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Ta có A = 2018.2020 + 2019.2021
= (2020 - 2).2020 + 2019.(2019 + 2)
= 20202 - 2.2020 + 20192 + 2.2019
= 20202 + 20192 - 2(2020 - 2019) = 20202 + 20192 - 2 = B
=> A = B
b) Ta có B = 964 - 1= (932)2 - 12
= (932 + 1)(932 - 1) = (932 + 1)(916 + 1)(916 - 1) = (932 + 1)(916 + 1)(98 + 1)(98 - 1)
= (932 + 1)(916 + 1)(98 + 1)(94 + 1)(94 - 1)
= (932 + 1)(916 + 1)(98 + 1)(94 + 1)(92 + 1)(92 - 1)
(932 + 1)(916 + 1)(98 + 1)(94 + 1)(92 + 1).80
mà A = (932 + 1)(916 + 1)(98 + 1)(94 + 1)(92 + 1).10
=> A < B
c) Ta có A = \(\frac{x-y}{x+y}=\frac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)^2}=\frac{x^2-y^2}{x^2+2xy+y^2}< \frac{x^2-y^2}{x^2+xy+y^2}=B\)
=> A < B
d) \(A=\frac{\left(x+y\right)^3}{x^2-y^2}=\frac{\left(x+y\right)^3}{\left(x+y\right)\left(x-y\right)}=\frac{\left(x+y\right)^2}{x-y}=\frac{x^2+2xy+y^2}{x-y}< \frac{x^2-xy+y^2}{x-y}=B\)
=> A < B
a) \(A=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=.............................................................\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)=2^{32}-1=B-1\)
Suy ra A < B
b) \(A=2015.2017=\left(2016-1\right)\left(2016+1\right)=2016^2-1=B-1\)
Suy ra A < B
Phần a bạn nhân thêm ở A là (2-1) là ra hằng đẳng thức, cứ thế mà triển. (Kết quả: A<B)
Phần b: phân tích A, ta có:
2015.2017= (2016-1).(2016+1)= 2016^2 -1 <2016^2
Suy ra: A<B
Ta có : \(\frac{x}{a}\)+\(\frac{y}{b}\)+\(\frac{z}{c}\)=0 => \(\frac{abz+acy+bcx}{xyz}\)=0=> abz+acy+bcz= 0
Lại có \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2\Rightarrow\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}-2\left(\frac{abz+acy+bcx}{xyz}\right)=4\)
=> \(\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}\)=4
y=\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
=>y=\(\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
=>y=\(\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
=>y=\(\left(2^8-1\right)\left(2^8+1\right)\)
=>y=\(2^{16}-1\)<\(2^{16}\)=x
=>x>y.
Vậy x>y
bài 1
a) 299992=(20000+9999)2=4.100002+40000.9999+99992
19999.39999+(10000+9999).(30000+9999)=3.100002+99992+40000.9999
ta có 4.100002>3.100002=>299992>19999.39999
b) chịu mình ko giỏi so sánh
bài 2
a) x2+8y2+9y=4y(x+3)
<=>x2-4xy+42+4y2+122+9=0
<=>(x-2y)2+(2y+3)2=0
xét (x-27)2\(\ge\)0 với mọi giá trị x,y
(2y+3)2\(\ge\)0 với mọi giá trị y
=>đồng thời xảy ra x-2y=0;2y-3=0
từ đó tìm ra y sau đó thay vào x-2y tìm nốt x
b)x2+2y2+5z2+1=2(xy+2yz+z)
<=>x2-2xy+y2+y2-4yz+4z2+z2-2z+1=0
<=>(x-y)2+(y-2z)2+(z-1)2=0
sau đó xm tyơng tự câu trên
c) câu này mình chịu
chào, hiện tại tôi đang ở tương lai năm 2024, 2017 và 2018 vui lắm, cố lên nhé!
Ta có : \(\hept{\begin{cases}A=1999.2001\\B=2000^2\end{cases}}\)
\(< =>\hept{\begin{cases}A=1999.2000+1999\\B=2000\cdot2000\end{cases}}\)
\(< =>\hept{\begin{cases}A=1999.2000+2000+1\\B=1999.2000+2000\end{cases}}\)
\(< =>\hept{\begin{cases}A=2000.2000+1\\B=2000.2000\end{cases}}\)
\(< =>A>B\)
a. Ta có : \(A=1999.2021=\left(2000-1\right)\left(2000+1\right)=2020^2-1< 2020\)
\(\Rightarrow A< B\)
b. Ta có : \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
...
\(=\left(2^8-1\right)\left(2^8+1\right)=2^{16}-1< 2^{16}\)
\(\Rightarrow A>B\)
c,d tương tự
\(a.\)
Ta sẽ biến đổi biểu thức \(B\) quy về dạng có thể dùng được hằng đẳng thức \(\left(x-y\right)\left(x+y\right)=x^2-y^2\), khi đó:
\(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)=2^{16}-1\)
Vì \(2^{16}>2^{26}-1\) nên \(2^{16}>\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
Vậy, \(A>B\)
Tương tự với câu \(b\) kết hợp với phương pháp tách hạng tử, khi đó xuất hiện hằng đẳng thức mới và dễ dàng đơn giản hóa biểu thức \(A\). Ta có:
\(A=4\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(=\frac{1}{2}\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(=\frac{1}{2}\left(3^{64}-1\right)\left(3^{64}+1\right)=\frac{1}{2}\left(3^{128}-1\right)\)
Mặt khác, do \(\frac{1}{2}<1\) nên \(\frac{1}{2}\left(3^{128}-1\right)<3^{128}-1\)
Vậy, \(B>A\)
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
b) A = 2010 . 2012
= ( 2011 - 1 )( 2011 + 1 )
= 20112 - 12 = 20112 - 1
20112 - 1 < 20112 => A < B