1)] = 1

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

<=> (3x + 1).2 = 1

<=> 3x + 1 = 1/2

<=> 3x = -1/2

<=> x = -1/6

Vậy S = {-1/6}.

36x² - 25 - x.(6x - 5) = 0

<=> (36x² - 25) - x.(6x - 5) = 0

<=> [(6x)² - 5²] - x.(6x - 5) = 0

<=> (6x - 5).(6x + 5) - x.(6x - 5) = 0

<=> (6x - 5).(6x + 5 - x) = 0

<=> (6x - 5).(5x + 5) = 0

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

<=> 6x - 5 = 0 hoặc x + 1 = 0

<=> x = 5/6 hoặc x = -1

Vậy S = {-1; 5/6}.

a)

\(\left(3x+1\right)^2-\left(3x-1\right)\left(3x+1\right)=1\)

\(\Rightarrow\left(9x^2+6x+1\right)-\left(9x^2-1\right)=1\)

\(\Rightarrow6x+2=1\)

\(\Rightarrow x=-\frac{1}{6}\)

Vậy pt có nghiệm là x = - 1 / 6

b)

\(36x^2-25-x\left(6x-5\right)=0\)

\(\Rightarrow\left(36x^2-25\right)-x\left(6x-5\right)=0\)

\(\Rightarrow\left(6x-5\right)\left(6x+5\right)-x\left(6x-5\right)=0\)

\(\Rightarrow\left(6x-5\right)\left(6x+5-x\right)=0\)

\(\Rightarrow\left(6x-5\right)\left(5x+5\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{5}{6}\\x=-1\end{array}\right.\)

Vậy pt có nghiệm là x = 5 / 6 ; x = - 1

a) ( 3.x + 1 ) . ( 7.x + 3 ) = (5.x-7 ) . ( 3.x + 1 )  

<=> ( 3.x + 1 ) . ( 7.x + 3 ) - ( 5.x - 7) . ( 3.x + 1 ) = 0

<=> ( 3.x + 1 ) . ( 7.x + 3 - 5.x + 7 ) = 0

<=> ( 3.x + 1 ) . ( 2.x + 10 ) = 0

<=> \(\orbr{\begin{cases}3.x+1=0\\2.x+10=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=-5\end{cases}}}\)

Vậy x = { \(\frac{-1}{3};-5\)} 

b) x² + 10.x + 25 - 4.x . ( x + 5 ) = 0 

<=> ( x + 5 )² -4.x . (x + 5 ) = 0

<=> ( x+ 5 ) . ( x + 5 - 4.x ) = 0

<=> ( x + 5 ) . ( 5 - 3.x )  = 0

<=> \(\orbr{\begin{cases}x+5=0\\5-3.x\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=\frac{5}{3}\end{cases}}}\)

Vậy x = \(\left\{\frac{5}{3};-5\right\}\)

c) (4.x - 5 )² - 2. ( 16.x² -25 ) = 0 

<=> ( 4.x-5)² -2 .( 4.x-5) .( 4.x + 5 ) = 0

<=> (  4.x -5 )² - ( 8.x+ 10 ) . ( 4.x -5 ) = 0

<=> ( 4.x -5 ) . ( 4.x-5 - 8.x - 10 ) = 0

<=> ( 4.x - 5 ) . ( -4.x - 15 ) = 0

<=> \(\orbr{\begin{cases}4.x-5=0\\-4.x-15=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{4}\\x=\frac{-15}{4}\end{cases}}}\)

Vậy x = \(\left\{\frac{5}{4};\frac{-15}{4}\right\}\)

d) ( 4.x + 3 )² = 4. ( x² - 2.x + 1 ) 

<=> 16.x² + 24.x + 9 - 4.x²  + 8.x - 4 = 0

<=> 12.x² + 32.x + 5 =0 

<=> 12. ( x +\(\frac{1}{8}\) ) . ( x + \(\frac{5}{2}\)) = 0 

<=> \(\orbr{\begin{cases}x+\frac{1}{6}=0\\x+\frac{5}{2}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{6}\\x=\frac{-5}{2}\end{cases}}}\)

Vậy x = \(\left\{\frac{-1}{6};\frac{-5}{2}\right\}\)

e) x² -11.x + 28 = 0

<=> x² -4.x  - 7.x + 28 = 0

<=> ( x - 7 ) . ( x - 4 ) = 0

<=> \(\orbr{\begin{cases}x-7=0\\x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=4\end{cases}}}\)

Vậy x = { 4 ; 7 } 

f ) 3.x.3 - 3.x² - 6.x = 0

<=> 3.x. ( x² -x - 2 ) = 0 

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

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

        \([x=0\)                \([x=0\)

( Lưu ý :Lưu ý này không cần ghi vào vở :  Chị nối 2 ý đó làm 1 nha cj ! ) 

Vậy x = { 2 ; -1 ; 0 } 

a, (3x+1)(7x+3)=(5x-7)(3x+1)

<=> (3x+1)(7x+3)-(5x-7)(3x+1)=0

<=> (3x+1)(7x+3-5x+7)=0

<=> (3x+1)(2x+10)=0

<=> 2(3x+1)(x+5)=0

=> 3x+1=0 hoặc x+5=0

=> x= -1/3 hoặc x=-5

Vậy...

a) (3x - 2)(4x + 5) = 0

⇔ 3x - 2 = 0 hoặc 4x + 5 = 0

1) 3x - 2 = 0 ⇔ 3x = 2 ⇔ x = 2/3

2) 4x + 5 = 0 ⇔ 4x = -5 ⇔ x = -5/4

Vậy phương trình có tập nghiệm S = {2/3;−5/4}

b) (2,3x - 6,9)(0,1x + 2) = 0

⇔ 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0

1) 2,3x - 6,9 = 0 ⇔ 2,3x = 6,9 ⇔ x = 3

2) 0,1x + 2 = 0 ⇔ 0,1x = -2 ⇔ x = -20.

