b.

\(B=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{\left(n-1\right)n\left(n+1\right)}\\ =\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+....+\dfrac{2}{\left(n-1\right).n.\left(n+1\right)}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right).n}-\dfrac{1}{n\left(n+1\right)}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n\left(n+1\right)}\right)=\dfrac{1}{4}-\dfrac{1}{2n\left(n+1\right)}\)

Bài 1:

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) Sửa đề \(8\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)-3^{64}\)

\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)-3^{64}\)

\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)-3^{64}\)

\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)-3^{64}\)

\(=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)-3^{64}\)

\(=\left(3^{32}-1\right)\left(3^{32}+1\right)-3^{64}\)

\(=3^{64}-1-3^{64}\)

\(=-1\)

Bài 2:

Ta có:

\(A=2009.2009\)

\(A=2009\left(2008+1\right)\)

\(A=2009.2008+2009\)

Ta lại có:

\(B=2008.2010\)

\(B=2008\left(2009+1\right)\)

\(B=2008.2009+2008\)

Vì 2008.2009 = 2009.2008

2009 > 2008

=> 2008.2009 + 2009 > 2009.2008 + 2008

=> A > B

1,a,(2-1)(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

2,

A=2009.2009=2009²

B=2008.2010=(2009-1)(2009+1)=2009²-1

Do2009²>2009²-1\(\Rightarrow A>B\)

\(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{16}{x^2-1}\)

\(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\frac{16}{x^2-1}\)

\(\Rightarrow\left(x+1\right)^2-\left(x-1\right)^2=16\)

\(\Rightarrow\left(x+1-x+1\right)\left(x+1+x-1\right)=16\)

\(\Rightarrow2\left(2x\right)=16\)

\(\Rightarrow4x=16\)

\(\Rightarrow x=4\)

vậy \(x=4\)

\(\frac{6x+1}{x^2-7x+10}+\frac{5}{x-2}=\frac{3}{x-5}\)

\(\frac{6x+1}{\left(x-2\right)\left(x-5\right)}+\frac{5\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}=\frac{3\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}\)

\(\Rightarrow6x+1+5x-5=3x-6\)

\(\Rightarrow11x-3x=-6+4\)

\(\Rightarrow8x=-2\)

\(\Rightarrow x=\frac{-1}{4}\)

3) \(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)

\(\frac{x^2+x+1}{x^3-1}+\frac{\left(2x^2-5\right)}{x^3-1}=\frac{4\left(x-1\right)}{x^3-1}\)

\(\Rightarrow x^2+x+1+2x^2-5=4x-4\)

\(\Rightarrow3x^2-3x=-4+4\)

\(\Rightarrow3x\left(x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

Bài 1: Thực hiện phép tính

a) Ta có: \(3x^2\left(5x^2-2x+4\right)\)

\(=15x^4-6x^3+12x^2\)

b) Ta có: \(\left(2x^2-4\right)\left(x^2-3\right)\)

\(=2x^4-6x^2-4x^2+12\)

\(=2x^4-10x^2+12\)

c) Ta có: \(\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right)\cdot\left(1-\frac{1}{x^2}\right)\)

\(=\frac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\cdot\frac{1-x^2}{x^2}\)

\(=\frac{x^2+2x+1-\left(x^2-2x+1\right)}{\left(x-1\right)\left(x+1\right)}\cdot\frac{-\left(x-1\right)\left(x+1\right)}{x^2}\)

\(=\frac{x^2+2x+1-x^2+2x-1}{-x^2}\)

\(=\frac{4x}{-x^2}=\frac{-4x}{x^2}=\frac{-4}{x}\)

d) Ta có: \(\frac{3x+1}{\left(x-1\right)^2}-\frac{1}{x+1}+\frac{x+3}{1-x^2}\)

\(=\frac{\left(3x+1\right)\left(x+1\right)}{\left(x-1\right)^2\cdot\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)^2\cdot\left(x+1\right)}-\frac{\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\cdot\left(x+1\right)}\)

\(=\frac{3x^2+3x+x+1-\left(x^2-2x+1\right)-\left(x^2-x+3x-3\right)}{\left(x-1\right)^2\cdot\left(x+1\right)}\)

\(=\frac{3x^2+4x+1-x^2+2x-1-x^2-2x+3}{\left(x-1\right)^2\cdot\left(x+1\right)}\)

\(=\frac{x^2+4x+3}{\left(x-1\right)^2\cdot\left(x+1\right)}=\frac{\left(x+1\right)\left(x+3\right)}{\left(x+1\right)\left(x-1\right)^2}\)

\(=\frac{x+3}{x^2-2x+1}\)

cảm ơn nhoa~~

Xai HDT

Xin lỗi, vì mình lớp 7 mà tò mò thôi, bạn giải chi tiết giùm mình nha!

2. \(\frac{1}{x-1}-\frac{7}{x-2}=\frac{1}{\left(x-1\right)\left(2-x\right)}\) (ĐKXĐ:\(x\ne1,x\ne2\))

\(\Leftrightarrow\frac{1}{x-1}+\frac{7}{2-x}=\frac{1}{\left(x-1\right)\left(2-x\right)}\)

\(\Leftrightarrow\frac{2-x+7\left(x-1\right)}{\left(x-1\right)\left(2-x\right)}=\frac{1}{\left(x-1\right)\left(2-x\right)}\)

\(\Rightarrow2-x+7\left(x-1\right)=1\)

\(\Leftrightarrow2-x+7x-7=1\)

\(\Leftrightarrow-x+7x=1-2+7\)

\(\Leftrightarrow6x=6\)

\(\Leftrightarrow x=1\) (Không thỏa mãn ĐKXĐ)

Vậy phương trình trên vô nghiệm

ko phan tich duoc nha bn

chuc bn hoc gioi

happy new year