So sánh

\(\text{a, 2014,2014 và 2013,2015}\)

\(2014,2014>2013,2015\)

\(\text{b, 2015,2015 và 2013,2017}\)

\(2015,2015>2013,2017\)

a , 2014, 2014  > 2013 ,2015

b. 2015,2015 > 2013, 2017

hok tốt !!!

a)2014.2014=2013.2014+2014

2013.2015=2013.2014+2013

vì hai thừa số thứ nhất của hai tổng giống nhau mà 2014>2013 => 2014.2014>2013.2015

b)

2015.2015=2013.2015+2015.2

2013.2017=2013.2015+2013.2

vì hai thừa số thứ nhất của hai tổng giống nhau mà 2015.2 >2013.2 => 2015.2015>2013.2017

chúc bạn học tốt nha

A = 2012.2016 = (2013 - 1)(2015 + 1)

= 2013.2015 + 2013 - 2015 - 1

= 2013.2015 - 3 < 2013.2015

=> A < B

Dễ c/m đẳng thức: \(\left(n-1\right)\left(n+1\right)=n^2-1\)

Lúc đó: \(A=2014^2-1+2015^2-1=2014^2+2015^2-2=B\)

Vậy A = B

\(A=2013.2015+2014.2016\)

   \(=\left(2015-2\right).2015+2014\left(2014+2\right)\)

   \(=(2015^2-4030)+(2014^2+4028)\)

   \(=\left(2015^2+2014^2\right)-\left(4030-4028\right)\)

  \(=2014^2+2015^2-2\)

\(\Rightarrow A=B\)

A=2012.2016

=(2013-1).2016

=2013.2016-2016

=2013.(2015+1)-2016

=2013.2015+2013-2016

=2013.2015-3 < 2013.2015

=> A < B

Nice

1)

a)\(A=2013.2015=2013.\left(2014+1\right)=2013.2014+2013\)

\(B=2014^2=2014.\left(2013+1\right)=2014.2013+2014\)

Ta có: \(2014.2013+2014>2013.2014+2013\)

\(\Rightarrow2014^2>2013.2015\)

\(\Rightarrow B>A\)

Vậy \(B>A\)

b) \(A=4.\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)

\(\Rightarrow2A=2.4\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3-1\right).\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3^8-1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3^{16}-1\right)...\left(3^{64}+1\right)\)

\(\Rightarrow2A=3^{128}-1\)

\(\Rightarrow A=\frac{3^{128}-1}{2}< 3^{128}-1=B\)

\(\Rightarrow A< B\)

Vậy \(A< B\)

2)

a)\(9x^2-6x+3=\left(3x\right)^2-2.3x.1+1^2+2\)

                           \(=\left(3x-1\right)^2+2\)

Ta có: \(\left(3x-1\right)^2\ge0\forall x\)

\(\Rightarrow\left(3x-1\right)^2+2\ge2\forall x\)

\(\Rightarrow\left(3x-1\right)^2+2>0\forall x\)

                                đpcm

b)\(x^2+y^2+2x+6y+16\)

\(=\left(x^2+2x+1\right)+\left(y^2+2.y.3+3^2\right)+6\)

\(=\left(x+1\right)^2+\left(y+3\right)^2+6\)

Ta có: \(\hept{\begin{cases}\left(x+1\right)^2\ge0\forall x\\\left(y+3\right)^2\ge0\forall y\end{cases}\Rightarrow}\left(x+1\right)^2+\left(y+3\right)^2+6\ge6\forall x;y\)

\(\Rightarrow\left(x+1\right)^2+\left(y+3\right)^2+6>0\)

                                         đpcm

Tham khảo nhé~

1.

a) A = 2013.2015 = (2014 - 1)(2014 + 1) = 2014² - 1

Vì 2014² - 1 < 2014² => A < B

Vậy A < B

b) \(A=4\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3^8-1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)

\(\Rightarrow2A=3^{128}-1\Leftrightarrow A=\frac{3^{128}-1}{2}\)

\(\Rightarrow A< B\)

Vậy A < B

Bài 2:

a) \(9x^2-6x+2=\left(3x\right)^2-2.3x+1+2=\left(3x-1\right)^2+2\)

Vì \(\left(3x-1\right)^2\ge0\Rightarrow\left(3x-1\right)^2+2>0\)

=> 9x² - 6x + 2 luôn nhận giá trị dương với mọi x

b) \(x^2+y^2+2x+6y+16=\left(x^2+2x+1\right)+\left(y^2+6y+9\right)+6=\left(x+1\right)^2+\left(y+3\right)^2+6\)

Vì \(\left(x+1\right)^2\ge0;\left(y+3\right)^2\ge0\Rightarrow\left(x+1\right)^2+\left(y+3\right)^2\ge0\Rightarrow\left(x+1\right)^2+\left(y+3\right)^2+6>0\)

=> x² + y² + 2x + 6y + 16 luôn nhận giá trị dương với mọi x

a) ta có:  \(1-\frac{2012}{2013}=\frac{1}{2013}\)

                 \(1-\frac{2013}{2014}=\frac{1}{2014}\)

mà \(\frac{1}{2013}>\frac{1}{2014}\) nên   \(\frac{2013}{2014}>\frac{2012}{2013}\)

sao giống lớp 4 thế ta