7326+3627:9

=7326+403

=7729

Đặt \(2x^2+3x+1=y\).Ta có:

\(\left(y-2\right)^2-5\left(y+2\right)+24=0\)

\(\Leftrightarrow y^2-4y+4-5y-10+24=0\)

\(\Leftrightarrow y^2-9y+18=0\)

\(\Leftrightarrow\left(y-3\right)\left(y-6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}y=3\\y=6\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x^2+3x+1=3\\2x^2+3x+1=6\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(2x-1\right)\left(x+2\right)=0\\\left(2x+5\right)\left(x-1\right)=0\end{cases}}\)

Vậy PT có 4 nghiệm là:\(\frac{1}{2}\)\(,\)\(-2,-\frac{5}{2},1\)

1) 3x = 45 + 15 = 60

x = 60 : 3 = 20

2) 5x = 50 - 35 = 15

x = 15 : 5 = 3

3) (2x - 5) + 17 = 6

2x - 5 = 6 - 17

2x - 5 = -11

2x = -11 + 5

2x = -6

x = -6 : 2 = -3

4) 10 - 2(4 -3x) = -4

2(4 - 3x) = -4 - 10 = -14

4 - 3x = -14 : 2 = -7

3x = -7 - 4 = -18

x = -18 : 3 = -6

\(\left\{{}\begin{matrix}\left\{{}\begin{matrix}3x-15=45\Leftrightarrow3x=60\Leftrightarrow x=20\\35-5x=50\Leftrightarrow5x=-15\Leftrightarrow x=-3\end{matrix}\right.\\\left\{{}\begin{matrix}\left(2x-5\right)+17=6\Leftrightarrow2x+5=-11\Leftrightarrow2x=-16\Leftrightarrow x=-8\\10-2\left(4-3x\right)=-4\Leftrightarrow8-6x=14\Leftrightarrow6x=-6\Leftrightarrow x=-1\end{matrix}\right.\\\left\{{}\begin{matrix}-12+3\left(-x+7\right)=-18\Leftrightarrow-3x+21=-6\Leftrightarrow-3x=-27\Leftrightarrow x=9\\24:\left(3x-2\right)=-3\Leftrightarrow3x-2=-8\Leftrightarrow3x=-6\Leftrightarrow x=-2\end{matrix}\right.\\-45:5\left(-3-2x\right)=3\Leftrightarrow-15-10x=-15\Leftrightarrow10x=0\Leftrightarrow x=0\end{matrix}\right.\)

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