Tháng 5 nhà Hà dùng hết số điện là:

      124 x 2  = 248 (số điện)

Tháng 5 có số ngày là: 31 ngày

Trung bình mỗi ngày trong tháng 5 nhà Hà dùng hết số điện là:

   248 : 31 = 8 (số)

Đáp số: 8 số 

Chiều cao của bể cá là:

\(48:\left(6+4\right)\times2=2,4\left(dm\right)\)

Chọn D 

@Huỳnh Thanh Phong: thanks ạ

\(\dfrac{1}{2}+\dfrac{5}{6}+\dfrac{11}{12}+...+\dfrac{89}{90}+\dfrac{109}{110}+\dfrac{10}{11}\)

\(=1-\dfrac{1}{2}+1-\dfrac{1}{6}+...+1-\dfrac{1}{90}+1-\dfrac{1}{110}+\dfrac{10}{11}\)

\(=\left(1+1+...+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{90}+\dfrac{1}{110}\right)+\dfrac{10}{11}\)

\(=10-\left(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{10\times11}\right)+\dfrac{10}{11}\)

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

\(=10-\left(1-\dfrac{1}{11}\right)+\dfrac{10}{11}\)

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

\(=10\)

Em bổ sung đề cho chính xác nhé

Đề bài là gì vậy 

\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)

\(\dfrac{1}{a}+\dfrac{1}{c}\ge\dfrac{4}{a+c}\)

\(\dfrac{4}{a+b}+\dfrac{4}{a+c}\ge4\left(\dfrac{4}{a+b+a+c}\right)=\dfrac{16}{2a+b+c}\)

\(\Rightarrow\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{16}{2a+b+c}\)

Tương tự ta có:

\(\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{1}{c}\ge\dfrac{16}{a+2b+c}\) ; \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{c}\ge\dfrac{16}{a+b+2c}\)

Cộng vế:

\(4\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{16}{2a+b+c}+\dfrac{16}{a+2b+c}+\dfrac{16}{a+b+2c}\)

\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{2a+b+c}+\dfrac{4}{a+2b+c}+\dfrac{4}{a+b+2c}\)

Dấu "=" xảy ra khi \(a=b=c\)

Ta có:

\(VP=\dfrac{4}{2a+b+c}+\dfrac{4}{2b+a+c}+\dfrac{4}{2c+a+b}\)

\(\le\dfrac{1}{2a}+\dfrac{1}{b+c}+\dfrac{1}{2b}+\dfrac{1}{c+a}+\dfrac{1}{2c}+\dfrac{1}{a+b}\)

\(=\dfrac{1}{2a}+\dfrac{1}{4}\left(\dfrac{4}{b+c}\right)+\dfrac{1}{2b}+\dfrac{1}{4}\left(\dfrac{4}{c+a}\right)+\dfrac{1}{2c}+\dfrac{1}{4}\left(\dfrac{4}{a+b}\right)\)

\(\le\dfrac{1}{2a}+\dfrac{1}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)+\dfrac{1}{2b}+\dfrac{1}{4}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)+\dfrac{1}{2c}+\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)

\(=\dfrac{1}{2a}+\dfrac{1}{4b}+\dfrac{1}{4c}+\dfrac{1}{2b}+\dfrac{1}{4c}+\dfrac{1}{4a}+\dfrac{1}{2c}+\dfrac{1}{4a}+\dfrac{1}{4b}\)

\(=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)

\(=VT\)

 Ta có đpcm. Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)

 Chú ý: Trong bài ta đã sử dụng bất đẳng thức \(\dfrac{4}{x+y}\le\dfrac{1}{x}+\dfrac{1}{y}\) với \(x,y>0\) hai lần

A = \(\dfrac{7}{18}\).[ - \(\dfrac{12}{23}\) - \(\dfrac{4}{15}\)] + \(\dfrac{7}{18}\).[\(\dfrac{8}{30}\)+ \(\dfrac{25}{23}\)]

A =- \(\dfrac{7}{18}\).\(\dfrac{12}{23}\) - \(\dfrac{7}{18}.\dfrac{4}{15}\) + \(\dfrac{7}{18}\).\(\dfrac{8}{30}\) + \(\dfrac{7}{18}\).\(\dfrac{25}{23}\)

A =  \(\dfrac{7}{18}\).(\(\dfrac{25}{23}\) - \(\dfrac{12}{23}\))  - (\(\dfrac{7}{18}\).\(\dfrac{4}{15}\) - \(\dfrac{7}{18}\).\(\dfrac{4}{15}\))

= \(\dfrac{7}{18}\).\(\dfrac{13}{23}\) - 0

= \(\dfrac{91}{414}\)

Số học sinh nam là:

\(\left(230+40\right):2=135\left(hs\right)\)

Số học sinh nữ là:

\(135-40=95\left(hs\right)\)

Đáp số: ... 

Ta có \(VT=\dfrac{1}{a}+\dfrac{1}{4b}\)

\(=\dfrac{1}{a}+\dfrac{\dfrac{1}{4}}{b}\)

\(=\dfrac{1^2}{a}+\dfrac{\left(\dfrac{1}{2}\right)^2}{b}\)

\(\ge\dfrac{\left(1+\dfrac{1}{2}\right)^2}{a+b}\) (áp dụng BĐT \(\dfrac{x^2}{m}+\dfrac{y^2}{n}\ge\dfrac{\left(x+y\right)^2}{m+n}\))

\(=\dfrac{\left(\dfrac{3}{2}\right)^2}{1}\) (vì \(a+b=1\))

\(=\dfrac{9}{4}\)

Ta có đpcm. Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}a+b=1\\\dfrac{1}{a}=\dfrac{1}{2b}\end{matrix}\right.\) \(\Leftrightarrow\left(a,b\right)=\left(\dfrac{2}{3},\dfrac{1}{3}\right)\)

Ta có:

\(\dfrac{1}{a}+\dfrac{1}{4b}=\dfrac{1}{a}+\dfrac{\left(\dfrac{1}{2}\right)^2}{b}\ge\dfrac{\left(1+\dfrac{1}{2}\right)^2}{a+b}=\dfrac{9}{4}\)

Dấu "=" xảy ra khi \(\left(a;b\right)=\left(\dfrac{2}{3};\dfrac{1}{3}\right)\)