post từng câu một thôi bn nhìn mệt quá

\(\text{méo biết}\)

= căn xy + căn x + căn y còn lại tự tính

a) +) Điều kiện : x \(\ge\) 0 ; y \(\ge\) 0 ; y \(\ne\) 1 ; x; y không đồng thời bằng 0

+) \(P=\frac{x\left(\sqrt{x}+1\right)-y\left(1-\sqrt{y}\right)-xy\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}=\frac{x\sqrt{x}+x-y+y\sqrt{y}-xy\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}\)

\(P=\frac{\left(x\sqrt{x}+y\sqrt{y}\right)+\left(x-y\right)-xy\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x+y-\sqrt{xy}\right)+\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)-xy\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}\)

\(P=\frac{x+y-\sqrt{xy}+\sqrt{x}-\sqrt{y}-xy}{\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}=\frac{\left(x+\sqrt{x}\right)+\left(y-xy\right)-\left(\sqrt{xy}+\sqrt{y}\right)}{\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}=\frac{\left(1+\sqrt{x}\right)\sqrt{x}+y\left(1-x\right)-\sqrt{y}\left(\sqrt{x}+1\right)}{\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}\)

\(P=\frac{\left(1+\sqrt{x}\right)\left(\sqrt{x}+y-y\sqrt{x}-\sqrt{y}\right)}{\left(1-\sqrt{y}\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-y\sqrt{x}\right)+\left(y-\sqrt{y}\right)}{\left(1-\sqrt{y}\right)}=\frac{\sqrt{x}\left(1-\sqrt{y}\right)\left(1+\sqrt{y}\right)-\sqrt{y}\left(1-\sqrt{y}\right)}{\left(1-\sqrt{y}\right)}\)

\(P=\sqrt{x}\left(1+\sqrt{y}\right)-\sqrt{y}=\sqrt{x}-\sqrt{y}+\sqrt{xy}\)

b) Để P = 2 <=> \(\sqrt{x}-\sqrt{y}+\sqrt{xy}=2\) <=> \(\sqrt{x}+\sqrt{xy}=\sqrt{y}+2\)

<=>  \(\left(\sqrt{x}+\sqrt{xy}\right)^2=\left(\sqrt{y}+2\right)^2\)

<=> \(x+xy+2x\sqrt{y}=y+4+4\sqrt{y}\)

<=> \(x+xy-y+\left(2x-4\right)\sqrt{y}=4\)(*)

P = 2 <=> (x; y) thỏa mãn (*)

\(P\le\sqrt{3\left(\sum\dfrac{1}{\left(x+y\right)^2+\left(x+1\right)^2+4}\right)}\le\sqrt{3\left(\sum\dfrac{1}{4xy+4x+4}\right)}\)

\(P\le\sqrt{\dfrac{3}{4}\sum\left(\dfrac{1}{xy+x+1}\right)}=\dfrac{\sqrt{3}}{2}\)

\(P_{max}=\dfrac{\sqrt{3}}{2}\) khi \(x=y=z=1\)

Có cách khác nè:

P=x4(x−1)3+y4(y−1)3≥2√x4y4(x−1)3(y−1)3x4(x−1)3+y4(y−1)3≥2x4y4(x−1)3(y−1)3

⇒P≥2x2y2√(x−1)3(y−1)3=2.x2x−1.y2y−1.1√(x−1)(y−1)⇒P≥2x2y2(x−1)3(y−1)3=2.x2x−1.y2y−1.1(x−1)(y−1)

Ta dễ dàng chứng minh được a2a−1≥4a2a−1≥4

⇒P≥2.4.4.1√(x−1)(y−1)≥32.1x−1+y−12≥32⇒P≥2.4.4.1(x−1)(y−1)≥32.1x−1+y−12≥32

Dấu "=" khi x=y=2

x4(x−1)3+16(x−1)≥8.x2(x−1)x4(x−1)3+16(x−1)≥8.x2(x−1)

Tương tự và cộng hai BĐT lại : 

p+16(x−1)+16(y−1)≥8.(x2x−1+y2y−1)p+16(x−1)+16(y−1)≥8.(x2x−1+y2y−1)

Ta xét A=x2x−1+y2y−1A=x2x−1+y2y−1

Đặt x - 1 = a và y - 1 = b, ta có A=(a+1)2a+(b+1)2b=a+2+1a+b+2+1b≥(a+b)+4a+b+4≥2√4+4=8⇒A≥8A=(a+1)2a+(b+1)2b=a+2+1a+b+2+1b≥(a+b)+4a+b+4≥24+4=8⇒A≥8

Do đó P≥8A−16(x+y)+32≥8.8−16.4+32=32P≥8A−16(x+y)+32≥8.8−16.4+32=32

Min P = 32 <=> x = y = 2 

bạn vào trang này nhé có bài như thến này đấy 

//123doc.org//document/3173507-ren-luyen-chuyen-de-tim-maxmin-on-thi-thpt-quoc-gia.htm

tính diện tích hình vẽ dưới đây

Đặt \(\left\{{}\begin{matrix}x-4=a\\y-3=b\end{matrix}\right.\) \(\Rightarrow a^2+b^2=5\)

\(Q=\sqrt{\left(a+5\right)^2+b^2}+\sqrt{\left(a+3\right)^2+\left(b+4\right)^2}\)

\(\Rightarrow Q\le\sqrt{2\left[\left(a+5\right)^2+b^2+\left(a+3\right)^2+\left(b+4\right)^2\right]}\) (Bunhiacopxki)

\(\Rightarrow Q\le\sqrt{4\left(a^2+8a+b^2+4b+25\right)}\)

\(\Rightarrow Q\le\sqrt{4\left(a^2+2.4a+b^2+2.2b+25\right)}\)

\(\Rightarrow Q\le\sqrt{4\left(a^2+2\left(a^2+4\right)+b^2+2\left(b^2+1\right)+25\right)}\)

\(\Rightarrow Q\le\sqrt{4\left(3a^2+3b^2+35\right)}\le\sqrt{4\left(3.5+35\right)}=10\sqrt{2}\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=6\\y=4\end{matrix}\right.\)