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I don't now
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a) ĐK : \(a\ne\pm1\); \(a\ne\frac{-1}{2}\)
\(P=[\frac{\left(x-1\right)\left(1-x\right)}{1-x^2}+\frac{x\left(1+x\right)}{1-x^2}-\frac{3x+1}{1-x^2}]:\frac{2x+1}{x^2-1}\)
\(=\left(\frac{-x^2+2x-1+x^2+x-3x-1}{1-x^2}\right):\frac{2x+1}{x^2+1}\)
\(=\left(\frac{-2}{1-x^2}\right):\frac{-2x-1}{1-x^2}\)
\(=\frac{2}{2x+1}\)
b)
\(\frac{2}{2x+1}=\frac{3}{x-1}\)
\(\Leftrightarrow2\left(x-1\right)=3\left(2x+1\right)\)
<=> x=-5/4 (nhận)
c) P>1
\(\Leftrightarrow\frac{2}{2x+1}>1\)
\(\Leftrightarrow2x+1>0\)
Khi đó : 2 > 2x+1
<=> x < 1/2
mà x thuộc Z nên
\(P>1\Leftrightarrow x\hept{\begin{cases}x\in Z\\x\ne-1\\x\le0\end{cases}}\)
a/ \(P=\left(\frac{x-1}{x+1}-\frac{x}{x-1}-\frac{3x+1}{1-x^2}\right):\frac{2x+1}{x^2-1}\)
\(P=\left(\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{3x+1}{x^2-1}\right):\frac{2x+1}{x^2-1}\)
\(P=\left(\frac{x^2-2x+1}{x^2-1}-\frac{x^2+x}{x^2-1}+\frac{3x+1}{x^2-1}\right).\frac{x^2-1}{2x+1}\)
\(P=\frac{x^2-2x+1-x^2-x+3x+1}{x^2-1}.\frac{x^2-1}{2x+1}\)
\(P=\frac{2}{2x+1}\)
b/ để \(P=\frac{3}{x-1}\)
<=> \(\frac{2}{2x+1}=\frac{3}{x-1}\)
=> \(2x-2=6x+3\)
<=> \(2x-6x=3+2\)
<=> \(-4x=5\)
<=> \(x=\frac{-5}{4}\)
c/ để \(P>1\)
<=> \(\frac{2}{2x+1}\)\(>1\)
<=> \(\frac{2}{2x+1}-1>0\)
<=> \(\frac{2}{2x+1}-\frac{2x+1}{2x+1}>0\)
<=> \(\frac{3-2x}{2x+1}>0\)
<=> \(\hept{\begin{cases}3-2x>0\\2x+1>0\end{cases}}\)hoặc \(\hept{\begin{cases}3-2x< 0\\2x+1< 0\end{cases}}\)
<=> \(\hept{\begin{cases}x< \frac{3}{2}\\x>\frac{-1}{2}\end{cases}}\)hoặc \(\hept{\begin{cases}x>\frac{3}{2}\\x< \frac{-1}{2}\end{cases}}\)
<=> \(\frac{-1}{2}< x< \frac{3}{2}\)hoặc \(x\in\varnothing\)
vậy \(\frac{-1}{2}< x< \frac{3}{2}\)thì \(P< 1\)
học tốt
\(A=\left(\frac{x+1}{x^3+1}-\frac{1}{x-x^2-1}-\frac{2}{x+1}\right)\div\left(\frac{x^2-2x}{x^3-x^2+x}\right)\)
a) ĐKXĐ : \(\hept{\begin{cases}x\ne-1\\x\ne2\end{cases}}\)
\(=\left(\frac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{1}{x^2-x+1}-\frac{2}{x+1}\right)\div\left(\frac{x\left(x-2\right)}{x\left(x^2-x+1\right)}\right)\)
\(=\left(\frac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{1\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}-\frac{2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\right)\div\frac{x-2}{x^2-x+1}\)
\(=\left(\frac{x+1+x+1-2x^2+2x-2}{\left(x+1\right)\left(x^2-x+1\right)}\right)\times\frac{x^2-x+1}{x-2}\)
