Tính tổng:
1/5 + 1/5+10 + 1/5+10+15 + .... + 1/5+10+15+....+100
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a)
=> Ta có : \(\widehat{A}+\widehat{B}+\widehat{C}\) = 180o
100o + \(\widehat{B}+\widehat{C}\) = 180o
\(\widehat{B}+\widehat{C}\) = 180o - 100o
\(\widehat{B}+\widehat{C}\) = 80o
Góc B = (80o+50o):2 = 65o
=> \(\widehat{C}\) = 65o - 50o = 15o
Vậy \(\widehat{B}\) = 65o ; \(\widehat{C}\) = 15o
b)
Ta có : \(\widehat{3A}+\widehat{B}+\widehat{2C}\) = 180o
\(\widehat{3A}+\widehat{2C}\) = 180o - 80o
\(\widehat{3A}+\widehat{2C}\) = 100o
=> \(\widehat{A}\) = 100o:(3+2).3 = 60o
\(\widehat{C}\) = 100o - 60o = 40o
Vậy \(\widehat{A}\) = 60o ; \(\widehat{C}\) = 40o
Áp dụng tính chất của dãy tỉ số bằng nhau,ta có :
\(\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b}{c}=\frac{\left(b+c\right)+\left(c+a\right)+\left(a+b\right)}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)
\(\frac{b+c}{a}=2\Rightarrow b+c=2a\)( 1 )
\(\frac{c+a}{b}=2\Rightarrow c+a=2b\)( 2 )
\(\frac{a+b}{c}=2\Rightarrow a+b=2c\)( 3 )
Từ ( 1 ),(2) và ( 3 ) \(\Rightarrow a=b=c\)
\(x-4xy+y=0\Leftrightarrow4x-16xy+4y=0\)
\(\Leftrightarrow4x-4y\left(4x-1\right)=0\)
\(\Leftrightarrow4x-1-4y\left(4x-1\right)=-1\)
\(\Leftrightarrow\left(4x-1\right)\left(1-4y\right)=-1\)(1)
Ta có \(-1=1.\left(-1\right)\) để pt (1) có nghiệm nghuyên khi 4x - 1 và 1 - 4y là ước nguyên của - 1
+) Nếu \(4x-1=1\) thì \(1-4y=-1\) => \(x=\frac{1}{2}\) thì \(y=\frac{1}{2}\) (loại)
+) Nếu \(4x-1=-1\) thì \(1-4y=1\) => x = 0 thì y = 0 (TM)
Vậy (x;y) = (0;0)
A,B,C tỉ lệ với a,b,c
\(\Rightarrow\frac{A}{a}=\frac{B}{b}=\frac{C}{c}\)
đặt \(\frac{A}{a}=\frac{B}{b}=\frac{C}{c}=k\)\(\Rightarrow\)A = ka ; B = kb ; C = kc
Q = \(\frac{Ax+By+C}{ax+by+c}=\frac{kax+kbx+kc}{ax+by+c}=\frac{k\left(ax+bx+c\right)}{ax+by+c}=k\)
\(\frac{a}{b}=\frac{c}{d}\)
\(\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{2c^2}{2d^2}=\frac{ac}{bd}\)
Ta có : \(\frac{2c^2}{2d^2}=\frac{ac}{bd}=\frac{2c^2-ac}{2d^2-bd}\)
Vậy \(\frac{a^2}{b^2}=\frac{2c^2-ac}{2d^2-bd}\)
`Answer:`
\dfrac15+\dfrac1{5+10}+\dfrac1{5+10+15}\ +\,.\!.\!.+\ \dfrac1{5+10+15\ +\,.\!.\!.+\ 100}\\=\dfrac15+\dfrac1{5.(1+2)}+\dfrac1{5.(1+2+3)}\ +\,.\!.\!.+\ \dfrac1{5.(1+2+3\ +\,.\!.\!.+\ 20)}\\=\dfrac15\left(1+\dfrac1{1+2}+\dfrac1{1+2+3}\ +\,.\!.\!.+\ \dfrac1{1+2+3\ +\,.\!.\!.+\ 20}\right)\\=\dfrac15\bigg(\dfrac22+\dfrac26+\dfrac2{12}\ +\,.\!.\!.+\ \dfrac2{20.21}\bigg)\\=\dfrac25\left(\dfrac1{1.2}+\dfrac1{2.3}+\dfrac1{3.4}\ +\,.\!.\!.+\ \dfrac1{20.21}\right)\\=\dfrac25\left(1-\dfrac12+\dfrac12-\dfrac13+\dfrac13-\dfrac14\ +\,.\!.\!.+\ \dfrac1{20}-\dfrac1{21}\right)\\=\dfrac25\left(1-\dfrac1{21}\right)\\=\dfrac25\!\cdot\!\dfrac{20}{21}\\=\dfrac8{21}
`Answer:`
Mình gửi lại bài nhé. Mong lần này không bị lỗi như lần trước.
\(\dfrac15+\dfrac1{5+10}+\dfrac1{5+10+15}\ +\,.\!.\!.+\ \dfrac1{5+10+15\ +\,.\!.\!.+\ 100}\\=\dfrac15+\dfrac1{5.(1+2)}+\dfrac1{5.(1+2+3)}\ +\,.\!.\!.+\ \dfrac1{5.(1+2+3\ +\,.\!.\!.+\ 20)}\\=\dfrac15\left(1+\dfrac1{1+2}+\dfrac1{1+2+3}\ +\,.\!.\!.+\ \dfrac1{1+2+3\ +\,.\!.\!.+\ 20}\right)\\=\dfrac15\bigg(\dfrac22+\dfrac26+\dfrac2{12}\ +\,.\!.\!.+\ \dfrac2{20.21}\bigg)\\=\dfrac25\left(\dfrac1{1.2}+\dfrac1{2.3}+\dfrac1{3.4}\ +\,.\!.\!.+\ \dfrac1{20.21}\right)\\=\dfrac25\left(1-\dfrac12+\dfrac12-\dfrac13+\dfrac13-\dfrac14\ +\,.\!.\!.+\ \dfrac1{20}-\dfrac1{21}\right)\\=\dfrac25\left(1-\dfrac1{21}\right)\\=\dfrac25\!\cdot\!\dfrac{20}{21}\\=\dfrac8{21}\)