C=2^0-2^1+2^2+...-2^99+2^100
D=5-5^2+5^3-...+5^99-5^100
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\(1-\left\{\dfrac{49}{9}+x-\dfrac{113}{18}\right\}:\dfrac{63}{4}=0\)
\(\Leftrightarrow\left\{\dfrac{49}{9}+x-\dfrac{113}{18}\right\}:\dfrac{63}{4}=1\)
\(\Leftrightarrow\dfrac{49}{9}+x-\dfrac{113}{18}=\dfrac{63}{4}\)
\(\Leftrightarrow x-\dfrac{5}{6}=\dfrac{63}{4}\)
\(\Leftrightarrow x=\dfrac{199}{12}\)
`1 - (49/9 + x - 113/18) : 63/4 = 0`
`(49/9 + x - 113/18) : 63/4 = 1-0`
`(49/9 + x - 113/18) : 63/4 = 1`
`49/9 + x - 113/18 = 1 xx 63/4`
`49/9 + x - 113/18 = 63/4`
`(49/9 - 113/18) + x =63/4`
`-5/6 + x =63/4`
`x-5/6 = 63/4`
`x=63/4 + 5/6`
`x=189/12 + 10/12`
`x=199/12`
Vậy `x=199/12`
\(\left[\left(8x-12\right):4\right].27=729\)
\(\Leftrightarrow\left(8x-12\right):4=27\)
\(\Leftrightarrow8x-12=108\)
\(\Leftrightarrow8x=120\)
\(\Leftrightarrow x=15\)
\(\dfrac{30\left(x-5\right)}{100}=\dfrac{20x}{100}+5\\ =>\dfrac{30\left(x-5\right)}{100}=\dfrac{20x+500}{100}\\ =>30\left(x-5\right)=20x+500\\ =>30x-150=20x+500\\ =>10x=650\\ =>x=65\)
\(4\left(x+1\right)-2x=-2\left(5-x\right)\)
\(\Leftrightarrow4x+4-2x=-10+2x\)
\(\Leftrightarrow2x+4-2x=-10\)
\(\Leftrightarrow0x+4=-10\) ( vô lí )
Vậy không có x thỏa mãn yêu cầu đề bài
a) \(-x^2+4x+4\)
\(=-\left(x^2-4x-4\right)\)
\(=-\left(x^2-4x+4-8\right)\)
\(=-\left(x-2\right)^2+8\le8\)
Dấu bằng xảy ra \(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
b) \(9x^2-6x+5\)
\(=\left(3x\right)^2-2.3x+1+4\)
\(=\left(3x+1\right)^2+4\ge4\)
Dấu bằng xảy ra \(\Leftrightarrow3x+1=0\Leftrightarrow x=-\dfrac{1}{3}\)
a) 13,45 + 7,98 + 8,55
= (13,45 + 8,55 ) + 7 ,98
= 22 + 7 ,98
= 29,98
b) 9,72 + 8,38 +. 3,62
= 9,72 + 12
=21,72
c) 7,29 + 9,11 + 0,89
= 7,29 + 10
= 17,29