\(=\frac{-2x^2+4x}{\left(x+1\right)\left(x^2-x+1\right)}\times\frac{x^2-x+1}{x-2}\)
\(=\frac{-2x\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}=\frac{-2x}{x+1}\)
b) \(\left|x-\frac{3}{4}\right|=\frac{5}{4}\)
<=> \(\orbr{\begin{cases}x-\frac{3}{4}=\frac{5}{4}\\x-\frac{3}{4}=-\frac{5}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\left(loai\right)\\x=-\frac{1}{2}\left(nhan\right)\end{cases}}\)
Với x = -1/2 => \(A=\frac{-2\cdot\left(-\frac{1}{2}\right)}{-\frac{1}{2}+1}=2\)
c) Để A ∈ Z thì \(\frac{-2x}{x+1}\)∈ Z
=> -2x ⋮ x + 1
=> -2x - 2 + 2 ⋮ x + 1
=> -2( x + 1 ) + 2 ⋮ x + 1
Vì -2( x + 1 ) ⋮ ( x + 1 )
=> 2 ⋮ x + 1
=> x + 1 ∈ Ư(2) = { ±1 ; ±2 }
| x+1 | 1 | -1 | 2 | -2 |
| x | 0 | -2 | 1 | -3 |
Các giá trị trên đều tm \(\hept{\begin{cases}x\ne-1\\x\ne2\end{cases}}\)
Vậy x ∈ { -3 ; -2 ; 0 ; 1 }
b. Sử dụng các hằng đẳng thức
\(a^3+b^3+c^2-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(=3\left(a^2+b^2+c^2-ab-bc-ca\right)\)
và \(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
nên \(A=\frac{a^2+b^2+c^2-ab-bc-ca}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{1}{2}.\frac{\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Do (a - b) + (b - c) + (c - a) = 0 nên áp dụng hđt \(X^2+Y^2+Z^2=-2\left(XY+YZ+ZX\right)\)khi X + Y + Z = 0, ta có:
\(A=-2\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right).\)
Bài 1 :
\(b,ax^2+3ax+9=a^2\)
\(\Leftrightarrow a^2x+3ax+9-a^2=0\)
\(\Leftrightarrow ax\left(a+3\right)+\left(a+3\right)\left(3-a\right)=0\)
\(\Leftrightarrow\left(a+3\right)\left(ax+3-a\right)=0\)
Vì \(a\ne3\Rightarrow\left(a+3\right)\ne0\Rightarrow ax+3-a=0\)
\(\Leftrightarrow ax=a-3\)
Vì \(a\ne0\Rightarrow x=\frac{a-3}{a}\)
\(B=\left(\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{x+2}\right):\frac{x^2-3x}{2x^2-x^3}\left(ĐKXĐ:x\ne2;-2;0\right)\)
a)\(B=\left(-\frac{\left(x+2\right)^2}{x^2-4}-\frac{4x^2}{x^2-4}+\frac{\left(x-2\right)^2}{x^2-4}\right):\frac{x\left(x-3\right)}{x^2\left(2-x\right)}\)
\(B=\left(\frac{-\left(x+2\right)^2-4x^2+\left(x-2\right)^2}{x^2-4}\right).\frac{-x\left(x-2\right)}{\left(x-3\right)}\)
\(B=\left(\frac{-x^2-4x-4-4x^2+x-4x+4}{\left(x-2\right)\left(x+2\right)}\right).-\frac{x\left(x-2\right)}{x-3}\)
\(B=\frac{-5x^2-7x}{\left(x+2\right)}.\frac{-x}{x-3}\)
\(B=\frac{\left(-5x^2-7x\right)-x}{\left(x+2\right)\left(x-3\right)}\)
\(B=\frac{5x^3+7x^2}{\left(x+2\right)\left(x+3\right)}\)
Đ e o o o o o = Đéo
Mk chỉ cần câu trả lời thôi nên đừng vào cmt bậy nha các bác ;)))