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\(P=\frac{1}{a^2-a}+\frac{1}{a^2-3a+2}+\frac{1}{a^2-5a+6}+\frac{1}{a^2-7a+12}+\frac{1}{a^2-9a+20}\)
\(=\frac{1}{a.\left(a-1\right)}+\frac{1}{\left(a-1\right).\left(a-2\right)}+\frac{1}{\left(a-2\right).\left(a-3\right)}+\frac{1}{\left(a-3\right).\left(a-4\right)}+\frac{1}{\left(a-4\right).\left(a-5\right)}\)
a) ĐKXĐ: \(a\ne0;1;2;3;4;5;6\)
b) \(P=\frac{1}{a-1}-\frac{1}{a}+\frac{1}{a-2}-\frac{1}{a-1}+\frac{1}{a-3}-\frac{1}{a-2}+\frac{1}{a-4}-\frac{1}{a-3}+\frac{1}{a-5}-\frac{1}{a-4}\)
\(A=\frac{1}{a-5}-\frac{1}{a}=\frac{a-\left(a-5\right)}{a.\left(a-5\right)}=\frac{5}{a.\left(a-5\right)}\)
c) \(a^3-a^2+2=0\)
\(\Leftrightarrow a^3+a^2-2a^2-2a+2a+2=0\)
\(\Leftrightarrow a^2.\left(a+1\right)-2a.\left(a+1\right)+2.\left(a+1\right)=0\)
\(\Leftrightarrow\left(a+1\right).\left(a^2-2a+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a+1=0\\a^2-2a+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}a=-1\\\left(a-1\right)^2=-1\left(loai\right)\end{cases}}}\)
Thay a=-1 vào P
\(P=\frac{5}{a.\left(a-5\right)}=\frac{5}{-1.\left(-1-5\right)}=\frac{5}{6}\)
Mình làm câu c trước để bạn hình dung ra nhé, câu a tương tự:
c) \(7\left(2^3+1\right)\left(2^6+1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\)
\(=\left(8-1\right)\left(2^3+1\right)\left(2^6+1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\)
\(=\left[\left(2^3-1\right)\left(2^3+1\right)\right]\left(2^6+1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\)
\(=\left(2^6-1\right)\left(2^6+1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\)
\(=\left(2^{12}-1\right)\left(2^{12}+1\right)\left(2^{24}+1\right)\)
\(=\left(2^{12}-1\right)\left(2^{24}+1\right)\)
\(=2^{36}-1\)
b) \(\left(x^2-x+4\right)\left(x^2+x+1\right)\left(x^2-1\right)\)
\(=\left(x^2.x^2.x^2\right).\left(-x+4+x+1+\left(-1\right)\right)\)
\(=x^8.\left(-4\right)\)
\(a,\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\)
\(=2^{16}-1\)
ta có : \(\frac{a^3-4a^2-a+4}{a^3-7a^2+14a-8}\)
= \(\frac{\left(a-1\right)\left(a+1\right)\left(a-4\right)}{\left(a-4\right)\left(a-2\right)\left(a-1\right)}\)
= \(\frac{a+1}{a-2}\)
nhớ nha
A = (22 - 1) (22 +1)(24 +1)...(264 +1) + 1 = (24 - 1)(24 +1)...(264 +1) + 1 = (28 -1)...(264 +1) + 1 = 2128 -1 + 1 = 2128
a) A = (8x3 - 4x2) : (2x2) - (4x2 - 3x) : x + 2x
= 8x3 : (2x2) - 4x2 : (2x)2 - 4x2 : x + 3x : x + 2x
= 4x - 2 - 4x + 3 + 2x
= 1 + 2x
Thay x = -1 vào biểu thức A, ta có:
A = 1 + 2.(-1)
= -1
Vậy giá trị của biểu thức A tại x = -1 là -1
b) B = (18a4 - 27a3) : (9a2) - 10a3 : (5a)
= 18a4 : (9a2) - 27a3 : (9a2) - 2a2
= 2a2 - 3a - 2a2
= -3a
Thay a = -8 vào biểu thức B, ta có:
B = -3.(-8)
= 24
Vậy giá trị của biểu thức B tại a = -8 là 24
a) (x - 1)(x + 1)(x2 + 1)(x4 + 1)(x8 + 1)
= (x2 - 1)(x2 + 1)(x4 + 1)(x8 + 1)
= (x4 - 1)(x4 + 1)(x8 + 1)
= (x8 - 1)(x8 + 1)
= x16 - 1
b) (a2 - 2b)(a2 + 2b)(a4 + 4b2)(a8 + 16b4)
= (a4 - 4b2)(a4 + 4b2)(a8 + 16b4)
= (a8 - 16b4)(a8 + 16b4)
= a16 - 256b8
\(M=\frac{1}{a^2-5a+6}+\frac{1}{a^2-7a+12}+\frac{1}{a^2-9a+20}+\frac{1}{a^2-11a+30}\)
\(M=\frac{1}{\left(a-2\right)\left(a-3\right)}+\frac{1}{\left(a-3\right)\left(a-4\right)}+\frac{1}{\left(a-4\right)\left(a-5\right)}+\frac{1}{\left(a-5\right)\left(a-6\right)}\)
\(M=\frac{1}{a-2}-\frac{1}{a-3}+\frac{1}{a-3}-\frac{1}{a-4}+\frac{1}{a-4}-\frac{1}{a-5}+\frac{1}{a-5}-\frac{1}{a-6}\)
\(M=\frac{1}{a-2}-\frac{1}{a-6}\)