in 

I think .-.

Bạn cho mình xem đề đi bạn

19.

\(\left(a+b\right)^2\le2\left(a^2+b^2\right)=4\Rightarrow-2\le a+b\le2\)

\(P=3\left(a+b\right)+ab=3\left(a+b\right)+\dfrac{\left(a+b\right)^2-\left(a^2+b^2\right)}{2}=\dfrac{1}{2}\left(a+b\right)^2+3\left(a+b\right)-1\)

Đặt \(a+b=x\Rightarrow-2\le x\le2\)

\(P=\dfrac{1}{2}x^2+3x-1=\dfrac{1}{2}\left(x+2\right)\left(x+4\right)-5\ge-5\) (đpcm)

Dấu "=" xảy ra khi \(x=-2\) hay \(a=b=-1\)

20.

Đặt \(P=2a+2ab+abc\)

\(P=2a+ab\left(2+c\right)\le2a+\dfrac{a}{4}\left(b+2+c\right)^2=2a+\dfrac{a}{4}\left(7-a\right)^2\)

\(P\le\dfrac{1}{4}\left(a^3-14a^2+57a-72\right)+18=18-\dfrac{1}{4}\left(8-a\right)\left(a-3\right)^2\le18\) (đpcm)

Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(3;2;0\right)\)

Câu 3:

\(n_{CuO}=\dfrac{6,4}{80}=0,08(mol)\\ n_{HCl}=\dfrac{36,5.20}{100.36,5}=0,2(mol)\\ CuO+2HCl\to CuCl_2+H_2\\ \dfrac{n_{CuO}}{1}<\dfrac{n_{HCl}}{2}\Rightarrow HCl\text{ dư}\\ \Rightarrow n_{HCl(dư)}=0,2-0,08.2=0,04(mol)\\ n_{CuCl_2}=n_{H_2}=0,08(mol);n_{HCl(p/ứ)}=0,16(mol)\\ \Rightarrow C\%_{CuCl_2}=\dfrac{0,08.135}{6,4+36,5-0,08.2}.100\%=25,27\%\\ C\%_{HCl(dư)}=\dfrac{0,04.36,5}{6,4+36,5-0,08.2}.100\%=3,42\%\)

đâu

\(\left(4xy^2-6x^2y+8x^2y^2\right):\left(2xy\right)-3y\\ =\left(4xy^2:2xy-6x^2y:2xy+8x^2y^2:2xy\right)-3y\\ =2y-3x+4xy-3y=-y-3x+4xy\)

5:

d: \(A=\dfrac{9\left(x_1+x_2\right)+10-3m}{18\left(x_1x_2+2\right)^2+1}\)

\(=\dfrac{9\cdot\dfrac{m-2}{3}+10-3m}{18\cdot\left(\dfrac{m-6}{3}+2\right)^2+1}=\dfrac{3m-6+10-3m}{18\cdot\left(\dfrac{m-6+6}{3}\right)^2+1}\)

\(=\dfrac{4}{18\cdot\dfrac{m^2}{9}+1}=\dfrac{4}{2m^2+1}< =\dfrac{4}{1}=4\)

Dấu = xảy ra khi m=0