A = \(\left|x-2,5\right|+\left|x-1,7\right|\)|
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A=-x+2.5+x-1.7
A=0.8
B=-x+1.5+x-2/5
B=1.5-2/5
B=11/10hayB=1.1
\(=x^6-6x^4+12x^2-8-x^3+x+6x^2-18x\\ =x^6-6x^4-x^3+18x^2-17x-8\)
A = 5(x + 3)(x - 3) + (2x + 3)3 + (x - 6)2
A = 5(x + 3)(x - 3) + 4x2 + 12x + 9 + x2 - 12x + 36
A = 5x2 - 45x + 4x2 + 12x + 9 + x2 - 12x + 36
A = 10x2 (1)
Thay x = -1/5 vào (1), ta có:
A = 10x2 = 10.(-1/5)2 = 2/5
A = 2/5
Vậy:...
\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
Ta có :
\(2P=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(5^{32}-1\)
\(\Rightarrow P=\frac{5^{32}-1}{2}\)
ta có : P=\(12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)=\frac{12\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{\left(5^2-1\right)}\)
=> P=\(\frac{5^{32}-1}{2}\)
\(=\left(x-\dfrac{1}{3}\right)\left(\dfrac{4}{3}x+\dfrac{1}{9}-x+\dfrac{1}{3}\right)\\ =\left(x-\dfrac{1}{3}\right)\left(\dfrac{1}{3}x+\dfrac{4}{9}\right)\\ =\dfrac{1}{3}x^2+\dfrac{4}{9}x-\dfrac{1}{9}x-\dfrac{4}{27}\\ =\dfrac{1}{3}x^2+\dfrac{1}{3}x-\dfrac{4}{27}\)
+, Nếu x < 1,7
=> A = 2,5-x+1,7-x = 4,2-2x
+, Nếu 1,7 < = x < = 2,5
=> A = 2,5-x+x-1,7 = 0,8
+, Nếu x > 2,5
=> A = x-2,5+x-1,7 = 2x-4,2
Tk mk nha