2016 x 2016-2000 x 2032

=2016 x (2000+16)-2000 x (2016+16)

=2016 x 2000+2016 x 16-2000 x 2016-2000 x 16

=2016 x 16-2000 x 16

=(2016-2000) x 16

=16 x 16

=256

2016 x 2016 - 2000 x 2032

= 2016 x 2016 - 2000 x ( 2016 + 16 )

= 2016 x 2016 - 2000 x 2016 - 2000 x 16 

= 2016 x ( 2016 - 2000 ) - 2000 x 16 

= 2016 x 16 - 2000 x 16

= 16 x ( 2016 - 2000 )

= 16 x 16

= 256

Cấm ai chép nhé

= 64000 + 256

= 64256

2016 x 2016 - 2000 x 2032 = 2016 x 2016 - (2016 - 16) x (2016 + 16) = 2016 x 2016 - (2016 x 2016 - 16 x 16)

= 2016 x 2016 - 2016 x 2016 + 16 x 16 = 256

Đáp án C

a) x=1 hoặc x=0

b) x=5 hoặc x=6

c) x=64,8

d) x=1 hoặc x=0

a) 2016x = 2017x

=> 2016x - 2017x =0

=> x(2016 - 2017) =0

=> x(-1)=0

=>x=0:(-1)=0

b) (x-5)2015=(x-5)2014

=> (x-5)2015 - (x-5)2014=0

=> (x-5)(2015-2014)=0

=> x-5=0

=>x=5

c)5x + 5x +2 =650

=> 10x + 2 =650

=> 10x =648

=> x = \(\frac{648}{10}=64,8\)

d) 2017x =2x

=> 2017x -2x =0

=> 2015x=0

=>x=0

Ta có : \(x^2+y^2\ge2xy\)

\(\Leftrightarrow2\left(x^2+y^2\right)\ge\left(x+y\right)^2\)

\(\Leftrightarrow x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\)

Áp dụng vào bài toán có :

\(P\le\frac{x+y}{\frac{\left(x+y\right)^2}{2}}+\frac{y+z}{\frac{\left(y+z\right)^2}{2}}+\frac{z+x}{\frac{\left(z+x\right)^2}{2}}\) \(=\frac{2}{x+y}+\frac{2}{y+z}+\frac{2}{z+x}=\frac{1}{2}\left(\frac{4}{x+y}+\frac{4}{y+z}+\frac{4}{z+x}\right)\)

Áp dụng BĐT Svacxo ta có :

\(\frac{4}{x+y}\le\frac{1}{x}+\frac{1}{y}\), \(\frac{4}{y+z}\le\frac{1}{y}+\frac{1}{z}\), \(\frac{4}{z+x}\le\frac{1}{z}+\frac{1}{x}\)

Do đó : \(P\le\frac{1}{2}\left[2.\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\right]=2016\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\frac{1}{672}\)

P/s : Dấu "=" không chắc lắm :))

thanks bạn mình hiểu sương sương rồi:))

1.

a.

\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}=1-\frac{1}{6}=\frac{5}{6}\)

b.

Tích có 100 thừa số 

=> n = 100

\(\left(100-1\right)\times\left(100-2\right)\times\left(100-3\right)\times...\times\left(100-99\right)\times\left(100-100\right)\)

\(=\left(100-1\right)\times\left(100-2\right)\times\left(100-3\right)\times...\times\left(100-99\right)\times0\)

\(=0\)

2.

a.

\(135\times789789-789\times135135=1001\times\left(135\times789-789\times135\right)=1001\times0=0\)

b.

\(\left(28\times9696-96\times2828\right)\div\left(1\times2\times3\times...\times2015\times2016\right)\)

\(=\left[101\times\left(28\times96-96\times28\right)\right]\div\left(1\times2\times3\times...\times2015\times2016\right)\)

\(=\left(101\times0\right)\div\left(1\times2\times3\times...\times2015\times2016\right)\)

\(=0\div\left(1\times2\times3\times...\times2015\times2016\right)\)

\(=0\)

3.

a.

\(\left[\left(x+32\right)-17\right]\times2=42\)

\(\left(x+32\right)-17=\frac{42}{2}\)

\(\left(x+32\right)-17=21\)

\(x+32=21+17\)

\(x+32=38\)

\(x=38-32\)

\(x=6\)

b.

\(125+\left(145-x\right)=175\)

\(145-x=175-125\)

\(145-x=50\)

\(x=145-50\)

\(x=95\)

A=1/1.2+1/2.3+1/3.4+1/4.5+1/5.6

A=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6

A=1-1/6

A=5/6

Vậy: A=5/6

B=(100-1).(100-2).(100-3). ... .(100-n)

B=99.98.97. ... .0

B=0

Vậy B=0

              Chắc Vậy

theo cảm tính là 5