\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}\\ \dfrac{1}{3^2}< \dfrac{1}{2\cdot3}\\ \dfrac{1}{4^2}< \dfrac{1}{3\cdot4}\\ ...\\ \dfrac{1}{50^2}< \dfrac{1}{49\cdot50}\\ \Rightarrow A=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}< 1+\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{49\cdot50}\\ \Rightarrow A< 1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\\ \Rightarrow A< 2-\dfrac{1}{50}< 2\)
`A = 1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + ... + 1/50^2`
`=> A = 1 + 1/2^2 + 1/3^2 + 1/4^2+ ... + 1/50^2`
ta giữ nguyên `1`:
ta thấy :
`1/2^2 = 1/(2.2) < 1/(1.2)`
`1/3^2 = 1/(3.3) < 1/(2.3)`
`1/4^2 = 1/(4.4) < 1/(3.4)`
`............................................`
`1/50^2 = 1/(50.50) < 1/(49.50)`
`=> A < 1 + 1/(1.2) + 1/(2.3) + .... + 1/(49.50)`
`=> A < 1 + 1 - 1/2 + 1/2 - 1/3 + ... + 1/49 - 1/50`
`=> A < 1 + 1 - 1/50`
`=> A < 2 - 1/50 < 2`
`=> A < 2`
Phần 9) \(\dfrac{-5}{6}x\left(\dfrac{2}{3}x^2y+\dfrac{3}{4}xy^2-\dfrac{1}{2}xy\right)\)= \(\dfrac{-5}{9}x^3y-\dfrac{5}{8}x^2y^2+\dfrac{5}{12}x^2y\)
Phần 10) \(x^4.y^5\left(\dfrac{7}{3}x^2.y^4-\dfrac{1}{7}x^3.y\right)=\dfrac{7}{3}x^6.y^9-\dfrac{1}{7}.x^7.y^6\)
Phần 11) \(5x^2-3x.\left(3x+2\right)=5x^2-9x^2-6x=-4x^2-6x\)
Phần 12) \(-4x^2+2x-4x\left(x-5\right)=-4x^2+2x-4x^2+20x=-8x^2+22x\)
\(CuongDoraemon123\)
