b) \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}+\frac{x+1}{3}=\frac{x+7}{12}\)

<=> \(\frac{13\left(x+1\right)}{12}-\frac{5x+3}{6}=\frac{x+7}{12}\)

<=> 13(x + 1) - 2(5x + 3) = x + 7

<=> 13x + 13 - 10x - 6 = x + 7

<=> 3x + 7 = x + 7

<=> 3x + 7 - x = 7

<=> 2x + 7 = 7

<=> 2x = 7 - 7

<=> 2x = 0

<=> x = 0

c) 2x + 4(x - 2) = 5

<=> 2x + 4x - 8 = 5

<=> 6x - 8 = 5

<=> 6x = 5 + 8

<=> 6x = 13

<=> x = 13/6

a) 7x + 21 = 0 <=> 7x = -21 <=> x=-3

b) 5x - 2 = 0 <=> 5x =2 <=> x= 2/5

c) 12 - 6x =0 <=>6x = -12 <=> x= -2 

d) -2x + 14 = 0 <=> 2x = 14 <=> x = 7

a) 7x + 21 = 0

<=> 7x = -21

<=> x = -3

Vậy S = {-3}

b) 5x - 2 = 0

<=> 5x = 2

\(\Leftrightarrow x=\dfrac{2}{5}\)

Vậy:....

c) 12 - 6x =0 

<=> 6x = 12

<=> x = 2

Vậy S = {2}

d) -2x + 14 = 0

<=> -2x = -14

<=> x = 7

Vậy S = {7}

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

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

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

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

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

\(\Rightarrow\left[\begin{array}{nghiempt}-2\\-3\end{array}\right.\)

Vậy x = -2 và x = -3

Ta có: \(x^2+5x+6=0\)

<=> \(\left(x^2+2x\right)+\left(3x+6\right)=0\)

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

<=> \(\left[\begin{array}{nghiempt}x+2=0\\x+3=0\end{array}\right.\)

<=> \(\left[\begin{array}{nghiempt}x=-2\\x=-3\end{array}\right.\)

Vậy x\(\in\left\{-3;-2\right\}\)

x² + 5x + 6 = 0

⇒x²+ 2x + 3x + 6 = 0

⇒x ( x + 2 ) + 3 ( x + 2 ) = 0

⇒( x + 2 ) ( x + 3 ) = 0

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

Vậy x = -2 và x = -3

Điều kiện x khác 0

     \(\left(5x^4-3x^3\right):2x^3=\frac{1}{2}\) 

\(\Rightarrow\frac{5}{2}x-\frac{3}{2}=\frac{1}{2}\)

\(\Rightarrow\frac{5}{2}x=2\Rightarrow x=\frac{4}{5}\)

       (3x - 1 )² - 16 =0

<=>(3x - 1 )² = 16

<=>3x-1  =4

<=>x=5/3

(3x - 1)² - 16 = 0

<=> (3x - 1)² - 4² = 0

<=> (3x - 1 - 4)(3x - 1 + 4) = 0

<=> (3x - 5)(3x + 3)  = 0

<=> 3x - 5 = 0 hoặc 3x + 3 = 0

<=> x = 5/3 hoặc x = - 1

(3x-1)^2-16=0

(3x-1)^2=16  

Th1: 3x-1=4

        3x=5  

        x=5/3

Th2: 3x-1=-4    

        3x=-3            

          x=-1 

a\(^n+2a^n\)+1+5a\(^n\)-4a\(^n\)+1

=(a\(^n-2a^n+5a^n-4a^n\))+(1+1)

=(-a\(^n+a^n\))+2

=2

vậy \(a^n-2a^n+1+5a^n-4a^n+1=2\)

mơn bạn

1+1+2+0+0+6=10 nha bạn

Trả lời :

1 + 1 + 2 + 0 + 0 + 6 = 10

Hk tốt

1+1+2+0+0+6=10