Ta có

+) Xét với x ≥ 1  thì ta có

hay *  vô nghiệm.

+) Xét với x < 1  thì 

Vậy phương trình đã cho vô nghiệm.

Chọn D.

nghiệm của đa thưc là : - 1 và - 12/25

 vì 2.5x²+3.7x+1.2 = 2.5x²+2.5x+1.2x+1.2 = 2.5x(x+1)+1.2(x+1) = (x+1)(2.5x+1.2)

2.5x²+3.7x+1.2 = 0  <=> (x+1)(2.5x+1.2)=0 <=> x+1 =0 hoặc 2.5x+1.2=0  <=> x= -1 hoặc x = -12/25

<=>\(1.x^2-1=3=>x^2=4=\left(+-2\right)^2=>x=+-2\)

Trả lời:

1, \(\left(x-1\right)\left(x+1\right)=3\)

\(\Leftrightarrow x^2-1=3\)

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)

Vậy S = { 2 ; - 2 }

2, \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x-2\right)\left(x+2\right)=5\)

\(\Leftrightarrow x^3-1-x\left(x^2-4\right)=5\)

\(\Leftrightarrow x^3-1-x^3+4x=5\)

\(\Leftrightarrow4x-1=5\)

\(\Leftrightarrow4x=6\)

\(\Leftrightarrow x=\frac{3}{2}\)

Vậy \(S=\left\{\frac{3}{2}\right\}\)

3, 7x + 4 ( 5 - x ) = 3

<=> 7x + 20 - 4x = 3

<=> 3x + 20 = 3

<=> 3x = - 17

<=> x = - 17/3

Vậy S = { - 17/3 }

a) 1/5.6 + 1/6.7 + 1/7.8 + ... + 1/24.25

= 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + ... + 1/24 - 1/25

= 1/5 - 1/25

= 4/25

b) 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/99.101

= 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/99 -1/101

= 1 - 1/101

= 100/101

c) 3/1.4 + 3/4.7 + ... + 3/2002.2005

= 1 - 1/4 + 1/4 - 1/7 + ... + 1/2002 - 1/2005

= 1 - 1/2005

= 2004/2005

d) 5/2.7 + 5/7.12 + ... + 5/1997.2002

= 1/2 - 1/7 + 1/7 - 1/12 + ... + 1/1997 - 1/2002

= 1/2 - 1/2002

= 500/1001

a,A =  \(\frac{1}{5\times6}+\frac{1}{6\times7}+\frac{1}{7\times8}+...+\frac{1}{24\times25}\)

A\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)

A\(=\frac{1}{5}-\frac{1}{25}=\frac{5}{25}-\frac{1}{25}=\frac{4}{25}\)

b, B=\(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{99\times101}\)

B= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

B=\(1-\frac{1}{101}=\frac{100}{101}\)

c, \(C=\frac{3}{1\times4}+\frac{3}{4\times7}+...+\frac{3}{2002\times2005}\)

C= \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{2002}-\frac{1}{2005}\)

C= \(1-\frac{1}{2005}=\frac{2004}{2005}\)

d, D= \(\frac{5}{2\times7}+\frac{5}{7\times12}+...+\frac{5}{1997\times2002}\)

D= \(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+...+\frac{1}{1997}-\frac{1}{2002}\)

D= \(\frac{1}{2}-\frac{1}{2002}=\frac{1001}{2002}-\frac{1}{2002}=\frac{1000}{2002}=\frac{500}{1001}\)