\(P=\dfrac{5a+10b+15c}{4}+\left(\dfrac{3}{a}+\dfrac{3a}{4}\right)+\left(\dfrac{9}{2b}+\dfrac{b}{2}\right)+\left(\dfrac{4}{c}+\dfrac{c}{4}\right)\)

\(\ge\dfrac{5\left(a+2b+3c\right)}{4}+2\sqrt{\dfrac{3}{a}.\dfrac{3a}{4}}+2\sqrt{\dfrac{9}{2b}.\dfrac{b}{2}}+2\sqrt{\dfrac{4}{c}.\dfrac{c}{4}}\)

\(\Leftrightarrow P\ge\dfrac{5.20}{4}+3+3+2=33\)

Dấu "=" xảy ra khi a=2;b=3;c=4

Vậy \(P_{min}=33\)

3a)\(\left\{{}\begin{matrix}\dfrac{1}{x-2}+\dfrac{1}{2y-1}=2\\\dfrac{2}{x-2}-\dfrac{3}{2y-1}=1\end{matrix}\right.\) (ĐK: x≠2;y≠\(\dfrac{1}{2}\))

Đặt \(\dfrac{1}{x-2}=a;\dfrac{1}{2y-1}=b\) (ĐK: a>0; b>0)

Hệ phương trình đã cho trở thành

\(\left\{{}\begin{matrix}a+b=2\\2a-3b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2-b\\2\left(2-b\right)-3b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2-b\\4-2b-3b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2-b\\b=\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{7}{5}\left(TM\text{Đ}K\right)\\b=\dfrac{3}{5}\left(TM\text{Đ}K\right)\end{matrix}\right.\) Khi đó \(\left\{{}\begin{matrix}\dfrac{1}{x-2}=\dfrac{7}{5}\\\dfrac{1}{2y-1}=\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7\left(x-2\right)=5\\3\left(2y-1\right)=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7x-14=5\\6y-3=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{19}{7}\left(TM\text{Đ}K\right)\\y=\dfrac{4}{3}\left(TM\text{Đ}K\right)\end{matrix}\right.\) Vậy hệ phương trình đã cho có nghiệm duy nhất (x;y)=\(\left(\dfrac{19}{7};\dfrac{4}{3}\right)\)

b) Bạn làm tương tự như câu a kết quả là (x;y)=\(\left(\dfrac{12}{5};\dfrac{-14}{5}\right)\)

c)\(\left\{{}\begin{matrix}3\sqrt{x-1}+2\sqrt{y}=13\\2\sqrt{x-1}-\sqrt{y}=4\end{matrix}\right.\)(ĐK: x≥1;y≥0)

\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-1}+2\sqrt{y}=13\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-1}+4\sqrt{x-1}=13\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7\sqrt{x-1}=13\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}49\left(x-1\right)=169\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}49x-49=169\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{218}{49}\\y=\dfrac{4}{49}\end{matrix}\right.\left(TM\text{Đ}K\right)\)

Bài 4:

Theo đề, ta có hệ:

\(\left\{{}\begin{matrix}3\left(3a-2\right)-2\left(2b+1\right)=30\\3\left(a+2\right)+2\left(3b-1\right)=-20\end{matrix}\right.\)

=>9a-6-4b-2=30 và 3a+6+6b-2=-20

=>9a-4b=38 và 3a+6b=-20+2-6=-24

=>a=2; b=-5

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=\dfrac{5}{6}\\\dfrac{6a+14b}{9}=\dfrac{9}{9}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=\dfrac{5}{6}\\6a+14b=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{5}{6}-b\\6\left(\dfrac{5}{6}-b\right)+14b=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{5}{6}-b\\5-6b+14b=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{5}{6}-b\\8b=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{5}{6}-\dfrac{1}{2}=\dfrac{1}{3}\\b=\dfrac{1}{2}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}a+b=\dfrac{5}{6}\\\dfrac{2a}{3}+\dfrac{14b}{9}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{14a}{9}+\dfrac{14b}{9}=\dfrac{35}{27}\\\dfrac{2a}{3}+\dfrac{14b}{9}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{8}{9}a=\dfrac{8}{27}\\\dfrac{2a}{3}+\dfrac{14b}{9}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{3}\\\dfrac{2a}{3}+\dfrac{14b}{9}=1\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{2.\dfrac{1}{3}}{3}+\dfrac{14b}{9}=1\)

\(\Leftrightarrow\dfrac{14b}{9}=1-\dfrac{2}{9}=\dfrac{7}{9}\)

\(\Leftrightarrow b=\dfrac{7}{9}:\dfrac{14}{9}=\dfrac{1}{2}\)

Vậy ...

