\(P=\left(2x-5y\right)^2-\left(15y-6x\right)^2-\left|xy-90\right|\)

\(\Leftrightarrow P=\left(2x-5y\right)^2-\left(6x-15y\right)^2-\left|xy-90\right|\)

\(\Leftrightarrow P=\left(2x-5y\right)^2-3\left(2x-3y\right)^2-\left|xy-90\right|\)

\(\Leftrightarrow P=\left(2x-5y\right)^2.\left(1-3\right)-\left|xy-90\right|\)

\(\Leftrightarrow P=-4\left(2x-5y\right)^2-\left|xy-90\right|\)

\(\Leftrightarrow P=-\left[4\left(2x-5y\right)^2-\left|xy-90\right|\right]\)

Ta có \(\hept{\begin{cases}\left(2x-5y\right)^2\ge0\\\left|xy-90\right|\ge0\end{cases}}\forall xy\)

\(\Rightarrow\hept{\begin{cases}4\left(2x-5y\right)^2\ge0\\\left|xy-90\right|\ge0\end{cases}}\forall xy\)

\(\Rightarrow P=-\left[4\left(2x-5y\right)^2+\left|xy-90\right|\right]\le0\forall xy\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}4\left(2x-5y\right)^2=0\\\left|xy-90\right|=0\end{cases}}\)

                      \(\Leftrightarrow\hept{\begin{cases}\left(2x-5y\right)^2=0\\xy-90=0\end{cases}}\)

                     \(\Leftrightarrow\hept{\begin{cases}2x-5y=0\\xy=90\end{cases}}\)

                     \(\Leftrightarrow\hept{\begin{cases}2x=5y\\xy=90\end{cases}}\)

               \(\Leftrightarrow2xy=5y^2\)\(\Leftrightarrow2.90=5y^2\Leftrightarrow5y^2=180\Leftrightarrow y^2=36\)

                                                                                                              \(\Rightarrow\orbr{\begin{cases}y=6\\y=-6\end{cases}}\Rightarrow\orbr{\begin{cases}x=90:6=15\\x=90:\left(-6\right)=-15\end{cases}}\)

Vậy \(P_{max}=0\Leftrightarrow x=15;y=6\)  hoặc x=-15; y=-6

Có 1 vài chỗ ko ok cho lắm bạn thông cảm

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Trả lời : 

Bn tham khảo link này :  

https://olm.vn/hoi-dap/detail/216085412740.html 

( Vào thống kê hỏi đáp của mk sẽ thấy ) 

\(log_{\sqrt{3}}\left(2x+y\right)-log_{\sqrt{3}}\left(4x^2+y^2+2xy+2\right)=\left(4x^2+y^2+2xy+2\right)-3\left(2x+y\right)-2\)

\(\Leftrightarrow log_{\sqrt{3}}\left(2x+y\right)+2+3\left(2x+y\right)=log_{\sqrt{3}}\left(4x^2+y^2+2xy+2\right)+\left(4x^2+y^2+2xy+2\right)\)

\(\Leftrightarrow log_{\sqrt{3}}\left(6x+3y\right)+\left(6x+3y\right)=log_{\sqrt{3}}\left(4x^2+y^2+2xy+2\right)+\left(4x^2+y^2+2xy+2\right)\)

Xét hàm \(f\left(t\right)=log_{\sqrt{3}}t+t\) với \(t>0\)

\(f'\left(t\right)=\dfrac{1}{t.ln\sqrt{3}}+1>0\Rightarrow f\left(t\right)\) đồng biến

\(\Rightarrow6x+3y=4x^2+y^2+2xy+2\)

\(\Leftrightarrow4x+y=\left(x+y-1\right)^2+1+3\left(x^2+1\right)-3\ge2\left(x+y-1\right)+6x-3\)

\(\Leftrightarrow4x+y\ge2\left(4x+y\right)-5\)

\(\Leftrightarrow4x+y\le5\)

