a) ( x - 2)mũ 2 + ( 1/2y + 2) mũ 2 =0
b) (5x +1) mũ 2 = 36/49
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x2 + xy + 5x + 5y = ( x2 + xy ) + ( 5x + 5y ) = x( x + y ) + 5( x + y ) = ( x + y )( x + 5 )
x2 - y2 + 3x - 3y = ( x2 - y2 ) + ( 3x - 3y ) = ( x - y )( x + y ) + 3( x - y ) = ( x - y )( x + y + 3 )
x² + xy + 5x + 5y
= (x²+ xy) + ( 5x+5y)
= x(x+y) + 5(x+y)
= (x+y)(x+5)
x² - y² + 3x - 3y
= (x² - y²) + ( 3x -3y)
= (x-y)(x+y) + 3(x-y)
= (x-y)(x+y+3)
chúc bạn học tốt ^^
a) 1 + 3 + 5 + ... + 13
= (13 + 1).[(13 - 1) : 2 + 1] : 2
= 14 . 7 : 2
= 49
= 7²
b) 3² + 4² + 12²
= 9 + 16 + 144
= 169
= 13²
\(A=2+2^2+2^3+2^4+...+2^{100}\)
\(=2+\left(2^2+2^3+2^4\right)+...+\left(2^{98}+2^{99}+2^{100}\right)\)
\(=2+2^2\left(1+2+2^2\right)+...+2^{98}\left(1+2+2^2\right)\)
\(=2+7\cdot\left(2^2+2^5+...+2^{98}\right)\)
=>A không chia hết cho 7 mà là chia 7 dư 2 nha bạn
Mình làm ngắn gọn nhé.
\(A=1+2+2^2+...+2^{50}\)
\(\Rightarrow2A=2+2^2+...+2^{51}\)
\(\Rightarrow2A-A=2+2^2+...+2^{51}-1-2-2^2-...-2^{50}\)
\(\Rightarrow A=2^{51}-1\)
\(B=1+3+...+3^{66}\)
\(3B=3+3^2+...+3^{67}\)
\(2B=3+3^2+...+3^{67}-1-3-...-3^{66}\)
\(2B=3^{67}-1\)
\(B=\frac{3^{67}-1}{2}\)
\(\left(\dfrac{1}{4}\right)^{2n}=\left(\dfrac{1}{8}\right)^2\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{2.2n}=\left(\dfrac{1}{2}\right)^{3.2}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{4n}=\left(\dfrac{1}{2}\right)^6\)
\(\Rightarrow4n=6\)
\(\Rightarrow n=\dfrac{6}{4}=\dfrac{3}{2}\)
`(2x - 1)^2 = 36`
`(2x - 1)^2 = 6^2` hoặc `(2x - 1)^2 = (-6)^2`
`2x - 1 = 6` hoặc `2x - 1 = -6`
`2x = 6 + 1` hoặc `2x = -6 + 1`
`2x = 7` hoặc `2x = -5`
`x = 7 : 2` hoặc `x = -5 : 2`
`x = 3,5` hoặc `x = -2,5`
\(\left(2x-1\right)^2=36\)
\(\Rightarrow\left(2x-1\right)^2=6^2\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=6\\2x-1=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=7\\2x=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
\(\left(\dfrac{1}{2}\right)^n=\left(\dfrac{1}{8}\right)^2\)
\(=>\left(\dfrac{1}{2}\right)^n=\left[\left(\dfrac{1}{2}\right)^3\right]^2\)
\(=>\left(\dfrac{1}{2}\right)^n=\left(\dfrac{1}{2}\right)^6\)
\(\Rightarrow n=6\)
\(^{3^2}\).\(^{3^3}\)+\(2^3\).\(2^2\)
(\(^{2^3}\).\(^{3^3}\))+(\(2^2\).\(^{3^2}\)
=275
b, \(\left(5x+1\right)^2=\frac{36}{49}\)
\(\Rightarrow\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\)
\(\Rightarrow5x+1=\frac{6}{7}\)
\(\Rightarrow5x=\frac{-1}{7}\)
\(\Rightarrow x=\frac{-1}{35}\)