3) 3(2 − 𝑥) − 14 = 21
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a: \(x\in\left\{25;30;35\right\}\)
b: \(x\in\left\{24;32;40;48;56;64\right\}\)
c: \(x\in\left\{3;4;6\right\}\)
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\(a,\frac{33\cdot21\cdot18}{9\cdot22\cdot14}=\frac{11\cdot3\cdot7\cdot3\cdot9\cdot2}{9\cdot11\cdot2\cdot7\cdot2}=\frac{9}{2}\)
\(b,\frac{36\cdot7\cdot12}{6\cdot49\cdot4}=\frac{9\cdot4\cdot7\cdot6\cdot2}{6\cdot7\cdot7\cdot4}=\frac{18}{7}\)
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\(\left(x^3-x^2+3x-3\right):\left(x^2+3\right)\\ =\left[\left(x^3-x^2\right)+\left(3x-3\right)\right]:\left(x^2+3\right)\\ =\left[x^2\left(x-1\right)+3\left(x-1\right)\right]:\left(x^2+3\right)\\ =\left[\left(x^2+3\right)\left(x-1\right)\right]:\left(x^2+3\right)\\ =x-1\)
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Bài 1:
a. $x(x^2-5)=x^3-5x$
b. $3xy(x^2-2x^2y+3)=3x^3y-6x^3y^2+9xy$
c. $(2x-6)(3x+6)=6x^2+12x-18x-36=6x^2-6x-36$
d.
$(x+3y)(x^2-xy)=x^3-x^2y+3x^2y-3xy^2=x^3+2x^2y-3xy^2$
Bài 2:
a.
\((2x+5)(2x-5)=(2x)^2-5^2=4x^2-25\)
b.
\((x-3)^2=x^2-6x+9\)
c.
\((4+3x)^2=9x^2+24x+16\)
d.
\((x-2y)^3=x^3-6x^2y+12xy^2-8y^3\)
e.
\((5x+3y)^3=(5x)^3+3.(5x)^2.3y+3.5x(3y)^2+(3y)^3\)
\(=125x^3+225x^2y+135xy^2+27y^3\)
f.
\((5-x)(25+5x+x^2)=5^3-x^3=125-x^3\)
3.(2-x)-14=21
(2-x)-14=21:3
(2-x)-14=7
2-x=7-14
2-x=-7
x=7+2
x=9
\(3\left(2-x\right)-14=21=>3\left(2-x\right)=35=>2-x=\frac{35}{3}=>x=2-\frac{35}{3}=\frac{-29}{3}\)