ĐK: x \(\ge\)\(\frac{8}{3}\)

pt <=> \(4.\left(x-3\right)+9-3.\sqrt{5x-6}=\sqrt{3x-8}-1\)

<=> \(4.\left(x-3\right)+3.\left(3-\sqrt{5x-6}\right)=\sqrt{3x-8}-1\)

<=>  \(4.\left(x-3\right)+3.\frac{\left(3-\sqrt{5x-6}\right)\left(3+\sqrt{5x-6}\right)}{3+\sqrt{5x-6}}=\frac{\left(\sqrt{3x-8}-1\right)\left(\sqrt{3x-8}+1\right)}{\sqrt{3x-8}+1}\)

<=> \(4.\left(x-3\right)+3.\frac{9-5x+6}{3+\sqrt{5x-6}}=\frac{3x-8-1}{\sqrt{3x-8}+1}\)

<=> \(4.\left(x-3\right)+15.\frac{3-x}{3+\sqrt{5x-6}}-3.\frac{x-3}{\sqrt{3x-8}+1}=0\)

<=> \(\left(x-3\right)\left(4-\frac{15}{3+\sqrt{5x-6}}-\frac{3}{\sqrt{3x-8}+1}\right)=0\)

<=> x = 3 (thoả mãn) hoặc \(4-\frac{15}{3+\sqrt{5x-6}}-\frac{3}{\sqrt{3x-8}+1}=0\) (2)

Giải (2):  (2) <=> \(\frac{15}{6}-\frac{15}{3+\sqrt{5x-6}}+\frac{3}{2}-\frac{3}{\sqrt{3x-8}+1}=0\)

<=> \(15\left(\frac{1}{6}-\frac{1}{3+\sqrt{5x-6}}\right)+3.\left(\frac{1}{2}-\frac{1}{\sqrt{3x-8}+1}\right)=0\)

<=>  \(15.\frac{\sqrt{5x-6}-3}{6.\left(3+\sqrt{5x-6}\right)}+3.\frac{\sqrt{3x-8}-1}{2.\left(\sqrt{3x-8}+1\right)}=0\)

<=> \(15.\frac{5.\left(x-3\right)}{6.\left(3+\sqrt{5x-6}\right)^2}+3.\frac{3.\left(x-3\right)}{2.\left(\sqrt{3x-8}+1\right)^2}=0\)

<=> \(\left(x-3\right).\left(\frac{75}{6.\left(3+\sqrt{5x-6}\right)^2}+\frac{9}{2.\left(\sqrt{3x-8}+1\right)^2}\right)=0\)

<=> x = 3 Vì \(\frac{75}{6.\left(3+\sqrt{5x-6}\right)^2}+\frac{9}{2.\left(\sqrt{3x-8}+1\right)^2}>0\) với mọi x \(\ge\frac{8}{3}\)

Vậy pt có 1 nghiệm duy nhất x = 3

Áp dụng BĐT Cô-si,ta có :

x⁴ + yz \(\ge\)\(2\sqrt{x^4yz}=2x^2\sqrt{yz}\); \(y^4+xz\ge2y^2\sqrt{xz}\); \(z^4+xy\ge2z^2\sqrt{xy}\)

\(\Rightarrow\frac{x^2}{x^4+yz}+\frac{y^2}{y^4+xz}+\frac{z^2}{z^4+xy}\le\frac{x^2}{2x^2\sqrt{yz}}+\frac{y^2}{2y^2\sqrt{xz}}+\frac{z^2}{2z^2\sqrt{xy}}=\frac{1}{2\sqrt{yz}}+\frac{1}{2\sqrt{xz}}+\frac{1}{2\sqrt{xy}}\)

CM : x + y + z \(\ge\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\)

\(\frac{x^2}{x^4+yz}+\frac{y^2}{y^4+xz}+\frac{z^2}{z^4+xy}\le\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{1}{2}.\frac{yz+xz+xy}{xyz}=\frac{1}{2}.\frac{3xyz}{xyz}=\frac{3}{2}\)

Áp dụng BĐT Cauchy cho các cặp số dương, ta có: \(\Sigma\frac{x^2}{x^4+yz}\le\Sigma\frac{x^2}{2x^2\sqrt{yz}}=\Sigma\frac{1}{2\sqrt{yz}}\)

\(\le\frac{1}{4}\Sigma\left(\frac{1}{y}+\frac{1}{z}\right)=\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

\(=\frac{1}{2}.\frac{xy+yz+zx}{xyz}\le\frac{1}{2}.\frac{x^2+y^2+z^2}{xyz}=\frac{1}{2}.\frac{3xyz}{xyz}=\frac{3}{2}\)

Đẳng thức xảy ra khi x = y = z = 1

chak là p kẻ hình oy nên t trả lời lời giải thui!
phần a , Xét t/g ABD và t/g AEB có:
               Góc A chung  (1)
               góc ABD = 1/2 Sđ cung BD  ( vì là góc tạo bởi tia tiếp tuyến và dây cung )
               góc AEB = 1/2 Sđ cung BD  ( vì là góc nội tiếp)
               Từ đó ta có Góc ABD= Góc AEB (2)
Từ (1) và (2) ta có t/g ABD= t/g AEB (g.g)
=> AB/AD = AE/AB
=> AB mũ2 = AD.AE
 còn phần b thì chịu. đúng hộ nha!

\(y=\frac{\left(x^2-2x+1\right)+2.\left(x-1\right)+8}{\left(x-1\right)^2}=\frac{\left(x-1\right)^2}{\left(x-1\right)^2}+\frac{2\left(x-1\right)}{\left(x-1\right)^2}+\frac{8}{\left(x-1\right)^2}=1+\frac{2}{x-1}+\frac{8}{\left(x-1\right)^2}\)

Đặt t = \(\frac{1}{x-1}\)

=> y = 1 + 2t + 8t² = 8.(t² + 2.\(\frac{1}{8}\). t + \(\left(\frac{1}{8}\right)^2\) ) - \(\frac{1}{8}\) + 1 = 8. (t + \(\frac{1}{8}\))² + \(\frac{7}{8}\) \(\ge\) 8.0 + \(\frac{7}{8}\) = \(\frac{7}{8}\) với mọi t

=> Min y = \(\frac{7}{8}\) khi t + \(\frac{1}{8}\) = 0 <=> t = -\(\frac{1}{8}\)<=>\(\frac{1}{x-1}\)  = \(\frac{1}{8}\) <=> x - 1 = 8 <=> x = 9

\(\frac{100\left(x+20\right)}{x\left(x+20\right)}-\frac{100x}{x\left(x+20\right)}=\frac{1}{3}\)

\(\frac{100x+2000-100x}{x\left(x+20\right)}=\frac{1}{3}\)

\(\frac{2000}{x\left(x+20\right)}=\frac{1}{3}\)

\(\Rightarrow x^2+20x=3.2000\)

\(\Rightarrow x^2+20x-6000=0\)

ĐKXĐ: \(x\ne0;x\ne-2\)

Ta có: \(\frac{100x+2000-100x}{x\left(x+20\right)}=\frac{1}{3}\)

\(\Leftrightarrow\frac{2000}{x^2+20x}=\frac{1}{3}\)

\(\Leftrightarrow x^2+20x=6000\)

\(\Leftrightarrow x^2+2.10x+100=6100\)

\(\Leftrightarrow\left(x+10\right)^2=6100\)

\(\Leftrightarrow\orbr{\begin{cases}x=10\sqrt{61}-10\left(TM\right)\\x=-10\sqrt{61}-10\left(TM\right)\end{cases}}\)

Vậy...