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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 \(S=\left\{\dfrac{-1}{2}\right\}\)
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)\(\Rightarrow\dfrac{BM}{AB}=\dfrac{BC}{AC}\Rightarrow BM=\dfrac{AB.BC}{AC}\Rightarrow\dfrac{1}{BM}=\dfrac{AC}{AB.BC}\Rightarrow\dfrac{1}{BM^2}=\dfrac{AC^2}{AB^2.BC^2}=\dfrac{AB^2+BC^2}{AB^2.BC^2}=\dfrac{1}{AB^2}+\dfrac{1}{BC^2}\)
\(a,=2x\left(2x+3y\right)\\ b,=3x\left(x^2-9\right)=3x\left(x-3\right)\left(x+3\right)\\ c,=x\left(x-7y\right)+\left(x-7y\right)=\left(x+1\right)\left(x-7y\right)\\ d,=\left(x+y\right)^2-9z^2=\left(x+y-3z\right)\left(x+y+3z\right)\)
a. 5x + 10y = 5(x + 2y)
b. 3x3 - 12x
= 3x(x2 - 4)
= 3x(x - 2)(x + 2)
c. 4x2 + 9x - 4xy - 9y
= 4x2 - 4xy + 9x - 9y
= 4x(x - y) + 9(x - y)
= (4x + 9)(x - y)
d. 3x2 + 5y - 3xy - 5x
= 3x2 - 3xy - 5x + 5y
= 3x(x - y) - 5(x - y)
= (3x - 5)(x - y)
e. 3y2 - 3z2 + 3x2 + 6xy
= 3x2 + 6xy + 3y2 - 3z2
= 3(x2 + 2xy + y2 - z2)
= \(3\left[\left(x+y\right)^2-z^2\right]\)
= \(3\left(x+y+z\right)\left(x+y-z\right)\)
a) \(=5\left(x+2y\right)\)
b) \(=3x\left(x^2-4\right)=3x\left(x-2\right)\left(x+2\right)\)
c) \(=4x\left(x-y\right)+9\left(x-y\right)=\left(x-y\right)\left(4x+9\right)\)
d) \(=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)
e) \(=3\left[\left(x^2+2xy+y^2\right)-z^2\right]=3\left[\left(x+y\right)^2-z^2\right]\)
\(=3\left(x+y-z\right)\left(x+y+z\right)\)
a, Vì ABCD là hbh nên AD//BC hay AM//BC và AD=AM=BC(tính chất đối xứng)
Do đó AMBC là hbh
b, Ta có AMBC là hbh
Mà N là giao điểm AB và MC nên N là trung điểm AB và MC
Vậy MN=NC
1) \(2\left(x-9\right)=12\Rightarrow x-9=12:2=6\)
\(\Rightarrow x=6+9=15\)
2) \(5x\left(x-3\right)-2\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(5x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{2}{5}\end{matrix}\right.\)
3) \(\left(x-4\right)\left(7x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{2}{7}\end{matrix}\right.\)
4) \(\left(x-1\right)\left(x+1\right)-x\left(x+3\right)=0\)
\(\Rightarrow x^2-1-x^2-3x=0\Rightarrow3x=-1\Rightarrow x=-\dfrac{1}{3}\)