\(Q=\dfrac{2002}{a}+\dfrac{2017}{b}+2996a-5501b=\left(\dfrac{2002}{a}+8008a\right)+\left(\dfrac{2017}{b}+2017b\right)-\left(5012a+7518b\right)\)

\(=\left(\dfrac{2002}{a}+8008a\right)+\left(\dfrac{2017}{b}+2017b\right)-2506\left(2a+3b\right)\)

Áp dụng bất đẳng thức Cosi cho 2 số dương:

\(\left\{{}\begin{matrix}\dfrac{2002}{a}+8008\ge2\sqrt{\dfrac{2002}{a}.8008}=8008\\\dfrac{2017}{b}+2017b\ge2\sqrt{\dfrac{2017}{b}.2017b}=4034\end{matrix}\right.\)

Ta có: \(2a+3b=4\Rightarrow-\left(2a+3b\right)=-4\Leftrightarrow-2506\left(2a+3b\right)=-10024\)

\(\Rightarrow Q\ge8008+4034-10024=2018\)

\(ĐTXR\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=1\end{matrix}\right.\)

a: \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+2\right)\left(y+3\right)-xy=100\\xy-\left(x-2\right)\left(y-2\right)=64\end{matrix}\right.\)

=>xy+3x+2y+6-xy=100 và xy-xy+2x+2y-4=64

=>3x+2y=94 và 2x+2y=68

=>x=26 và x+y=34

=>x=26 và y=8

b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3x+3+2}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x+2-2}{x+1}-\dfrac{5y+20-11}{y+4}=9\end{matrix}\right.\)

=>\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x+1}-\dfrac{2}{y+4}=4-3=1\\\dfrac{-2}{x+1}+\dfrac{11}{y+4}=9+5-2=12\end{matrix}\right.\)

=>x+1=18/35; y+4=9/13

=>x=-17/35; y=-43/18

\(a) \begin{cases}x=y+4\\2x+3=0\end{cases}\Leftrightarrow\begin{cases}x = y + 4\\2x = -3\end{cases}\Leftrightarrow\begin{cases}\dfrac{-3}{2} = y + 4\\x = \dfrac{-3}{2}\end{cases}\Leftrightarrow\begin{cases}y = \dfrac{-11}{2}\\x = \dfrac{-3}{2}\end{cases}\\b) \begin{cases}2x + y = 7\\3y - x = 7\end{cases}\Leftrightarrow\begin{cases}2x + y = 7\\6y - 2x = 14\end{cases}\Leftrightarrow\begin{cases}2x + y = 7\\7y = 21\end{cases}\Leftrightarrow\begin{cases}2x + 3 = 7\\y = 3\end{cases}\Leftrightarrow\begin{cases}x=2\\y=3\end{cases}\\ c) \begin{cases} 5x + y = 3 \\ -x - \dfrac{1}{5}y=\dfrac{-3}{5} \end{cases} \Leftrightarrow \begin{cases} 5x + y = 3 \\ 5x + y = 3 \end{cases} (luôn\ đúng) \Leftrightarrow Phương\ trình\ vô\ số\ nghiệm \\d) \begin{cases} 3x - 5y = -18 \\ x - 5 = 2y \end{cases} \Leftrightarrow \begin{cases} 3x - 5y = -18 \\ 3x - 6y = 15 \end{cases} \Leftrightarrow \begin{cases} x - 5 = 2.(-33)\\ y = -13 \end{cases} \Leftrightarrow \begin{cases}x = -61\\y=-33 \end{cases} \)

