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bài 1.( x 3 + 1 x 2 ) 1 0 = ( x 3 + x − 2 ) 1 0 (x^3+\dfrac{1}{x^2})^10 = (x^3+x^{-2})^10 ( x 3 + x 2 1 ) 1 0 = ( x 3 + x − 2 ) 1 0 T k + 1 = C 10 k . x 30 − 3 k . x − 2 k = C 10 k . x 30 − 5 k T_{k+1}=C_{10}^k.x^{30-3k}.x^{-2k}=C_{10}^k.x^{30-5k} T k + 1 = C 1 0 k . x 3 0 − 3 k . x − 2 k = C 1 0 k . x 3 0 − 5 k k + 1 k+1 k + 1 x 15 ⇔ 30 − 5 k = 15 ⇔ k = 3 x^{15} \Leftrightarrow 30-5k=15 \Leftrightarrow k=3 x 1 5 ⇔ 3 0 − 5 k = 1 5 ⇔ k = 3 x 15 x^{15} x 1 5 C 10 4 . x 15 = 210. x 15 C_{10}^4.x^{15}=210.x^{15} C 1 0 4 . x 1 5 = 2 1 0 . x 1 5 4. C n 3 = C n + 1 2 4.C_n^3=C_{n+1}^2 4 . C n 3 = C n + 1 2 ⇔ 4. n ! 6. ( n − 3 ) ! = ( n + 1 ) ! ( n − 1 ) ! \Leftrightarrow \dfrac{4.n!}{6.(n-3)!} = \dfrac{(n+1)!}{(n-1)!} ⇔ 6 . ( n − 3 ) ! 4 . n ! = ( n − 1 ) ! ( n + 1 ) ! ⇔ 2 3 . n . ( n − 1 ) ( n − 2 ) = n . ( n + 1 ) \Leftrightarrow \dfrac{2}{3}.n.(n-1)(n-2)=n.(n+1) ⇔ 3 2 . n . ( n − 1 ) ( n − 2 ) = n . ( n + 1 ) n ≥ 3 n \geq 3 n ≥ 3 n n n 2 3 . ( n − 1 ) ( n − 2 ) = n + 1 \dfrac{2}{3}.(n-1)(n-2)=n+1 3 2 . ( n − 1 ) ( n − 2 ) = n + 1 ( x 5 − 5. x − 1 ) 12 (\dfrac{x}{5} - 5.x^{-1})^{12} ( 5 x − 5 . x − 1 ) 1 2 T k + 1 = 1 5 12 − k . ( − 5 ) k . C 12 k . x 12 − k . x − k T_{k+1}=\dfrac{1}{5^{12-k}}.(-5)^k .C_{12}^k.x^{12-k}.x^{-k} T k + 1 = 5 1 2 − k 1 . ( − 5 ) k . C 1 2 k . x 1 2 − k . x − k = 1 5 12 − k . ( − 5 ) k . C 12 k . x 12 − 2 k =\dfrac{1}{5^{12-k}}.(-5)^k .C_{12}^k.x^{12-2k} = 5 1 2 − k 1 . ( − 5 ) k . C 1 2 k . x 1 2 − 2 k T k + 1 T_{k+1} T k + 1 x 4 ⇔ 12 − 2 k = 4 ⇔ k = 4 x^4 \Leftrightarrow 12-2k=4 \Leftrightarrow k=4 x 4 ⇔ 1 2 − 2 k = 4 ⇔ k = 4
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn