Ta có:

`P=a+b+1/a+1/b`

`=16/16a+25/25b+1/a+1/b`

`=(1/16a+1/a)+(1/25b+1/b)+15/16a+24/25b`

\(\left\{{}\begin{matrix}a\ge4\\b\ge5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{16}a>0;\dfrac{1}{a}>0\\\dfrac{1}{25}b>0;\dfrac{1}{b}>0\end{matrix}\right.\)

Áp dụng bđt cô-si ta có: 

\(P\ge2\sqrt{\dfrac{1}{16}a\cdot\dfrac{1}{a}}+2\sqrt{\dfrac{1}{25}b\cdot\dfrac{1}{b}}+\dfrac{15}{16}a+\dfrac{24}{25}b\\ \ge2\cdot\dfrac{1}{4}+2\cdot\dfrac{1}{5}+\dfrac{15}{16}\cdot4+\dfrac{24}{25}\cdot5=\dfrac{189}{20}\)

Dấu "=" xảy ra khi: `a=4` và `b=5` 

Ta có:

`P=a+b+1/a+1/b`

`=16/16a+25/25b+1/a+1/b`

`=(1/16a+1/a)+(1/25b+1/b)+15/16a+24/25b`

\(\left\{{}\begin{matrix}a\ge4\\b\ge5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{16}a>0;\dfrac{1}{a}>0\\\dfrac{1}{25}b>0;\dfrac{1}{b}>0\end{matrix}\right.\)

Áp dụng bđt cô-si ta có: 

\(P\ge2\sqrt{\dfrac{1}{16}a\cdot\dfrac{1}{a}}+2\sqrt{\dfrac{1}{25}b\cdot\dfrac{1}{b}}+\dfrac{15}{16}a+\dfrac{24}{25}b\\ \ge2\cdot\dfrac{1}{4}+2\cdot\dfrac{1}{5}+\dfrac{15}{16}\cdot4+\dfrac{24}{25}\cdot5=\dfrac{189}{20}\)

Dấu "=" xảy ra khi: `a=4` và `b=5` 

`25%x+x=-1,25`

`=>0,25x+1*x=-1,25`

`=>x*(0,25+1)=-1,25`

`=>x*1,25=-1,25`

`=>x=(-1,25)/(1,25)`

`=>x=-1`

Vậy: ...

a) 

\(\left\{{}\begin{matrix}\dfrac{1}{x+y}-\dfrac{2}{x-y}=2\\\dfrac{5}{x+y}-\dfrac{4}{x-y}=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x+y}-\dfrac{4}{x-y}=4\\\dfrac{5}{x+y}-\dfrac{4}{x-y}=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x+y}=-1\\\dfrac{1}{x+y}-\dfrac{2}{x-y}=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+y=-3\\-\dfrac{1}{3}-\dfrac{2}{x-y}=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+y=-3\\\dfrac{2}{x-y}=-\dfrac{7}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=-3\\x-y=-\dfrac{7}{6}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-\dfrac{16}{3}\\x+y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{8}{3}\\y=-3+\dfrac{8}{3}=-\dfrac{1}{3}\end{matrix}\right.\) 

`1)x/z^2*(xz)/y^3:x^3/(yz)`

`=x/z^2*(xz)/y^3*(yz)/x^3`

`=(x*xz*yz)/(z^2*y^3*x^3)`

`=1/(xy^2)`

`2)2/x-2/x:1/x+4/x*x^2/2`

`=2/x-2/x*x+2x`

`=2/x-2+2x`

`=(2x^2-2x+2)/x`

`3)((1-x)/x+x^2-1):(x-1)/x`

`=((1-x)/x+x^2-1)*x/(x-1)`

`=(1-x)/x*x/(x-1)+(x^2-1)*x/(x-1)`

`=-1+x(x+1)`

`=x^2+x-1` 

`(39*17+39*12):29`

`=[39*(17+12)]:29`

`=39*29:29`

`=39*1`

`=39`

`1)(1-4x^2)/(x^2+4x):(2-4x)/(3x)`

`=(1-4x^2)/(x^2+4x)*(3x)/(2-4x)`

`=((1-2x)(1+2x))/(x(x+4))*(3x)/(2(1-2x))`
`=(3(1+2x))/(2x)`

`=(3+6x)/(2x)` 

`2)x^2/y*(xz)/y^2:x^2/y^2`

`=x^2/y*(xz)/y^2*y^2/x^2`

`=x^2/y*z/x`

`=(xz)/y` 

`3)(-25a^3b^5)/(3cd^2):(15ab^2)`

`=(-25a^3b^5)/(3cd^2)*1/(15ab^2)`

`=(-5a^2b^3)/(3cd^2)`

`1)x^3/(x+1975)*(2x+1954)/(x+1)+x^3/(x+1975)*(21-x)/(x+1)`

`=(x^3(2x+1954))/((x+1975)(x+1))+(x^3(21-x))/((x+1975)(x+1))`

`=(2x^4+1954x^3+21x^3-x^4)/((x+1975)(x+1))`

`=(x^4+1975x^3)/((x+1975)(x+1))`

`=(x^3(x+1975))/((x+1975)(x+1))`

`=x^3/(x+1)`

`2)(19x+8)/(x-7)*(5x-9)/(x+1945)-(19+8)(x-7)*(4x-2)/(x+1945)`

`=(19x+8)/(x-7)*((5x-9)/(x+1945)-(4x-2)/(x+1945))`

`=(19x+8)/(x-7)*((5x-9-4x+2)/(x+1945))`

`=(19x+8)/(x-7)*(x-7)/(x+1945)`

`=(19x+8)/(x+1945)` 

`a)2^x-15=17`

`=>2^x=15+17`

`=>2^x=32`

`=>2^x=2^5`

`=>x=5`

`b)(7x-11)^3=2^5*5^2+200`

`=>(7x-11)^3=32*25+200`

`=>(7x-11)^3=1000`

`=>(7x-11)^3=10^3`

`=>7x-11=10`

`=>7x=10+11`

`=>x=21/7=3`

`c)x^3=4^2*2^2`

`=>x^3=16*4`

`=>x^3=64`

`=>x^3=4^3`

`=>x=4`

`d)x^2021=x^10`

`=>x^2021-x^10=0`

`=>x^10*(x^2011-1)=0`

`=>x^10=0` hoặc `x^2011-1=0`

`=>x=0` hoặc `x^2011=1^2011`

`=>x=0` hoặc `x=1`