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\(A=2^5.5^2-8^2-7=800-64-7=729=27^2\)
\(B=2^3.4^2+3^2.3^2-40=128+81-40=169=13^2\)
\(C=11.2^4+6^2.19+40=176+684+40=900=30^2\)
\(D=4^3+6^3+7^3+2=64+216+343+2=625=25^2\)
a: \(2^5\cdot5^2-8^2-7=27^2\)
b: \(2^3+3^2\cdot3^2-40=7^2\)
d: Ta có: \(D=5^3+6^3+59\)
\(=125+216+59\)
\(=400=20^2\)
Bài 8:
Ta có: \(A=-x^2+2x+4\)
\(=-\left(x^2-2x-4\right)\)
\(=-\left(x^2-2x+1-5\right)\)
\(=-\left(x-1\right)^2+5\le5\forall x\)
Dấu '=' xảy ra khi x=1
\(\left(x-2\right)\left(x-3\right)\left(x-4\right)\left(x-5\right)+1\)(2)
\(=\left(x-2\right)\left(x-5\right)\left(x-3\right)\left(x-4\right)+1\)
\(=\left(x^2-7x+10\right)\left(x^2-7x+12\right)+1\)(1)
Đặt \(x^2-7x+10=t\)
\(\Rightarrow\left(1\right)=t\left(t+2\right)+1=t^2+2t+1=\left(t+1\right)^2\)
Mà \(x^2-7x+10=t\)nên \(\left(2\right)=\left(x^2-7x+11\right)^2\)
Vậy \(\left(x-2\right)\left(x-3\right)\left(x-4\right)\left(x-5\right)+1\)\(=\left(x^2-7x+11\right)^2\)
\(\left(x+1\right)^3-x\left(x-3\right)\left(x+3\right)-6\left(x-1\right)\left(x+2\right)=13\)
\(\Leftrightarrow x^3+3x^2+3x+1-x\left(x^2-9\right)-6\left(x^2+x-2\right)=13\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+9x-6x^2-6x+12=13\)
\(\Leftrightarrow-3x^2+6x=0\)
\(\Leftrightarrow-3\left(x^2-2\right)=0\)
\(\Leftrightarrow x^2-2=0\Leftrightarrow x^2=2\)
\(\Leftrightarrow x=\pm\sqrt{2}\)
A = 2⁵.(-5)² - 8² - 7
= 32.25 - 64 - 7
= 729
= 27²
B = 2³.(-4)² + (-3)².3² - 40
= 8.16 + 9.9 - 40
= 169
= 13²
C = (1/4 - 1/2 - 1)³ . (2 - 2/5)³
= (-5/4)³ . (8/5)³
= (-5/4 . 8/5)³
= (-2)³
D = (-1/4)² : (1/2 - 1/3)
= 1/16 : 1/6
= 3/8
E = 4 . (1/4)² + 25 . [(3/4)³ : (5/4)³] : (3/2)³
= 1/4 + 25 . (3/4 . 5/4)³ : (3/2)³
= 1/4 + 25 . (15/16)³ : 27/8
= 1/4 + 25 . 3375/4096 : 27/8
= 1/4 + 84375/4096 : 27/8
= 1/4 + 3125/512
= 3253/512
F = 2³ + 3.(1/2)⁰ - 1 + [(-2)² : 1/2] - 8
= 8 + 3.1 - 1 + (4 : 1/2) - 8
= 8 + 3 - 1 + 8 - 8
= 10