\(\left\{{}\begin{matrix}x^2+y^4+xy=2xy^2+7\\xy^3-x^2y+4xy+11x=28+11y^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-y^2\right)^2+xy-7=0\\\left(x^{ }-y^2\right)\left(11-xy\right)+4\left(xy-7\right)=0\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x-y^2=a\\xy-7=b\end{matrix}\right.\) hệ trở thành \(\left\{{}\begin{matrix}a^2+b=0\\a\left(4-b\right)+4b=0\end{matrix}\right.\)\(\Rightarrow a\left(4+a^2\right)-4a^2=0\Leftrightarrow a\left(a^2-4a+4\right)=0\Leftrightarrow a\left(a-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}a=0;b=0\\a=2;b=-4\end{matrix}\right.\)

Giải từng trường hợp rồi kết hợp nghiệm

a) 11x + 11y + x² + xy

= 11.(x+y) + x.(x+y)

= (x+y).(11+x)

b) 255 + x² - 4xy + y²

= 255 + 2xy + x² -2xy + y²

= 255 + 2xy + (x-y)²

...

a) Ta có: \(11x+11y+x^2+xy\)

\(=11\left(x+y\right)+x\left(x+y\right)\)

\(=\left(x+y\right)\left(11+x\right)\)

b) Ta có: \(225-4x^2-4xy-y^2\)

\(=225-\left(4x^2+4xy+y^2\right)\)

\(=15^2-\left(2x+y\right)^2\)

\(=\left(15-2x-y\right)\left(15+2x+y\right)\)

a, sai đề ko zậy ak

b, 225 - 4x² - 4xy - y²

= 225 - (4x² + 4xy + y²)

= 225 - [(2x)² + 2*2xy+y²]

= 15² - (2x + y)²

^{= [15 - (2x + y)] * [15 + (2x+y)]}

^{= (15 - 2x - y) * (15 + 2x + y)}

Chúc bạn học có hiệu quả!!

Dự đoán dấu "=" xảy ra khi \(x=y=z=1\) ta tìm được \(P=9\)

Ta sẽ chứng minh nó là \(GTLN\) của \(P\)

Thật vậy, ta cần chứng minh 

\(Σ\frac{11x+4y}{4x^2-xy+2y^2}\le\frac{3\left(xy+yz+xz\right)}{xyz}\)

\(\Leftrightarrow\left(\frac{3}{x}-\frac{11x+4y}{4x^2-xy+2y^2}\right)\ge0\)

\(\LeftrightarrowΣ\frac{\left(x-y\right)\left(x-6y\right)}{x\left(4x^2-xy+2y^2\right)}\ge0\)

\(\LeftrightarrowΣ\left(\frac{\left(x-y\right)\left(x-6y\right)}{x\left(4x^2-xy+2y^2\right)}+\frac{1}{y}-\frac{1}{x}\right)\ge0\)

\(\LeftrightarrowΣ\frac{\left(x-y\right)^2\left(x+y\right)}{xy\left(4x^2-xy+2y^2\right)}\ge0\) (luôn đúng)

Vậy \(P_{Max}=9\) khi \(x=y=z=1\)

ggvcgfdsx

Cách giải khác:

Ta chứng minh bổ đề:

\(\dfrac{11x+4y}{4x^2-xy+2y^2}\le\dfrac{2}{x}+\dfrac{1}{y}\Leftrightarrow\left(x-y\right)^2\left(x+y\right)\ge0\)(Đúng)

Tương tự ta cho 2 BĐT còn lại ta cũng có:

\(\dfrac{11y+4z}{4y^2-yz+2z^2}\le\dfrac{2}{y}+\dfrac{1}{z};\dfrac{11z+4x}{4z^2-xz+2x^2}\le\dfrac{2}{z}+\dfrac{1}{x}\)

Cộng theo vế 3 BĐT trên ta có:

\(P\le\dfrac{3}{x}+\dfrac{3}{y}+\dfrac{3}{z}=\dfrac{3\left(xy+yz+xz\right)}{xyz}=9\)

Đẳng thức xảy ra khi \(x=y=z=1\)

Câu hỏi của Neet - Toán lớp 10 | Học trực tuyến đổi biến (a,b,c)->(x,y,z) là y nhau

Bạn cần viết đề bằng công thức toán để được hỗ trợ tốt hơn.

