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126-2.(x-1)=20 120+3.(x-3)=180
2.(x-1)=126-20 3.(x-3)=180-120
2.(x-1)=106 3.(x-3)=60
x-1=106:2 x-3=60:3
x-1=53 x-3=20
x=53+1 x=20+3
x=54 x=23
5)
để \(\frac{5x-3}{x+1}\)là số nguyên
\(5x-3⋮x+1\)
\(x+1⋮x+1\)
\(\Rightarrow5\left(x+1\right)⋮x+1\)
\(5x-3-\left(5x-5\right)⋮x+1\)
\(-2⋮x+1\)
\(\Rightarrow x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
| x+1 | 1 | -1 | 2 | -2 |
| x | 0 | -2 | 1 | -3 |
Vậy \(x\in\left\{0;-2;1;-3\right\}\)
\(a,-12\left(x-5\right)+7\left(3-x\right)=5\)
\(-12x+60+21-7x=5\)
\(-12x-7x=5-60-21\)
\(-19x=-76\Leftrightarrow x=4\)
\(b,30\left(x+2\right)-6\left(x-5\right)-24x=100\)
\(30x+60-6x+30-24x=100\)
\(30x-6x-24x=100-60-30\)
\(0x=10\left(vl\right)\)
Vậy pt vô nghiệm
\(a,\left(-5\right).\left|x\right|=-75\)
\(\left|x\right|=\frac{-75}{-5}=15\)
\(\Rightarrow\orbr{\begin{cases}x=15\\x=-15\end{cases}}\)
Vậy....
\(b,\left(-6\right)^3.x^2=-1944\)
\(-216.x^2=-1944\)
\(x^2=9\)
\(\Rightarrow x=\pm3\)
Vậy....
\(d,\left|9-x\right|=-7+64\)
\(\left|9-x\right|=57\)
\(\Rightarrow\orbr{\begin{cases}9-x=57\\9-x=-57\end{cases}\Rightarrow\orbr{\begin{cases}x=-48\\x=66\end{cases}}}\)
Vậy...
\(e,\left|x+101\right|-\left(-16\right)=\left(-43\right).\left(-5\right)\)
\(\left|x+101\right|+16=215\)
\(\left|x+101\right|=199\)
\(\Rightarrow\orbr{\begin{cases}x+101=199\\x+101=-199\end{cases}\Rightarrow\orbr{\begin{cases}x=98\\x=-300\end{cases}}}\)
Vậy..
hok tốt!!
a,\(\left(-5\right).\left|x\right|=-75\)
\(=>\left|x\right|=-75:\left(-5\right)=15\)
\(=>\orbr{\begin{cases}x=15\\x=-15\end{cases}}\)
b,\(\left(-6\right)^3.x^2=-1944\)
\(=>\frac{1944}{216}=x^2\)
\(=>x=\sqrt{\frac{1944}{216}}=3\)
1, \(\left(3x-6\right)\left(2x-10\right)=0\)
\(\Leftrightarrow3x-6=0or2x-10=0\Leftrightarrow x=3orx=5\)
or là từ '' hoặc ''
2, \(7\left(x+5\right)+10=5x-11\)
\(\Leftrightarrow7x+35+10=5x-11\)
\(\Leftrightarrow7x-5x=-11-10-35\)
\(\Leftrightarrow2x=-56\Leftrightarrow x=-28\)
1) \(\Rightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)
2) \(\Rightarrow5\left(x-2\right).3\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
3) \(\Rightarrow2\left(x-4\right)\left(x-7\right)=0\Rightarrow\left[{}\begin{matrix}x=4\\x=7\end{matrix}\right.\)
a)
x+3=0⇔x=-3
x-5=0⇔x=5
b
5x-10=0⇔x=2
3x+6=0⇔x=-2
c)
x-4=0⇔x=4
2x-14=0⇔x=7