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Câu 1:
\(Tacó\)
\(\frac{2}{2x-1}+\frac{4x^2+1}{4x^2-1}-\frac{1}{2x+1}=\frac{2}{2x-1}+\frac{4x^2+1}{\left(2x+1\right)\left(2x-1\right)}-\frac{1}{2x+1}\)
\(=\frac{4x+2}{\left(2x+1\right)\left(2x-1\right)}+\frac{4x^2+1}{\left(2x+1\right)\left(2x-1\right)}-\frac{2x-1}{\left(2x+1\right)\left(2x-1\right)}\)
\(=\frac{4x+2+4x^2+1-2x+1}{\left(2x+1\right)\left(2x-1\right)}=\frac{2x\left(2x+1\right)+4}{\left(2x+1\right)\left(2x-1\right)}=\frac{2x+4}{2x-1}\)
\(b,x=\frac{1}{2}\Rightarrow2x-1=0\left(loại\right)\)
..... 2 câu sau easy
Câu 2:
a) \(ĐKXĐ:x\ne1\)
\(A=\left(\frac{1}{x-1}-\frac{2x}{x^3+x-x^2-1}\right)\div\left(1-\frac{2x}{x^2+1}\right)\)
\(\Leftrightarrow A=\left(\frac{1}{x-1}-\frac{2x}{\left(x-1\right)\left(x^2+1\right)}\right)\div\frac{x^2-2x+1}{x^2+1}\)
\(\Leftrightarrow A=\frac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}\div\frac{\left(x-1\right)^2}{x^2+1}\)
\(\Leftrightarrow A=\frac{\left(x-1\right)^2\left(x^2+1\right)}{\left(x-1\right)\left(x^2+1\right)\left(x-1\right)^2}\)
\(\Leftrightarrow A=\frac{1}{x-1}\)
b) Để A > 0
\(\Leftrightarrow x-1>0\)(Vì\(1>0\))
\(\Leftrightarrow x>1\)
3.
- Gọi x(km) là quãng đường AB ( x > 0 )
- Thời gian đi:\(\dfrac{x}{60}\) (h)
- Thời gia về:\(\dfrac{x}{50}\) (h)
Ta có phương trình
\(\dfrac{x}{50}\) - \(\dfrac{x}{60}\) = \(\dfrac{4}{5}\)
\(\Leftrightarrow\) 6x - 5x = 240
\(\Leftrightarrow\) x = 240 (TM)
Vậy quãng đường AB là 240km
Bài 1 :
( x2 + 1 )2 - 4x2
= ( x2 + 1 - 2x ) ( x2 + 1 + 2x )
Bài 2:
Sửa đề: \(A=\left(\dfrac{1}{x-1}-\dfrac{1}{x+1}\right):\left(1+\dfrac{x}{1-x}\right)\)
a: \(A=\dfrac{x+1-x+1}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{1-x+x}{1-x}\)
\(=\dfrac{2}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{-\left(x-1\right)}{1}=\dfrac{-2}{x+1}\)
b: Thay x=-2 vào A, ta được:
\(A=\dfrac{-2}{-2+1}=\dfrac{-2}{-1}=2\)
c: Để A là số nguyên thì \(x+1\in\left\{1;-1;2;-2\right\}\)
hay \(x\in\left\{0;-2;-3\right\}\)
Bài 1:
a: \(A=\dfrac{x+1+x}{x+1}:\dfrac{3x^2+x^2-1}{x^2-1}\)
\(=\dfrac{2x+1}{x+1}\cdot\dfrac{\left(x+1\right)\left(x-1\right)}{\left(2x+1\right)\left(2x-1\right)}=\dfrac{x-1}{2x-1}\)
b: Thay x=1/3 vào A, ta được:
\(A=\left(\dfrac{1}{3}-1\right):\left(\dfrac{2}{3}-1\right)=\dfrac{-2}{3}:\dfrac{-1}{3}=2\)
a:\(A=\left(\dfrac{1}{x-1}+\dfrac{x}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+x+1}{x+1}\right)\cdot\dfrac{\left(x+1\right)^2}{2x+1}\)
\(=\left(\dfrac{x+1+x}{\left(x-1\right)\left(x+1\right)}\right)\cdot\dfrac{\left(x+1\right)^2}{2x+1}=\dfrac{x+1}{x-1}\)
b: Khi x=1/2 thì \(A=\left(\dfrac{1}{2}+1\right):\left(\dfrac{1}{2}-1\right)=\dfrac{3}{2}:\dfrac{-1}{2}=-3\)
a: \(Q=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\dfrac{x\left(x-1\right)}{x+1}=\dfrac{x^2}{x-1}\)
b: |x|=1/3 thì x=1/3 hoặc x=-1/3
Khi x=1/3 thì \(Q=\left(\dfrac{1}{3}\right)^2:\left(\dfrac{1}{3}-1\right)=-\dfrac{1}{6}\)
Khi x=-1/3 thì \(Q=\left(-\dfrac{1}{3}\right)^2:\left(-\dfrac{1}{3}-1\right)=-\dfrac{1}{12}\)
c: Để Q là số nguyên thì \(x^2-1+1⋮x-1\)
=>\(x-1\in\left\{1;-1\right\}\)
=>x=2
d: Để Q=4 thì x^2=4x-4
=>x=2
Bài 3 :
a) Xét \(\Delta ABDvà\Delta CDB\) có :
\(\left\{{}\begin{matrix}\widehat{DAB}=\widehat{BCD}=90^o\\\dfrac{AD}{AB}=\dfrac{CB}{CD}\left(=\dfrac{3}{4}\right)\end{matrix}\right.\)
=> \(\Delta ABD\sim\Delta CBD\left(c.g.c\right)\) (1)
Xét \(\Delta ABDvà\Delta HBA\) có :
\(\left\{{}\begin{matrix}\widehat{DAB}=\widehat{AHB}=90^o\\\widehat{B}:chung\end{matrix}\right.\)
=> \(\Delta ABD\sim\Delta HBA\left(g.g\right)\) (2)
Từ (1) và (2) => \(\Delta AHB\sim\Delta BCD\left(\sim DAB\right)\)
b) Xét \(\Delta ADHvà\Delta BDA\) có :
\(\left\{{}\begin{matrix}\widehat{D}:Chung\\\widehat{DHA}=\widehat{DAB}=90^o\end{matrix}\right.\)
=> \(\Delta ADH\sim\Delta BDA\left(g.g\right)\)
\(=>\dfrac{AD}{BD}=\dfrac{DH}{DA}\)
=> \(AD^2=DH.BD\) (đpcm)
Bài 1:
a: =>5x-10=3x+3
=>2x=13
hay x=13/2
b: \(\Leftrightarrow2x\left(x-2\right)+3\left(x+1\right)=2\left(x+1\right)\left(x-2\right)\)
\(\Leftrightarrow2x^2-4x+3x+3=2x^2-2x-4\)
=>-x+3=-2x-4
=>x=-7
c: =>2x+7=3 hoặc 2x+7=-3
=>2x=-4 hoặc 2x=-10
=>x=-2 hoặc x=-5
Câu 3:
a: \(A=\left(\dfrac{1}{x+1}-\dfrac{2}{x-1}+\dfrac{x}{\left(x-1\right)\left(x+1\right)}\right)\cdot\dfrac{x+1}{1}\)
\(=\dfrac{x-1-2x-2+x}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{1}\)
\(=\dfrac{-3}{x-1}\)
b: Khi x=1 thì A không xác định
Khi x=2 thì \(A=\dfrac{-3}{2-1}=-3\)