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C =[125.(73-75)]:(-25)2
=125.(-2):(-25).2
=-250:(-50)
=5
E=125.9.(-4).(-8).25.7
=125.(-8).25.(-4).7.9
=-1000.(-100).63
=6300000
C = ( 125 . 73 - 125 . 75 ) : ( -50)
C = 125 .( -2 ) : (-50)
C = -250 : -50
C = 50
G = ( -3 )2+( -5)3: l -5 l
G = 9 + ( -125) : 5
G = 9 + ( -25 )
G= ( -16 )
E = 125 .9.*( -4 ) . ( -8) .25.7
E = 25.5 .( -4 ) .9 .2. ( -4 ) .25 .7
E =( -100 ).10.( -100).63
E = 10000.630
E = 630000
**** mha
\(9-25=\left(-7-x\right)-\left(25-7\right)\)
\(-7-x-18=-16\)
\(-x=-16+18+7\)
\(-x=9\)
\(x=-9\)
Vậy \(x=-9\).
9-25=(-7-x)-(25-7)
Ta có (-7-x)-(25-7)=9-25
(-7-x)-18=-16
-7-x=-16+18
-7-x=2
x=(-7)-2
x=-9
Vậy x=-9
\(125^3.25^4=25^9.25^4=25^{13}\)
\(16^2.64^5=4^4.4^{15}=4^{19}\)
a) \(-65-\left(x+15\right)+105=0\)
\(-65-\left(x+15\right)=-105\)
\(x+15=-65-\left(-105\right)\)
\(x+15=40\)
\(x=25\)
b) |-6| + (-9) - (x+1) = 7
-3- (x+1) =7
x+1 = -10
x = -11
c)\(\left|5-x\right|+\left(-25+7\right)=-3-\left(-10\right)\)
\(\left|5-x\right|+\left(-18\right)=7\)
\(\left|5-x\right|=25\)
\(\Rightarrow\orbr{\begin{cases}5-x=25\\5-x=-25\end{cases}\Rightarrow\orbr{\begin{cases}x=-20\\x=30\end{cases}}}\)
d) \(28-\left|x+6\right|+\left(-2\right)=0\)
\(28-\left|x+6\right|=2\)
\(\left|x+6\right|=26\)
\(\Rightarrow\orbr{\begin{cases}x+6=26\\x+6=-26\end{cases}\Rightarrow\orbr{\begin{cases}x=20\\x=-32\end{cases}}}\)
e) \(x-\left(13-15\right)=5+\left(10-x\right)-\left(-1\right)\)
\(x+2=5+10-x+1\)
\(x+2=16-x\)
\(x+x=16-2\)
\(2x=14\)
\(x=7\)
f) \(-120-\left(x-5\right)=125\)
\(x-5=-120-125\)
\(x-5=-245\)
\(x=-240\)
g) \(10-\left(-5+2\right)+\left(-9\right)=\left(-20+7\right)-x\)
\(-16=-13-x\)
\(x=-13+16\)
\(x=3\)
\(a,\)\(9-25=\left(7-x\right)-\left(25+7\right)\)
\(-16=\left(7-x\right)-32\)
\(7-x=-16+32\)
\(7-x=16\)
\(x=7-16=-9\)
\(b,-7624+\left(1543+7624\right)-x=25\)
\(-7624+9167-x=25\)
\(1543-x=25\)
\(x=1543-25\)
\(x=1518\)
\(c,\left(27-514\right)-\left(486-73\right)+x=7\)
\(-487-413+x=7\)
\(-900+x=7\)
\(x=7+900\)
\(x=907\)
a)\(-17+\left|5-x\right|=10\)
\(\Leftrightarrow\left|5-x\right|=10-\left(-17\right)\)
\(\Leftrightarrow\left|5-x\right|=10+17\)
\(\Leftrightarrow\left|5-x\right|=27\)
\(\Leftrightarrow\orbr{\begin{cases}5-x=27\\5-x=-27\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-22\\x=32\end{cases}}\)
b) \(45-5\left|12-x\right|=125\div\left(-25\right)\)
\(\Leftrightarrow45-5\left|12-x\right|=-5\)
\(\Leftrightarrow5\left|12-x\right|=45-\left(-5\right)\)
\(\Leftrightarrow5\left|12-x\right|=45+5\)
\(\Leftrightarrow5\left|12-x\right|=50\)
\(\Leftrightarrow\left|12-x\right|=10\)
\(\Leftrightarrow\orbr{\begin{cases}12-x=10\\12-x=-10\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=22\end{cases}}\)
c) \(2< \left|3-x\right|\le5\)
\(\Leftrightarrow\left|3-x\right|\in\left\{3;4;5\right\}\)
* \(\left|3-x\right|=3\Leftrightarrow\orbr{\begin{cases}3-x=3\\3-x=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}}\)
* \(\left|3-x\right|=4\Leftrightarrow\orbr{\begin{cases}3-x=4\\3-x=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=7\end{cases}}}\)
* \(\left|3-x\right|=5\Leftrightarrow\orbr{\begin{cases}3-x=5\\3-x=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=8\end{cases}}}\)
d) \(\left|x+4\right|< 3\)
mà \(\left|x+4\right|\ge0\)
\(\Rightarrow\left|x+4\right|\in\left\{0;1;2\right\}\)
* \(\left|x+4\right|=0\Leftrightarrow x+4=0\Leftrightarrow x=-4\)
* \(\left|x+4\right|=1\Leftrightarrow\orbr{\begin{cases}x+4=1\\x+4=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=-5\end{cases}}}\)
* \(\left|x+4\right|=2\Leftrightarrow\orbr{\begin{cases}x+4=2\\x+4=-2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-6\end{cases}}}\)
a) \(2.\left(x+\frac{2}{5}\right)+1\frac{1}{4}=\frac{11}{20}\)
\(2.\left(x+\frac{2}{5}\right)+\frac{5}{4}=\frac{11}{20}\)
\(2.\left(x+\frac{2}{5}\right)=\frac{-7}{10}\)
\(x+\frac{2}{5}=\frac{-7}{20}\)
\(x=\frac{-13}{20}\)
Vậy \(x=\frac{-13}{20}\)
b)\(x-1\frac{1}{8}-\frac{2}{3}x-\frac{5}{6}x=75\%\)
\(\left(x-\frac{2}{3}x-\frac{5}{6}x\right)-\frac{9}{8}=\frac{3}{4}\)
\(\frac{-1}{2}x-\frac{9}{8}=\frac{3}{4}\)
\(\frac{-1}{2}x=\frac{15}{8}\)
\(x=\frac{-15}{4}\)
Vậy \(x=\frac{-15}{4}\)
Ta có: 25(7-x)=-125
=>\(7-x=-\dfrac{125}{25}=-5\)
=>x=7+5=12
\(25\cdot\left(7-x\right)=-125\)
\(7-x=\left(-125\right)\div25\)
\(7-x=-5\)
\(x=7-\left(-5\right)\)
\(x=7+5\)
\(x=12\)
Vậy \(x=12\)