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Đặt \(A=\dfrac{\left(1^4+\dfrac{1}{4}\right)\left(3^4+\dfrac{1}{4}\right)...\left(19^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right)\left(4^4+\dfrac{1}{4}\right)...\left(20^4+\dfrac{1}{4}\right)}\)
\(=\dfrac{\left[\left(1^4+\dfrac{1}{4}\right).2^4\right]\left[\left(3^4+\dfrac{1}{4}\right).2^4\right]...\left[\left(19^4+\dfrac{1}{4}\right).2^4\right]}{\left[\left(2^4+\dfrac{1}{4}\right).2^4\right]\left[\left(4^4+\dfrac{1}{4}\right).2^4\right]...\left[\left(20^4+\dfrac{1}{4}\right).2^4\right]}\)
\(=\dfrac{\left(2^4+4\right)\left(6^4+4\right)...\left(38^4+4\right)}{\left(4^4+4\right)\left(8^4+4\right)...\left(40^4+4\right)}\)
Lưu ý: \(a^4+4=\left(a^4+4a^2+4\right)-4a^2=\left(a^2+2\right)^2-\left(2a\right)^2\)
\(=\left(a^2-2a+2\right)\left(a^2+2a+2\right)\)
Áp dụng vào biểu thức A, ta có:
\(A=\dfrac{\left(2^4+4\right)\left(6^4+4\right)...\left(38^4+4\right)}{\left(4^4+4\right)\left(8^4+4\right)...\left(40^4+4\right)}\)
\(=\dfrac{\left(2^2-2.2+2\right)\left(2^2+2.2+2\right)...\left(38^2-38.2+2\right)\left(38^2+38.2+2\right)}{\left(4^2-2.4+2\right)\left(4^2+2.4+2\right)...\left(40^2-2.40+2\right)\left(40^2+2.40+2\right)}\)
\(=\dfrac{2.10.26..1370.1522}{10.26.50...1522.1682}=\dfrac{2}{1682}=\dfrac{1}{841}\)
Vậy \(A=\dfrac{1}{841}\)
a) Ta có : (x - 5)2 - 16
= (x - 5)2 - 42
= (x - 5 - 4)(x - 5 + 4)
= (x - 1)(x - 9)
b) 25 - (3 - x)2
= 52 - (3 - x)2
= (5 - 3 + x)(5 + 3 - x)
= (x + 2)(8 - x)
c) (7x - 4)2 - (2x + 1)2
= (7x - 4 - 2x - 1)(7x - 4 + 2x + 1)
= (5x - 5)(9x - 3)
= 5(x - 1)3(3x - 1)
= 15(x - 1)(3x - 1)
\(=x^6-6x^4+12x^2-8-x^3+x+6x^2-18x\\ =x^6-6x^4-x^3+18x^2-17x-8\)
e) Ta có: \(2\left|x-\dfrac{1}{2}\right|\ge0\forall x\)
\(\Leftrightarrow2\left|x-\dfrac{1}{2}\right|+2021\ge2021\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)
đặt \(x^2+4x+8=a\)
=> \(A=a^2+3ax+2x^2=a^2+ax+2ax+2x^2=a\left(a+x\right)+2x\left(a+x\right)\)
\(=\left(a+x\right)\left(a+2x\right)\)
b) ta có
\(B=\left(x+1\right)\left(x+7\right)\left(x+3\right)\left(x+5\right)+15=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
đặt \(x^2+8x+11=a\)
=> \(B=\left(a-4\right)\left(a+4\right)+15=a^2-16+15=a^2-1=\left(a-1\right)\left(a+1\right)\)
\(=\left(x^2+8x+10\right)\left(x^2+8x+12\right)=\left(x^2+8x+10\right)\left(x^2+6x+2x+12\right)\)
\(=\left(x^2+8x+10\right)\left[x\left(x+6\right)+2\left(x+6\right)\right]=\left(x^2+8x+10\right)\left(x+6\right)\left(x+2\right)\)
a) \(=x^3-\dfrac{1}{27}-x^2+\dfrac{2}{3}x-\dfrac{1}{9}=x^3-x^2+\dfrac{2}{3}x-\dfrac{2}{27}\)
b) \(=x^6-6x^4+12x^2-8-x^3+x+x^2-3x=x^6-6x^4-x^3+13x^2-2x-8\)
\(x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\)
\(=x^2\left(y-z\right)+y^2z-xy^2+xz^2-yz^2\)
\(=x^2\left(y-z\right)+yz\left(y-z\right)-x\left(y^2-z^2\right)\)
\(=\left(y-z\right)\left(x^2+yz-xy-xz\right)\)
\(=\left(y-z\right)\left[x\left(x-y\right)-z\left(x-y\right)\right]\)
\(=\left(y-z\right)\left(x-y\right)\left(x-z\right)\)
\(=\left(x-\dfrac{1}{3}\right)\left(\dfrac{4}{3}x+\dfrac{1}{9}-x+\dfrac{1}{3}\right)\\ =\left(x-\dfrac{1}{3}\right)\left(\dfrac{1}{3}x+\dfrac{4}{9}\right)\\ =\dfrac{1}{3}x^2+\dfrac{4}{9}x-\dfrac{1}{9}x-\dfrac{4}{27}\\ =\dfrac{1}{3}x^2+\dfrac{1}{3}x-\dfrac{4}{27}\)
(x+1)(x+3)(x+5)(x+7) + 15
= [ (x+1)(x+7) ].[ (x+3)(x+5) ] + 15
= (x² + 7x + x + 7).(x² + 5x + 3x + 15) + 15
= (x² + 8x + 7).(x² + 8x + 15) + 15
= (x² + 8x + 11 - 4)(x² + 8x + 11 + 4) + 15.
Đặt x² + 8x + 11 = y (1) ta được :
(t - 4)(t + 4) + 15 = t² - 16 + 15 = t² - 1 = (t+1)(t-1) (2).
Thay (1) vào (2) ta được: đa thức trên được phân tích thành:
(x² + 8x + 11 + 1)(x² + 8x + 11 - 1)
= (x² + 8x + 12)(x² + 8x + 10).
Chúc bn học tốt!
\(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)
\(=x^4+9x^3+23x^2+15x+7x^3+63x^2+161x+105+15\)
\(=x^4+16x^3+86x^2+176x+120\)
\(=\left(x+2\right)\left(x+6\right)\left(x^2+8x+10\right)\)