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a, x - 3 : 2 = 5 14 : 5 12
=> x - 3 : 2 = 5 2
=> x - 3 : 2 = 25
=> x – 3 = 25
=> x = 53
b, 30 : x - 7 = 15 19 : 15 18
=> 30 : x - 7 = 15
=> x – 7 = 2
=> x = 9
c, x 70 = x
=> x 70 - x = 0
=> x ( x 69 - 1 ) = 0
=> 
d, 2 x + 1 3 = 9 . 81
=> 2 x + 1 3 = 9 3
=> 2x + 1 = 9
=> x = 4
e, 5 x + 5 x + 2 = 650
=> 5 x 1 + 5 2 = 650
=> 5 x . 26 = 650
=> 5 x = 25
=> x = 2
f, 4 x - 1 2 = 25 . 9
=> 4 x - 1 2 = 5 2 . 3 2
=> 4 x - 1 2 = 15 2
=> 4x – 1 = 15
=> x = 4
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:x=3/5-7/8=24/40-35/40=-11/40
b: =>1/3:(2x-1)=-1/6
=>2x-1=-2
=>2x=-1
=>x=-1/2
c: =>x-3/4=17/2+7/4=34/4+7/4=41/4
=>x=11
d: =>3x+2=0 hoặc 2/5x+7=0
=>x=-2/3 hoặc x=-7:2/5=-35/2
c) \(5^{x+2}+5^x=650\)
\(\Leftrightarrow 5^x(5^2+1)=650\)
\(\Leftrightarrow 5^x.26=650\)
\(\Rightarrow 5^x=25=5^2\Rightarrow x=2\)
d) \(81^x=(-3)^7\)
Ta thấy \(81^x>0, \forall x\in\mathbb{R}\)
\((-3)^7<0\)
Do đó pt đã cho vô nghiệm.
Lời giải:
a) \((2x-1)^3=(2x-1)^4\)
\(\Leftrightarrow (2x-1)^4-(2x-1)^3=0\)
\(\Leftrightarrow (2x-1)^3[(2x-1)-1]=0\)
\(\Leftrightarrow (2x-1)^3(2x-2)=0\)
\(\Rightarrow \left[\begin{matrix} 2x-1=0\\ 2x-2=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{1}{2}\\ x=1\end{matrix}\right.\)
b) \(2017^{x+2}=(2018-5^3)^{x+2}\)
\(\Rightarrow \left[\begin{matrix} x+2=0(1)\\ 2017=2018-5^3(2)\end{matrix}\right.\)
(1)\(\Rightarrow x=-2\)
(2): hiển nhiên vô lý
Vậy pt có nghiệm $x=-2$