Question

Giúp em làm Bài 3 với, Làm ơn ạ !!!

Answer 1

Bài 4:

\(28x^3+6x^2+12x+8=0\)

\(\Leftrightarrow28x^3+14x^2-8x^2-4x+16x+8=0\)

\(\Leftrightarrow14x^2\left(2x+1\right)-4x\left(2x+1\right)+8\left(2x+1\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left(14x^2-4x+8\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left(x^2-\dfrac{2}{7}x+\dfrac{4}{7}\right)=0\)

\(\Leftrightarrow2x+1=0\) hay \(\left(x^2-\dfrac{2}{7}x+\dfrac{4}{7}\right)=0\)

\(\Leftrightarrow x=\dfrac{-1}{2}\) hay \(x^2-2.\dfrac{1}{7}x+\dfrac{1}{49}+\dfrac{27}{49}=0\)

\(\Leftrightarrow x=\dfrac{-1}{2}\) hay  \(\left(x-\dfrac{1}{7}\right)^2+\dfrac{27}{49}=0\) (vô nghiệm vì \(\left(x-\dfrac{1}{7}\right)^2+\dfrac{27}{49}\ge\dfrac{27}{49}\))

-Vậy \(S=\left\{\dfrac{-1}{2}\right\}\)

Answer 2

Bài 3: 

a) AB//CD \(\Rightarrow\widehat{BAM}=\widehat{ACD}\) (so le trong)

\(\widehat{AMB}=\widehat{ADC}=90^0\)

\(\Rightarrow\)△ABM∼△CAD (g-g).

b) △ADC vuông tại D \(\Rightarrow AD^2+DC^2=AC^2\Rightarrow AD^2+AB^2=AC^2\Rightarrow AC=\sqrt{AD^2+AB^2}=\sqrt{9^2+12^2}=15\left(cm\right)\)△ADC có DN phân giác \(\Rightarrow\dfrac{NA}{NC}=\dfrac{DA}{DC}\)

\(\Rightarrow\dfrac{NA}{DA}=\dfrac{NC}{DC}=\dfrac{NA+NC}{DA+DC}=\dfrac{AC}{DA+DC}\)

\(\Rightarrow NC=\dfrac{AC.DC}{DA+DC}=\dfrac{15.12}{9+12}=\dfrac{60}{7}\left(cm\right)\)

△ADC có NK//AD (cùng vuông góc với DC) \(\Rightarrow\dfrac{NK}{AD}=\dfrac{NC}{AC}\)

\(\Rightarrow NK=\dfrac{NC}{AC}.AD=\dfrac{\dfrac{60}{7}}{15}.9=\dfrac{36}{7}\left(cm\right)\)

c) △ABM∼△CAD \(\Rightarrow\dfrac{BM}{AD}=\dfrac{AM}{CD}\Rightarrow\dfrac{BM}{AM}=\dfrac{AD}{CD}\Rightarrow\dfrac{BM}{AM}=\dfrac{AN}{CN}\)

\(\Rightarrow BM.CN=AM.AN\)

△BMC∼△ABC (g-g)\(\Rightarrow\dfrac{BM}{AB}=\dfrac{BC}{AC}\Rightarrow BM=\dfrac{AB.BC}{AC}\Rightarrow\dfrac{1}{BM}=\dfrac{AC}{AB.BC}\Rightarrow\dfrac{1}{BM^2}=\dfrac{AC^2}{AB^2.BC^2}=\dfrac{AB^2+BC^2}{AB^2.BC^2}=\dfrac{1}{AB^2}+\dfrac{1}{BC^2}\)