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Bài 1.
a) Do hai phân thức bằng nhau , ta có :
( x +2)P( x2 - 22) = ( x - 1)Q( x -2)
=( x + 2)P( x - 2)( x + 2) = ( x - 1)Q( x - 2)
Suy ra : P = x - 1 ; Q = ( x + 2)2
b) Do hai phân thức bằng nhau , ta có :
( x + 2)P(x2 - 2x + 1) = ( x - 2)Q( x2 - 1)
= ( x + 2)P( x - 1)2 = ( x - 2)Q( x - 1)( x + 1)
Suy ra : P = ( x - 2)( x + 1) = x2 - x - 2
Q = ( x + 2)( x - 1) = x2 + x + 2
Bài 2. a) Do : \(\dfrac{P}{Q}=\dfrac{R}{S}=>PS=QR\)
Xét : ( P + Q)S= PS + QS = QR + QS = Q( R + S)
-> \(\dfrac{P+Q}{Q}=\dfrac{R+S}{S}\)
b) Do : \(\dfrac{P}{Q}=\dfrac{R}{S}=>PS=QR\)
Xét : ( S - R)P = PS - PR = QR - PR = R( Q - P)
-> \(\dfrac{R-S}{R}=\dfrac{Q-P}{P}\)
- > \(\dfrac{R}{R-S}=\dfrac{P}{Q-P}\)
\(A=\left(\dfrac{6}{1.4}\right)\left(\dfrac{12}{2.5}\right)\left(\dfrac{20}{3.6}\right)\left(\dfrac{x^2+3x+2}{x\left(x+3\right)}\right)\)
\(A=\dfrac{2.3}{1.4}.\dfrac{3.4}{2.5}.\dfrac{4.5}{3.6}...\dfrac{\left(x+1\right)\left(x+2\right)}{x\left(x+3\right)}\)
\(A=\dfrac{2.3.4...\left(x+1\right)}{1.2.3...x}.\dfrac{3.4.5...\left(x+2\right)}{4.5.6...\left(x+3\right)}=\left(x+1\right)\dfrac{3}{x+3}=\dfrac{3\left(x+1\right)}{x+3}\)
\(\dfrac{1}{x+5}+\dfrac{1}{x-5}-\dfrac{2x+10}{\left(x+5\right)\cdot\left(x-5\right)}=\dfrac{x-5+x+5-2\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}\)
=\(\dfrac{2x-2}{x+5}=\dfrac{2\left(x-1\right)}{x+5}\)
b, Khi A=-3
thì ta có
\(\dfrac{2\left(x-1\right)}{x+5}=-3\)
\(\Leftrightarrow\) \(\dfrac{2\left(x-1\right)}{x+5}=\dfrac{-3\left(x+5\right)}{x+5}\Leftrightarrow2x-2=-3x-15\Leftrightarrow2x+3x=-15+2\Leftrightarrow5x=-13\Rightarrow x=-\dfrac{13}{5}\)
a ) Rút gọn : \(A=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2x+10}{\left(x+5\right)\left(x-5\right)}\)
\(\Leftrightarrow A=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}\)
\(\Leftrightarrow A=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2}{x-5}\)
\(\Leftrightarrow A=\dfrac{1}{x+5}.\)
Khi \(A=-3\),thì :
\(\dfrac{1}{x+5}=-3\Leftrightarrow x=-\dfrac{16}{3}\)
Ta có : \(9x^2-42x+49\)
\(=\left(3x\right)^2-2.3x.7+49\)
\(=\left(3x-7\right)^2\)
Thay \(x=-\dfrac{16}{3},\) ta có :
\(\left(3.\dfrac{-16}{3}-7\right)^2=\left(-16-7\right)^2=\left(-23\right)^2=529\)
\(=\left(\dfrac{x+1-1}{x+1}\right)\left(\dfrac{x+2-1}{x+2}\right)...\left(\dfrac{x+2014-1}{x+2014}\right)\)
\(=\dfrac{x\left(x+1\right)\left(x+2\right)...\left(x+2013\right)}{\left(x+1\right)\left(x+2\right)...\left(x+2013\right)\left(x+2014\right)}\)
\(=\dfrac{x}{x+2014}\)
\(a,\dfrac{x^2-2x}{x^2-4}=\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x}{x+2}\)
b) \(\dfrac{x^2+5x+4}{x^2-1}=\dfrac{x^2+x+4x+4}{x^2-1}=\dfrac{\left(x+1\right)\left(x+4\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+4}{x-1}\)
