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Bài 1:
a, \(x^2+10x+26+y^2+2y\)
\(=x^2+2.x.5+5^2+y^2+2.y.1+1^2\)
\(=\left(x+5\right)^2+\left(y+1\right)^2\)
b, \(x^2-2xy+2y^2+2y+1\)
\(=x^2-2.x.y+y^2+y^2+2.y.1+1^2\)
\(=\left(x-y\right)^2+\left(y+1\right)^2\)
c, \(4x^2+2z^2-4xz-2z+1\)
\(=\left(2x\right)^2-2.2x.z+z^2+z^2-2.z.1+1^2\)
\(=\left(2x-z\right)^2+\left(z-1\right)^2\)
Chúc bạn học tốt!!!
Bài1:
Bn kia giải r nhé
Bài 2:
a)\(127^2+146.127+73^2=127^2+2.73.127+73^2\)
=\(\left(127+73\right)^2=200^2=40000\)
b)\(31,8^2-63,6.21,8+21,8^2=\left(31,8-21,8\right)^2=10^2=100\)
c)\(2018^2-2017^2+2016^2-2015^2+...+2^2-1\)
=\(\left(2018+2017\right)+\left(2015+2016\right)+...+\left(2+1\right)\)
=4025+4031+...+3
=...(bn tự tính)
d)\(2017^2-2016.2018=2017^2-\left(2017^2-1\right)=1\)
20182 - 20172 + 20162 - 20152 + ... + 22 - 12
= (2018+2017)(2018-2017) + (2016+2015)(2016-2015) + ... + (2+1)(2-1)
= 2018 + 2017 + 2016 + 2015 + ... + 2 + 1
= \(\dfrac{\left(1+2018\right).2018}{2}=2037171\)
\(2018^2-2017^2+2016^2-2015^2+...+2^2-1^2\)
\(=\left(2018+2017\right)\left(2018-2017\right)+\left(2016+2015\right)\left(2016-2015\right)+...+\left(2+1\right)\left(2-1\right)\)
\(=4035+4031+...+3\)
Từ 3 đến 4035 có số lượng số hạng là:
\(\left(4035-3\right):4+1=1009\)
Ta có:
\(4035+4031+....+3\)
\(=\dfrac{\left(4035+3\right).1009}{2}=2037171\)
Chúc bạn học tốt!!!
\(2018^2-2017.2019\)
\(=2018^2-\left(2018-1\right)\left(2018+1\right)\)
\(=2018^2-\left(2018^2-1\right)=1\)
\(56^2+56.88+44^2\)
\(=56^2+2.56.44+44^2\)
\(=\left(56+44\right)^2\)
\(=100^2=10000\)
\(\frac{2018^3+1}{2018^2-2017}\)
\(=\frac{\left(2018+1\right)\left(2018^2-2018+1\right)}{2018^2-2017}\)
\(=\frac{2019\left(2018^2-2017\right)}{2018^2-2017}=2019\)
Chúc bạn học tốt.
b) \(x,y\ge1\Rightarrow xy\ge1\)
BĐT đã cho tương đương với:
\(\left(\dfrac{1}{1+x^2}-\dfrac{1}{1+xy}\right)+\left(\dfrac{1}{1+y^2}-\dfrac{1}{1+xy}\right)\ge0\)
\(\Leftrightarrow\dfrac{xy-x^2}{\left(1+x^2\right)\left(1+xy\right)}+\dfrac{xy-y^2}{\left(1+y^2\right)\left(1+xy\right)}\ge0\)
\(\Leftrightarrow+\dfrac{x\left(y-x\right)}{\left(1+x^2\right)\left(1+xy\right)}+\dfrac{y\left(x-y\right)}{\left(1+y^2\right)\left(1+xy\right)}\ge0\)
\(\Leftrightarrow\dfrac{\left(y-x\right)^2\left(xy-1\right)}{\left(1+x^2\right)\left(1+y^2\right)\left(1+xy\right)}\ge0\)
BĐT cuối luôn đúng nên ta có đpcm
Đẳng thức xảy ra khi x=y hoặc xy=1
Ta có \(A=\frac{2017-2018}{2017+2018}=\frac{\left(2017-2018\right)\left(2017+2018\right)}{\left(2017+2018\right)^2}=\frac{2017^2-2018^2}{2017^2+2018^2+2.2017.2018}< \frac{2017^2-2018^2}{2017^2+2018^2}=B\)
Vậy A<B
Ta thấy \(A=\frac{2018-2017}{2018+2017}=\frac{2018^2-2017^2}{\left(2018+2017\right)^2}=\frac{2018^2-2017^2}{2018^2+2.2018.2017+2017^2}\)
Mà \(2018^2+2.2018.2017+2017^2>2018^2+2017^2\)
\(\Rightarrow\frac{2018^2-2017^2}{2018^2+2.2018.2017+2017^2}< \frac{2018^2-2017^2}{2018^2+2017^2}\)
Vậy A<B
2017.2019 = (2018-1)(2018+1) = 20182 -1 => a =1
b= 20183 +1 (???)
a) A= 20182 -20172 = (2018-2017)(2018+2017) = 1.4035=4035
b) B = 20182 -20172 + 20162 - 20152 + ... + 22 -12
= (2018-2017)(2018+2017)+(2016-2015)(2016+2015)...(2-1)(2+1)
=2018+2017+2016+2015+...+2+1
=(2018+1).1004=2027076