b)

\(3\left(2x^2-7\right)=33\)

\(\Leftrightarrow2x^2-7=11\)

\(\Leftrightarrow2x^2=18\)

\(\Leftrightarrow x^2=9\)

\(\Leftrightarrow x=\pm3\)

a) -2(2x - 8) + 3(4 - 2x) = -72 - 5(3x - 7)

=> -4x + 16 + 12 - 6x = -72 - 15x + 35

=> -10x + 28 = -37 - 15x

=> -10x + 15x = -37 - 28

=> 5x = -65

=> x = -65 : 5

=> x = -13

b) 3(2x² - 7) = 33

=> 2x² - 7 = 33 : 3

=> 2x² - 7 = 11

=> 2x² = 11 + 7

=> 2x² = 18

=> x² = 18 : 2

=> x² = 9

=> \(\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)

Vậy ...

ai giúp minh đi cần quá gấp

dễ mà

a) -2(2x - 8) + 3(4 - 2x) = -72  - 5(3x - 7)

=> -4x + 18 + 12 - 6x = -72 - 15x + 35

=> -10x + 15x = -37 - 30

=> 5x = -37

=> x = -7,4

b) 3|2x² - 7| = 33

=> |2x² - 7| = 11

=> \(\orbr{\begin{cases}2x^2-7=11\\2x^2-7=-11\end{cases}}\)

=> \(\orbr{\begin{cases}2x^2=18\\2x^2=-4\left(loại\right)\end{cases}}\)

=> \(\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)

tự làm nha !!!

chúc bạn học tốt

a) -2(2x-8)+3(4-2x)=-72-5(3x-7)

<=> -4x+16+12-6x=-72-15x+35

<=> -10x+28=-37-15x

<=> -10x+28+37+15x=0

<=> 5x+65=0

<=> 5x=-65

<=> x=-13

b) 3I2x²-7I=33

<=> I2x²-7I=11

<=> \(\orbr{\begin{cases}2x^2-7=11\\2x^2-7=-11\end{cases}\Leftrightarrow\orbr{\begin{cases}2x^2=18\\2x^2=-3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x^2=9\\x^2=\frac{-3}{2}\left(ktm\right)\end{cases}\Leftrightarrow}x=\pm3}\)

\(a,-2.\left(2x-8\right)+3.\left(4-2x\right)=-72-5\left(3x-7\right)\)

\(< =>-4x+16+12-6x=-72-15x+35\)

\(< =>-10x+15x=-72+35-16-12=-65\)

\(< =>5x=-65< =>x=\frac{-65}{5}=-12\)

\(b,3.\left|2x^2-7\right|=33\)

\(< =>\left|2x^2-7\right|=\frac{33}{3}=11\)

\(< =>\orbr{\begin{cases}2x^2-7=11\\2x^2-7=-11\end{cases}}\)

\(< =>\orbr{\begin{cases}2x^2=11+7=18\\2x^2=-11+7=-4\end{cases}}\)

\(< =>\orbr{\begin{cases}x^2=9\\x^2=-2\end{cases}< =>\orbr{\begin{cases}x=3or-3\\x=\varnothing\end{cases}}}\)

Bài 1: a) \(-2.\left(2x-8\right)+3.\left(4-2x\right)=\left(-72\right)-5.\left(3x-7\right)\)

\(-4x+16+12-6x=-72-15x+35\)

\(-4x-6x+15x=-72+35-16-12\)

\(5x=-65\)

\(x=-\frac{65}{5}\)

\(x=-13\)

b) \(3.\left|2x^2-7\right|=33\)

\(\left|2x^2-7\right|=\frac{33}{3}=11\)

\(\Rightarrow\orbr{\begin{cases}2x^2-7=11\\2x^2-7=-11\end{cases}\Rightarrow\orbr{\begin{cases}2x^2=18\\2x^2=-4\end{cases}\Rightarrow}\orbr{\begin{cases}x^2=9\\x^2=-2\left(vl\right)\end{cases}\Rightarrow}\orbr{\begin{cases}x=\pm3\\\end{cases}}}\)

Bài 2:

Ta có: \(2n+1⋮n-3\)

\(2n-6+7⋮n-3\)

\(2\left(n-3\right)+7⋮n-3\)

Vì \(2\left(n-3\right)⋮n-3\)

Để \(2\left(n-3\right)+7⋮n-3\)

Thì \(7⋮n-3\Rightarrow n-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

Vậy.....

hok tốt!!