ĐKXĐ: \(0\le x,y\le1\)

ÁP DỤNG BẤT ĐẲNG THỨC BU-NHI-A-CỐP-XKI ta có:\(x\sqrt{1-y^2}+y\sqrt{1-x^2}\le\sqrt{\left(x^2+1-x^2\right)\left(1-y^2+y^2\right)}=1\)

DẤU "=" XẢY RA KHI VÀ CHỈ KHI: \(\frac{x}{\sqrt{1-y^2}}=\frac{\sqrt{1-x^2}}{y}\)\(\Leftrightarrow xy=\sqrt{\left(1-x^2\right)\left(1-y^2\right)}\)

                                                                                                            \(\Leftrightarrow x^2y^2=\left(1-x^2\right)\left(1-y^2\right)\)

                                                                                                             \(\Leftrightarrow x^2y^2=1-x^2-y^2+x^2y^2\)

                                                                                                               \(\Leftrightarrow x^2+y^2=1\)

Ta có  

\(\left(\sqrt{27+10\sqrt{2}}-\sqrt{27-10\sqrt{2}}\right)^2\)

\(=27+10\sqrt{2}+27-10\sqrt{2}-2\sqrt{\left(27+10\sqrt{2}\right)\left(27-10\sqrt{2}\right)}\)

\(=54-2\sqrt{529}=8\)

\(\Rightarrow\)  \(\sqrt{27+10\sqrt{2}}-\sqrt{27-10\sqrt{2}}=\sqrt{8}=2\sqrt{2}\)

Xét tử số

\(\left(27+10\sqrt{2}\right)\sqrt{27-10\sqrt{2}}-\left(27-10\sqrt{2}\right)\sqrt{27+10\sqrt{2}}\)

\(=\left(\sqrt{27+10\sqrt{2}}.\sqrt{27-10\sqrt{2}}\right)\left(\sqrt{27+10\sqrt{2}}-\sqrt{27-10\sqrt{2}}\right)\)

\(=23\left(\sqrt{27+10\sqrt{2}}-\sqrt{27-10\sqrt{2}}\right)\)

\(=23.2\sqrt{2}=46\sqrt{2}\)

Lại có  \(\left(\sqrt{\sqrt{13}-3}+\sqrt{\sqrt{13}+3}\right)^2\)

\(=\sqrt{13}-3+\sqrt{13}+3+2\sqrt{\left(\sqrt{13}-3\right)\left(\sqrt{13}+3\right)}\)

\(=2\sqrt{13}+2\sqrt{4}=2\sqrt{13}+4\)

ta bình phương mẫu số

\(\left(\frac{\sqrt{\sqrt{13}-3}+\sqrt{\sqrt{13}+3}}{\sqrt{\sqrt{13}+2}}\right)^2=\frac{\left(\sqrt{\sqrt{13}-3}+\sqrt{\sqrt{13}+3}\right)^2}{\sqrt{13}+2}\)

\(=\frac{2\sqrt{13}+4}{\sqrt{13}+2}=2\)

Vậy mẫu  \(=\sqrt{2}\)

Vậy  \(x=\frac{46\sqrt{2}}{\sqrt{2}}=46\)  thay vào ta đc A = 92880

a. Xét \(\Delta AHC\)có \(AH^2+HC^2=AC^2\)(1)

Xét \(\Delta AHB\) có \(AH^2+HB^2=AB^2\)(2)

Từ (1) và (2) \(\Rightarrow HC^2-HB^2=AC^2-AB^2\left(đpcm\right)\)

b. Ta có \(HC=20-HB\Rightarrow\left(20-HB\right)^2-HB^2=AC^2-AB^2\)

\(\Rightarrow400-40HB=15^2-11^2=104\)\(\Rightarrow HB=7,4\Rightarrow HC=12,6\left(cm\right)\)

\(AH=\sqrt{AC^2-HC^2}=\sqrt{15^2-\left(12,6\right)^2}=\frac{6\sqrt{46}}{5}\left(cm\right)\)

Hình vẽ:

ĐKXĐ:\(x\ge0,y\ge1\)

\(x-4\sqrt{x}+y-6\sqrt{y-1}+z=0\)

\(\Leftrightarrow x-4\sqrt{x}+4+y-1-6\sqrt{y-1}+9+z-12=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+\left(\sqrt{y-1}-3\right)^2+z-12=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(\sqrt{x}-2\right)^2=0\\\left(\sqrt{y-1}-3\right)^2=0\\z-12=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=4\\y=10\\z=12\end{cases}}\)(THỎA MÃN ĐKXĐ)

Ta chứng minh bất đẳng thức sau  

Với x, y, z > 0 ta luôn có  \(x^4+y^4+z^4\ge xyz\left(x+y+z\right)\)  (1)

Theo BĐT Cô-si

\(x^4+x^4+y^4+z^4\ge4\sqrt[4]{x^8y^4z^4}=4x^2yz\)

\(y^4+y^4+z^4+x^4\ge4\sqrt[4]{y^8z^4x^4}=4y^2zx\)

\(z^4+z^4+x^4+y^4\ge4\sqrt[4]{z^8x^4y^4}=4z^2xy\)

Cộng vế theo vế ta được:  \(4\left(x^4+y^4+z^4\right)\ge4\left(x^2yz+y^2zx+z^2xy\right)\)

\(\Leftrightarrow\)  \(x^4+y^4+z^4\ge xyz\left(x+y+z\right)\)

Vậy (1) đc c/m

Bất đẳng thức cần c/m có thể viết lại thành

\(\frac{abcd}{a^4+b^4+c^4+abcd}+\frac{abcd}{b^4+c^4+d^4+abcd}+\frac{abcd}{c^4+d^4+a^4+abcd}+\frac{abcd}{d^4+a^4+b^4+abcd}\le1\)

Áp dụng (1) ta có  

\(\frac{abcd}{a^4+b^4+c^4+abcd}\le\frac{abcd}{abc\left(a+b+c\right)+abcd}=\frac{abcd}{abc\left(a+b+c+d\right)}=\frac{d}{a+b+c+d}\)

Tương tự  

\(\frac{abcd}{b^4+c^4+d^4+abcd}\le\frac{a}{a+b+c+d}\)

\(\frac{abcd}{c^4+d^4+a^4+abcd}\le\frac{b}{a+b+c+d}\)

\(\frac{abcd}{d^4+a^4+b^4+abcd}\le\frac{c}{a+b+c+d}\)

Cộng theo vế suy ra đpcm.

\(x+y+z+35=2\left(2\sqrt{x+1}+3\sqrt{y+2}+4\sqrt{z+3}\right)\)

\(\Leftrightarrow x+y+z+35-4\sqrt{x+1}-6\sqrt{y+2}-8\sqrt{z+3}=0\)

\(\Leftrightarrow\left(x+1-4\sqrt{x+1}+4\right)+\left(y+2-6\sqrt{y+2}+9\right)+\left(z+3-8\sqrt{z+3}+16\right)=0\)

\(\Leftrightarrow\left(\sqrt{x+1}-2\right)^2+\left(\sqrt{y+2}-3\right)^2+\left(\sqrt{z+3}-4\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(\sqrt{x+1}-2\right)^2=0\\\left(\sqrt{y+2}-3\right)^2=0\\\left(\sqrt{z+3}-4\right)^2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{x+1}=2\\\sqrt{y+2}=3\\\sqrt{z+3}=4\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=3\\y=7\\z=13\end{cases}}\)