a: \(\left\{{}\begin{matrix}4\sqrt{5}-y=3\sqrt{2}\\10x+\sqrt{2}\cdot y=-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=4\sqrt{5}-3\sqrt{2}\\10x+\sqrt{2}\left(4\sqrt{5}-3\sqrt{2}\right)=-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=4\sqrt{5}-3\sqrt{2}\\10x=-1-4\sqrt{10}+6=5-4\sqrt{10}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=4\sqrt{5}-3\sqrt{2}\\x=\dfrac{1}{2}-\dfrac{2\sqrt{10}}{5}\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}\dfrac{3}{4}x+\dfrac{2}{5}y=2,3\\x-\dfrac{3}{5}y=0,8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{9}{4}x+\dfrac{6}{5}y=6,9\\2x-\dfrac{6}{5}y=1,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{17}{4}x=8,5\\x-0,6y=0,8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=8,5:\dfrac{17}{4}=8,5\cdot\dfrac{4}{17}=2\\0,6y=x-0,8=2-0,8=1,2\end{matrix}\right.\)

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

c: ĐKXĐ: y>2

\(\left\{{}\begin{matrix}\left|x-1\right|-\dfrac{3}{\sqrt{y-2}}=-1\\2\left|1-x\right|+\dfrac{1}{\sqrt{y-2}}=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2\left|x-1\right|-\dfrac{6}{\sqrt{y-2}}=-2\\2\left|x-1\right|+\dfrac{1}{\sqrt{y-2}}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{7}{\sqrt{y-2}}=-7\\2\left|1-x\right|+\dfrac{1}{\sqrt{y-2}}=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\sqrt{y-2}=1\\2\left|x-1\right|=5-1=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-2=1\\\left|x-1\right|=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=3\\x-1\in\left\{2;-2\right\}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=3\\x\in\left\{3;-1\right\}\end{matrix}\right.\left(nhận\right)\)

Tất cả các hpt này đều giải bằng PP đặt ẩn phụ

a) \(\begin{cases}2\left(x+y\right)+3\left(x-y\right)=4\\\left(x+y\right)+2\left(x-y\right)=5\end{cases}\)

Đặt \(x+y=a\) ; \(x-y=b\) ta được:

\(\begin{cases}2a+3b=4\\a+2b=5\end{cases}\) \(\Leftrightarrow\) \(\begin{cases}2a+3b=4\\2a+4b=10\end{cases}\)\(\Leftrightarrow\) \(\begin{cases}-b=-6\\2a+4b=10\end{cases}\)

\(\Leftrightarrow\) \(\begin{cases}b=6\\2a+4.6=10\end{cases}\) \(\Leftrightarrow\) \(\begin{cases}a=-7\\b=6\end{cases}\) \(\Leftrightarrow\) \(\begin{cases}x+y=6-7\\x-y=6-7\end{cases}\)

\(\Leftrightarrow\) \(\begin{cases}x-7=-1\\6-y=-1\end{cases}\) \(\Leftrightarrow\) \(\begin{cases}x=6\\y=-7\end{cases}\)

Lúc khác mình làm tiếp mấy câu kia

Tiếp nào!

b) \(\begin{cases}\dfrac{3}{x}-\dfrac{4}{y}=2\\\dfrac{4}{x}-\dfrac{5}{y}=3\end{cases}\) Đặt \(\dfrac{1}{x}=a\) ; \(\dfrac{1}{y}=b\) ta được:

\(\begin{cases}3a-4b=2\\4a-5b=3\end{cases}\) \(\Leftrightarrow\) \(\begin{cases}12a-16b=8\\12a-15b=9\end{cases}\) \(\Leftrightarrow\) \(\begin{cases}-1b=-1\\12a-15b=9\end{cases}\)

\(\Leftrightarrow\) \(\begin{cases}b=1\\a=2\end{cases}\)\(\Leftrightarrow\) \(\begin{cases}a=2\\b=1\end{cases}\) \(\Leftrightarrow\) \(\begin{cases}\dfrac{1}{a}=2\\\dfrac{1}{b}=1\end{cases}\) \(\Leftrightarrow\) \(\begin{cases}a=\dfrac{1}{2}\\b=1\end{cases}\)

c) Làm tương tự thay \(\dfrac{1}{2x-y}=a\) ; \(\dfrac{1}{x+y}=b\)