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$\frac{2}{x+\frac{1}{1+\frac{x+1}{x-2}}}=\frac{6}{3x-1}$

Trần Thùy Linh · Accepted Answer

ĐK: $x\ne\left\{2,\frac{1}{3},\pm1\right\}$

$PT\Leftrightarrow\frac{2}{x+\frac{x-2}{2x-1}}=\frac{6}{3x-1}$

$\Leftrightarrow\frac{2\left(2x-1\right)}{2x^2-2}=\frac{6}{3x-1}$

$\Leftrightarrow\frac{2x-1}{x^2-1}=\frac{6}{3x-1}$

$\Leftrightarrow\left(2x-1\right)\left(3x-1\right)=6\left(x^2-1\right)$

$\Leftrightarrow-5x+7=0\Leftrightarrow x=\frac{7}{5}$

Vậy........Câu trả lời: ĐK: $x\ne\left\{2,\frac{1}{3},\pm1\right\}$

$PT\Leftrightarrow\frac{2}{x+\frac{x-2}{2x-1}}=\frac{6}{3x-1}$

$\Leftrightarrow\frac{2\left(2x-1\right)}{2x^2-2}=\frac{6}{3x-1}$

$\Leftrightarrow\frac{2x-1}{x^2-1}=\frac{6}{3x-1}$

$\Leftrightarrow\left(2x-1\right)\left(3x-1\right)=6\left(x^2-1\right)$

$\Leftrightarrow-5x+7=0\Leftrightarrow x=\frac{7}{5}$

Vậy........

Hiếu Văn Huỳnh · Answer

ĐK: x≠{2,13,±1}x≠{2,13,±1}

PT⇔2x+x−22x−1=63x−1PT⇔2x+x−22x−1=63x−1

⇔2(2x−1)2x2−2=63x−1⇔2(2x−1)2x2−2=63x−1

⇔2x−1x2−1=63x−1⇔2x−1x2−1=63x−1

⇔(2x−1)(3x−1)=6(x2−1)⇔(2x−1)(3x−1)=6(x2−1)

⇔−5x+7=0⇔x=7Câu trả lời: ĐK: x≠{2,13,±1}x≠{2,13,±1}

PT⇔2x+x−22x−1=63x−1PT⇔2x+x−22x−1=63x−1

⇔2(2x−1)2x2−2=63x−1⇔2(2x−1)2x2−2=63x−1

⇔2x−1x2−1=63x−1⇔2x−1x2−1=63x−1

⇔(2x−1)(3x−1)=6(x2−1)⇔(2x−1)(3x−1)=6(x2−1)

⇔−5x+7=0⇔x=7

Bài 5: Phương trình chứa ẩn ở mẫu