5 . 2x + 4 . 2x = 72
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-2(2x-8)+3(4-2x)=-72-5(3x-7)
-2.2x-(-2).8+3.4-3.2x=-72-5.3x+5.7
-4x+16+12-6x=-72-15x+35
-4x-6x+15x=-72+35-16-12
5x=-65
x=-65:5
x=-13
![](https://rs.olm.vn/images/avt/0.png?1311)
\(-2\left(2x-8\right)+3\left(4-2x\right)=-72-5\left(3x-7\right)\)
\(\Leftrightarrow-4x-16+12-6x=-72-15x+35\)
\(\Leftrightarrow-10x-4=-37-15x\)
\(\Leftrightarrow-10x+15x=-37+4\)
\(\Leftrightarrow5x=-33\)
\(\Leftrightarrow x=\frac{-33}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài giải
a, \(-2\left(2x-8\right)+3\left(4-2x\right)=-72-5\left(3x-7\right)\)
\(-4x+8+12-6x=-72-15x+7\)
\(-10x+20=-65-15x\)
\(-10x+15x=-65-20\)
\(5x=-85\)
\(x=-85\text{ : }5\)
\(x=-17\)
b, \(3\left|2x^2-7\right|=33\)
\(\left|2x^2-7\right|=33\text{ : }3\)
\(\left|2x^2-7\right|=11\)
\(\Rightarrow\orbr{\begin{cases}2x^2-7=-11\\2x^2-7=11\end{cases}}\Rightarrow\orbr{\begin{cases}2x^2=-4\text{ ( loại ) }\\2x^2=18\end{cases}}\Rightarrow\text{ }x^2=9\text{ }\Rightarrow\text{ }x=\pm3\)
\(\Rightarrow\text{ }x=\pm3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
=> \(\orbr{\begin{cases}2x^2-7=11\\2x^2-7=-11\end{cases}}\)
=> \(\orbr{\begin{cases}2x^2=18\\2x^2=-4\left(loại\right)\end{cases}}\)
=> \(\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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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 ...
![](https://rs.olm.vn/images/avt/0.png?1311)
1e) Để \(\frac{2x-1}{x-3}\) nguyên thì \(2x-1⋮x-3\)
\(\Leftrightarrow2x-6+5⋮x-3\)
\(\Leftrightarrow2\left(x-3\right)+5⋮x-3\)
Do \(2\left(x-3\right)⋮x-3\) \(\Rightarrow5⋮x-3\)
\(\Rightarrow x-3\in\left\{-5;-1;1;5\right\}\)
\(\Leftrightarrow x\in\left\{-2;2;4;8\right\}\)
Vậy:...................
5 . 2x + 4 . 2x = 72
( 5 + 4 ) . 2x = 72
9 .2x = 72
2x = 72 : 9
2x = 8
2x = 23
\(\Rightarrow\)x = 3
Vậy x = 3
5 . 2x + 4 . 2x = 72
<=> 2x . (5 + 4) = 72
<=> 2x . 9 = 72
<=> 2x = \(\frac{72}{9}\)
<=> 2x = 8
<=> 2x = 23
=> x = 3.
Vậy x = 3.