Question

Bài 1:giải các phương trình sau

a)3x+5=14

b)(x+3)(2x-5)=0

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

d)\(\left|2x-1\right|=x+4\)

Answer 1

Giải:

a) \(3x+5=14\)

\(\Leftrightarrow3x=14-5=9\)

\(\Leftrightarrow x=\dfrac{9}{3}=3\)

Vậy ...

b) \(\left(x+3\right)\left(2x-5\right)=0\)

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

Vậy ...

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

ĐKXĐ: \(x\ne2;x\ne4\)

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

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

\(\Rightarrow\left(x-1\right)\left(4-x\right)+\left(-x-3\right)\left(x-2\right)=2\)

\(\Leftrightarrow\left(x-1\right)\left(4-x\right)-\left(x+3\right)\left(x-2\right)=2\)

\(\Leftrightarrow4x-4-x^2+x-\left(x^2+3x-2x-6\right)=2\)

\(\Leftrightarrow4x-4-x^2+x-x^2+3x-2x-6=2\)

Tự triển khai và tìm ra nghiệm của phương trình.

d) \(\left|2x-1\right|=x+4\) (2)

TH1: \(2x-1\ge0\Leftrightarrow x\ge\dfrac{1}{2}\)

\(\left(2\right)\Leftrightarrow2x-1=x+4\)

\(\Leftrightarrow2x-x=4+1\)

\(\Leftrightarrow x=5\) (thỏa mãn)

TH2: \(2x-1< 0\Leftrightarrow x< \dfrac{1}{2}\)

\(\left(2\right)\Leftrightarrow-2x+1=x+4\)

\(\Leftrightarrow-2x-x=4-1\)

\(\Leftrightarrow-3x=3\)

\(\Leftrightarrow x=-1\) (thỏa mãn)

Vậy ...

Answer 2

$a) 3x + 5 = 14$

$\Leftrightarrow 3x = 14 - 5$

$\Leftrightarrow 3x = 9$

$\Leftrightarrow x = \frac{9}{3}$

$\Leftrightarrow x = 3$

Vậy tập nghiệm của pt: S = {3}

$b) (x + 3)(2x - 5) = 0$

$\Leftrightarrow x + 3 = 0 hoặc 2x - 5 = 0$

$\Leftrightarrow x = - 3 hoặc 2x = 5$

$\Leftrightarrow x = - 3 hoặc x = \frac{5}{2}$

Vậy tập nghiệm của pt: S = {$- 3$; $\frac{5}{2}$}

Answer 3

a) \(3x+5=14\)

⇔\(3x=14-5\)

⇔\(3x=9\)

⇔\(x=\dfrac{9}{3}\)

⇔\(x=3\)

Vậy S={3}

b)\(\left(x+3\right)\left(2x-5\right)=0\)

⇔\(\left\{{}\begin{matrix}x+3=0\\2x-5=0\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=0-3\\2x=5\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=-3\\x=\dfrac{5}{2}\end{matrix}\right.\)

Vậy S={\(-3;\dfrac{5}{2}\)}

b) \(\dfrac{x-1}{x-2}+\dfrac{x+3}{x-4}=\dfrac{2}{\left(x-2\right)\left(4-x\right)}\) (\(x\ne2;x\ne4\))

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

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

⇔\(\left(x-1\right)\left(4-x\right)-\left(x+3\right)\left(x-2\right)=2\)

⇔\(4x-x^2-4+x-x^2+2x-3x+6=2\)

⇔\(-2x^2+4x+2=2\)

⇔\(-2x\left(x-2\right)=0\)

⇔\(\left\{{}\begin{matrix}-2x=0\\x-2=0\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=0\\x=2\left(loại\right)\end{matrix}\right.\)

Vậy S={0}

d) \(\left|2x-1\right|=x+4\)

*Nếu 2x-1≥0 ⇔ x≥\(\dfrac{1}{2}\) thì phương trình trở thành:

\(2x-1=x+4\)

⇔\(2x-x=1+4\)

⇔\(x=5\) (thỏa mãn)

*Nếu 2x-1<0 ⇔ x<\(\dfrac{1}{2}\) thì phương trình trở thành:

\(-2x+1=x+4\)

⇔\(-2x-x=-1+4\)

⇔\(-3x=3\)

⇔\(x=-1\) (thỏa mãn)

Vậy S={-1;5}