Bài 2: Phương trình bậc nhất một ẩn và cách giải

a: \(m^2+m+1=m^2+2\cdot m\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(m+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)

Do đó: Phương trình \(\left(m^2+m+1\right)x-3=0\) luôn là pt bậc nhất 1 ẩn

b: \(m^2+2m+3=\left(m+1\right)^2+2>0\)

Do đó: Phương trình \(\left(m^2+2m+3\right)x-m+1=0\) luôn là pt bậc nhất 1 ẩn

a, Ta có : \(m^2+m+1=m^2+m+\dfrac{1}{4}+\dfrac{3}{4}=\left(m+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)

Vậy ta có đpcm 

b, Ta có : \(m^2+2m+3=m^2+2m+1+2=\left(m+1\right)^2+2>0\)

Vậy ta có đpcm 

a: =>3x=-12

hay x=-4

b: =>2x=8

hay x=4

c: =>-2x=2

hay x=-1

a. \(2x+x+12=0\\ \Leftrightarrow2x+x=0-12\\ \Leftrightarrow3x=-12\\ \Leftrightarrow x=-4\)

Vậy S = { -4 }

b. \(x-5=3-x\\ \Leftrightarrow x+x=3+5\\ \Leftrightarrow2x=8\\ \Leftrightarrow x=4\)

Vậy S = { 4 }

c. \(7-3x=9-x\\ \Leftrightarrow-3x+x=9-7\\ \Leftrightarrow-2x=2\\ \Leftrightarrow x=-1\)

Vậy S = { -1 }

Các pt a,c,d và pt bậc nhất 1 ẩn

a: a=1; b=2

c: a=-2; b=1

d: a=3; b=0

a,c

pt bậc nhất : a ; c ; d ; d 

a, x + 2 = 0 hệ số a = 1 ; b = 2 

c, -2t + 1 = 0 hệ số a = -2 ; b = 1

d, 3y = 0 hệ số a = 3 ; b = 0 

Bài 3: 

a: \(=\dfrac{\left(x+1\right)^2}{x^2}:\left(\dfrac{x^2+1}{x^2}+\dfrac{2}{x+1}\cdot\dfrac{x+1}{x}\right)\)

\(=\dfrac{\left(x+1\right)^2}{x^2}:\dfrac{x^2+1+2x}{x^2}=1\)

b: \(=\dfrac{x}{x-3}-\dfrac{x\left(x+3\right)}{2x+3}\cdot\left(\dfrac{x+3}{x\left(x-3\right)}-\dfrac{x}{\left(x-3\right)\left(x+3\right)}\right)\)

\(=\dfrac{x}{x-3}-\dfrac{x\left(x+3\right)}{2x+3}\cdot\dfrac{x^2+6x+9-x^2-3x}{x\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{2x^2+3x}{\left(2x+3\right)\left(x-3\right)}-\dfrac{3x+9}{\left(x-3\right)\left(2x+3\right)}\)

\(=\dfrac{2x^2-9}{\left(2x+3\right)\left(x-3\right)}\)

lỗi ạ

lỗi

lỗi r bn

a) \(\dfrac{x+1}{4}-\dfrac{5+2x}{8}=\dfrac{3-4x}{2}\)

⇔\(\dfrac{2\left(x+1\right)}{8}-\dfrac{5+2x}{8}=\dfrac{4\left(3-4x\right)}{8}\) 

⇔ 2x + 2 - 5 - 2x = 12 -16x

⇔ 16x = 15 

⇔ x = 15/16

b) \(\dfrac{4-3x}{5}-\dfrac{4-x}{10}=\dfrac{x+2}{2}\)

⇔\(\dfrac{2\left(4-3x\right)}{10}-\dfrac{4-x}{10}=\dfrac{5\left(x+2\right)}{10}\)

⇔ 8 - 6x - 4 + x = 5x + 10

⇔ 10x = -6

⇔ x = -6/10

Câu 1:

x + 1/4 - 5 + 2x/8 = 3 - 4x/2

<=> 2x + 2/8 - 5 + 2x/8 = 12 - 16x/8

<=> 2x + 2 - 5 - 2x = 12 - 16x

<=> -3 = 12 - 16x <=> 15 = 16x <=> x = 15/16

Câu 2:

4 - 3x/5 - 4 - x/10 = x + 2/2

<=> 8 - 6x/10 - 4 - x/10 = 5x + 10/10

<=> 8 - 6x - 4 + x = 5x + 10

<=> 4 - 5x = 5x + 10

<=> 4 = 10x + 10 <=> 10x = -6 <=> x = -3/5

\(\left(m^2-16\right)\cdot x=3m+12\)

\(\left(m-4\right)\left(m+4\right)\cdot x-3m-12=0\)

\(\left(m-4\right)\left(m+4\right)\cdot x-3\left(m+4\right)=0\)

\(\left(m+4\right)\left(xm-4x-3\right)=0\)

\(\Rightarrow m+4=0\Rightarrow m=-4\)

\(xm-4x-3=0\)

\(\Leftrightarrow-4x-4x-3=0\Leftrightarrow-8x-3=0\Leftrightarrow x=\dfrac{-3}{8}\)

\(\text{2x - (x - 3)(5 - x) = (x+4)}^2.\)

\(\Leftrightarrow2x-\left(5x-x^2-15+3x\right)=x^2+8x+16.\)

\(\Leftrightarrow2x-5x+x^2+15-3x-x^2-8x-16=0.\)

\(\Leftrightarrow-14x-1=0.\Leftrightarrow x=\dfrac{-1}{14}.\)

\(\text{(4x + 1)(x - 2) + 25 = (2x+3)}^2-4x.\)

\(\Leftrightarrow4x^2-8x+x-2+25=4x^2+12x+9-4x.\)

\(\Leftrightarrow-15x+14=0.\Leftrightarrow x=\dfrac{14}{15}.\)

Bài 5 :

a) \(7-\left(2x+4\right)=-\left(x+4\right)\)

\(\Leftrightarrow7-2x-4=-x-4\)

\(\Leftrightarrow7-4+4=-x+2x\)

\(\Leftrightarrow7x\)

a: \(\Leftrightarrow x^2+x+1+2x^2-5=4x-4\)

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

=>x=0

b: \(\Leftrightarrow\dfrac{15x}{\left(x+4\right)\left(x-1\right)}-1=\dfrac{12}{x+4}+\dfrac{4}{x-1}\)

\(\Leftrightarrow15x-x^2-3x+4=12x-12+4x+16\)

\(\Leftrightarrow-x^2+12x+4-12x-4x-4=0\)

=>-x(x+4)=0

=>x=0