KHÓ QUÁ KO NHÌN ĐC

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SIÊU KHÓ

ah ko đọc đc luôn!

\(G=\left(a-b\right)^4+\left(b-c\right)^4+\left(c-a\right)^4\)

\(G=\left(a^2-2ab+b^2\right)^2+\left(b^2-2bc+c^2\right)^2+\left(c^2-2ac+a^2\right)^2\)

\(G\ge\frac{\left(2a^2+2b^2+2c^2-2ab-2bc-2ca\right)}{3}\)

\(2a^2+2b^2+2c^2\ge2ab+2bc+2ca\)(BĐT tương đương)

\(G\ge\frac{\left(2a^2+2b^2+2c^2-2ab-2bc-2ca\right)}{3}\ge\frac{0}{3}=0\)

\(< =>MIN:G=0\)dấu "=" xảy ra khi và chỉ khi \(a=b=c=1\)

Xét tam giác \(ABC\)vuông tại \(A\)đường cao \(AH\): 

\(\frac{1}{AH^2}=\frac{1}{AB^2}+\frac{1}{AC^2}=\frac{1}{6^2}+\frac{1}{8^2}\Rightarrow AH=4,8\left(cm\right)\).

\(BC^2=AB^2+AC^2\)(định lí Pythagore) 

\(=6^2+8^2=100\)

\(\Rightarrow BC=10\left(cm\right)\)

\(HC=\frac{AC^2}{BC}=\frac{8^2}{10}=6,4\left(cm\right)\)

\(HB=BC-HC=10-6,4=3,6\left(cm\right)\)

Xét tam giác \(AHB\)vuông tại \(H\)đường cao \(HQ\): 

\(AQ=\frac{AH^2}{AB}=\frac{4,8^2}{6}=3,84\left(cm\right)\)

Xét tam giác \(ACQ\)vuông tại \(A\): 

\(CQ^2=AC^2+AQ^2=8^2+3,84^2\Rightarrow CQ=\frac{8\sqrt{769}}{25}\left(cm\right)\)

\(a,A=\left(1-\frac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-3}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)

\(A=\frac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}:\frac{x-9-x+4+\sqrt{x}+2}{x-5\sqrt{x}+6}\)

\(A=\frac{1}{\sqrt{x}+1}:\frac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

\(A=\frac{\sqrt{x}-2}{\sqrt{x}+1}=\frac{\sqrt{x}+1-3}{\sqrt{x}+1}=1-\frac{3}{\sqrt{x}+1}\)

\(< =>b,3⋮\sqrt{x}+1\)

lập bảng thì ra đc

\(x=4;0\)

??? ở đâu

\(x=\sqrt{2+\sqrt{\frac{5+\sqrt{5}}{2}}}+\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}\)

\(\Leftrightarrow x^2=4+2\sqrt{\frac{3-\sqrt{5}}{2}}\)

\(\Leftrightarrow x^2=4+\sqrt{6-2\sqrt{5}}\)

\(\Leftrightarrow x^2=4+\sqrt{5}-1\)

\(\Leftrightarrow x^2=3+\sqrt{5}\)

\(\Leftrightarrow x^2=\frac{6+2\sqrt{5}}{2}\)

\(\Leftrightarrow x=\frac{\sqrt{5}+1}{\sqrt{2}}\)

Ta có:

\(P=\left(x-\sqrt{3-\sqrt{5}}-\sqrt{2}+1\right)^{2019}\)

\(=\left(\frac{\sqrt{5}+1}{\sqrt{2}}-\sqrt{3-\sqrt{5}}-\sqrt{2}+1\right)^{2019}\)

\(=\left(\frac{\sqrt{5}+1-\sqrt{6-2\sqrt{5}}-2+\sqrt{2}}{\sqrt{2}}\right)^{2019}\)

\(=\left(\frac{\sqrt{5}+1-\left(\sqrt{5}-1\right)-2+\sqrt{2}}{\sqrt{2}}\right)^{2019}\)

\(=\left(1\right)^{2019}=1\)