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14 tháng 8 2017
$A=\dfrac{2018.2017-1}{2016.2018+2017}$
$=>A={2018.2016+2018-1}{2016.2018+2017}$
$=>A={2018.2016+2017}{2016.2018+2017}$
$=>A=1$
DH
14 tháng 8 2017
\(A=\dfrac{2018.2017-1}{2018.2016+2017}\)
\(A=\dfrac{2018.\left(2016+1\right)-1}{2018.2016+2017}\)
\(A=\dfrac{2018.2016+2018-1}{2018.2016+2017}\)
\(A=\dfrac{2018.2016+2017}{2018.2016+2017}=1\)
\(B=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}+\dfrac{1}{2187}\)
\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^7}\)
\(\Rightarrow3B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^6}\)
\(\Rightarrow3B-B=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^6}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^7}\right)\)
\(\Rightarrow2B=1-\dfrac{1}{3^7}\Rightarrow B=\dfrac{1-\dfrac{1}{2187}}{2}=\dfrac{1093}{2187}\)
Chúc bạn học tốt!!!
16 tháng 7 2021
Bài 9:
a) Ta có: \(A=\left(2x+y\right)^2-\left(2x+y\right)\left(2x-y\right)+y\left(x-y\right)\)
\(=4x^2+4xy+y^2-4x^2+y^2-xy-y^2\)
\(=3xy-y^2\)
\(=3\cdot\left(-2\right)\cdot3-3^2=-18-9=-27\)
b) Ta có: \(B=\left(a-3b\right)^2-\left(a+3b\right)^2-\left(a-1\right)\left(b-2\right)\)
\(=a^2-6ab+9b^2-a^2-6ab-9b^2-ab+2a+b-2\)
\(=-13ab+2a+b-2\)
\(=-13\cdot\dfrac{1}{2}\cdot\left(-3\right)+2\cdot\dfrac{1}{2}+\left(-3\right)-2\)
\(=\dfrac{31}{2}\)
16 tháng 7 2021
Bài 7:
a) \(498^2=\left(500-2\right)^2=250000-2000+4=248004\)
b) \(93\cdot107=100^2-7^2=10000-49=9951\)
c) \(163^2+74\cdot163+37^2=\left(163+37\right)^2=200^2=40000\)
d) \(1995^2-1994\cdot1996=1995^2-1995^2+1=1\)
e) \(9^8\cdot2^8-\left(18^4-1\right)\left(18^4+1\right)\)
\(=18^8-18^8+1=1\)
f) \(125^2-2\cdot125\cdot25+25^2=\left(125-25\right)^2=100^2=10000\)
28 tháng 8 2021
Bài 1:
a: \(\left(3x+2\right)^2-4=3x\left(3x+4\right)\)
Bài 2:
a: \(4x^2+4x+1\)
b: \(9x^2+9x+\dfrac{9}{4}\)
HD
1
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 6 2019
1/
\(3\left(-1-4x^2+5x\right)+4\left(3x^2+7x-6\right)=-27\)
\(\Leftrightarrow-3-12x^2+15x+12x^2+28x-24=-27\)
\(\Leftrightarrow43x=0\Rightarrow x=0\)
2/
\(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2-1\right)=27\)
\(\Leftrightarrow x^3+27-x^3+x=27\)
\(\Leftrightarrow x=0\)
3 tháng 7 2017
a, \(\dfrac{27}{8x^3-1}:\dfrac{3}{2x-1}\)
\(=\dfrac{27}{\left(2x-1\right)\left(4x^2+2x+1\right)}.\dfrac{2x-1}{3}\)
\(=\dfrac{9}{4x^2+2x+1}\)
b, \(\dfrac{8x^3+36x^2+54x+27}{2x+3}=\dfrac{\left(2x+3\right)^3}{2x+3}=\left(2x+3\right)^2\)
18 tháng 7 2018
\(\dfrac{x^3}{27}+\dfrac{x^6}{729}-x^9=\dfrac{x}{3}+\dfrac{x}{3}-\dfrac{3x^9}{3}=\dfrac{2x-3x^9}{3}=\dfrac{x\left(2-3x^8\right)}{3}\)
Y
14 tháng 8 2018
a) \(\left(5x-1\right)^6=729\)
\(\Rightarrow\left[{}\begin{matrix}\left(5x-1\right)^6=3^6\\\left(5x-1\right)^6=\left(-3\right)^6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}5x-1=3\\5x-1=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}5x=4\\5x=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\)
b) \(\dfrac{8}{25}=\dfrac{2^x}{5^{x-1}}\)
\(\Rightarrow\left[{}\begin{matrix}2^x=2^3\\5^{x-1}=5^2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x-1=2\end{matrix}\right.\)
\(\Rightarrow x=3\)
Vậy x = 3
c) \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{2}\right)^{10}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{3x}=\left(\dfrac{1}{2}\right)^{10}\)
\(\Rightarrow3x=10\)
\(\Rightarrow x=\dfrac{10}{3}\)
d) \(9^x:3^x=3\)
\(\Rightarrow\left(9:3\right)^x=3\)
\(\Rightarrow3^x=3^1\)
\(\Rightarrow x=1\)
A = 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
A x 3 = 3 x (1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729)
= 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 A x 3 - A
= 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 - (1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729)
= 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 - 1/3 - 1/9 - 1/27 - 1/81 - 1/243 - 1/729
= 1 - 1/729 A x 2
= 728/729
A = 364/729