\(2x+12=3\left(x-7\right)\)

\(\Leftrightarrow2x+12=3x-21\)

\(\Leftrightarrow2x-3x=-12-21\)

\(\Leftrightarrow-x=-33\)

\(\Leftrightarrow x=33\)

 bỏ dấu\(\Leftrightarrow\)cx đc nhé,lp 6 chx hok đến phương trình

 #hoktot<3# 

\(2x+12=3\left(x-7\right)\)

\(\Leftrightarrow2x+12=3x-21\)

\(\Leftrightarrow2x+12-3x+21=0\)

\(\Leftrightarrow-x+33=0\Leftrightarrow33-x=0\Leftrightarrow x=33\)

\(2x+12=3\left(x-7\right)\)

=> \(2x+12=3x-21\)

=> \(3x-2x=12+21\)

=> \(x=33\)

1) \(\Rightarrow16x^2+24x+9+9x^2-24x+16+4-25x^2=x\)

\(\Rightarrow x=29\)

2)

a) \(=x^2-9-x^2+6x-9=6x-18\)

b) \(=\left(3x-1+2x+1\right)^2=\left(5x\right)^2=25x^2\)

a) 

$|3x-2|=2x\Rightarrow x\geq 0$.

Xét 2 TH:

TH1: $x\geq \frac{2}{3}$ thì pt trở thành:

$3x-2=2x\Leftrightarrow x=2$ (thỏa mãn)

TH2: $0\leq x< \frac{2}{3}$ thì pt trở thành:

$2-3x=2x\Leftrightarrow x=\frac{2}{5}$ (thỏa mãn)

b) 

PT $\Rightarrow x\geq 0$

$\Rightarrow |4+2x|=4+2x$. PT trở thành:

$4+2x=4x\Leftrightarrow x=2$ (thỏa mãn)

c) 

Xét các TH sau:

TH1: $x\geq \frac{3}{2}$. Khi đó, pt trở thành:

$2x-3=-x+21$

$\Leftrightarrow x=8$ (thỏa mãn)

TH2: $x< \frac{3}{2}$. Khi đó, pt trở thành:

$3-2x=-x+21$

$\Leftrightarrow x=-18$ (thỏa mãn)

d) 

Từ PT suy ra $x-2\geq 0\Leftrightarrow x\geq 2(*)$

Khi đó: $|3x-1|=3x-1$. PT trở thành:
$3x-1=x-2$

$\Leftrightarrow 2x=-1<0\Rightarrow x<0$ (mâu thuẫn với $(*)$)

Vậy PT vô nghiệm.

\(|-2x+1,5|=\dfrac{1}{4}\Rightarrow-2x+1,5=\pm\dfrac{1}{4}\)

\(-2x+1,5=\dfrac{1}{4}\Rightarrow-2x=1,5-0,25\Rightarrow-2x=1,25\Rightarrow x=1,25:\left(-2\right)\Rightarrow x=...\)

\(-2x+1,5=-\dfrac{1}{4}\Rightarrow-2x=-0,25-1,5\Rightarrow-2x=1,75\Rightarrow x=1,75:\left(-2\right)\Rightarrow x=...\)

\(\dfrac{3}{2}-|1.\dfrac{1}{4}+3x|=\dfrac{1}{4}\Rightarrow|1.\dfrac{1}{4}+3x|=\dfrac{3}{2}-\dfrac{1}{4}\Rightarrow|1.\dfrac{1}{4}+3x|=\dfrac{5}{4}\)

\(\Rightarrow1.\dfrac{1}{4}+3x=\pm\dfrac{5}{4}\)

\(1.\dfrac{1}{4}+3x=\dfrac{5}{4}\Rightarrow\dfrac{1}{4}+3x=\dfrac{5}{4}\Rightarrow3x=\dfrac{5}{4}-\dfrac{1}{4}\Rightarrow3x=1\Rightarrow x=3\)

