Bài 3: Phương trình đưa được về dạng ax + b = 0

=>(2x-1)(x-1)=(2x-3)(x-2)

=>2x^2-2x-x+1=2x^2-4x-3x+6

=>-3x+1=-7x+6

=>4x=5

=>x=5/4

Bài 5 :

Thay \(x=-3\) vào pt : \(3x+m-x-1=0\)

\(\Leftrightarrow3\left(-3\right)+m-\left(-3\right)-1=0\)

\(\Leftrightarrow-9+m+3-1=0\)

\(\Leftrightarrow m-7=0\)

\(\Leftrightarrow m=7\)

Vậy \(m=7\) để pt nhận \(x=-3\) là nghiệm

Bài 6 :

Thay \(x=1\) vào pt : \(\left(2m-4\right)x+6=0\)

\(\Leftrightarrow2mx-4x+6=0\)

\(\Leftrightarrow2m-4+6=0\)

\(\Leftrightarrow2m+2=0\)

\(\Leftrightarrow m=-1\)

Vậy \(m=-1\) để pt nhận \(x=1\) là nghiệm

k,\(\dfrac{x}{3}-\dfrac{2x+1}{2}=\dfrac{x}{6}-x\)

giúp mk câu k nhé đề bài như trên

b: \(\Leftrightarrow4x+8-9=4x-4\)

=>-1=-4(loại)

d: \(\Leftrightarrow3\left(x-2\right)+2\left(x+1\right)=8x\)

=>8x=3x-6+2x+2=5x-4

=>3x=-4

=>x=-4/3

f: \(\Leftrightarrow3\left(x+2\right)+4\left(2x-3\right)=2\left(x-12\right)\)

=>3x+6+8x-12=2x-24

=>11x-6=2x-24

=>9x=-18

=>x=-2

Lời giải:

b.

$\frac{2x}{3}=8$

$\Leftrightarrow 2x=3.8=24$

$\Leftrightarrow x=24:2=12$
d.

$\frac{6}{5}x=-9$

$\Leftrightarrow x=-9: \frac{6}{5}=\frac{-15}{2}$

f.

$\frac{2-3x}{4}=\frac{4x-5}{5}$

$\Leftrightarrow 5(2-3x)=4(4x-5)$

$\Leftrightarrow 10-15x=16x-20$

$\Leftrightarrow 30=31x$

$\Leftrightarrow x=\frac{30}{31}$

h.

$\frac{10-3x}{2}=\frac{6x+1}{3}$

$\Leftrightarrow 3(10-3x)=2(6x+1)$

$\Leftrightarrow 30-9x=12x+2$

$\Leftrightarrow 28=21x$

$\Leftrightarrow x=\frac{28}{21}=\frac{4}{3}$

b)x+3=4:2
 => x=-1
d)5x-15=3x-5
<=> 5x-3x=15-5
<=> 2x=10
<=> x=5
f) 35-7x=11-5x
<=> 35-11=-5x+7x
<=> 24=2x
<=> x=12

h) 6x-2-3x=10
<=> 3x=10+2
<=> x=4
j)3-2x=3x+3-x-2
<=> 3-2x=2x+1
<=>-4x=-2
<=> x=1/2
 

Lời giải:

b.

$3x+1=5x+2$

$\Leftrightarrow 1-2=5x-3x$

$\Leftrightarrow -1=2x\Leftrightarrow x=\frac{-1}{2}$

d.

$6x-4=7-5x$

$\Leftrightarrow 6x+5x=4+7$

$\Leftrightarrow 11x=11\Leftrightarrow x=1$

f.

$x-12+4x=25+2x-1$
$\Leftrightarrow 5x-12=24+2x$

$\Leftrightarrow 5x-2x=24+12$

$\Leftrightarrow 3x=36$

$\Leftrightarrow x=12$

a: \(\Leftrightarrow\left(\dfrac{x+2001}{5}+1\right)+\left(\dfrac{x+1999}{7}+1\right)+\left(\dfrac{x+1997}{9}+1\right)+\left(\dfrac{x+1995}{11}+1\right)=0\)

=>x+2006=0

=>x=-2006

b: \(\Leftrightarrow\left(\dfrac{x-15}{100}-1\right)+\left(\dfrac{x-10}{105}-1\right)+\left(\dfrac{x-100}{5}-1\right)=\left(\dfrac{x-100}{15}-1\right)+\left(\dfrac{x-105}{10}-1\right)+\left(\dfrac{x-110}{5}-1\right)\)

=>x-105=0

=>x=105

`(x^2-1)(x+3) =0`

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

`<=>` \(\left[{}\begin{matrix}x-1=0\\x+1=0\\x+3=0\end{matrix}\right.\)

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

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

hay \(x\in\left\{1;-1;-3\right\}\)

\(x^2+4x+8=0\)

\(\text{Δ}=4^2-4\cdot1\cdot8=-16< 0\)

Do đó: Phương trình vô nghiệm

Cho `x^2+4x+8=0`

`=>x^2+4x+4+4=0`

`=>(x+2)^2+4=0`

`=>(x+2)^2=-4` (Vô lí vì `(x+2)^2 >= 0` mà `-4 < 0`)

Vậy đa thức không có nghiệm

x² + 4x + 8 = x² + 4x + 4 + 4 = (x + 2)² + 4

(x + 2)² ≥ 0 ⩝ x; 4 > 0

=> (x + 2)² + 4 > 0 => x² + 4x + 8 > 0

=> pt vô nghiệm