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1: \(A=5x^5-5x^3+7x^2-2x+4\)
\(B\left(x\right)=-5x^6+2x^4+4x^3+4x^2-4x-1\)
2: \(A\left(x\right)+B\left(x\right)=5x^5-5x^3+7x^2-2x+4-5x^6+2x^4+4x^3+4x^2-4x-1\)
\(=-5x^6+5x^5+2x^4-x^3+11x^2-6x+3\)
\(A\left(x\right)-B\left(x\right)\)
\(=5x^5-5x^3+7x^2-2x+4+5x^6-2x^4-4x^3-4x^2+4x+1\)
\(=5x^6+5x^5-2x^4-9x^3+3x^2+2x+5\)
a) \(A=\left(x-2\right)x^2+3x\left(x-y\right)-8y\left(x+y\right)\)
\(A=x^3-2x^2+3x^2-3xy-8xy-8y^2\)
\(A=x^3+x^2-11xy-8y^2\)
b) Đây không phải là đa thức thuần nhất
a) \(A=2+2^2+2^3+...+2^{2017}\)
\(2A=2^2+2^3+2^4+...+2^{2018}\)
\(2A-A=\left(2^2+2^3+2^4+...+2^{2018}\right)-\left(2+2^2+2^3+...+2^{2017}\right)\)
\(A=2^{2018}-2\)
b) \(C=1+3^2+3^4+...+3^{2018}\)
\(3^2\cdot C=3^2+3^4+3^6+...+3^{2020}\)
\(9C-C=\left(3^2+3^4+3^6+...+3^{2020}\right)-\left(1+3^2+3^4+...+3^{2018}\right)\)
\(8C=3^{2020}-1\)
\(\Rightarrow C=\dfrac{3^{2020}-1}{8}\)
\(Toru\)
Ta có: A = 1 + 3 + 32 + 33 +.........+ 3100
=> 3A = 3 + 32 + 33 +.........+ 3101
=> 3A - A = 3101 - 1
=> 2A = 3101 - 1
=> A = 3101 - 1 / 2
A=1+3+32+33+.........+3100
=> 3A=3+32+33+.........+3101
=> 3A-A=2A=(3+32+33+.........+3101)-(1+3+32+33+.........+3100)
=> 2A=3101-1
=> A=(3101-1):2
Ta có: A = 1 + 3 + 32 + 33 +.........+ 3100
=> 3A = 3 + 32 + 33 +.........+ 3101
=> 3A - A = 3101 - 1
=> 2A = 3101 - 1
=> A = 3101 - 1/2
ta có : \(A=6.7+6.7^2+6.7^3+...+6.7^{100}\)
\(\Rightarrow7A=7.\left(6.7+6.7^2+6.7^3+...+6.7^{100}\right)\)
\(7A=6.7^2+6.7^3+6.7^4+...+6.7^{101}\)
\(\Rightarrow7A-A=6A=\left(6.7^2+6.7^3+6.7^4+...+6.7^{101}\right)-\left(6.7+6.7^2+6.7^3+...+6.7^{100}\right)\)
\(6A=6.7^{101}-6.7=6\left(7^{101}-7\right)\Leftrightarrow A=7^{101}-7\)
vậy \(A=7^{101}-7\)
ta có : \(B=6.5-6.5^2+6.5^3-...+6.5^{99}-6.5^{100}\)
\(\Rightarrow5B=5\left(6.5-6.5^2+6.5^3-...+6.5^{99}-6.5^{100}\right)\)
\(5B=6.5^2-6.5^3+6.5^4-...+6.5^{100}-6.5^{101}\)
\(\Rightarrow5B+B=6B=6.5^2-6.5^3+6.5^4-...+6.5^{100}-6.5^{101}+6.5-6.5^2+6.5^3-...+6.5^{99}-6.5^{100}\)
\(6B=6.5-6.5^{101}=6.\left(5-5^{101}\right)\Leftrightarrow B=5-5^{101}\)
vậy \(B=5-5^{101}\)
\(A=6\cdot7+6\cdot7^2+6\cdot7^3+...+6\cdot7^{100}\\ =6\cdot\left(7+7^2+7^3+...+7^{100}\right)\\ =\left(7-1\right)\cdot\left(7+7^2+7^3+...+7^{100}\right)\\ =\left(7-1\right)\cdot7+\left(7-1\right)\cdot7^2+\left(7-1\right)\cdot7^3+...+\left(7-1\right)\cdot7^{100}\\ =7^2-7+7^3-7^2+7^4-7^3+...+7^{101}-7^{100}\\ =7^{101}-7=7\cdot\left(7^{100}-1\right)\)
\(B=6\cdot5-6\cdot5^2+6\cdot5^3-...+6\cdot5^{99}-6\cdot5^{100}\\ =6\cdot\left(5-5^2+5^3-...+5^{99}-5^{100}\right)\\ =\left(5+1\right)\cdot\left(5-5^2+5^3-...+5^{99}-5^{100}\right)\\=\left(5+1\right)\cdot5-\left(5+1\right)\cdot5^2+\left(5+1\right)\cdot5^3-...+\left(5+1\right)\cdot5^{99}-5^{100}\\ =5^2+5-5^3-5^2+5^4+5^3+...+5^{100}+5^{99}-5^{101}-5^{100}\\ =5-5^{101}\\ =5\cdot\left(1-5^{100}\right)\)