a) Áp dụng hệ thức lượng trong tam giác vuông có:

\(AB^2=BH.BC\)

\(AC^2=CH.CB\)

\(\Rightarrow\dfrac{BH}{CH}=\dfrac{AB^2}{AC^2}=\dfrac{c^2}{b^2}\)

b) Áp dụng hệ thức lượng trong tam giác vuông có:

\(BH^2=BE.BA\)

\(CH^2=CF.CA\)

\(\Rightarrow\dfrac{BH^2}{CH^2}=\dfrac{BE}{CF}.\dfrac{BA}{CA}\)\(\Leftrightarrow\dfrac{c^4}{b^4}=\dfrac{BE}{CF}.\dfrac{c}{b}\)

\(\Leftrightarrow\dfrac{BE}{CF}=\dfrac{c^3}{b^3}\)

\(A=\dfrac{\sqrt{3x-9}}{x}=\sqrt{\dfrac{3x-9}{x^2}}=\sqrt{-\dfrac{9}{x^2}+\dfrac{3}{x}}\)\(=\sqrt{\dfrac{1}{4}-\left(\dfrac{9}{x^2}+2.\dfrac{3}{x}.\dfrac{1}{2}+\dfrac{1}{4}\right)}\)

\(=\sqrt{\dfrac{1}{4}-\left(\dfrac{3}{x}-\dfrac{1}{2}\right)^2}\le\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}\)

Dấu "=" xảy ra khi \(\dfrac{3}{x}-\dfrac{1}{2}=0\Leftrightarrow x=6\) (tm)

Vậy \(A_{max}=\dfrac{1}{2}\)

\(B=\dfrac{\sqrt{2x-4}}{x}=\sqrt{\dfrac{2x-4}{x^2}}=\sqrt{-\dfrac{4}{x^2}+\dfrac{2}{x}}=\sqrt{\dfrac{1}{4}-\left(\dfrac{4}{x^2}-2.\dfrac{2}{x}.\dfrac{1}{2}+\dfrac{1}{4}\right)}\)

\(=\sqrt{\dfrac{1}{4}-\left(\dfrac{2}{x}-\dfrac{1}{2}\right)^2}\le\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}\)

Dấu "=" xảy ra khi \(\dfrac{2}{x}-\dfrac{1}{2}=0\Leftrightarrow x=4\) (tm)

Vậy \(B_{max}=\dfrac{1}{2}\)

\(C=\dfrac{\sqrt{x-4}}{3x}=\sqrt{\dfrac{x-4}{9x^2}}=\sqrt{-\dfrac{4}{9x^2}+\dfrac{1}{9x}}=\sqrt{\dfrac{1}{144}-\left(\dfrac{4}{9x^2}-2.\dfrac{2}{3x}.\dfrac{1}{12}+\dfrac{1}{144}\right)}\)

\(=\sqrt{\dfrac{1}{144}-\left(\dfrac{2}{3x}-\dfrac{1}{12}\right)^2}\le\sqrt{\dfrac{1}{144}}=\dfrac{1}{12}\)

Dấu "=" xảy ra khi \(x=8\) (tm)

Vậy \(C_{max}=\dfrac{1}{12}\)

a)Áp dụng hệ thức lượng trong tam giác vuông có:

\(AH^2=AE.AB\)

\(AH^2=AF.AC\)

\(\Rightarrow AE.AB=AF.AC\)

b)(\(\dfrac{BE}{CF}\) chứ)

Áp dụng hệ thức lượng trong tam giác vuông có:

\(AB^2=BH.BC\)

\(AC^2=CH.BC\)

\(\Rightarrow\dfrac{AB^2}{AC^2}=\dfrac{BH}{CH}\)\(\Leftrightarrow\dfrac{AB^4}{AC^4}=\dfrac{BH^2}{CH^2}=\dfrac{BE.AB}{CF.AC}\)

\(\Leftrightarrow\dfrac{BE}{CF}=\left(\dfrac{AB}{AC}\right)^3\)

c)Áp dụng định lý Thales có:

\(\dfrac{BH}{BC}=\dfrac{BE}{BA}\Leftrightarrow BA.BH=BE.BC\)

\(\dfrac{CF}{CA}=\dfrac{CH}{BC}\Leftrightarrow CF.BC=CA.CH\)

\(\Rightarrow BA.CA.BH.CH=BE.CF.BC^2\)

\(\Leftrightarrow AH.BC.AH^2=BC^2.BE.BF\)

\(\Leftrightarrow BC^..BE.BF=AH^3\) 

Vậy ....

