\(\hept{\begin{cases}x^2+xy+x-y-2y^2=0\\x^2-y^2+x+y=6\end{cases}}\)

\(hpt\Leftrightarrow\hept{\begin{cases}\left(x-y\right)\left(x+2y+1\right)=0\left(1\right)\\\left(x+y\right)\left(x-y+1\right)=6\left(2\right)\end{cases}}\)

\(\left(1\right)\Leftrightarrow\orbr{\begin{cases}x-y=0\\x+2y+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=y\\x=-1-2y\end{cases}}\)

• Xét \(x=y\) thay vào \(\left(2\right)\) ta có:

\(\left(2\right)\Leftrightarrow2y=6\Leftrightarrow y=3\Leftrightarrow x=y=3\)

• Xét \(x=-1-2y\) thay vào \(\left(2\right)\) ta có:

\(\left(2\right)\Leftrightarrow3y^2+3y=6\Leftrightarrow3y^2+3y-6=0\)

\(\Leftrightarrow3\left(y-1\right)\left(y+2\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}y=1\\y=-2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}y=1\Rightarrow x=-1-2y=-3\\y=-2\Rightarrow x=-1-2y=3\end{cases}}\)

Vậy hpt có nghiệm là \(\left(x;y\right)=\left(3;3\right);\left(-3;1\right);\left(3;-2\right)\)

Mơn thắng nguyễn nhìu nha 

Ta đặt \(x=tanA;y=tanB;z=tanC\) với \(ABC\) là các góc tam giá từ đây cần c/m

\(sinA+sinB+sinC\le\frac{3\sqrt{3}}{2}\)

tài liệu c/m BĐT này đầy trên mạng bn có thể tham tham khảo

VD:Cm : sinA+sinB+sinC bé hơn hoặc bằng (3* căn3)/2? | Yahoo Hỏi & Đáp

Dự đoán khi \(x=y=z=\frac{1}{\sqrt{3}}\) thì ta tìm được \(P=\frac{3\sqrt{3}}{2}\)

Ta sẽ chứng minh nó là GTNN

Thật vậy, ta cần chứng minh 

\(Σ\frac{1}{\sqrt{x^2+xy+xz+yz}}\le\frac{3\sqrt{3}}{2\sqrt{xy+xz+yz}}\left(xy+yz+xz=1\right)\)

\(\LeftrightarrowΣ\sqrt{x+y}\le\frac{3\sqrt{3\left(x+y\right)\left(x+z\right)\left(y+z\right)}}{2\sqrt{xy+xz+yz}}\)

Nhưng theo BĐT Cauchy-Schwarz ta có: 

\(\left(Σ\sqrt{x+y}\right)^2\le\left(1+1+1\right)Σ\left(x+y\right)=6\left(x+y+z\right)\)

Như vậy, ta còn phải chứng minh :

\(\sqrt{6\left(x+y+z\right)}\le\frac{3\sqrt{3\left(x+y\right)\left(x+z\right)\left(y+z\right)}}{2\sqrt{xy+xz+yz}}\)

\(\Leftrightarrow9\left(x+y\right)\left(x+z\right)\left(y+z\right)\ge8\left(x+y+z\right)\left(xy+xz+yz\right)\)

\(\LeftrightarrowΣz\left(x-y\right)^2\ge0\) luôn đúng. Nên \(P_{Min}=\frac{3\sqrt{3}}{2}\Leftrightarrow x=y=z=\frac{1}{\sqrt{3}}\)

\(\frac{1}{6}\)nha bạn

Áp dụng bất đẳng thức Cauchy - Schwarz 

\(\Rightarrow\hept{\begin{cases}\frac{x^3}{\left(2x+y\right)\left(y+z\right)}+\frac{2x+y}{8}+\frac{y+z}{8}\ge3\sqrt[3]{\frac{x^3}{64}}=\frac{3x}{4}\\\frac{y^3}{\left(2y+z\right)\left(z+x\right)}+\frac{2y+z}{8}+\frac{x+z}{8}\ge3\sqrt[3]{\frac{y^3}{64}}=\frac{3y}{4}\\\frac{z^3}{\left(2z+x\right)\left(x+y\right)}+\frac{2z+x}{8}+\frac{x+y}{8}\ge3\sqrt[3]{\frac{z^3}{64}}=\frac{3z}{4}\end{cases}}\)

\(\Rightarrow\frac{x^3}{\left(2x+y\right)\left(y+z\right)}+\frac{y^3}{\left(2y+z\right)\left(x+z\right)}+\frac{z^3}{\left(2z+x\right)\left(x+y\right)}+\frac{5\left(x+y+z\right)}{8}\ge\frac{3\left(x+y+z\right)}{4}\)

\(\Rightarrow\frac{x^3}{\left(2x+y\right)\left(y+z\right)}+\frac{y^3}{\left(2y+z\right)\left(x+z\right)}+\frac{z^3}{\left(2z+x\right)\left(x+y\right)}+\frac{5}{8}\ge\frac{3}{4}\)

\(\Rightarrow\frac{x^3}{\left(2x+y\right)\left(y+z\right)}+\frac{y^3}{\left(2y+z\right)\left(x+z\right)}+\frac{z^3}{\left(2z+x\right)\left(x+y\right)}\ge\frac{1}{8}\)

\(\Leftrightarrow P_{min}=\frac{1}{8}\)

5000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

dối quá

chắc bằng 500000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000.

số gì mà khiếp vậy.................................

Đáp án = 130000000000000000

tuyển bạn trai

lương: tích

Đặt \(\hept{\begin{cases}x+y=a\\xy=b\end{cases}}\) thì ta có phương trình:

\(ab^2+a=3+b\Leftrightarrow a\left(b^2+1\right)=b+3\)

\(\Leftrightarrow a=\frac{b+3}{b^2+1}\). Nếu \(b=3\) vô nghiệm thì xét \(b\ne3\)

Khi đó: \(a=\frac{b+3}{b^2+1}\Leftrightarrow a\left(b-3\right)=\frac{b^2-9}{b^2+1}\)\(=\frac{b^2+1-10}{b^2+1}\)

\(=\frac{b^2+1}{b^2+1}-\frac{10}{b^2+1}=1-\frac{10}{b^2+1}\)

Suy ra \(b^2+1\inƯ\left(10\right)=....\)

Tự làm nốt nhá, trở thành bài lớp 6 r` :)

Mơn nhìu ạ