1/ 3-2x+4+6x=x+7+3x

⇔-2x+6x-x-3x=0

⇔0x=0 (Vô số nghiệm)

2/-6(1,5-2x)=3(-15+2x)

⇔-9+12x=-45+6x

⇔6x+36=0

⇔6(x+6)=0

⇔x+6=0

⇔x=-6

Vậy S ϵ {-6}

3/ 3(2x-5)+5(x-1)=4(x+1)

⇔6x-15+5x-5=4x+4

⇔7x=24

⇔x=\(\dfrac{24}{7}\) 

Vậy S ϵ {\(\dfrac{24}{7}\)}

1) Ta có: \(3-2x+4+6x=x+7+3x\)

\(\Leftrightarrow4x+7=4x+7\)

\(\Leftrightarrow4x+7-4x-7=0\)

\(\Leftrightarrow0x=0\)(luôn đúng)

Vậy: S={x|\(x\in R\)}

2) Ta có: \(-6\cdot\left(1.5-2x\right)=3\left(-15+2x\right)\)

\(\Leftrightarrow-9+12x=-45+6x\)

\(\Leftrightarrow12x-9+45-6x=0\)

\(\Leftrightarrow6x+36=0\)

\(\Leftrightarrow6x=-36\)

hay x=-6

Vậy: S={-6}

3) Ta có: \(3\left(2x-5\right)+5\left(x-1\right)=4\left(x+1\right)\)

\(\Leftrightarrow6x-15+5x-5=4x+4\)

\(\Leftrightarrow11x-20-4x-4=0\)

\(\Leftrightarrow7x-24=0\)

\(\Leftrightarrow7x=24\)

\(\Leftrightarrow x=\dfrac{24}{7}\)

Vậy: \(S=\left\{\dfrac{24}{7}\right\}\)

1) 3 - 2x + 4 + 6x = x + 7 + 3x

⇔-2x + 6x - x - 3x = 7 - 3 - 4

⇔0x = 0

⇔x ∈ R

Vậy ...

2) -6(1,5 - 2x) = 3(-15 + 2x)

⇔-9 + 12x = -45 + 6x

⇔6x = -36

⇔x = -6

Vậy ...

1) \(|5x-3|=|7-x|\)

\(\Leftrightarrow\orbr{\begin{cases}5x-3=7-x\\5x-3=x-7\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}6x=10\\4x=-4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}\)

Vậy...

2) \(2.|3x-1|-3x=7\)

\(\Leftrightarrow2.|3x-1|=7+3x\)

\(\Leftrightarrow\orbr{\begin{cases}2.\left(3x-1\right)=7+3x\\2.\left(3x-1\right)=-7-3x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}6x-2=7+3x\\6x-2=-7-3x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x=9\\9x=-5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{-5}{9}\end{cases}}\)

Vậy...

5) \(\text{1,6 - |x + 1,5| = 0}\)

\(|x + 1,5| = 1,6-0\)

\(\text{|x + 1,5| = 1,6}\)

\( |x | = 1,6-1,5\)

\(|x|=0,1\)

a: =>1/3:x=3/5-2/3=9/15-10/15=-1/15

=>x=-1/3:1/15=5

b: \(\Leftrightarrow x\cdot\dfrac{2}{3}-3=\dfrac{2}{5}\cdot\left(-10\right)=-4\)

=>x*2/3=-1

=>x=-3/2

c: =>2x+1=4 hoặc 2x+1=-4

=>x=3/2 hoặc x=-5/2

h: =>x-3=4

=>x=7

g: =>2x-1=3

=>2x=4

=>x=2

f: \(\Leftrightarrow x\cdot\left(\dfrac{3}{2}-\dfrac{7}{3}\right)=\dfrac{3}{2}-\dfrac{2}{3}\)

=>x*-5/6=5/6

=>x=-1

d: =>|2x-1|=3

=>2x-1=3 hoặc 2x-1=-3

=>x=-1 hoặc x=2

f: Ta có: \(16x^2-9\left(x+1\right)^2=0\)

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

\(\Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\)

\(\frac{2}{3}.x-\frac{3}{2}.\left(x-\frac{1}{2}\right)=\frac{5}{12}\)

\(\Leftrightarrow\frac{2}{3}.x-\frac{3}{2}.x+\frac{3}{4}=\frac{5}{12}\)

\(\Leftrightarrow x.\left(\frac{2}{3}-\frac{3}{2}\right)=\frac{5}{12}-\frac{3}{4}\)

\(\Leftrightarrow x.\frac{-5}{6}=\frac{-4}{12}\)

\(\Leftrightarrow x=\frac{-4}{12}:\frac{-5}{12}\)

\(\Leftrightarrow x=\frac{4}{5}\)

Vậy : \(x=\frac{4}{5}\)

Đề câu 2 bạn viết lại hộ mình nhé ! vũ thị diệu hương

a) \(\left(x+1\right)^3+\left(x-2\right)^3-2x^2\left(x-1,5\right)=5\)
\(\Leftrightarrow\left(x^3+3x^2+3x+1\right)+\left(x^3-6x^2+12x-8\right)-\left(2x^3-3x^2\right)=5\)
\(\Leftrightarrow x^3+3x^2+3x+1+x^3-6x^2+12x-8-2x^3+3x^2=5\)
\(\Leftrightarrow15x-12=0\)
\(\Leftrightarrow15x=12\)
\(\Leftrightarrow x=\dfrac{4}{5}\)
Vậy \(S=\left\{\dfrac{4}{5}\right\}\)
(bạn ơi xem lại đề câu b ik để mik giải lun)

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