Số m vải còn lại sau khi bán ngày 1 chiếm:

1 - 3/5 = 2/5

Số m vải còn lại sau khi bán ngày 2 chiếm:

2/5 - 2/5 . 2/7 = 2/7

Số mét vải cửa hàng đã bán:

40 : 2/7 = 140 (m)

\(\left(12\dfrac{1}{3}-10\dfrac{1}{4}\right):\left(2\dfrac{1}{2}+1\dfrac{1}{3}\right)\\ =\left(\dfrac{37}{3}-\dfrac{41}{4}\right):\left(\dfrac{5}{2}+\dfrac{4}{3}\right)\\ =\dfrac{25}{12}:\dfrac{23}{6}\\ =\dfrac{25}{12}\cdot\dfrac{6}{23}\\ =\dfrac{25}{46}\)

\(A=2\sqrt{2}\left(\dfrac{a}{2\sqrt{2b\left(a+b\right)}}+\dfrac{b}{2\sqrt{2c\left(b+c\right)}}+\dfrac{a}{2\sqrt{2a\left(c+a\right)}}\right)\)

\(A\ge2\sqrt{2}\left(\dfrac{a}{2b+a+b}+\dfrac{b}{2c+b+c}+\dfrac{a}{2a+c+a}\right)\)

\(A\ge2\sqrt{2}\left(\dfrac{a^2}{a^2+3ab}+\dfrac{b^2}{b^2+3bc}+\dfrac{c^2}{c^2+3ca}\right)\)

\(A\ge\dfrac{2\sqrt{2}\left(a+b+c\right)^2}{a^2+b^2+c^2+3\left(ab+bc+ca\right)}=\dfrac{2\sqrt{2}\left(a+b+c\right)^2}{\left(a+b+c\right)^2+ab+bc+ca}\)

\(A\ge\dfrac{2\sqrt{2}\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\dfrac{1}{3}\left(a+b+c\right)^2}=\dfrac{3\sqrt{2}}{2}\)

Dấu "=" xảy ra khi \(a=b=c\) 

Bổ sung các bđt được áp dụng trong bài thầy Lâm cho rõ ràng:

Áp dụng Bđt Cauchy và Bunhiacopxki : 

\(a+3b=2b+\left(a+b\right)\ge2\sqrt[]{2b\left(a+b\right)}\)

\(ab+bc+ca\le\sqrt[]{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)}=a^2+b^2+c^2\)

AB//CD

=>\(\widehat{B}+\widehat{C}=180^0\)

=>\(\widehat{C}+\widehat{C}+40^0=180^0\)

=>\(2\cdot\widehat{C}=180^0-40^0=140^0\)

=>\(\widehat{C}=70^0\)

\(\widehat{B}=70^0+40^0=110^0\)

ABCD là hình thang có AB//CD
=>\(\widehat{A}+\widehat{D}=180^0\)(hai góc trong cùng phía)

=>\(2\cdot\widehat{D}+\widehat{D}=180^0\)

=>\(3\widehat{D}=180^0\)

=>\(\widehat{D}=60^0\)

\(\widehat{A}=2\cdot60^0=120^0\)

xét ΔABC vuông tại A có \(cosB=\dfrac{AB}{BC}\)

=>\(\dfrac{6}{BC}=cos30=\dfrac{\sqrt{3}}{2}\)

=>\(BC=6\cdot\dfrac{2}{\sqrt{3}}=4\sqrt{3}\left(cm\right)\)

\(\left|A+B\right|< =\left|A\right|+\left|B\right|\)

=>\(\left(\left|A+B\right|\right)^2< =\left(\left|A\right|+\left|B\right|\right)^2\)

=>\(A^2+B^2+2AB< =A^2+B^2+2\left|AB\right|\)

=>2AB<=2|AB|

=>AB<=|AB|(luôn đúng)

Dấu '=' xảy ra khi AB>=0

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{10\cdot11}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{10}-\dfrac{1}{11}\)

\(=1-\dfrac{1}{11}=\dfrac{10}{11}\)

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{10\cdot11}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}\\ =1-\dfrac{1}{11}\\ =\dfrac{11}{11}-\dfrac{1}{11}=\dfrac{10}{11}\)

Xét ΔABC vuông tại A có \(sinB=\dfrac{AC}{BC}\)

=>\(\dfrac{6}{BC}=sin30=\dfrac{1}{2}\)

=>\(BC=6\cdot2=12\left(cm\right)\)

Ta có: \(\widehat{xOy}+\widehat{yOz}=180^0\)(hai góc kề bù)

=>\(\widehat{yOz}+125^0=180^0\)

=>\(\widehat{yOz}=55^0\)

\(a+\dfrac{2}{b}=b+\dfrac{2}{c}\Rightarrow a-b=\dfrac{2}{c}-\dfrac{2}{b}=2\left(\dfrac{b-c}{bc}\right)\)

\(\Rightarrow\dfrac{a-b}{b-c}=\dfrac{2}{bc}\)

Tương tự: \(a+\dfrac{2}{b}=c+\dfrac{2}{a}\Rightarrow\dfrac{a-c}{b-a}=\dfrac{2}{ab}\)

\(b+\dfrac{2}{c}=c+\dfrac{2}{a}\Rightarrow\dfrac{b-c}{c-a}=\dfrac{2}{ca}\)

Nhân vế với vế:

\(\left(\dfrac{a-b}{b-c}\right)\left(\dfrac{a-c}{b-a}\right)\left(\dfrac{b-c}{c-a}\right)=\dfrac{8}{\left(abc\right)^2}\)

\(\Rightarrow\left(abc\right)^2=8\)

\(\Rightarrow\left|abc\right|=2\sqrt{2}\)