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=>2x+2+6.5x=50
=>2x+2+30x=50
=>2x+30x=50-2
=>32x=48
=>x=48:32
=>x=3/2
\(a,12\left(x-1\right)=0\\ x-1=0\\ x=1\\ b,45+5\left(x-3\right)=70\\ 5\left(x-3\right)=25\\ x-3=5\\ x=8\\ c,3.x-18:2=12\\ 3.x-9=12\\ 3.x=21\\ x=7\)
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
1) \(xy-2x-y=-6\Rightarrow x\left(y-2\right)-y=-6\Rightarrow x\left(y-2\right)-y+2=-6+2\)
\(\Rightarrow x\left(y-2\right)-\left(y-2\right)=-4\Rightarrow\left(y-2\right)\left(x-1\right)=-4\)
\(\Rightarrow x-1\inƯ\left(-4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
Ta có bảng sau:
x - 1 | 1 | -1 | 2 | -2 | 4 | -4 |
y - 2 | -4 | 4 | -2 | 2 | -1 | 1 |
Suy ra ta có các cặp (x,y) sau:
x | 2 | 0 | 3 | -1 | 5 | -3 |
y | -2 | 6 | 0 | 4 | 1 | 3 |
2) \(|x+1|+|x-2|+|x+7|=5x-1\)
Ta thấy: \(|x+1|\ge0,|x-2|\ge0,|x+7|\ge0\) với \(\forall x\inℤ\)
Mà \(|x+1|+|x-2|+|x+7|=5x-10\Rightarrow5x-10\ge0\Rightarrow5x\ge10\Rightarrow x\ge2>0\)
\(\Rightarrow|x+1|=x+1,|x-2|=x-2,|x+7|=x+7\)
\(\Rightarrow|x+1|+|x-2|+|x+7|=x+1+x-2+x+7=5x-10\)
\(\Rightarrow\left(x+x+x\right)+\left(1-2+7\right)=5x-10\Rightarrow3x+6=5x-10\)
\(\Rightarrow3x-5x=-10-6\Rightarrow-2x=-16\Rightarrow x=\frac{-16}{-2}=8\)
a, 2+4+6+...+2x = 156
=> 2(1+2+3+....+x) = 156
=> 1+2+3+...+x = 78
=> (x+1).x : 2 = 78
=> (x+1)x = 156
=> x(x+1) = 13.12
=> x = 12
b, |x + 1| + |x - 2| + |x + 7| = 5x - 10
Vì \(\hept{\begin{cases}\left|x+1\right|\ge0\\\left|x-2\right|\ge0\\\left|x+7\right|\ge0\end{cases}}\Rightarrow\left|x+1\right|+\left|x-2\right|+\left|x+7\right|\ge0\)
\(\Rightarrow5x-10\ge0\Rightarrow x\ge0\)
\(\Rightarrow x+1+x-2+x+7=5x-10\)
\(\Rightarrow3x+6=5x-10\)
\(\Rightarrow3x-5x=-10-6\)
\(\Rightarrow-2x=-16\)
\(\Rightarrow x=8\)
1. \(3-|2x+1|=-5\)
\(\Rightarrow|2x+1|=8\)
\(\Rightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=7\\2x=-9\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{9}{2}\end{cases}}\)
Vậy \(x\in\left\{\frac{7}{2};-\frac{9}{2}\right\}\)
2.\(12+|3-x|=9\)
\(\Rightarrow|3-x|=-3\)
Mà \(|3-x|\ge0\forall x\)
\(\Rightarrow\)Vô lí
Vậy không có x
3.\(|x+9|=12+\left(-9\right)+2\)
\(\Rightarrow|x+9|=5\)
\(\Rightarrow\orbr{\begin{cases}x+9=5\\x+9=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-4\\x=-14\end{cases}}\)
Vậy \(x\in\left\{-4;-14\right\}\)
4.\(5x-16=40+x\)
\(\Rightarrow5x-x=40+16\)
\(\Rightarrow4x=56\)
\(\Rightarrow x=14\)
Vậy \(x=14\)
5.\(5x-7=-21-2x\)
\(\Rightarrow5x+2x=-21+7\)
\(\Rightarrow7x=-14\)
\(\Rightarrow x=-2\)
Vậy \(x=-2\)
6.\(\left(2x-1\right)\left(y-2\right)=12\)
Vì \(x,y\inℤ\)nên \(2x-1;y-2\inℤ\)
\(\Rightarrow2x-1;y-2\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
Ta có bảng : (em tự xét bảng nhé)