A

D

\(\left\{{}\begin{matrix}SO\perp BC\\SO\perp CA\end{matrix}\right.\) \(\Rightarrow SO\perp\left(ABC\right)\)

\(AA'=\dfrac{a\sqrt{3}}{2}\) (trung tuyến tam giác đều) \(\Rightarrow AO=\dfrac{2}{3}AA'=\dfrac{a\sqrt{3}}{3}\)

\(\Rightarrow M\) nằm trên đoạn thẳng OA'

Qua M kẻ đường thẳng song song BC cắt AB và AC lần lượt tại D và E

Trong mp (SAA'), qua M kẻ đường thẳng song song SO cắt SA' tại F

Trong mp (SBC), qua F kẻ đường thẳng song song BC cắt SB và SC lần lượt tại G và H

\(\Rightarrow\) Hình thang DEHG là thiết diện của (P) và chóp

\(FM||SO\Rightarrow FM\perp\left(ABC\right)\Rightarrow FM\perp ED\)

Áp dụng định lý Talet cho tam giác ABC:

\(\dfrac{DE}{BC}=\dfrac{AM}{AA'}\Rightarrow DE=\dfrac{BC.AM}{AA'}=\dfrac{a.x}{\dfrac{a\sqrt{3}}{2}}=\dfrac{2x\sqrt{3}}{3}\)

Talet tam giác SOA':

\(\dfrac{FM}{SO}=\dfrac{MA'}{OA'}\Rightarrow FM=\dfrac{SO.MA'}{OA'}=\dfrac{2a.\left(\dfrac{a\sqrt{3}}{2}-x\right)}{\dfrac{a\sqrt{3}}{6}}=6a-4\sqrt{3}x\)

Talet tam giác SBC:

\(\dfrac{GH}{BC}=\dfrac{SF}{SA'}=1-\dfrac{FA'}{SA'}=1-\dfrac{FM}{SO}=1-\dfrac{6a-4\sqrt{3}x}{2a}=\dfrac{2\sqrt{3}x-2a}{a}\)

\(\Rightarrow GH=2\sqrt{3}x-2a\)

\(S_{DEHG}=\dfrac{1}{2}\left(DE+GH\right).FM=\dfrac{1}{2}\left(\dfrac{2x\sqrt{3}}{3}+2\sqrt{3}x-2a\right)\left(6a-4\sqrt{3}x\right)\)

\(=\dfrac{1}{3}\left(4\sqrt{3}x-3a\right)\left(6a-4\sqrt{3}x\right)\le\dfrac{1}{12}\left(4\sqrt{3}x-3a+6a-4\sqrt{3}x\right)^2=\dfrac{9a^2}{12}\)

Dấu "=" xảy ra khi \(4\sqrt{3}x-3a=6a-4\sqrt{3}x\Leftrightarrow x=\dfrac{9a}{8\sqrt{3}}=\dfrac{3a\sqrt{3}}{8}\)

Ta có: (u.v)' = u'.v + u.v'

\(Q=80K^{\dfrac{1}{3}}\left(100-K\right)^{\dfrac{1}{2}}\)

\(Q'=80.\left(K^{\dfrac{1}{3}}\right)'.\left(100-K\right)^{\dfrac{1}{2}}+80.K^{\dfrac{1}{3}}.\left(\left(100-K\right)^{\dfrac{1}{2}}\right)'\)= \(80.\dfrac{1}{3}.K^{-\dfrac{2}{3}}.\left(100-K\right)^{\dfrac{1}{2}}+80.K^{\dfrac{1}{3}}.\dfrac{1}{2}.\left(100-K\right)^{-\dfrac{1}{2}}.\left(-1\right)\) = \(80.\left(\dfrac{\left(100-K\right)^{\dfrac{1}{2}}}{3K^{\dfrac{2}{3}}}-\dfrac{K^{\dfrac{1}{3}}}{2\left(100-K\right)^{\dfrac{1}{2}}}\right)\)= \(80.\left(\dfrac{2\left(100-K\right)^{\dfrac{1}{2}}\left(100-K\right)^{\dfrac{1}{2}}-3K^{\dfrac{2}{3}}K^{\dfrac{1}{3}}}{6K^{\dfrac{2}{3}}\left(100-K\right)^{\dfrac{1}{2}}}\right)\) = \(80.\left(\dfrac{2\left(100-K\right)-3K}{6K^{\dfrac{2}{3}}\left(100-K\right)^{\dfrac{1}{2}}}\right)\) = \(80.\left(\dfrac{200-5K}{6K^{\dfrac{2}{3}}\left(100-K\right)^{\dfrac{1}{2}}}\right)\) = \(\dfrac{400\left(40-K\right)}{6K^{\dfrac{2}{3}}\left(100-K\right)^{\dfrac{1}{2}}}\) = \(\dfrac{200\left(40-K\right)}{3K^{\dfrac{2}{3}}\left(100-K\right)^{\dfrac{1}{2}}}\).

