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a: góc DAC=90-40=50 độ
b: góc ADB=90 độ
c: góc DAB=90-80=10 độ
=>góc BAE=10+50=60 độ
góc AED=180-60=120 độ
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a) Để f(x) có bậc là 1 thì m=0
b) Thay x=-1 vào f(x), ta được:
\(m\cdot\left(-1\right)^2+3\cdot\left(-1\right)+6=0\)
\(\Leftrightarrow m+3=0\)
hay m=-3
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a) (1,75 : \(\dfrac{7}{2}\)).\(\dfrac{8}{5}\)=(\(\dfrac{7}{4}\) : \(\dfrac{7}{2}\)).\(\dfrac{8}{5}\)=(\(\dfrac{7}{4}\).\(\dfrac{2}{7}\)).\(\dfrac{8}{5}\)=\(\dfrac{1}{2}\).\(\dfrac{8}{5}\)=\(\dfrac{4}{5}\)
b) \(\dfrac{7}{2}\).\(4\dfrac{5}{3}\)-\(2\dfrac{5}{3}\).\(\dfrac{7}{2}\)=(\(4\dfrac{5}{3}\)-\(2\dfrac{5}{3}\)).\(\dfrac{7}{2}\)=2.\(\dfrac{7}{2}\)=7
c)\(\dfrac{-5}{9}\).(\(\dfrac{3}{10}-\dfrac{1}{5}\))=\(\dfrac{-5}{9}\).(\(\dfrac{3}{10}-\dfrac{2}{10}\))=\(\dfrac{-5}{9}\).\(\dfrac{1}{10}\)=\(\dfrac{-1}{18}\)
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\(a,1-\left(\dfrac{\dfrac{5}{3}}{8}+x-\dfrac{\dfrac{7}{5}}{24}\right)-\dfrac{\dfrac{16}{2}}{3}=0\\ \Leftrightarrow1-\left(\dfrac{5}{24}+x-\dfrac{7}{120}\right)=\dfrac{8}{3}\\ \Leftrightarrow\dfrac{3}{20}+x=1-\dfrac{8}{3}=-\dfrac{5}{3}\\ \Leftrightarrow x=-\dfrac{5}{3}-\dfrac{3}{20}=-\dfrac{109}{60}\)
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\(\dfrac{\sqrt{5^2}-\sqrt{19^2}}{\sqrt{8^2}-\sqrt{22^2}}=\dfrac{5-19}{8-22}=\dfrac{-14}{-14}=1\)
Bài 1:
a) Ta có: \(\dfrac{-10}{21}x^2y\cdot\dfrac{14}{-15}xy^3\)
\(=\left(\dfrac{10}{21}\cdot\dfrac{14}{15}\right)\cdot\left(x^2\cdot x\right)\cdot\left(y\cdot y^3\right)\)
\(=\dfrac{4}{9}x^3y^4\)
Hệ số là \(\dfrac{4}{9}\)
Phần biến là \(x^3;y^4\)
Bậc là 7
b) Ta có: \(\dfrac{-32}{3}x^2y\cdot\left(\dfrac{-6}{8}xy^4\right)^3\)
\(=\dfrac{-32}{3}\cdot\dfrac{-27}{64}\cdot\left(x^2\cdot x^3\right)\cdot\left(y\cdot y^{12}\right)\)
\(=\dfrac{9}{2}x^5y^{13}\)
Hệ số là \(\dfrac{9}{2}\)
Phần biến là \(x^5;y^{13}\)
Bậc là 18
cam mon ạ