Bài 1:
a) \(\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=17\)
\(\Rightarrow x^3-3x^2+3x-1+2^3-x^3+3x^2+6x=17\)
\(\Rightarrow9x+7=17\)
\(\Rightarrow9x=17-7=10\)
\(\Rightarrow x=\dfrac{10}{9}\)
b) \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-2\right)=15\)
\(\Rightarrow x^3+2^3-x^3+2x=15\)
\(\Rightarrow8+2x=15\)
\(\Rightarrow2x=15-8=7\)
\(\Rightarrow x=\dfrac{7}{2}\)
c) \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\)
\(\Rightarrow x^3-3x^2.3+3x.3^2-3^3-x^3+3^3+9\left(x^2+2x+1\right)=15\)
\(\Rightarrow-9x^2+27x+9x^2+18x+9=15\)
\(\Rightarrow45x+9=15\)
\(\Rightarrow45x=6\)
\(\Rightarrow x=\dfrac{6}{45}=\dfrac{2}{15}\)
d) \(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=3\)
\(\Rightarrow x\left(x^2-5^2\right)-x^3-2^3=3\)
\(\Rightarrow x^3-25x-x^3-8=3\)
\(\Rightarrow-25x-8=3\)
\(\Rightarrow-25x=3+8=11\)
\(\Rightarrow x=-\dfrac{11}{25}\)
Bài 2:
a) Ta có:
\(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(B=\left(2^8-1\right)\left(2^8+1\right)\)
\(B=2^{16}-1\)
Vì 216 - 1 < 216
=> B < A
b) Ta có:
\(A=4\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(A=\dfrac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(A=\dfrac{1}{2}\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(A=\dfrac{1}{2}\left(3^8-1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)
\(A=\dfrac{1}{2}\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\left(3^{64}+1\right)\)
\(A=\dfrac{1}{2}\left(3^{32}-1\right)\left(3^{32}+1\right)\left(3^{64}+1\right)\)
\(A=\dfrac{1}{2}\left(3^{64}-1\right)\left(3^{64}+1\right)\)
\(A=\dfrac{1}{2}\left(3^{128}-1\right)\)
Vì 1/2( 3128 - 1) < 3128 - 1
=> A < B