Vậy phương trình có tập hợp nghiệm S = {3;-20}

c) (4x + 2)(x² +  1) = 0 ⇔ 4x + 2 = 0 hoặc x² +  1 = 0

1) 4x + 2 = 0 ⇔ 4x = -2 ⇔ x = −1/2

2) x² +  1 = 0 ⇔ x² = -1 (vô lí vì x² ≥ 0)

Vậy phương trình có tập hợp nghiệm S = {−1/2}

d) (2x + 7)(x - 5)(5x + 1) = 0

⇔ 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0

1) 2x + 7 = 0 ⇔ 2x = -7 ⇔ x = −7/2

2) x - 5 = 0 ⇔ x = 5

3) 5x + 1 = 0 ⇔ 5x = -1 ⇔ x = −1/5

Vậy phương trình có tập nghiệm là S = {−7/2;5;−1/5}

a: \(2x^3-50x=0\)

=>\(2x\left(x^2-25\right)=0\)

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

=>x∈{0;5;-5}

b: \(2x\left(3x-5\right)-\left(5-3x\right)=0\)

=>2x(3x-5)+(3x-5)=0

=>(3x-5)(2x+1)=0

=>\(\left[\begin{array}{l}3x-5=0\\ 2x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac53\\ x=-\frac12\end{array}\right.\)

c: \(9\left(3x-2\right)=x\left(2-3x\right)\)

=>9(3x-2)-x(2-3x)=0

=>9(3x-2)+x(3x-2)=0

=>(3x-2)(x+9)=0

=>\(\left[\begin{array}{l}3x-2=0\\ x+9=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac23\\ x=-9\end{array}\right.\)

d: \(\left(2x-1\right)^2-25=0\)

=>(2x-1-5)(2x-1+5)=0

=>(2x-6)(2x+4)=0

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

=>\(\left[\begin{array}{l}x-3=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=-2\end{array}\right.\)

e: \(25x^2-2=0\)

=>\(25x^2=2\)

=>\(x^2=\frac{2}{25}\)

=>\(\left[\begin{array}{l}x=\frac{\sqrt2}{5}\\ x=-\frac{\sqrt2}{5}\end{array}\right.\)

f: \(x^2-25=6x-9\)

=>\(x^2-6x-16=0\)

=>(x-8)(x+2)=0

=>\(\left[\begin{array}{l}x-8=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=8\\ x=-2\end{array}\right.\)

g: 5x(x-3)-2x+6=0

=>5x(x-3)-2(x-3)=0

=>(x-3)(5x-2)=0

=>\(\left[\begin{array}{l}x-3=0\\ 5x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=\frac25\end{array}\right.\)

h: 3x(x-7)-2(x-7)=0

=>(x-7)(3x-2)=0

=>\(\left[\begin{array}{l}x-7=0\\ 3x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=7\\ x=\frac23\end{array}\right.\)

i: \(7x^2-28=0\)

=>\(7x^2=28\)

=>\(x^2=4\)

=>x=2 hoặc x=-2

j: 2x+1+x(2x+1)=0

=>(2x+1)(x+1)=0

=>\(\left[\begin{array}{l}2x+1=0\\ x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac12\\ x=-1\end{array}\right.\)

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

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

=>4(x+2)=0

=>x+2=0

=>x=-2

l: \(x^3+5x^2-4x-20=0\)

=>\(x^2\left(x+5\right)-4\left(x+5\right)=0\)

=>\(\left(x+5\right)\left(x^2-4\right)=0\)

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

=>x∈{-5;2;-2}

m: \(x^2-25+2\left(x+5\right)=0\)

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

=>(x+5)(x-3)=0

=>\(\left[\begin{array}{l}x+5=0\\ x-3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-5\\ x=3\end{array}\right.\)

n: \(x^2-3x+2=0\)

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

=>x(x-1)-2(x-1)=0

=>(x-1)(x-2)=0

=>\(\left[\begin{array}{l}x-1=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=2\end{array}\right.\)

o: \(x^2-6x+8=0\)

=>\(\left(x-2\right)\left(x-4\right)=0\)

=>\(\left[\begin{array}{l}x-2=0\\ x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=4\end{array}\right.\)

p: \(x^2-5x-14=0\)

=>\(x^2-7x+2x-14=0\)

=>(x-7)(x+2)=0

=>\(\left[\begin{array}{l}x-7=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=7\\ x=-2\end{array}\right.\)

q: \(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)

=>\(x^2-4x+4-x^2+9=6\)

=>-4x+13=6

=>-4x=6-13=-7

=>x=7/4

r: \(\left(2x-1\right)^2-\left(2x-5\right)\left(2x+5\right)=18\)

=>\(4x^2-4x+1-\left(4x^2-25\right)=18\)

=>-4x+26=18