Question

Giải phương trình

a)\(x-\dfrac{5x+2}{6}=\dfrac{7-3x}{4}\)

b)\(\dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\)

c)\(\left(\dfrac{3}{2x+1}+2\right)\left(5x-2\right)=\dfrac{5x-2}{2x+1}\)

d)\(\dfrac{2x-1}{x^3+1}+\dfrac{1}{x+1}=\dfrac{2}{x^2-x+1}\)

GIÚP MÌNH VỚI , ĐANG CẦN GẤP

Answer 1

a.x-\(\dfrac{5x+2}{6}=\dfrac{7-3x}{4}\)

⇔\(x=\dfrac{7-3x}{4}+\dfrac{5x+2}{6}\)

⇔\(x=\dfrac{21-9x+10x+4}{12}\)

⇔x=\(\dfrac{x+25}{12}\)

⇔12x=x+25

⇔x=\(\dfrac{25}{11}\)

Vậy pt đã cho có n₀ là S=\(\left\{\dfrac{25}{11}\right\}\)

Answer 2

b.ĐKXĐ:x≠-2;x≠2

\(\dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\)

⇔\(\dfrac{\left(x-2\right)\cdot\left(x-2\right)-3\cdot\left(x+2\right)}{\left(x-2\right)\cdot\left(x+2\right)}\)=\(\dfrac{2x-22}{\left(x-2\right)\cdot\left(x+2\right)}\)

⇔\(\dfrac{x^2-7x-2}{\left(x-2\right)\cdot\left(x+2\right)}=\dfrac{2x-22}{\left(x-2\right)\cdot\left(x+2\right)}\)

⇒\(\left(x^2-7x-2\right)\cdot\left(x-2\right)\cdot\left(x+2\right)=\left(2x-22\right)\cdot\left(x-2\right)\cdot\left(x+2\right)\)

⇔x²-7x-2=2x-22

⇔x²-9x+20=0

⇔(x-4)(x-5)=0

⇔\(\left\{{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

Vậy pt đã cho có n₀ là S={4;5}

Answer 3

\(\left(\dfrac{3}{2x+1}+\dfrac{4x+2}{2x+1}\right)\cdot\left(5x-2\right)=\dfrac{5x-2}{2x+1}\)

⇔\(\dfrac{\left(4x+5\right)\cdot\left(5x-2\right)}{2x+1}=\dfrac{5x-2}{2x+1}\)

⇒(4x+5)(5x-2)(2x+1)=(5x-2)(2x+1)

⇔4x+5=1

⇔x=-1

Vậy pt đã cho có n₀ là S={-1}

Answer 4

d.\(\dfrac{2x-1}{\left(x+1\right)\cdot\left(x^2-x+1\right)}+\dfrac{1}{x+1}=\dfrac{2}{x^2-x+1}\)

⇔\(\dfrac{2x-1+x^2-x+1}{\left(x+1\right)\cdot\left(x^2-x+1\right)}=\dfrac{2}{x^2-x+1}\)

⇔\(\dfrac{x\cdot\left(x+1\right)}{\left(x+1\right)\cdot\left(x^2-x+1\right)}=\dfrac{2}{x^2-x+1}\)

⇔\(\dfrac{x}{x^2-x+1}=\dfrac{2}{x^2-x+1}\)

⇒x=2

Vậy pt đã cho có n₀ là S={2}