The number of.illiterate.people used to be as great as the number of hungry people after war.(literate) ( illiterate people : người mù chữ ) 

a) \(\Delta ABC\perp\) tại A ; \(AH\perp BC\) ; ta có : 

\(BC^2=AB^2+AC^2=5^2+12^2=169\Rightarrow BC=13\left(cm\right)\)

\(BH=\dfrac{AB^2}{BC}=\dfrac{5^2}{13}=\dfrac{25}{13}\left(cm\right)\)

\(AH=\dfrac{AB.AC}{BC}=\dfrac{5.12}{13}=\dfrac{60}{13}\left(cm\right)\)

b) \(\Delta AHB\perp\) tại H ; \(HE\perp AB\)  ta có : \(AH^2=AE.AB\)

CMTT : \(AH^2=AF.AC\)

Suy ra : \(AE.AB=AF.AC\left(đpcm\right)\)

\(\Delta ABC\perp\) tại A ; \(AH\perp BC\) . Áp dụng HTL ; ta có : 

\(\dfrac{AB^2}{AC^2}=\dfrac{BH.BC}{CH.BC}=\dfrac{BH}{CH}\Rightarrow\left(\dfrac{5}{7}\right)^2=\dfrac{BH}{CH}\)  \(\Rightarrow\dfrac{BH}{CH}=\dfrac{25}{49}\)  \(\Rightarrow\dfrac{x}{y}=\dfrac{25}{49}\)

\(BH.CH=AH^2=15^2\Rightarrow xy=225\) 

Ta có : \(\left\{{}\begin{matrix}49x=25y\\xy=225\end{matrix}\right.\)  \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{25}{49}y\\\dfrac{25}{49}y^2=225\end{matrix}\right.\)   \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{25}{49}y\\y^2=441\end{matrix}\right.\)  \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{25}{49}y\\y=21\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{75}{7}\\y=21\end{matrix}\right.\) (cm)

Vậy ... 

A = \(\varnothing\) \(\Leftrightarrow x^2-2mx+m^2-7m+1>0\)  \(\forall x\)

\(\Leftrightarrow\Delta'< 0\) \(\Leftrightarrow m^2-\left(m^2-7m+1\right)< 0\Leftrightarrow7m-1< 0\)

\(\Leftrightarrow m< \dfrac{1}{7}\)

n với x thì làm sao mà xét được bn ? 

Dựa vào đáp án ; ta suy ra : X là ancol đơn chức 

Đặt \(n_{C_3H_6\left(OH\right)_2}=a=n_X\)

Ta có : \(a+\dfrac{a}{2}=0,075\Rightarrow a=0,05\)

Khi đó :  \(6,1=76.0,05+M_X.0,05\) \(\Rightarrow M_X=46\Rightarrow X:C_2H_5OH\)

a) Đặt : \(a=37;b=12\) ; ta có : 

\(D=\dfrac{a^3+b^3}{a+b}-ab=a^2-ab+b^2-ab=a^2-2ab+b^2=\left(a-b\right)^2\)

\(=\left(37-12\right)^2=25^2=625\)

b) Đặt : \(a=52;b=48\) ; ta có : 

\(E=\dfrac{a^3-b^3}{a-b}+ab=a^2+ab+b^2+ab=\left(a+b\right)^2=\left(52+48\right)^2=100^2=10000\) 

8) \(49x^2-\left(3x+2\right)^2=\left(7x-3x-2\right)\left(7x+3x+2\right)\)

\(=4\left(2x-1\right)\left(5x+1\right)\)

9) \(x^4-10x^2+9=x^2\left(x^2-9\right)-\left(x^2-9\right)\)

\(=\left(x^2-1\right)\left(x^2-9\right)=\left(x-1\right)\left(x+1\right)\left(x-3\right)\left(x+3\right)\)

5) \(x^6-7x^3-8=x^3\left(x^3-8\right)+x^3-8=\left(x^3+1\right)\left(x^3-8\right)\)

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

6) \(\left(x^2-x\right)^2+\left(x^2-x\right)-6\)  

Đặt t = \(x^2-x\)  ; ta có : \(t^2+t-6=t\left(t+3\right)-2\left(t+3\right)=\left(t-2\right)\left(t+3\right)\)

\(=\left(x^2-x-2\right)\left(x^2-x+3\right)=\left[x\left(x-2\right)+x-2\right]\left(x^2-x+3\right)\)

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

7) \(x^2\left(x^2-7\right)^2-36=\left[x\left(x^2-7\right)-6\right]\left[x\left(x^2-7\right)+6\right]\)

\(=\left[x^3-7x-6\right]\left[x^3-7x+6\right]\) 

\(=\left[x^2\left(x+1\right)-x\left(x+1\right)-6\left(x+1\right)\right]\left[x^2\left(x-1\right)+x\left(x-1\right)-6\left(x-1\right)\right]\)

\(=\left(x^2-x-6\right)\left(x+1\right)\left(x^2+x-6\right)\left(x-1\right)\)

\(=\left[x\left(x-3\right)+2\left(x-3\right)\right]\left(x+1\right)\left[x\left(x+3\right)-2\left(x+3\right)\right]\left(x-1\right)\)

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