\(1.\dfrac{1}{4}+3x=-\dfrac{5}{4}\Rightarrow\dfrac{1}{4}+3x=-\dfrac{5}{4}\Rightarrow3x=-\dfrac{5}{4}-\dfrac{1}{4}\Rightarrow3x=-\dfrac{3}{2}x=...\)

a: ta có: \(\left|-2x+\dfrac{3}{2}\right|=\dfrac{1}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x+\dfrac{3}{2}=\dfrac{1}{4}\\-2x+\dfrac{3}{2}=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-2x=-\dfrac{5}{4}\\-2x=-\dfrac{7}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{8}\\x=\dfrac{7}{8}\end{matrix}\right.\)

b: Ta có: \(\dfrac{3}{2}-\left|\dfrac{5}{4}+3x\right|=\dfrac{1}{4}\)

\(\Leftrightarrow\left|3x+\dfrac{5}{4}\right|=\dfrac{5}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{5}{4}=\dfrac{5}{4}\\3x+\dfrac{5}{4}=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=0\\3x=-\dfrac{5}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{6}\end{matrix}\right.\)

a, \(\dfrac{x}{3}=\dfrac{8}{12};12:3=4\)\(;8:4=2\)

⇒\(x=\dfrac{2}{3}\)

a: \(\dfrac{x}{3}=\dfrac{8}{12}\)

nên x/3=2/3

hay x=2

b: \(x+\dfrac{1}{15}=\dfrac{1}{3}\)

\(\Leftrightarrow x=\dfrac{1}{3}-\dfrac{1}{15}=\dfrac{5}{15}-\dfrac{1}{15}=\dfrac{4}{15}\)

c: \(x\cdot\dfrac{4}{7}-\dfrac{2}{7}=\dfrac{5}{7}\)

\(\Leftrightarrow x\cdot\dfrac{4}{7}=\dfrac{5}{7}+\dfrac{2}{7}=1\)

hay x=7/4

d: \(\Leftrightarrow x\cdot\dfrac{3}{4}-\dfrac{5}{4}=-\dfrac{1}{2}\)

\(\Leftrightarrow x\cdot\dfrac{3}{4}=-\dfrac{2}{4}+\dfrac{5}{4}=\dfrac{3}{4}\)

hay x=1

Đặt t = 2x^2 +x pt trở thành

t^2 - 4t + 3=0

=>t^2 -t -3t +3 =0

=>t( t - 1) -3( t - 1)=0 

=>(t - 3)(t - 1 )=0 

*)Với t-3=0 <=> 2x^2 + x -3=0

=>2x^2 +3x -2x - 3 =0

=>x(2x + 3) - (2x + 3)=0

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

*)Với t-1=0 <=> 2x^2 + x -1=0

=>2x^2 - x + 2x -1=0

=>x(2x - 1) + (2x - 1) =0

=>(x + 1)(2x - 1)=0 <=> x=-1 hoặc x=1/2

3 . (-x + 1) - 2.(x + 7) - 3² = -(-2x - 9) - 1⁵⁸

=> -3x + 3 - 2x - 14 - 9 = 2x + 9 - 1

=> -3x - 2x - 2x = 9 - 1 + 9 + 14 - 3

=> -7x = 28

=> x = 28 : (-7)

=> x = -4

Vậy ...

=>(-3x+3)-(2x+14)-9=2x+9-1

=>(-3x-2x)-(14-3+9)=2x+8

=>-5x-20=2x+8

=>2x+5x=-8-20

=>-7x=-28

=>x=-28/7

\(\left(x+3\right)^2+\left(x-2\right)^2=2x^2\\ \Leftrightarrow x^2+6x+9+x^2-4x+4=2x^2\\ \Leftrightarrow2x=-13\Leftrightarrow x=-\dfrac{13}{2}\)

<=> x²+6x+9+x²+4x+4-2x² =0

<=>10x+13=0

<=> x= -13/10

Chúc bạn học tốt nha!

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

`<=>x^2+6x+9 +x^2-4x+4-2x^2=0`

`<=> 2x +13=0`

`<=>2x=-13`

`<=>x=(-13)/2`

vậy `x=(-13)/2`