\(\dfrac{x+4}{3}=\dfrac{x-11}{-6}\)

\(\dfrac{2x+8}{6}=\dfrac{-x+11}{6}\)

\(\Leftrightarrow2x+8=-x+11\)

\(\Leftrightarrow3x=3\)

\(\Leftrightarrow x=1\)

Nhân chéo ta được\(-6(x+4)=3(x-11)=>-6x-24=3x-33=>6x-3x-24+33=0=>3x+9=0=>3x=-9=>x=-3\)

mn giúp em vs ạ em cần gấp cảm ơn ạ 

Bài 3: 

Gọi số học sinh là x

Theo đề, ta có: \(x\in BC\left(12;18;21\right)\)

hay x=504

CÂU 1: 

\(\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x}{4y^3}\)

CÂU 2: 

\(\dfrac{12x^3y^2}{18xy^5}=\dfrac{2x^2}{3y^3}\)

CÂU 3: 

\(\dfrac{15x\left(x+5\right)^3}{20x^2\left(x+5\right)}=\dfrac{3\left(x+5\right)^2}{4x}\)

CÂU 4: 

\(\dfrac{3xy+x}{9y+3}=\dfrac{x\left(3y+1\right)}{3\left(3y+1\right)}=\dfrac{x}{3}\)

CÂU 5: 

\(\dfrac{3xy+3x}{9y+9}=\dfrac{3x\left(y+1\right)}{9\left(y+1\right)}=\dfrac{x}{3}\)

CÂU 6: 

\(\dfrac{x^2-xy}{5y^2-5xy}=\dfrac{x\left(x-y\right)}{5y\left(y-x\right)}=\dfrac{-x\left(y-x\right)}{5y\left(y-x\right)}=\dfrac{-x}{5y}\)

CÂU 7:

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

CÂU 8: 

\(\dfrac{7x^2+14x+7}{3x^2+3x}=\dfrac{7\left(x^2+2x+1\right)}{3x\left(x+1\right)}\\ =\dfrac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\dfrac{7\left(x+1\right)}{3x}\)

CÂU 9: 

\(\dfrac{10xy^2\left(x+y\right)}{15xy\left(x+y\right)^3}=\dfrac{2y}{3\left(x+y\right)^2}\)

Bài 10:

a: \(x^2-36=0\)

=>\(x^2=36\)

=>\(\left[\begin{array}{l}x=6\\ x=-6\end{array}\right.\)

b: \(x^3-0,25x=0\)

=>\(x\left(x^2-0,25\right)=0\)

=>x(x-0,5)(x+0,5)=0

=>\(\left[\begin{array}{l}x=0\\ x-0,5=0\\ x+0,5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=0,5\\ x=-0,5\end{array}\right.\)

c: \(\left(3x+2\right)^2-\left(x+1\right)^2=0\)

=>(3x+2-x-1)(3x+2+x+1)=0

=>(2x+1)(4x+3)=0

=>\(\left[\begin{array}{l}2x+1=0\\ 4x+3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac12\\ x=-\frac34\end{array}\right.\)

d: \(49x^2-\left(5x-3\right)^2=0\)

=>\(\left(7x\right)^2-\left(5x-3\right)^2=0\)

=>(7x-5x+3)(7x+5x-3)=0

=>(2x+3)(12x-3)=0

=>\(\left[\begin{array}{l}2x+3=0\\ 12x-3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=-3\\ 12x=3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac32\\ x=\frac{3}{12}=\frac14\end{array}\right.\)

Bài 9:

a: \(x^4+4\)

