Xét tam giác ABC vuông tại A

a, Theo Pytago ta có \(c=\sqrt{a^2-b^2}=3\sqrt{13}\)

sinB = AC/BC = 18/21 = 6/7 => ^B = \(\approx\)59⁰

Do ^B ; ^C phụ nhau => ^C \(\approx\)31⁰

b, Do ^B ; ^C phụ nhau => ^B = 60⁰

tanC = AB/AC = c/b => c = b.tanC = \(\dfrac{10\sqrt{3}}{3}\)

cosC = AC/BC = b/a => a = b/cosC = \(\dfrac{20\sqrt{3}}{3}\)

c, Theo Pytago \(a=\sqrt{b^2+c^2}=\sqrt{34}\)

tanB = AC/AB => ^B  \(\approx\)31⁰ 

Do ^B ; ^C phụ nhau ^C \(\approx\)59⁰

Cho pt đt có dạng y = ax + b 

Do đt đi qua A(-4;5) ; B(-1;3 ) 

=> \(\left\{{}\begin{matrix}-4a+b=5\\-a+b=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{2}{3}\\b=\dfrac{7}{3}\end{matrix}\right.\)

=> pt đt có dạng y = -2/3x + 7/3 

đk x khác -1/3 ; 3/2 

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

\(\Leftrightarrow4x^2-12x+9=9x^2-1\Leftrightarrow5x^2+12x-10=0\Leftrightarrow x=\dfrac{-6\pm\sqrt{86}}{5}\)

Xét tam giác ABC vuông tại B

ADHT \(\dfrac{1}{BH^2}=\dfrac{1}{AB^2}+\dfrac{1}{BC^2}\Rightarrow\dfrac{1}{BH^2}=\dfrac{1}{\dfrac{9BC^2}{16}}+\dfrac{1}{BC^2}\)

\(\Leftrightarrow\dfrac{1}{BH^2}=\dfrac{16}{9BC^2}+\dfrac{1}{BC^2}=\dfrac{25}{9BC^2}\Rightarrow BH^2=\dfrac{9BC^2}{25}\)

\(\Leftrightarrow BH=\dfrac{3}{5}BC\Rightarrow BC=\dfrac{5}{3}BH=\dfrac{5}{3}.\dfrac{12}{5}=4\)cm 

=> AB = 3cm 

Theo Pytago \(AC=\sqrt{AB^2+BC^2}=5cm\)

adht \(AB^2=AH.AC\Rightarrow AH=\dfrac{AB^2}{AC}=\dfrac{9}{5}cm\)

=> CH = AC - AH = 16/5 cm

\(4x^3+2x^2-4x-2=2x^2\left(2x+1\right)-2\left(2x+1\right)=\left(2x^2-2\right)\left(2x+1\right)\)

\(=2\left(x^2-1\right)\left(2x-1\right)=2\left(x-1\right)\left(x+1\right)\left(2x-1\right)\)

\(37-\left(x-2\right)^2=28\Leftrightarrow\left(x-2\right)^2=9\Leftrightarrow\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

\(\left(3x-6\right)^3.3=81\Leftrightarrow\left(3x-6\right)^3=27\Rightarrow3x-6=3\Leftrightarrow x=3\)

\(M=\left(x-2\right)\left(x^2+2x+4\right)-\left(x+2\right)\left(x^2+2x+4\right)\)

\(=x^3-8-\left(x^3+2x^2+4x+2x^2+4x+8\right)\)

\(=x^3-8-x^3-4x^2-8x-8=-4x^2-8x-16\)

Bài 3 

a, \(\left(x-y\right)\left(x^2+xy+y^2\right)=x^3+x^2y+xy^2-x^2y-xy^2-y^3=x^3-y^3\)

b, \(\left(\dfrac{1}{2}xy-1\right)\left(x^3-2x-6\right)=\dfrac{x^4y}{2}-x^2y-3xy-x^3+2x+6\)

c, \(\left(x-2y^2\right)\left(x^2+2xy+4y^4\right)=x^3+2x^2y+4xy^4-2x^2y^2-4xy^3-8y^6\)

d, \(\left(x-1\right)\left(x-2\right)x=\left(x^2-3x+2\right)x=x^3-3x^2+2x\)

Bài 2 

a, \(-2x^3y\left(2x^2-3y+5xy\right)=-4x^5y+6x^3y-10x^4y^2\)

b, \(-2y\left(x^2y^2-xy+2y\right)=-2x^2y^3+2xy^2-4y^2\)

c, \(\dfrac{2}{5}x^2y\left(x^2y-5x+10y\right)=\dfrac{2}{5}x^4y^2-2x^3y+4x^2y^2\)

d, \(-\dfrac{1}{3}xy\left(\dfrac{3}{2}x^2y-xy+\dfrac{6}{5}x\right)=-\dfrac{1}{2}x^3y^2+\dfrac{1}{3}x^2y^2-\dfrac{2}{5}x^2y\)

\(\dfrac{2}{15}\left(\dfrac{5}{3}-\dfrac{2}{7}\right)-\left(\dfrac{7}{3}-\dfrac{3}{7}\right)\dfrac{2}{15}=\dfrac{2}{15}\left(\dfrac{5}{3}-\dfrac{2}{7}-\dfrac{7}{3}+\dfrac{3}{7}\right)=\dfrac{2}{15}\left(-\dfrac{2}{3}+\dfrac{1}{7}\right)=-\dfrac{4}{45}+\dfrac{2}{105}=-\dfrac{22}{315}\)