\(\frac{a^4}{a^3+2b^3}=a-\frac{2ab^3}{a^3+b^3+b^3}\ge a-\frac{2ab^3}{3ab^2}=a-\frac{2}{3}b\)

tương tự cộng lại ta có đpcm 

Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=d\)

Ta có: 3A=3+\(^{3^2+3^3+3^4+3^5+...+3^{2012}+3^{2013}}\)

\(\Rightarrow\)3A-A=2A=(\(3+3^2+3^3+3^4+...+3^{2013}\)) - (\(1-3^{ }-3^2-3^3-3^4-...-3^{2012}\))

\(\Rightarrow\)2A=\(3^{2013}-1\)\(\Rightarrow\)A=\(\left(3^{2013}-1\right):2\)\(\Rightarrow\)B-A=(\(^{\left(3^{2013}:2\right)-\left(\left(3^{2013}-1\right):2\right)\Rightarrow}\)

A = 1 + 3 + 3² +...+ 3²⁰¹²

3A = 3 + 3² + 3³ +...+ 3²⁰¹³

3A - A = (3 + 3² + 3³ +...+ 3²⁰¹³) - (1 + 3 + 3² +...+ 3²⁰¹²)

2A = 3²⁰¹³ - 1

A = \(\frac{3^{2013}-1}{2}\)

=> B - A = \(\frac{3^{2013}}{2}-\frac{3^{2013}-1}{2}=\frac{3^{2013}-\left(3^{2013}-1\right)}{2}=\frac{3^{2013}-3^{2013}+1}{2}=\frac{1}{2}\)

Of course , I can help you .

A = 1 + 3 + 3^2 + 3^3 + ... + 3^2012

3A = 3 + 3^2 + 3^3 + ... + 3^2013

3A - A = 3 + 3^2 + 3^3 + ... + 3^2013 - 1 - 3 - 3^2 - ... - 3*2012

2A = ( 3 - 3 ) + ( 3^2 - 3^2 ) + ... + ( 3^2012 - 3^2012 ) + ( 3^2013 - 1)

2A = 0 + 0 + 0 + 0 +...+ 0 + 3^2013 - 1

2A = 3^2013 - 1

A = 3^2013 : 2 - 1/2

=> B - A  = 3^2013 : 2 - ( 3^2013 : 2 - 1/2 )

=> B - A = 3^2013 : 2 - 3^2013 + 0,5

=> B - A = 0,5

Bài 1 

ta có a+3+b-3 =a +b chia hết cho 4

nên (b-a )(a+b) cũng chia hết cho 4

bài 2.

ta có: \(M=6x^2-5x-6-12xy+6y^2+6y-3x+2y+2027\)

\(=6\left(x-y\right)^2-8\left(x-y\right)+2021=24-16+2021=2029\)

A=1*2+2*3+3*4+...+2017*2018

3A=1*2*3+2*3*(4-1)+...+2017*2018*(2019-2016)

3A=1*2*3+2*3*4-1*2*3+...+2017*2018*2019-2016*2017*2018

3A=2017*2018*2019

A=\(\frac{2017.2018.2019}{3}\)

mk chỉ biết tính a thôi

thank you 

\(cosa=\frac{1}{3}\Rightarrow sina=\pm\sqrt{1-cos^2a}=\pm\frac{2\sqrt{2}}{3}\)

Thay giá trị vào M và bấm máy

Lời giải:

a)

$x^3+y^3+2x^2-2xy+2y^2=(x^3+y^3)+2(x^2-xy+y^2)$

$=(x+y)(x^2-xy+y^2)+2(x^2-xy+y^2)=(x^2-xy+y^2)(x+y+2)$

b)

$a^4+ab^3-a^3b-b^4=(a^4-a^3b)+(ab^3-b^4)$

$=a^3(a-b)+b^3(a-b)=(a-b)(a^3+b^3)=(a-b)(a+b)(a^2-ab+b^2)$

c)

\(a^3-b^3+3a^2+3ab+3b^2=(a^3-b^3)+3(a^2+ab+b^2)\)

\(=(a-b)(a^2+ab+b^2)+3(a^2+ab+b^2)=(a^2+ab+b^2)(a-b+3)\)

d)

\(x^4+x^3y-xy^3-y^4=x^3(x+y)-y^3(x+y)=(x+y)(x^3-y^3)=(x+y)(x-y)(x^2+xy+y^2)\)

dễ

hết

chỗ

chê

\(1-\)\(a,\frac{7}{8}-\frac{2}{3}=\frac{21}{24}-\frac{16}{24}=\frac{5}{24}\)

\(b,\frac{7}{8}-\frac{2}{3}\times\frac{3}{4}=\frac{7}{8}-\frac{1}{2}=\frac{7}{8}-\frac{4}{8}=\frac{3}{8}\)

\(2-\)\(a,\frac{5}{8}-x=\frac{5}{9}\)

\(x=\frac{5}{8}-\frac{5}{9}\)

\(x=\frac{5}{72}\)

\(b,\frac{3}{4}:x=\frac{2}{5}\)

\(x=\frac{3}{4}:\frac{2}{5}\)

\(x=\frac{15}{8}\)

1, Tính:

\(\frac{7}{8}\) - \(\frac{2}{3}\) = \(\frac{21}{24}\) - \(\frac{16}{24}\) = \(\frac{5}{24}\)

\(\frac{7}{8}\) - \(\frac{2}{3}\) x \(\frac{3}{4}\) = \(\frac{7}{8}\) - \(\frac{1}{2}\) = \(\frac{7}{8}\) - \(\frac{4}{8}\) = \(\frac{3}{8}\)

2, Tìm x

\(\frac{5}{8}\)- x = \(\frac{5}{9}\)

           x = \(\frac{5}{8}\)- \(\frac{5}{9}\)

           x = \(\frac{5}{72}\)

\(\frac{3}{4}\): x = \(\frac{2}{5}\)

           x = \(\frac{3}{4}\) : \(\frac{2}{5}\)

           x = \(\frac{15}{8}\)

a.(-2/3+3/7) : 4/5 + (-1/3+4/7) : 4/5

= [(-2/3 + 3/7) + (-1/3 + 4/7)] : 4/5

= [(-2/3 + (-1/3) + (3/7 + 4/7)] : 4/5

= [-1 + 1] : 4/5

= 0 : 4/5

= 0   

a) \(\left(\frac{-2}{3}+\frac{3}{7}\right).\frac{5}{4}+\left(\frac{-1}{3}+\frac{4}{7}\right).\frac{5}{4}\)

=\(\left(\frac{-2}{3}+\frac{-1}{3}+\frac{3}{7}+\frac{4}{7}\right).\frac{5}{4}\)

= \(0.\frac{5}{4}=0\)

b) \(\frac{5}{9}:\left(\frac{1}{11}-\frac{5}{22}+\frac{1}{15}-\frac{2}{3}\right)\)

=\(\frac{5}{9}:\frac{-81}{110}=\frac{-550}{729}\)