Khối 2 có 120-12=108 bạn

Khối 3 có 120x1/2=60 bạn

Gọi số học sinh khối 4 là x

Theo đề. ta có: \(\dfrac{120+108+60+x}{4}-x=6\)

=>288+x-4x=24

=>3x=288-24=264

=>x=88

a: \(\Leftrightarrow\left[{}\begin{matrix}cos\left(x+\dfrac{\Pi}{2}\right)=-\dfrac{\sqrt{3}}{2}\\sinx=-\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\Pi}{2}=\dfrac{5}{6}\Pi+k2\Pi\\x+\dfrac{\Pi}{2}=-\dfrac{5}{6}\Pi+k2\Pi\\x=-\dfrac{\Pi}{2}+k2\Pi\\x=\dfrac{3}{2}\Pi+k2\Pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\Pi+k2\Pi\\x=-\dfrac{4}{3}\Pi+k2\Pi\\x=-\dfrac{\Pi}{2}+k2\Pi\\x=\dfrac{3}{2}\Pi+k2\Pi\end{matrix}\right.\)

b: \(\Leftrightarrow2\cdot sinx\cdot cosx+\sqrt{3}\cdot sinx=0\)

\(\Leftrightarrow sinx\left(2cosx+\sqrt{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\Pi\\cosx=-\dfrac{\sqrt{3}}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=k\Pi\\x=\dfrac{5}{6}\Pi+k2\Pi\\x=-\dfrac{5}{6}\Pi+k2\Pi\end{matrix}\right.\)

- Sửa đề: CMR \(8\left(a^4+b^4+\dfrac{1}{ab}\right)\ge33\)

- Áp dụng BĐT Caushy, ta có:

\(\left\{{}\begin{matrix}a^4+\dfrac{1}{16}+\dfrac{1}{16}+\dfrac{1}{16}\ge\dfrac{1}{2}a\left(1\right)\\b^4+\dfrac{1}{16}+\dfrac{1}{16}+\dfrac{1}{16}\ge\dfrac{1}{2}b\left(2\right)\end{matrix}\right.\)

- Lấy \(\left(1\right)+\left(2\right)\), ta được:

\(a^4+b^4+\dfrac{3}{8}\ge\dfrac{1}{2}\left(a+b\right)=\dfrac{1}{2}\)

\(\Leftrightarrow a^4+b^4\ge\dfrac{1}{8}\left(3\right)\)

- Cũng theo BĐT Caushy, ta có:

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

- Lấy \(\left(3\right)+\left(4\right)\), ta có:

\(a^4+b^4+\dfrac{1}{ab}=\dfrac{1}{8}+4=\dfrac{33}{8}\)

\(\Leftrightarrow8\left(a^4+b^4+\dfrac{1}{ab}\right)\ge33\)

- Dấu "=" xảy ra khi \(a=b=\dfrac{1}{2}\)

d: DH=căn 4*9=6cm

=>IK=DH=6cm

b: tan DEF=DH/HE=6/4=3/2

=>góc DEF=56 độ

Với 0 < x < 1 ; ta có : \(A=\dfrac{3}{1-x}+\dfrac{4}{x}\ge\dfrac{\left(\sqrt{3}+2\right)^2}{1-x+x}=\left(\sqrt{3}+2\right)^2\)

" = " \(\Leftrightarrow\dfrac{\sqrt{3}}{1-x}=\dfrac{2}{x}\)  \(\Leftrightarrow x=\dfrac{2}{\sqrt{3}+2}=4-2\sqrt{3}\) (t/m) 

- Với \(0< x< 1\), ta có:

\(A=\dfrac{3}{1-x}+\dfrac{4}{x}\)

\(=\dfrac{3+3x-3x}{1-x}+\dfrac{4-4x+4x}{x}\)

\(=\dfrac{3\left(1-x\right)+3x}{1-x}+\dfrac{4\left(1-x\right)+4x}{x}\)

\(=3+\dfrac{3x}{1-x}+\dfrac{4\left(1-x\right)}{x}+4\)

\(=7+\left[\dfrac{3x}{1-x}+\dfrac{4\left(1-x\right)}{x}\right]\)

\(\ge7+2\sqrt{\dfrac{3x}{1-x}.\dfrac{4\left(1-x\right)}{x}}\)

\(=7+4\sqrt{3}\)

- Dấu "=" xảy ra khi \(\sqrt{\dfrac{3x}{1-x}}=\sqrt{\dfrac{4\left(1-x\right)}{x}}\Leftrightarrow x=4-2\sqrt{3}\left(tm\right)\)

- Vậy \(MinA=7+4\sqrt{3}\), đạt tại \(x=4-2\sqrt{3}\).

a: \(\Leftrightarrow sin\left(x+\dfrac{\Pi}{3}\right)=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\Pi}{3}=\dfrac{\Pi}{6}+k2\Pi\\x+\dfrac{\Pi}{3}=\dfrac{5}{6}\Pi+k2\Pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\Pi}{2}+k2\Pi\\x=\dfrac{1}{2}\Pi+k2\Pi\end{matrix}\right.\)

b: \(\Leftrightarrow cos\left(2x-\dfrac{\Pi}{4}\right)=-\dfrac{\sqrt{3}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\Pi}{4}=\dfrac{5}{6}\Pi+k2\Pi\\2x-\dfrac{\Pi}{4}=-\dfrac{5}{6}\Pi+k2\Pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{24}\Pi+k\Pi\\x=-\dfrac{7}{24}\Pi+k\Pi\end{matrix}\right.\)

`2/5 xx 3/4 xx 5/6 : 3/4 `
`= 2/5 xx 3/4 xx 5/6 xx 4/3`
`= (2/5 xx 5/6) xx (3/4 xx 4/3)`
`= 2/6 xx 1`
= `1/3`

\(\dfrac{2}{5}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot\dfrac{4}{3}=\dfrac{2}{5}\cdot\dfrac{5}{6}=\dfrac{1}{3}\)

Ta có :

`-x_1^2+4x+12 = -(x_1^2 -4x-12) = -(x_1^2 -4x +4 -16) = -(x_1^2 -4x+4)+16 = -(x_1-2)^2+16 <= 16 AA x`

`-> \sqrt{-x_1^2+4x+12} <= \sqrt{16} =4 AA x`
Tương tự :
`-x_2^2 -2x+3 = -(x_2^2 -2x-3) = -(x_2^2 -2x+1-4) =-(x_2-1)^2+4 <= 4 AA x`
`-> \sqrt{-x_2^2 -2x+3} <= 2 AA x`
`-> C <= 4 -2 = 2`
Dấu `=` xảy ra : $x_1 =2$ và $x_2 =1$

đó là các số -7; -2;1;2;4;5;8;13

\(ĐK:n\in Z;n\ne3\)

\(n+7⋮\left(n-3\right)=>\left(n-3\right)+10⋮\left(n-3\right)\\ =>10⋮\left(n-3\right)\\ =>n-3\inƯ\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\\ =>n\in\left\{4;5;8;13;2;1;-2;-7\right\}\left(TMDK\right)\)