Câu hỏi của quynh nhu nguyen

Question

7. Giải các phương trình

a/$\dfrac{2x-1}{x-1}+1=\dfrac{1}{x-1}$

b/$\dfrac{5x}{2x+2}+1=-\dfrac{6}{x+1}$

c/$x+\dfrac{1}{x}=x^2+\dfrac{1}{x^2}$

Sáng · Accepted Answer

a. $\dfrac{2x-1}{x-1}+1=\dfrac{1}{x-1}$ (ĐKXĐ: $x\ne1$)

$\Leftrightarrow2x-1+x-1=1$

$\Leftrightarrow3x-2=1\Leftrightarrow x=1$(KTM)

$\Rightarrow S=\left\{\varnothing\right\}$

b. $\dfrac{5x}{2x+2}+1=\dfrac{-6}{x+1}$ (ĐKXĐ: $x\ne-1$)

$\Leftrightarrow5x+2\left(x+1\right)=-12$

$\Leftrightarrow5x+2x+2=-12\Leftrightarrow7x=-14\Leftrightarrow x=-2$ (TM)

$\Rightarrow S=\left\{-2\right\}$

c. $x+\dfrac{1}{x}=x^2+\dfrac{1}{x^2}$ (ĐKXĐ: $x\ne0$)

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

$\Leftrightarrow-x^3-x=-x^4-1\Leftrightarrow x^4-x^3-x+1=0$

$\Leftrightarrow x^3\left(x-1\right)-\left(x-1\right)=0\Leftrightarrow\left(x-1\right)^2\left(x^2+x+1\right)=0$

$\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1$ (TM)

$\Rightarrow S=\left\{1\right\}$Câu trả lời: a. $\dfrac{2x-1}{x-1}+1=\dfrac{1}{x-1}$ (ĐKXĐ: $x\ne1$)

$\Leftrightarrow2x-1+x-1=1$

$\Leftrightarrow3x-2=1\Leftrightarrow x=1$(KTM)

$\Rightarrow S=\left\{\varnothing\right\}$

b. $\dfrac{5x}{2x+2}+1=\dfrac{-6}{x+1}$ (ĐKXĐ: $x\ne-1$)

$\Leftrightarrow5x+2\left(x+1\right)=-12$

$\Leftrightarrow5x+2x+2=-12\Leftrightarrow7x=-14\Leftrightarrow x=-2$ (TM)

$\Rightarrow S=\left\{-2\right\}$

c. $x+\dfrac{1}{x}=x^2+\dfrac{1}{x^2}$ (ĐKXĐ: $x\ne0$)

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

$\Leftrightarrow-x^3-x=-x^4-1\Leftrightarrow x^4-x^3-x+1=0$

$\Leftrightarrow x^3\left(x-1\right)-\left(x-1\right)=0\Leftrightarrow\left(x-1\right)^2\left(x^2+x+1\right)=0$

$\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1$ (TM)

$\Rightarrow S=\left\{1\right\}$

Lê Thị Hồng Vân · Answer

https://i.imgur.com/jL8Ia6a.jpgCâu trả lời: https://i.imgur.com/jL8Ia6a.jpg

Nguyễn Huyền Trâm · Answer

a. 2x−1x−1+1=1x−12x−1x−1+1=1x−1 (ĐKXĐ: x≠1x≠1)

⇔2x−1+x−1=1⇔2x−1+x−1=1

⇔3x−2=1⇔x=1⇔3x−2=1⇔x=1(không thỏa mãn )

⇒S={∅}⇒S={∅}

b. 5x2x+2+1=−6x+15x2x+2+1=−6x+1 (ĐKXĐ: x≠−1x≠−1)

⇔5x+2(x+1)=−12⇔5x+2(x+1)=−12

⇔5x+2x+2=−12⇔7x=−14⇔x=−2⇔5x+2x+2=−12⇔7x=−14⇔x=−2 (thỏa mãn )

⇒S={−2}⇒S={−2}

c. x+1x=x2+1x2x+1x=x2+1x2 (ĐKXĐ: x≠0x≠0)

⇔x3+x=x4+1⇔−(x3+x)=−(x4+1)⇔x3+x=x4+1⇔−(x3+x)=−(x4+1)

⇔−x3−x=−x4−1⇔x4−x3−x+1=0⇔−x3−x=−x4−1⇔x4−x3−x+1=0

⇔x3(x−1)−(x−1)=0⇔(x−1)2(x2+x+1)=0⇔x3(x−1)−(x−1)=0⇔(x−1)2(x2+x+1)=0

⇔(x−1)2=0⇔x−1=0⇔x=1⇔(x−1)2=0⇔x−1=0⇔x=1 (thỏa mãn )

⇒S={1}Câu trả lời: a. 2x−1x−1+1=1x−12x−1x−1+1=1x−1 (ĐKXĐ: x≠1x≠1)