\(\frac{30+40+46+51}{4}=41,75\)

có \(\infty\) số

\(3\left(x+1\right)\left(x+3\right)-2\left(x-1\right)\left(x+2\right)-x\left(x-1\right)\)
\(=\left(3x+3\right)\left(x-3\right)-\left(2x+2\right)\left(x+2\right)-x^2+x\)
\(=3x^2-9x-3x+9-\left(2x^2+4x+2x+4\right)-x^2+x\)
\(=3x^2-9x-3x+9-2x^2-4x-2x-4-x^2+x\)
\(=-17x+5\)
\(=5-17x\)
Sai thì thông cảm cho mk nhé

\(36+42-26+4653-65-42-15+46\)
\(=\left(42-42\right)-\left(26-46\right)-\left(65+15\right)+36+4653\)
\(=0+20-80+36+4653\)
\(=20-80+36+4653\)
\(=\left(-60\right)+36+4653\)
\(=\left(-24\right)+4653\)
\(=4629\)

tất nhiên là hơn thế nữa hihi=))

e nha

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9389

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