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14 tháng 10 2018

Bo may la binh day k di hieu ashdbfgbgygygggydfsghuyfhdguuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu3

22 tháng 5 2023

Áp dụng BĐT Bunhiacopxki:

\(\sqrt{\left(a^2+c^2\right)\left(b^2+c^2\right)}\ge\sqrt{\left(ac+bc\right)^2}=ac+bc\)

CMTT : \(\sqrt{\left(a^2+d^2\right)\left(b^2+d^2\right)}\ge ad+bd\)

Ta có :\(\sqrt{\left(a^2+c^2\right)\left(b^2+c^2\right)}+\sqrt{\left(a^2+d^2\right)\left(b^2+d^2\right)}\ge ac+bc+ad+bd=\left(a+b\right)\left(c+d\right)\)

22 tháng 5 2023

Áp dụng BĐT Bunhiacopxki:

(�2+�2)(�2+�2)≥(��+��)2=��+��

CMTT : (�2+�2)(�2+�2)≥��+��

Ta có :(�2+�2)(�2+�2)+(�2+�2)(�2+�2)≥��+��+��+��=(�+�)(�+�)

4 tháng 11 2019

\(\sqrt[3]{3x+1}+\sqrt[3]{5-x}+\sqrt[3]{2x-9}-\sqrt[3]{4x-3}=0\)

Đây nè @Võ Hồng Phúc(Phúc bím)

4 tháng 11 2019

nè nè chi

a: \(=3\sqrt{5}-\left(\sqrt{5}-2\right)=2\sqrt{5}+2\)

b: \(=\left|a-b\right|-\left|b-c\right|-\left|c-d\right|\)

\(=b-a-\left(c-b\right)-\left(d-c\right)\)

=b-a-c+b-d+c

=2b-d-a

26 tháng 6 2021

\(A=2.\left|\left(-3\right)\right|^3+2.\left(-2\right)^2-4\left|\left(-2\right)^3\right|\)

\(=54+8-32=30\)

\(B=\left|\sqrt{2}-2\right|+\left|\sqrt{2}-3\right|=2-\sqrt{2}+3-\sqrt{2}\)

\(=5-2\sqrt{2}\)

\(C=\left|3-\sqrt{3}\right|-\left|1+\sqrt{3}\right|=3-\sqrt{3}-1-\sqrt{3}\)

\(=2-2\sqrt{3}\)

\(D=\left|5+\sqrt{6}\right|-\left|\sqrt{6}-5\right|=5+\sqrt{6}-5+\sqrt{6}\)

\(=2\sqrt{6}\)

\(E=\sqrt{15^2}-\sqrt{5^2}=15-5=10\)

26 tháng 6 2021

`A=2sqrt{(-3)^6}+2sqrt{(-2)^4}-4sqrt{(-2)^6}=2|(-3)^3|+2|(-2)^2|-4|(-2)^3|=54+8-32=30` $\\$ `B=sqrt{(sqrt2-2)^2}+sqrt{(sqrt2-3)^2}=2-sqrt2+3-sqrt2=5-2sqrt2` $\\$ `C=sqrt{(3-sqrt3)^2}-sqrt{(1+sqrt3)^2}=3-sqrt3-sqrt3-1=2-2sqrt3` $\\$ `D=sqrt{(5+sqrt6)^2}-sqrt{(sqrt6-sqrt5)^2}=5+sqrt6-5+sqrt6=2sqrt6` $\\$ `E=sqrt{17^2-8^2}-sqrt{3^2+4^2}=sqrt{289-64}-sqrt{9+16}=sqrt(225)-sqrt{25}=15-5=10`

5 tháng 8 2016

Ta có \(\left(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\right)^2\)\(\ge\)\(\left(a+c\right)^2+\left(b+d\right)^2\)

       \(\Leftrightarrow\)\(a^2+b^2+c^2+d^2+2\sqrt{\left(a^2+b^2\right)\left(c^2+d^2\right)}\)\(\ge\)\(a^2+b^2+c^2+d^2\)\(+2\left(ac+bd\right)\)

      \(\Leftrightarrow\)\(\sqrt{\left(a^2+b^2\right)\left(c^2+d^2\right)}\)\(\ge\)\(ac+bd\)

      \(\Leftrightarrow\)\(\left(a^2+b^2\right)\left(c^2+d^2\right)\)\(\ge\)\(\left(ac+bd\right)^2\)(*)

   Vì (*) luôn đúng theo bđt bunhia copxki \(\Rightarrow\)đpcm

   dấu ''='' xảy ra khi a/c=b/d

24 tháng 5 2018

Cái này là Mincopxki rồi bạn. `

Mincopxki: \(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\ge\sqrt{\left(a+c\right)^2+\left(b+d\right)^2}\)