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6) c) x3 - x2 + x = 1
<=> x3 - x2 + x - 1 = 0
<=> (x3 - x2) + (x - 1) = 0
<=> x2 (x - 1) + (x - 1) = 0
<=> (x - 1) (x2 + 1) = 0
=> x - 1 = 0 hoặc x2 + 1 = 0
* x - 1 = 0 => x = 1
* x2 + 1 = 0 => x2 = -1 => x = -1
Vậy x = 1 hoặc x = -1
Bài 5:
a) Đặt \(A=\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=3^{32}-1\)
\(\Rightarrow A=\frac{3^{32}-1}{8}\)
b) (7x+6)2 + (5-6x)2 - (10-12x)(7x+6)
=(7x+6)2 + (5-6x)2 - 2(5-6x)(7x+6)
\(=\left(7x+6-5+6x\right)^2\)
\(=\left(13x+1\right)^2\)
Bài 2: Rút gọn biểu thức sau một cách nhanh nhất:
a, A=(6x-2)2+(2-5x)2+2.(6x-2)(2-5x)
\(=\left(6x-2\right)^2+2\left(6x-2\right)\left(2-5x\right)+\left(2-5x\right)^2\)
\(\text{(Hằng đẳng thức số 2)}\)
\(=\left(6x-2+2-5x\right)\)
\(=x\)
\(B=\left(2a^2+2a+1\right)\left(2a^2-2a+1\right)-\left(2a^2+1\right)^2\)
\(=\left(2a^2+1+2a\right)\left(2a^2+1-2a\right)-\left(2a^2+1\right)^2\)
\(=\left(2a^2+1\right)^2-4a^2-\left(2a^2+1\right)^2\)
\(=-4a^2\)
\(C=10^2+8^2+6^2...+2^2-\left(9^2+7^2+5^2+...+1^2\right)\)
\(\Rightarrow C=\left(10^2-9^2\right)+\left(8^2-7^2\right)+...+\left(2^2-1^2\right)\)
\(\Rightarrow C=\left(10-9\right)\left(10+9\right)+\left(8-7\right)\left(8+7\right)+...+\left(2-1\right)\left(2+1\right)\)
\(\Rightarrow C=19+15+...+3\)
Vậy C = {(19 + 3)[(19-3):4+1]} :2 = 60
\(C=10^2-9^2+8^2-...-3^2+2^2-1^2\)
\(=\left(10+9\right)\left(10-9\right)+\left(8+7\right)\left(8-7\right)+...+\left(2+1\right)\left(2-1\right)\)
\(=10+9+8+...+2+1\)
\(=\frac{\left(1+10\right)10}{2}=55\)
Vậy C=55
1: \(4a^2b^4-c^4d^2\)
\(=\left(2ab^2-c^2d\right)\left(2ab^2+c^2d\right)\)
4: \(\left(a+b\right)^3-\left(a-b\right)^3\)
\(=\left(a+b-a+b\right)\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(=2b\left(a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2\right)\)
\(=2b\left(3a^2+b^2\right)\)
5: \(\left(a+b\right)^3+\left(a-b\right)^3\)
\(=a^3+b^3+3a^2b+3ab^2+a^3-3a^2b+3ab^2-b^3\)
\(=2a^3+6ab^2\)
\(=2a\left(a^2+3b^2\right)\)
= \(10^2+8^2+6^2+4^2+2^2-9^2-7^2-5^2-3^2-1\)-1
=\(55\)
\(\left(10^2+8^2+6^2+4^2+2^2\right)-\left(9^2+7^2+5^2+3^2+1^2\right)\)
\(=10^2+8^2+6^2+4^2+2^2-9^2-7^2-5^2-3^2-1^2\)
\(=\left(10^2-9^2\right)+\left(8^2-7^2\right)+...+\left(2^2-1^2\right)\)
\(=\left(10-9\right)\left(10+9\right)+\left(8-7\right)\left(8+7\right)+...+\left(2-1\right)\left(1+2\right)\)
\(=10+9+8+7+...+2+1\)
\(=\frac{\left(1+10\right)\cdot10}{2}\)
\(=55\)