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Ta co´ :
a, 3 . ( x + 2 ) - 6 . ( x - 5 ) = 2 . ( 5 - 2x )
. 3x + 6 - 6x + 30 = 10 - 4x
3x - 6x + 4x = 10 - 6 - 30
x = - 26
b, ( - 2x ) . ( - 4x ) + 28 = 100
8x + 28 = 100
8x = 100 - 28
8x = 72
x = 72 : 8
x = 9
c, 5x . ( - x )2 + 1 = 6
5x . ( - x ) . ( - x ) + 1 = 6
5x . [ ( - x ) . ( - x ) ] + 1 = 6
5x . 2x + 1 = 6
5x . 2x = 6 - 1
5x . 2x = 5
Kẻ bảng . Co´ 4 trường hợp laˋ 5x = 1 ; 2x = 5
5x = 5. ; 2x = 1
5x = - 1 ; 2x = - 5
5x = - 5 ; 2x = - 1
4x-2x+x-27:9=33
2x+x-27:9=33
3x-27:9=33
3x-27=33×9=297
3x=297+27=324
x=324÷3=108
1.
a, \(x-14=3x+18\)
\(\Rightarrow x-3x=18+14\)
\(\Rightarrow-2x=32\Rightarrow x=\frac{32}{-2}=-16\)
b, \(\left(x+7\right).\left(x-9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+7=0\\x-9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-7\\x=9\end{cases}}}\)
c, \(\left|2x-5\right|-7=22\)
\(\Rightarrow\left|2x-5\right|=22+7\)
\(\Rightarrow\left|2x-5\right|=29\)
\(\Rightarrow\orbr{\begin{cases}2x+5=29\\2x-5=29\end{cases}}\Rightarrow\orbr{\begin{cases}2x=24\\2x=34\end{cases}\Rightarrow}\orbr{\begin{cases}x=12\\x=17\end{cases}}\)
d, \(\left(\left|2x\right|-5\right)-7=22\)
\(\Rightarrow\left(\left|2x\right|-5\right)=29\)
\(\Rightarrow\left|2x\right|=29+5\Rightarrow\left|2x\right|=34\Rightarrow x=\pm17\)
e, \(\left|x+3\right|+\left|x+9\right|+\left|x+5\right|=4x\)
Vì \(\left|x+3\right|\ge0;\left|x+9\right|\ge0;\left|x+5\right|\ge0;4x\ge0\)
Nên \(\left|x+3\right|+\left|x+9\right|+\left|x+5\right|=4x\ge0\)
\(\Rightarrow\left|x+3\right|>0\Rightarrow\left|x+3\right|=x+3\)
\(\left|x+9\right|>0\Rightarrow\left|x+9\right|=x+9\)
\(\left|x+5\right|>0\Rightarrow\left|x+5\right|=x+5\)
Ta có :
\(x+3+x+9+x+5=4x\)
\(\Rightarrow3x+\left(3+9+5\right)=4x\)
\(\Rightarrow4x-3x=17\)
\(\Rightarrow x=17\)
2. a , b sai đề bn
c, \(\left(5x+1\right).\left(y-1\right)=4\)
\(\Rightarrow\left(5x+1\right).\left(y-1\right)\inƯ\left(4\right)\)
\(\text{ }Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
Ta có bảng sau :
5x+1 | 1 | -1 | 2 | -2 | 4 | -4 |
y-1 | -4 | 4 | -2 | 2 | -1 | 1 |
x | 0 | -2/5 | 1/5 | -3/5 | 3/5 | -1 |
y | -3 | 5 | -1 | 3 | 0 | 2 |
d, \(5xy-5x+y=5\)
\(\Rightarrow\left(5xy-5x\right)+y=5\)
\(\Rightarrow5x.\left(y-1\right)+y=5\)
\(\Rightarrow\left(5x+1\right).\left(y-1\right)=4\)
\(\Rightarrow\left(5x+1\right).\left(y-1\right)\inƯ\left(4\right)\)
\(Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
Ta có bảng sau :
5x+1 | 1 | -1 | 2 | -2 | 4 | -4 |
y-1 | -4 | 4 | -2 | 2 | -1 | 1 |
x | 0 | -2 | 1/5 | -3/5 | 3/5 | -1 |
y | -3 | 5 | -1 | 3 | 0 | 2 |
a, 3(x+2)-6(x-5)=2(5-2x)
3x+6-6x+30=10-4x
3x+6-6x+30-10+4x=0
3x-6x+4x+6+30-10=0
x+26=0
x= -26
b, (-2x)(-4x)+28=100
8x2 + 28=100
8x2 = 72
x2 = 9
=> x=3 hoặc x= -3
a/ 3(x+2) - 6(x-5) = 2(5-2x)
<-> 3x+6-6x+30 = 10-4x
<-> x = -26
b/ (-2x)(-4x) + 28 = 100
<-> 8x = 72
<-> x = 9