\(\Rightarrow P=\dfrac{2x+y+6+\left(4x+y-5\right)}{2x+y+6}=1+\dfrac{4x+y-5}{2x+y+6}\le1\)

\(P_{max}=1\) khi \(x=y=1\)

a: \(3x^2y\left(2x^2-xy+5y^2\right)=6x^4y-3x^3y^2+15x^2y^3\)

b: \(\left(x+2\right)\left(x^2+3x-4\right)\)

\(=x^3+3x^2-4x+2x^2+6x-8\)

\(=x^3+5x^2+2x-8\)

Xét \(y=0\)\(\Rightarrow...\)

Xét \(y\ne0\). Ta có:

\(\left\{{}\begin{matrix}x^2+y^2+xy+2x=5y\\\left(x^2+2x\right)\left(x+y-3\right)=-3y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+2x=5y-y^2-xy\left(1\right)\\\left(x^2+2x\right)\left(x+y-3\right)=-3y\left(2\right)\end{matrix}\right.\)

Thay (1) vào (2), ta có:

\(\left(5y-y^2-xy\right)\left(x+y-3\right)=-3y\)

\(-y\left(x+y-5\right)\left(x+y-3\right)=-3y\)

\(\Leftrightarrow\left(x+y-5\right)\left(x+y-3\right)=3\left(\cdot\right)\)

Đặt \(x+y-5=t\), phương trình \(\left(\cdot\right)\) trở thành

\(t\left(t+2\right)=3\)\(\Leftrightarrow t^2+2t+1=4\Leftrightarrow\left(t+1\right)^2=4\)

\(\Leftrightarrow\left[{}\begin{matrix}t+1=2\\t+1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+y-5=1\\x+y-5=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+y=6\\x+y=2\end{matrix}\right.\)\(\Rightarrow...\)

1A,B,D

2 M=2

3 \(=\dfrac{3}{4x}\)

4 \(=\dfrac{4\left(x+y\right)}{x-y}=\dfrac{4x+4y}{x-y}\)

5 K rút gọn đc

6 \(=\dfrac{4\left(x-1\right)+2\left(x-1\right)}{6\left(x-1\right)}=\dfrac{6\left(x-1\right)}{6\left(x-1\right)}=1\)

a)

\(\begin{array}{l}\left( {2x - 5y} \right)\left( {2x + 5y} \right) + {\left( {2x + 5y} \right)^2}\\ = \left( {2x + 5y} \right)\left( {2x - 5y + 2x + 5y} \right)\\ = \left( {2x + 5y} \right).4x\\ = 2x.4x + 5y.4x\\ = 8{x^2} + 20xy\end{array}\)

b)

\(\begin{array}{l}\left( {x + 2y} \right)\left( {{x^2} - 2xy + 4{y^2}} \right) + \left( {2x - y} \right)\left( {4{x^2} + 2xy + {y^2}} \right)\\ = {x^3} + {\left( {2y} \right)^3} + {\left( {2x} \right)^3} - {y^3}\\ = {x^3} + 8{y^3} + 8{x^3} - {y^3}\\ = \left( {{x^3} + 8{x^3}} \right) + \left( {8{y^3} - {y^3}} \right)\\ = 9{x^3} + 7{y^3}\end{array}\)

\(=\left[7\left(4x^2-25y^2-4x+10y\right)\left(14x^2-3y^2\right):\left(-3y\right)\right]\)

\(=\dfrac{7\left(56x^2-362x^2y^2+75y^4-56x^3+12xy^2-140x^2y-30y^3\right)}{-3y}\)

\(=\dfrac{7\left(56x^2-362x^2y^2+75y^4-56x^3+12xy^2-140x^2y-30y^3\right)}{-3y}\)

\(=\dfrac{-392x^2}{3y}+\dfrac{2534}{3}x^2y-175y^3+\dfrac{392}{3}x^3:y-28xy+\dfrac{980}{3}x^2+70y^2\)