Chi tiết hả :)

\(4-\dfrac{2}{3}\left(x-3\right)=2-\dfrac{1}{2}+\dfrac{2}{3}\)

\(\Leftrightarrow4-\dfrac{2}{3}\left(x-3\right)=\dfrac{12}{6}-\dfrac{3}{6}+\dfrac{4}{6}\)

\(\Leftrightarrow4-\dfrac{2}{3}\left(x-3\right)=\dfrac{13}{6}\)

\(\Leftrightarrow\dfrac{2}{3}\left(x-3\right)=4-\dfrac{13}{6}=\dfrac{24}{6}-\dfrac{13}{6}=\dfrac{11}{6}\)

\(\Leftrightarrow x-3=\dfrac{11}{6}:\dfrac{2}{3}\)

\(\Leftrightarrow x=3+\dfrac{11}{4}\)\(\Leftrightarrow x=\dfrac{12}{4}+\dfrac{11}{4}\)\(\Leftrightarrow x=\dfrac{23}{4}\)

Vậy...

\(4-\dfrac{2}{3}\left(x-3\right)=2-\dfrac{1}{2}+\dfrac{2}{3}\)

\(\Leftrightarrow4-\dfrac{2}{3}\left(x-3\right)=\dfrac{13}{6}\)

\(\Leftrightarrow\dfrac{2}{3}\left(x-3\right)=4-\dfrac{13}{6}=\dfrac{11}{6}\)

\(\Leftrightarrow x-3=\dfrac{11}{4}\)

\(\Leftrightarrow x=\dfrac{23}{4}\)

Vậy...

Thay x=-8 và y=6 cào C ta được:

\(C=\dfrac{\left(-8\right)^3}{2}+\dfrac{\left(-8\right)^2.6}{4}+\dfrac{\left(-8\right).6^2}{6}+\dfrac{6^3}{27}\)\(=\dfrac{-512}{2}+\dfrac{384}{4}-\dfrac{288}{6}+\dfrac{216}{27}\)\(=-256+96-48+8=-200\)

B1)

b, Có \(x\ge0\Rightarrow B=x+\sqrt{x}+4\ge0+0+4=4\)

Dấu "=" xảy ra khi x=0

Vậy \(B_{min}=4\)

\(=\left(\sqrt{5}-\sqrt{3}\right).\sqrt{2}.\sqrt{8-\sqrt{15}}\)

\(=\left(\sqrt{5}-\sqrt{3}\right).\sqrt{16-2\sqrt{15}}\)

\(=\left(\sqrt{5}-\sqrt{3}\right)\sqrt{15-2\sqrt{15}+1}\)

\(=\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{15}-1\right)^2}\)

\(=\left(\sqrt{5}-\sqrt{3}\right)\left|\sqrt{15}-1\right|\)\(=\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{15}-1\right)=5\sqrt{3}-\sqrt{5}-3\sqrt{5}+\sqrt{3}\)

\(=6\sqrt{3}-4\sqrt{5}\)

Vậy...

Áp dụng định lí py-ta-go vào tam giác DFE vuông tại D có:

\(DE^2=EF^2-DF^2=729\)

\(\Rightarrow DE=27\) (cm)

Áp dụng ht lượng trong tam giác vuông có:

\(\dfrac{1}{DI^2}=\dfrac{1}{DE^2}+\dfrac{1}{DF^2}=\dfrac{1}{27^2}+\dfrac{1}{36^2}=\dfrac{2025}{27^2.36^2}\)

\(\Leftrightarrow DI^2=\dfrac{27^2.36^2}{45^2}\)\(\Leftrightarrow DI=\dfrac{27.36}{45}=21,6\) (cm)

\(DE^2=EI.EF\Leftrightarrow EI=\dfrac{DE^2}{EF}=\dfrac{27^2}{45}=16,2\) (cm)

\(DF^2=FI.EF\Leftrightarrow FI=\dfrac{DF^2}{EF}=\dfrac{36^2}{45}=28,8\) (cm)