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Ta có : A = (3 + 1) (32 + 1) (34 + 1) ... (364 + 1)
=> 8A = (32 - 1)(32 + 1)(34 + 1)......(364 + 1)
=> 8A = (34 - 1)(34 + 1)......(364 + 1)
=> 8A = (364 - 1)(364 + 1)
=> A = \(\frac{3^{64}-1}{8}\)
Ta có \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\cdot\cdot\cdot\left(3^{64}+1\right)\)
\(\Rightarrow2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\cdot\cdot\cdot\left(3^{64}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\cdot\cdot\cdot\left(3^{64}+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right)\cdot\cdot\cdot\left(3^{64}+1\right)\)
\(=\left(3^{64}-1\right)\left(3^{64}+1\right)=\left(3^{128}-1\right)\)
\(\Rightarrow A=\frac{3^{128}-1}{2}\)
Ta có
B = 2 a − 3 a + 1 − a − 4 2 − a a + 7 = 2 a 2 + 2 a – 3 a – 3 – ( a 2 – 8 a + 16 ) – ( a 2 + 7 a ) = 2 a 2 + 2 a – 3 a – 3 – a 2 + 8 a – 16 – a 2 – 7 a = - 19
Đáp án cần chọn là: D
a: \(A=4\cdot\dfrac{5}{2}\sqrt{x}-\dfrac{8}{3}\cdot\dfrac{3}{2}\sqrt{x}-\dfrac{4}{3x}\cdot\dfrac{3x}{8}\cdot\sqrt{x}\)
\(=10\sqrt{x}-4\sqrt{x}-\dfrac{1}{2}\sqrt{x}\)
\(=\dfrac{11}{2}\sqrt{x}\)
b: \(B=\dfrac{y}{2}+\dfrac{3}{4}\cdot\left|2y-1\right|-\dfrac{3}{2}\)
\(=\dfrac{y}{2}+\dfrac{3}{4}\left(1-2y\right)-\dfrac{3}{2}\)
=1/2y+3/4-3/2y-3/2
=-y-3/4
A = (3 + 1) (32 + 1) (34 + 1) ... (364 + 1)
2A = (3 - 1)(3 + 1) (32 + 1) (34 + 1) ... (364 + 1)
2A = (32 - 1)(32 + 1) (34 + 1) ... (364 + 1)
= (34 - 1)(34 + 1) ... (364 + 1)
= (38 - 1)(38 + 1)(316+1)(332+1)(364+1)
= (316-1)(316+1)(332+1)(364+1)
= (332-1)(332+1)(364+1)
= (364-1)(364+1)
= (3128-1)
=> A = \(\frac{3^{128}-1}{2}\)
\(2A=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
áp dụng hằng đẳng thức \(a^2-b^2\)
ta có 2A=\(3^{128}-1\)=>A=\(\frac{3^{128}-1}{2}\)
Đáp án cần chọn là: D
A = 3. − 4 .60 − 60 50.20 = 3. − 4 − 1 .60 50.20 = − 13.60 50.20 = − 13.3 50 = − 39 50
