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\(B=3+3^3+3^5+...+3^{101}\)
\(3^2.B=3^3+3^5+3^7+...+3^{103}\)
\(\left(3^2-1\right)B=\left(3^3+3^5+3^7+...+3^{103}\right)-\left(3+3^3+3^5+...+3^{101}\right)\)
\(8B=3^{103}-3\)
\(B=\frac{3^{103}-3}{8}\)
a) A + x2 - 4xy2 + 2xz - 3y2 = 0
=> A = -x2 + 4xy2 - 2xz + 3y2
b) B + 5x2 - 2xy = 6x2 + 9xy - y2
=> B = 6x2 + 9xy - y2 - 5x2 + 2xy= x2 + 11xy - y2
c) 3xy - 4y2 - A = x2 - 7xy + 8y2
=> A = 3xy - 4y2 - x2 + 7xy - 8y2 = -12y2 + 10xy - x2
Trả lời:
a, A + ( x2 - 4xy2 + 2xz - 3y2 ) = 0
=> A = - ( x2 - 4xy2 + 2xz - 3y2 ) = - x2 + 4xy2 - 2xz + 3y2
b, B + ( 5x2 - 2xy ) = 6x2 + 9xy - y2
=> B = 6x2 + 9xy - y2 - ( 5x2 - 2xy ) = 6x2 + 9xy - y2 - 5x2 + 2xy = x2 + 11xy - y2
c, ( 3xy - 4y2 ) - A = x2 - 7xy + 8y2
=> A = 3xy - 4y2 - ( x2 - 7xy + 8y2 ) = 3xy - 4y2 - x2 + 7xy - 8y2 = 10xy - 12y2 - x2
d, B + ( 4x2y + 5y2 - 3xz + z2 ) = x2 + 11xy - y2 + 4x2y + 5y2 - 3xz + z2 = x2 + 11xy + 4y2 + 4x2y - 3xz + z2
Bài 2:
a) 2(5x -8) –( x2 +10x) = -17
=> 10x - 16 – x2 - 10x = -17
=> - 16 – x2 = -17
=> x2 = - 16 +17
=> x2 = 1
=> \(\orbr{\begin{cases}x=-1\\x=1\end{cases}}\)
b) x2 -3x - 4 = 0
=> x2 - 4x + x - 4 = 0
=> ( x2 - 4x ) + ( x - 4 ) = 0
=> x ( x - 4 ) + ( x - 4 ) = 0
=>( x - 4 )( x + 1 ) = 0
=> \(\orbr{\begin{cases}x-4=0\\x+1=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=4\\x=-1\end{cases}}\)
c) x(2x -1) + 2 x2 = 3
=> 2x2 - x + 2 x2 = 3
=> 4x2 - x - 3 = 0
=> 4x2 - 4x +3x - 3 = 0
=> ( 4x2 - 4x ) + ( 3x - 3 ) = 0
=> 4x( x - 1 ) + 3( x - 1 ) = 0
=> ( x - 1 )( 4x + 3) = 0
=> \(\orbr{\begin{cases}x-1=0\\4x+3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=1\\x=-\frac{3}{4}\end{cases}}\)
a) Xét tam giác vuông HAM và tam giác vuông KCM có :
\(\hept{\begin{cases}AM=MC\\\widehat{HMA}=\widehat{KMC}\end{cases}\Rightarrow\Delta HAM=\Delta KCM\left(ch-gn\right)}\)(ĐPCM)
=> HM = KM
b) Ta có \(\frac{BH+BK}{2}=\frac{BM-HM+BM+MK}{2}=\frac{2BM}{2}=BM\)(vì HM = KM)
Xét tam giác vuông BAM có AB2 + AM2 = BM2 (Định lý Py-ta-go)
=> AB2 < BM2
=> AB < BM
hay \(AB< \frac{BH+BK}{2}\left(\text{ĐPCM}\right)\)
`@` `\text {Ans}`
`\downarrow`
`a)`
`-(-18 + 45) - (18 + 55)`
`= 18 - 45 - 18 - 55`
`= (18 - 18) - (45 + 55)`
`= -100`
`b)`
`24. (5 - 178) + 178 . (10 + 24)`
`= 24.5 - 24.178 + 178. 10 + 178. 24`
`= 24.5 + 178.(-24 + 10 + 24)`
`= 24.5 + 178.10`
`=120 + 1780`
`=``1900`
`c)`
`29.(-101)`
`= -2929`
`d)`
\((- 56 + 130) – (43 – 56) – (- 20 – 43)\)
`= -56 + 130 - 43 + 56 + 20 + 43`
`= (56 - 56) + (130 + 20) + (-43+43)`
`= 150`