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a,\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)=\left(2^8-1\right)\left(2^8+1\right)=2^{16}-1\)
b,\(=\left(2^3-1\right)\left(2^3+1\right)\left(2^6+1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\)
tiếp tục giống bài a
c, \(=\left[x^2-\left(x-1\right)\right]\left[x^2+\left(x+1\right)\right]\left(x^2-1\right)=\left(x^2-x^2+1\right)\left(x^2-1\right)=x^2-1\)
\(2\left(x-2\right)\left(x+3\right)-x^2+4=0\)
\(2\left(x^2+3x-2x-6\right)-x^2+4=0\)
\(2x^2+6x-4x-12-x^2+4=0\)
\(x^2+2x-8=0\)
\(x^2+4x-2x-8=0\)
\(x\left(x+4\right)-2\left(x+4\right)=0\)
\(\left(x+4\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+4=0\rightarrow x=\left(-4\right)\\x-2=0\rightarrow x=2\end{cases}}\)
3/
a/ \(2\left(x+1\right)^2-3\left(x-1\right)^2+\left(x+2\right)\left(5-x\right)\)
\(=2\left(x^2+2x+1\right)-3\left(x^2-2x+1\right)+\left(5x-x^2+10-2x\right)\)
\(=2x^2+4x+2-3x^2+6x-3+5x-x^2+10-2x\)
\(=-2x^2+13x+9\)
b/ \(\left(3x-1\right)^3+\left(3x-1\right)^3-6x^2+9\)
\(=2\left(3x-1\right)^3-6x^2+9\)
\(=2\left(\left(3x\right)^3-3\left(3x\right)^2\cdot1+3\cdot3x\cdot1-1\right)-6x^2+9\)
\(=2\left(27x^3-27x^2+9x-1\right)-6x^2+9\)
\(=54x^3-54x^2+18x-2-6x^2+9\)
\(=54x^3-60x^2+18x+7\)
Số hơi dài, nên dễ tính sai -,- tính mik hay cẩu thả có j sai ibbb ạ
a: \(P=\left(\dfrac{-\left(x+1\right)}{x-1}+\dfrac{x-1}{x+1}-\dfrac{4x^2}{\left(x-1\right)\left(x-1\right)}\right)\cdot\dfrac{\left(x-1\right)^2}{4\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x^2-2x-1+x^2-2x+1-4x^2}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x-1}{4\left(x+1\right)}\)
\(=\dfrac{-4x^2-4x}{x+1}\cdot\dfrac{1}{4\left(x+1\right)}\)
\(=\dfrac{-4x\left(x+1\right)}{x+1}\cdot\dfrac{1}{4\left(x+1\right)}=\dfrac{-x}{x+1}\)
b: khi x=5/8 thì \(P=\left(-\dfrac{5}{8}\right):\dfrac{13}{8}=\dfrac{-5}{13}\)
c: Để P là số nguyên thì \(-x-1+1⋮x+1\)
\(\Leftrightarrow x+1\in\left\{1;-1\right\}\)
hay \(x\in\left\{0;-2\right\}\)
a, = -3/2
b, = x-z/2
c, = (x-4).(x+4)/-x.(x-4) = -(x+4)/x = -x-4/x
k mk nha
1.a) (2 + 1)(22 + 1)((24 + 1)(28 + 1) = (22 - 1)(22 + 1)(24 + 1)(28 + 1) = (24 - 1)(24 + 1)(28 + 1)
= (28 - 1)(28 + 1) = 216 - 1
b) 7(23 + 1)(26 + 1)(212 + 1)(224 + 1) = (23 - 1)(23 + 1)(26 + 1)(212 + 1)(224 + 1)
= (26 - 1)(26 + 1)(212 + 1)(224 + 1) = (212 - 1)(212 + 1)(224 + 1) = (224 - 1)(224 + 1) = 248 - 1
c) (x2 - x + 1)(x2 + x + 1)(x2 - 1) = [(x2 - x + 1)(x + 1)][(x2 + x + 1)(x - 1)] = (x3 + 1)(x3 - 1) = x6 - 1
2. Đặt A = 4x - x2 - 1 = -(x^2 - 4x + 4) + 3 = -(x - 2)2 + 3 \(\le\)3 \(\forall\)x
Dấu "=" xảy ra <=> x - 2 = 0 <=> x = 2
Vậy MaxA = 3 khi x = 2