Question

giải các phương trình sau 

1/ 2 ( x-5) =(-x-5)

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

3/ 4(x-5) -(3x-1) =x-19

4, 7-(x-2) =5(2-3x)

5, 2x-(5-3x)=7x+1

Answer 1

1/ \(2\left(x-5\right)=\left(-x-5\right)\)

\(\Leftrightarrow2x-10=-x-5\)

\(\Leftrightarrow3x=5\)

\(\Leftrightarrow x=\dfrac{5}{3}\)

Vậy: \(S=\left\{\dfrac{5}{3}\right\}\)

==========

2/ \(2\left(x+3\right)-3\left(x-1\right)=2\)

\(\Leftrightarrow2x+6-3x+3=2\)

\(\Leftrightarrow-x=-7\)

\(\Leftrightarrow x=7\)

Vậy: \(S=\left\{7\right\}\)

==========

3/ \(4\left(x-5\right)-\left(3x-1\right)=x-19\)

\(\Leftrightarrow4x-20-3x+1=x-19\)

\(\Leftrightarrow0x=0\)

Vậy: \(S=\left\{x|x\text{ ∈ }R\right\}\) 

===========

4/ \(7-\left(x-2\right)=5\left(2-3x\right)\)

\(\Leftrightarrow7-x+2=10-15x\)

\(\Leftrightarrow14x=1\)

\(\Leftrightarrow x=\dfrac{1}{14}\)

Vậy: \(S=\left\{\dfrac{1}{14}\right\}\)

==========

5/ \(2x-\left(5-3x\right)=7x+1\)

\(\Leftrightarrow2x-5+3x=7x+1\)

\(\Leftrightarrow-2x=6\)

\(\Leftrightarrow x=-3\)

Vậy: \(S=\left\{-3\right\}\)

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Chúc bạn học tốt.

Answer 2

1. \(2\left(x-5\right)=-x-5\)

\(\Leftrightarrow3x=5\)

\(\Leftrightarrow x=\dfrac{5}{3}\)

Vậy \(S=\left\{\dfrac{5}{3}\right\}\)

2. \(2\left(x+3\right)-3\left(x-1\right)=2\)

\(\Leftrightarrow2x+6-3x+3=2\)

\(\Leftrightarrow x=7\)

Vậy \(S=\left\{7\right\}\)

3. \(4\left(x-5\right)-\left(3x-1\right)=x-19\)

\(\Leftrightarrow4x-20-3x+1-x+19=0\)

\(\Leftrightarrow0x=0\)

Vậy \(S=\left\{x\in R\right\}\)

4. \(7-\left(x-2\right)=5\left(2-3x\right)\)

\(\Leftrightarrow7-x+2-10+15x=0\)

\(\Leftrightarrow14x-1=0\)

\(\Leftrightarrow x=\dfrac{1}{14}\)

Vậy \(S=\left\{\dfrac{1}{14}\right\}\)

4. \(2x-\left(5-3x\right)=7x+1\)

\(\Leftrightarrow2x-5+3x-7x-1=0\)

\(\Leftrightarrow-2x-6=0\)

\(\Leftrightarrow x=-3\)

Vậy \(S=\left\{-3\right\}\)

Answer 3

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

\(\Leftrightarrow2x-10+x+5=0\)

\(\Leftrightarrow3x=5\)

hay \(x=\dfrac{5}{3}\)

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

\(\Leftrightarrow2x+6-3x+3=2\)

\(\Leftrightarrow-x=-7\)

hay x=7

3: Ta có: \(4\left(x-5\right)-\left(3x-1\right)=x-19\)

\(\Leftrightarrow4x-20-3x+1-x+19=0\)

\(\Leftrightarrow0x=0\)(luôn đúng

Answer 4

4: Ta có: \(7-\left(x-2\right)=5\left(2-3x\right)\)

\(\Leftrightarrow7-x+2=10-15x\)

\(\Leftrightarrow14x=1\)

hay \(x=\dfrac{1}{14}\)

5: Ta có: \(2x-\left(5-3x\right)=7x+1\)

\(\Leftrightarrow5x-5-7x-1=0\)

\(\Leftrightarrow-2x=6\)

hay x=-3