c) \(\dfrac{x^4+4}{x\left(x^2+2\right)-2x^2-\left(x-1\right)^2-1}\)
\(=\dfrac{x^4+4x^2-4x^2+4}{x^3+2x-2x^2-x^2+2x-1-1}\)
\(=\dfrac{\left(x^2+2\right)^2-4x^2}{\left(x^3+2x-2x^2\right)-\left(x^2-2x+2\right)}\)
\(=\dfrac{\left(x^2+2-2x\right)\left(x^2+2+2x\right)}{x\left(x^2+2-2x\right)-\left(x^2+2-2x\right)}\)
\(=\dfrac{x^2+2+2x}{x-1}\)
Bài 2:
a) \(\left(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}\right):\dfrac{4x}{10x-5}\)
\(=\dfrac{\left(2x+1\right)^2-\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}.\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{8x}{\left(2x-1\right)\left(2x+1\right)}.\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{10}{2x+1}\)
b) \(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\dfrac{1-2x+x^2}{x\left(x+1\right)}:\dfrac{1+x^2-2x}{x}\)
\(=\dfrac{1}{x+1}\)
c) Trong ngoặc giữa hai phân số là dấu gì vậy ?
a ) \(D=\left(\dfrac{1}{1-x}+\dfrac{1}{1+x}\right):\left(\dfrac{1}{1-x}-\dfrac{1}{1+x}\right)+\dfrac{1}{x+1}\)
\(=\left(\dfrac{1+x+1-x}{\left(1-x\right)\left(1+x\right)}\right):\left(\dfrac{1+x-1+x}{\left(1-x\right)\left(1+x\right)}\right)+\dfrac{1}{x+1}\)
\(=\dfrac{2}{\left(1-x\right)\left(1+x\right)}:\dfrac{2x}{\left(1-x\right)\left(1+x\right)}+\dfrac{1}{x+1}\)
\(=\dfrac{2}{\left(1-x\right)\left(1+x\right)}.\dfrac{\left(1-x\right)\left(1+x\right)}{2x}+\dfrac{1}{x+1}\)
\(=\dfrac{1}{x}+\dfrac{1}{x+1}\)
\(=\dfrac{x+1+x}{x\left(x+1\right)}=\dfrac{2x+1}{x\left(x+1\right)}\)
b ) Khi \(x^2-x=0\Leftrightarrow x\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Thay 0,1 vào biểu thức D
Khi \(x=0\), ta có :
\(\dfrac{2.0+1}{0.\left(0+1\right)}\) ( ko được )
Khi \(x=1,\) ta có :
\(\dfrac{2.1+1}{1.\left(1+1\right)}=\dfrac{3}{2}\)
c ) Khi \(D=\dfrac{3}{2}\)
Ta có : \(\dfrac{2x+1}{x\left(x+1\right)}=\dfrac{3}{2}\)
\(\Leftrightarrow4x+2=3x^2+3x\)
\(\Leftrightarrow-3x^2+x+2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Vậy ...........
cái cuối là \(R\left(2023\right)\) hay 2.2023 vậy bạn ?
Sửa đề: 1/R(2023)
R(3)=1*3
R(4)=2*4
R(5)=3*5
...
R(2022)=2020*2022
R(2023)=2021*2023
=>\(S=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{2021\cdot2023}+\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{2020\cdot2022}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2021\cdot2023}+\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{2020\cdot2022}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2021}-\dfrac{1}{2023}+\dfrac{1}{2}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2022}\right)\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{2022}{2023}+\dfrac{505}{1011}\right)\simeq0.7496\)