a=1 b=2 c=3 d=4

đây nè

k mik đi

Ta có :

\(\frac{30}{43}=\frac{1}{\frac{43}{30}}=\frac{1}{1+\frac{13}{30}}=\frac{1}{1+\frac{1}{\frac{30}{13}}}=\frac{1}{1+\frac{1}{2+\frac{4}{13}}}=\frac{1}{1+\frac{1}{2+\frac{1}{\frac{13}{4}}}}=\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}}\)

Vậy \(a=1;b=2;c=3;d=4\)

Ta có: \(\frac{30}{43}=\frac{1}{\frac{43}{30}}=\frac{1}{1+\frac{13}{30}}=\frac{1}{1+\frac{1}{2+\frac{4}{13}}}=\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}}\)

\(\Rightarrow\)a = 1 ; b  = 2 ; c = 3 ; d = 4

Vậy: 

a = 1 ; b  = 2 ; c = 3 ; d = 4

Sao bạn kia đăng lên rồi tự giải vậy ??????????????????????????????

bài wá rễ

Ai là hs giỏi thì giúp mik vs mai nộp rồi

Ta có : \(\frac{30}{43}=\frac{1}{\frac{43}{30}}=\frac{1}{1+\frac{13}{30}}=\frac{1}{1+\frac{1}{2+\frac{4}{13}}}=\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}}\)

Vậy a = 1,b = 2,c = 3,d = 4

\(\frac{30}{43}=\frac{1}{\frac{43}{30}}=\frac{1}{1+\frac{13}{30}}=\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}}\)

Vậy a = 1;b=2;c=3;d=4 ( thỏa mãn )

a) 3x - 6y = 3 . x - 3 . 2y = 3(x - 2y)

b) 2525x² + 5x³ + x²y = x² (2525 + 5x + y)

c) 14x²y – 21xy² + 28x²y² = 7xy . 2x - 7xy . 3y + 7xy . 4xy = 7xy(2x - 3y + 4xy)

d) 2525x(y - 1) - 2525y(y - 1) = 2525(y - 1)(x - y)

e) 10x(x - y) - 8y(y - x) =10x(x - y) - 8y[-(x - y)]

                                    = 10x(x - y) + 8y(x - y)

                                    = 2(x - y)(5x + 4y)

oooooooooooooooooooooooooooooooooooooooooooooooolllllllllllllllllllllllllllllllllllllleeeeeeeeeeeeeeeeeeeeellllllllllllllllllllllllllleeeeeeeeeeeeeee

\(\frac{30}{43}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)\(\Rightarrow\frac{1}{\frac{43}{30}}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)

\(\Rightarrow\frac{1}{1+\frac{13}{30}}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)\(\Rightarrow\frac{1}{1+\frac{1}{\frac{30}{13}}}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)\(\Rightarrow\frac{1}{1+\frac{1}{2+\frac{4}{13}}}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)\(\Rightarrow\frac{1}{1+\frac{1}{2+\frac{1}{\frac{13}{4}}}}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)\(\Rightarrow\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)

\(\Rightarrow a=1,b=2,c=3,d=4\)

Ta có: \(\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}}=\frac{1}{1+\frac{1}{2+\frac{1}{\frac{13}{4}}}}=\frac{1}{1+\frac{1}{2+\frac{4}{13}}}=\frac{1}{1+\frac{1}{\frac{30}{13}}}=\frac{1}{1+\frac{13}{30}}=\frac{1}{\frac{43}{30}}=\frac{30}{43}\)

Vậy a=1

      b=2

      c=3

      d=4

\(\frac{30}{43}=\frac{1}{\frac{43}{30}}=\frac{1}{1+\frac{13}{30}}=\frac{1}{1+\frac{1}{\frac{30}{13}}}=\frac{1}{1+\frac{1}{2+\frac{4}{13}}}=\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}}\)

a=1;b=2;c=3;d=4