\(=x^4+4x^2+4-4x^2\)

\(=\left(x^2+2\right)^2-\left(2x\right)^2\)

\(=\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)

b: \(\left(x+y\right)^3-\left(x-y\right)^3\)

\(=\left(x+y-x+y\right)\left\lbrack\left(x+y\right)_{}^2+\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right\rbrack\)

\(=2y\cdot\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)=2y\left(3x^2+y^2\right)\)

c: \(x^4-x^2+1\)

\(=x^4+2x^2+1-3x^2\)

\(=\left(x^2+1\right)^2-\left(x\sqrt3\right)^2\)

\(=\left(x^2+1-x\sqrt3\right)\left(x^2+1+x\sqrt3\right)\)

Bài 8:

a: \(x^4-1\)

\(=\left(x^2-1\right)\left(x^2+1\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\)

b: \(1-y^3+6xy^2-12x^2y+8x^3\)

\(=8x^3-12x^2y+6xy^2-y^3+1\)

\(=\left(2x-y\right)^3+1\)

\(=\left(2x-y+1\right)\left\lbrack\left(2x-y\right)^2-\left(2x-y\right)+1\right\rbrack\)

\(=\left(2x-y+1\right)\left(4x^2-4xy+y^2-2x+y+1\right)\)

c: \(x^4-y^2\left(2x-y\right)^2\)

\(=\left(x^2\right)^2-\left(2xy-y^2\right)^2\)

\(=\left(x^2-2xy+y^2\right)\left(x^2+2xy-y^2\right)=\left(x-y\right)^2\cdot\left(x^2+2xy-y^2\right)\)

d: \(\left(x+a\right)^4-\left(x-a\right)^4\)

\(=\left\lbrack\left(x+a\right)^2-\left(x-a\right)^2\right\rbrack\cdot\left\lbrack\left(x+a\right)^2+\left(x-a\right)^2\right\rbrack\)

\(=\left(x+a-x+a\right)\left(x+a+x-a\right)\left(x^2+2xa+a^2+x^2-2xa+a^2\right)\)

\(=2a\cdot2x\cdot\left(2x^2+2a^2\right)=8ax\left(x^2+a^2\right)\)

Bài 7:

a: \(x^4+2x^2y+y^2\)

\(=\left(x^2\right)^2+2\cdot x^2\cdot y+y^2=\left(x^2+y\right)^2\)

b: \(\left(2x+y\right)^2-\left(x+2y\right)^2\)

=(2x+y+x+2y)(2x+y-x-2y)

=(3x+3y)(x-y)

=3(x+y)(x-y)

d: \(\left(8x^3-27y^3\right)-2x\left(4x^2-9y^2\right)\)

\(=8x^3-27y^3-8x^3+18xy^2=-27y^3+18xy^2\)

\(=-9y^2\left(3y-2x\right)\)

e: \(\left(x+1\right)^3+\left(x-2\right)^3\)

\(=\left(x+1+x-2\right)\left\lbrack\left(x+1\right)^2-\left(x+1\right)\left(x-2\right)+\left(x-2\right)^2\right\rbrack\)

\(=\left(2x-1\right)\left(x^2+2x+1-x^2+x-2+x^2-4x+4\right)=\left(2x-1\right)\left(x^2-x+3\right)\)

f: \(64x^3+125y^3+5y\cdot\left(16x^2-25y^2\right)\)

\(=64x^3+125y^3+80x^2y-125y^3=64x^3+80x^2y\)

\(=16x^2\left(4x+5y\right)\)

c) Ta có: \(\sqrt{\sqrt{x}+3}=3\)

\(\Leftrightarrow\sqrt{x}+3=9\)

\(\Leftrightarrow\sqrt{x}=6\)

hay x=36

Ta có: \(\sqrt{x-2\sqrt{x-1}}=2\)

\(\Leftrightarrow x-2\sqrt{x-1}-4=0\)

\(\Leftrightarrow x-1-2\cdot\sqrt{x-1}\cdot1+1=4\)

\(\Leftrightarrow\left(\sqrt{x-1}-1\right)^2=4\)

\(\Leftrightarrow\sqrt{x-1}-1=2\)

\(\Leftrightarrow\sqrt{x-1}=3\)

\(\Leftrightarrow x-1=9\)

hay x=10