Giải:

a) \(\left(x-5\right)^2-16\)

\(=\left(x-5-4\right)\left(x-5+4\right)\)

\(=\left(x-9\right)\left(x-1\right)\)

b) \(25-\left(3-x\right)^2\)

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

\(=\left(2+x\right)\left(8-x\right)\)

c) \(49\left(y-4\right)^2-9\left(y+2\right)^2\)

\(=\left[7\left(y-4\right)\right]^2-\left[3\left(y+2\right)\right]^2\)

\(=\left[7\left(y-4\right)-3\left(y+2\right)\right]\left[7\left(y-4\right)+3\left(y+2\right)\right]\)

\(=\left(7y-28-3y-6\right)\left(7y-28+3y+6\right)\)

\(=\left(4y-34\right)\left(10y-22\right)\)

d) \(11x+11y-x^2-xy\)

\(=11\left(x+y\right)-x\left(x+y\right)\)

\(=\left(11-x\right)\left(x+y\right)\)

e) \(x^2-xy-8x+8y\)

\(=x\left(x-y\right)-8\left(x-y\right)\)

\(=\left(x-8\right)\left(x-y\right)\)

Vậy ...

\(\left(x-5\right)^2-16\)

\(=\left(x-5\right)^2-4^2\)

\(=\left(x-5-4\right)\left(x-5+4\right)\)

\(=\left(x-9\right)\left(x-1\right)\)

\(25-\left(3-x\right)^2\)

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

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

\(=\left(8-x\right)\left(2+x\right)\)

\(49\left(y-4\right)^2-9\left(y+2\right)^2\)

\(=7^2\left(y-4\right)^2-3^2\left(y+2\right)^2\)

\(=\left[7\left(y-4\right)\right]^2-\left[3\left(y+2\right)\right]^2\)

\(=\left(7y-28\right)^2-\left(3y+6\right)^2\)

\(=\left(7y-28-3y-6\right)\left(7y-28+3y+6\right)\)

\(=\left(4y-34\right)\left(10y-22\right)\)

\(11x+11y-x^2-xy\)

\(=11\left(x+y\right)-x\left(x+y\right)\)

\(=\left(11-x\right)\left(x+y\right)\)

\(x^2-xy-8x+8y\)

\(=x\left(x-y\right)-8\left(x-y\right)\)

\(=\left(x-8\right)\left(x-y\right)\)

Mai cho bn đấy tui dg định off =))

a)\(11x+11y-x^2-xy\)

\(=\left(11x+11y\right)-\left(x^2+xy\right)\)

\(=11\left(x+y\right)-x\left(x+y\right)\)

\(=\left(11-x\right)\left(x+y\right)\)

b)\(x^2-xy-8x+8y\)

\(=\left(x^2-xy\right)-\left(8x-8y\right)\)

\(=x\left(x-y\right)-8\left(x-y\right)\)

\(=\left(x-8\right)\left(x-y\right)\)

c)\(x^2-6x-y^2+9\)

\(=\left(x^2-6x+9\right)-y^2\)

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

d)\(x^2+2xy+y^2-xz-yz\)

\(=\left(x^2+2xy+y^2\right)-\left(xz+yz\right)\)

\(=\left(x+y\right)^2-z\left(x+y\right)\)

\(=\left(x+y\right)\left(x+y-z\right)\)

a) \(11x+11y-x^2-xy\)

\(=11\left(x+y\right)-x\left(x+y\right)\)

\(=\left(x+y\right)\left(11-x\right)\)

b) \(x^2-xy-8x+8y\)

\(=x\left(x-y\right)-8\left(x-y\right)\)

\(=\left(x-y\right)\left(x-8\right)\)

c) \(x^2-6x-y^2+9\)

\(=\left(x^2-6x+9\right)-y^2\)

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

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

d) \(x^2+2xy+y^2-xz-yz\)

\(=\left(x+y\right)^2-z\left(x+y\right)\)

\(=\left(x+y\right)\left(x+y-z\right)\)