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Question

$\left(3x-1\right)\cdot\left(2x+7\right)-\left(x+1\right)\cdot\left(6x-5\right)=16$

DƯƠNG PHAN KHÁNH DƯƠNG · Accepted Answer

$\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16$

$\Leftrightarrow6x^2+19x-7-6x^2-x+5=16$

$\Leftrightarrow18x=18$

$\Leftrightarrow x=1$Câu trả lời: $\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16$

$\Leftrightarrow6x^2+19x-7-6x^2-x+5=16$

$\Leftrightarrow18x=18$

$\Leftrightarrow x=1$

Hắc Hường · Answer

Giải:

$\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16$

$\Leftrightarrow6x^2-2x+21x-7-\left(6x^2+6x-5x-5\right)=16$

$\Leftrightarrow6x^2+19x-7-\left(6x^2+x-5\right)=16$

$\Leftrightarrow6x^2+19x-7-6x^2-x+5=16$

$\Leftrightarrow18x-2=16$

$\Leftrightarrow18x=18$

$\Leftrightarrow x=1$

Vậy ...Câu trả lời: Giải:

$\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16$

$\Leftrightarrow6x^2-2x+21x-7-\left(6x^2+6x-5x-5\right)=16$

$\Leftrightarrow6x^2+19x-7-\left(6x^2+x-5\right)=16$

$\Leftrightarrow6x^2+19x-7-6x^2-x+5=16$

$\Leftrightarrow18x-2=16$

$\Leftrightarrow18x=18$

$\Leftrightarrow x=1$

Vậy ...

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