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C = (-15 + |x|) + (25 - |-x|)
= -15 + |x| + 25 - |-x|
= (25 - 15) + |x| - |-x|
= 10 + |x| - |-x|
Ta có: |x| = |-x| => |x| - |-x| = 0
=> C = 10 + |x|
C = (-15 + IxI ) + ( 25 - I-xI)
C = (-15 + x ) + ( 25 - x )
C = -15 + x + 25 - x
C = ( -15 + 25 ) + ( x - x )
C = 10 + 0
C = 10
Vậy C = 10
a)x+(-81)-(11-81)
=x-81-11+81
=x-11
b)(-1-x+2)+1
=-1-x+2+1
=-x+2
c)15-(11-x)+(11-15)
=15-11+x+11-15
=x
d)15-(15-x+202)+202
=15-15+x-202+202
=x
\(4^{18}.8^{15}=\left(2^2\right)^{18}.\left(2^3\right)^{15}\)
\(=2^{36}.2^{45}\)
\(=2^{81}\)
\(4^{15}.5^{30}=\left(2^2\right)^{15}.5^{30}\)
\(=2^{30}.5^{30}\)
\(=\left(2.5\right)^{30}\)
\(=10^{30}\)
\(\frac{3.72^2.54^2}{108^4}=\frac{3.\left(3^2.2^3\right)^2.\left(3^3.2\right)^2}{\left(3^3.2^2\right)^4}\)
\(=\frac{3.3^4.2^6.3^6.2^2}{3^{12}.2^8}\)
\(=\frac{3^{11}.2^8}{3^{12}.2^8}\)
\(=\frac{1}{3}\)
a) \(x+\left(-81\right)-\left(11-81\right)=x-81-11+81=x-11\)
b) \(\left(-1-x+2\right)+1=\left(-1\right)-x+2+1=-x+2=2-x\)
c) \(15-\left(11-x\right)+\left(11-15\right)=15-11-x+11-15\)
\(=\left(15-15\right)+\left(11-11\right)+x=0+0+x=x\)
d) \(15-\left(15-x+202\right)=15-15+x-202=x-202\)
a)x-81-11+81=x-11
b)(-1-x+2)+1=-1-x+2+1=2-x
c)15-(11-x)+(11-15)=15-11+x+11-15=x
d)15-(15-x+202)+202=15-15+x-202+202=x