Ta có :  \(A_i=P.h=850.5=4250\left(J\right)\)

\(A_{tp}=\dfrac{A_i}{H}=6071\left(J\right)\)

\(t=\dfrac{A_{tp}}{P}=4,2\left(s\right)\)

Vậy...

Gọi hiệu của hai số là \(a\)

Ta có: 

 Tổng của chúng là \(7a\)

Tích của chúng bằng \(192a\)

Vậy : Số nhỏ nhất là :  \(\left(7a-a\right):2=3a\)

Số lớn nhất là :  \(\left(7a+a\right):2=4a\)

\(\Leftrightarrow\) Số nhỏ nhất là:  \(\dfrac{192a}{3a}=64\)

Số lớn nhất là: \(\dfrac{192a}{4a}=48\)

Vậy...

\(\dfrac{25^5+25^7+25^9}{5^{11}+5^{13}+5^{15}+5^{17}+5^{19}+5^{21}}\)

\(=\dfrac{5^{10}+5^{14}+5^{18}}{5^{11}+5^{13}+5^{17}+5^{19}+5^{21}}\)

\(=\dfrac{5^{10}\left(1+5^4+5^8\right)}{5^{11}\left(1+5^2+5^4+5^6+5^8+5^{10}\right)}\)

\(=\dfrac{1+5^4+5^8}{5+5^3+5^5+5^7+5^9+5^{11}}\)

\(=\dfrac{1+5^4+5^8}{\left(5+5^5+5^9\right)+\left(5^3+5^7+5^{11}\right)}\)

\(=\dfrac{1+5^4+5^8}{\left(1+5^4+5^8\right)\left(5+5^3\right)}\)

\(=\dfrac{1}{5+5^3}\)

\(=\dfrac{1}{130}\)

\(\)

a/ Thay  \(m=1\) vào hpt ta có :

\(\left\{{}\begin{matrix}x+2y=4\\2x-3y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

Vậy...

b/ Ta có :

\(\left\{{}\begin{matrix}x+2y=m+3\\2x-3y=m\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{m+3}{2y}\\\dfrac{2\left(m+3\right)}{2y}-3y=m\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{m+3}{2y}\\\dfrac{m+3}{y}-3y=m\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{m+3}{2y}\\m-3y^2+3=my\end{matrix}\right.\)

a/ ĐKXĐ : \(x\ne1;-2\)

\(\dfrac{2x}{x-1}-\dfrac{1}{x+2}=2\)

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

\(\Leftrightarrow2x^2+3x-x+1=2x^2+4x-2x-4\)

\(\Leftrightarrow2x+1=2x-4\)

\(\Leftrightarrow1=-4\left(loại\right)\)

Vậy...

b/ĐKXĐ :  \(x\ne\pm5\)

\(\dfrac{x}{x^2-25}-\dfrac{1-x}{x-5}=\dfrac{1}{x+5}\)

\(\Leftrightarrow\dfrac{x}{\left(x-5\right)\left(x+5\right)}+\dfrac{\left(x-1\right)\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}=\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}\)

\(\Leftrightarrow x+x^2+5x-x-5=x-5\)

\(\Leftrightarrow x^2+4x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)

Vậy...

Gọi \(ƯCLN\left(5a+2b;7a+3b\right)=d\) \(\left(d\in N\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}5a+2b⋮d\\7a+3b⋮d\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}15a+6b⋮d\\14a+6b⋮d\end{matrix}\right.\)

\(\Leftrightarrow a⋮d\)

Mà \(5a+2b⋮d\) \(\Leftrightarrow b⋮d\)

\(\Leftrightarrow d⋮a,b\Leftrightarrow d⋮d'\left(1\right)\)

Gọi \(d'=ƯCLN\left(a,b\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}a⋮d'\\b⋮d'\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5a+2b⋮d'\\7a+3b⋮d'\end{matrix}\right.\)

\(\Leftrightarrow d'⋮d\) \(\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)

\(F_A=P-P_1=18-12=6\left(N\right)\)

\(\Leftrightarrow V=\dfrac{F_A}{d}=\dfrac{6}{136000}\left(m^3\right)\)

\(m=\dfrac{P}{10}=1,8\left(kg\right)\)

\(D=\dfrac{m}{V}=\dfrac{1,8}{\dfrac{6}{136000}}=40800\left(kg\backslash m^3\right)\)

@@ t từng phát ngôn thế này à