\(\left(x-2\right)^3-\left(x-2x^2+2x+4\right)+6\left(x-2\right)\left(x+2\right)=60\)

\(\Rightarrow x^3-6x^2+12x-8-x+2x^2-2x-4+6\left(x^2-4\right)=60\)

\(\Rightarrow x^3-6x^2+12x-8-x+2x^2-2x-4+6x^2-24=60\)

\(\Rightarrow x^3+2x^2-7x-36=60\)

\(\Rightarrow x^3+2x^2-7x-96=0\)

Sai đề không ???

k sai đề nha

a) \(\dfrac{6}{13}:\left(\dfrac{1}{2}-x\right)=\dfrac{15}{39}\)

\(\dfrac{1}{2}-x=\dfrac{6}{13}:\dfrac{15}{39}\)

\(\dfrac{1}{2}-x=\dfrac{6}{5}\)

\(x=\dfrac{1}{2}-\dfrac{6}{5}\)

\(x=-\dfrac{7}{10}\)

b) \(3\times\left(\dfrac{x}{4}+\dfrac{x}{28}+\dfrac{x}{70}+\dfrac{x}{130}\right)=\dfrac{60}{13}\)

\(3\times x\times\left(\dfrac{1}{4}+\dfrac{1}{28}+\dfrac{1}{70}+\dfrac{1}{130}\right)=\dfrac{60}{13}\)

\(x\times\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+\dfrac{3}{7\times13}\right)=\dfrac{60}{13}\)

\(x\times\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}\right)=\dfrac{60}{13}\)

\(x\times\left(1-\dfrac{1}{13}\right)=\dfrac{60}{13}\)

\(x\times\dfrac{12}{13}=\dfrac{60}{13}\)

\(x=\dfrac{60}{13}:\dfrac{12}{13}\)

\(x=5\)

2.

a, x-13=-46

=>x=(-46)+13

=>x=33

b, 4x-6=22

=>4x=22+6

=>4x=28

=>x=28:4

=>x=7

3.

a, 32=2⁵

48=2⁴.3

=>ƯCLN(32,48)=2⁴=16

16=2⁴

72=2³.3²

=>ƯCLN(16,72)=2³=8

b,

24=2³.3

60=2².3.5

=>BCNN(24,60)=2³.3.5=120

72=2³.3²

180=2².3².5

=>BCNN(72,180)=2³.3².5=360

a) x - 13  =  -   46 

    x         =  - ( 46 + 13 )                                

    x         =  - 59   

Chắc vậy

2a/ -33

2b/  7

3a/ 16

       8

3b/ 120

       360

Chọn A

Bài 2 :

a ) l x l < 3

=> l x l thuộc { 0 ; 1 ; 2 }

=> x thuộc { - 2 ; - 1 ; 0 ; 1 ; 2 }

Vậy x thuộc  { - 2 ; - 1 ; 0 ; 1 ; 2 }

\(a,\dfrac{12}{5}=\dfrac{x}{1,5}\Rightarrow x=\dfrac{12\cdot1,5}{5}=3,6\\ b,\dfrac{x}{5}=\dfrac{3}{20}\Rightarrow x=\dfrac{5\cdot3}{20}=\dfrac{3}{4}\\ c,\dfrac{4}{x}=\dfrac{10}{9}\Rightarrow x=\dfrac{4\cdot9}{10}=\dfrac{18}{5}\\ d,\Rightarrow\dfrac{x}{15}=\dfrac{60}{x}\Rightarrow x^2=60\cdot15=900\Rightarrow\left[{}\begin{matrix}x=30\\x=-30\end{matrix}\right.\\ 2,\)

a, Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{x+y-z}{3+5-6}=\dfrac{8}{2}=4\\ \Rightarrow\left\{{}\begin{matrix}x=12\\y=20\\z=24\end{matrix}\right.\)

b, Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{x-y+z}{3-5+6}=\dfrac{-4}{4}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-3\\y=-5\\z=-6\end{matrix}\right.\)

c, Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{2y}{10}=\dfrac{3z}{18}=\dfrac{x-2y+3z}{3-10+18}=\dfrac{-33}{11}=-3\\ \Rightarrow\left\{{}\begin{matrix}x=-9\\y=-15\\z=-18\end{matrix}\right.\)

d, Đặt \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=k\Rightarrow x=3k;y=5k;z=6k\)

\(x^2-4y^2+2z^2=-475\\ \Rightarrow9k^2-100k^2+72z^2=-475\\ \Rightarrow-19k^2=-475\\ \Rightarrow k^2=25\Rightarrow\left[{}\begin{matrix}k=5\\k=-5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=15;y=25;z=30\\x=-15;y=-25;z=-30\end{matrix}\right.\)