Ta có : 

\(1-\frac{1945}{1975}=\frac{6}{395}\)

\(1-\frac{1975}{2005}=\frac{6}{401}\) 

Vì \(\frac{6}{395}>\frac{6}{401}\) nên \(1-\frac{1945}{1975}>1-\frac{1975}{2005}\)

\(\Rightarrow\)\(1+\frac{-1945}{1975}-1>1+\frac{-1975}{2005}-1\) ( trừ hai vế cho 1 ) 

\(\Rightarrow\)\(\frac{-1945}{1975}>\frac{-1975}{2005}\)

\(\Rightarrow\)\(A>B\)

Vậy \(A>B\)

Chúc bạn học tốt ~ 

Ta có: \(\text{1 + A =}1+\frac{-1945}{1975}=\frac{6}{395}\)

         \(1+B=1+\frac{-1975}{2005}=\frac{6}{401}\)

Vì\(\frac{6}{395}>\frac{6}{401}\)nên\(1+A>1+B\)

Suy ra \(A>B\)

AB=2xAI=2x4,5=9(cm)

\(\frac{15}{16}:\frac{25}{24}=\frac{15}{16}x\frac{24}{25}=\frac{15x24}{16x25}=\frac{9}{10}\)

\(\frac{15}{19}.\frac{38}{5}< x< \frac{67}{15}+\frac{56}{15}\)

\(\frac{15.38}{19.5}< x< \frac{67+56}{15}\)

\(6< x< \frac{41}{5}\)

\(6< x< 8,2\)

\(6< x\le8\)

=> x = 7 ; 8 

Ta có :\(\frac{15}{19}\cdot\frac{38}{5}< x< \frac{67}{15}+\frac{56}{15}\)\(\Rightarrow\)\(\frac{15\cdot38}{19\cdot5}< x< \frac{67+56}{15}\)\(\Rightarrow\)\(6< x< \frac{41}{5}\)\(\Rightarrow\)\(6< x< 8,2\)

Vậy các giá trị của số tự nhiên X thỏa mãn là X=7

x là 7 và 8

a) \(\dfrac{2\left(x-2\right)}{x\left(x-2\right)}=\dfrac{2}{x}\)

\(a,=\dfrac{2\left(x-2\right)}{x\left(x-2\right)}=\dfrac{2}{x}\\ b,=\dfrac{\left(1-3x\right)\left(2x-1\right)+2x\left(3x-2\right)-\left(3x-2\right)}{2x\left(2x-1\right)}\\ =\dfrac{\left(1-3x\right)\left(2x-1\right)+\left(2x-1\right)\left(3x-2\right)}{2x\left(2x-1\right)}\\ =\dfrac{\left(2x-1\right)\left(1-3x+3x-2\right)}{2x}=\dfrac{-1}{2x}\)

a \(\dfrac{2x-4}{x^2-2x}\)= \(\dfrac{2\left(x-2\right)}{x\left(x-2\right)}\)=\(\dfrac{2}{x}\)

Lời giải:

$(2x-3)(x^2+1)=0$

\(\Leftrightarrow \left[\begin{matrix} 2x-3=0\\ x^2+1=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{3}{2}(\text{chọn})\\ x^2=-1<0(\text{vô lý})\end{matrix}\right.\)

Vậy pt có nghiệm $x=\frac{3}{2}$

\(\dfrac{2}{x+1}-\dfrac{1}{x-2}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\\ \Leftrightarrow\dfrac{2.\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}-\dfrac{1.\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\\ \Leftrightarrow2x-4-x+1=3x-11\\ \Leftrightarrow x-3=3x-11\\ \Leftrightarrow x-3x=-11+3\\ \Leftrightarrow-2x=-8\\ \Leftrightarrow x=4\)

Vậy tập nghiệm của phương trình là S = { 4 }

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

a: \(P=\dfrac{x}{2\left(x-1\right)}-\dfrac{x^2+1}{2\left(x-1\right)\left(x+1\right)}\)

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

b: Để P=-1/2 thì 1/2(x+1)=-1/2

=>x+1=-1

=>x=-2