Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a VT=.\(\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\left(\frac{1}{x+1}-\frac{x}{1-x}+\frac{2}{x^2-1}\right)\)
=\(\frac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}:\frac{x-1+x\left(x-1\right)+2}{\left(x+1\right)\left(x-1\right)}\)
\(=\frac{x^2+2x+1-x^2+2x-1}{\left(x+1\right)\left(x-1\right)}.\frac{\left(x+1\right)\left(x-1\right)}{x^2+2x+1}\)
\(=\frac{4x}{\left(x+1\right)^2}\)=VP
b.VT\(=\frac{2+x}{2-x}.\frac{\left(2-x\right)^2}{4x^2}.\left(\frac{2}{2-x}-\frac{4}{\left(x+2\right)\left(x^2-2x+4\right)}.\frac{4-2x+x^2}{2-x}\right)\)
=\(\frac{4-x^2}{4x^2}.\left(\frac{2}{2-x}-\frac{4}{4-x^2}\right)=\frac{4-x^2}{4x^2}.\frac{2\left(2+x\right)-4}{4-x^2}\)
=\(\frac{2x}{4x^2}=\frac{1}{2x}\)=VP
c VT=.\(\left[\left(\frac{3}{x-y}+\frac{3x}{x^2-y^2}\right).\frac{\left(x+y\right)^2}{2x+y}\right].\frac{x-y}{3}\)
\(=\left[\frac{3\left(x+y\right)+3x}{\left(x+y\right)\left(x-y\right)}.\frac{\left(x+y\right)^2}{2x+y}\right].\frac{x-y}{3}\)
\(=\frac{3\left(2x+y\right)\left(x+y\right)^2}{\left(x+y\right)\left(x-y\right)\left(2x+y\right)}.\frac{x-y}{3}\)
\(=x+y=\)VP
Vậy các đẳng thức được chứng minh
=
a) Đk \(x\ne\pm1\), sau khi rút gọn ta được: (bạn tư làm)
\(P=\frac{x}{x+1}\)
b) Khi \(\left|x-\frac{2}{3}\right|=\frac{1}{3}\) thì hoặc \(x-\frac{2}{3}=\frac{1}{3}\) hoặc \(x-\frac{2}{3}=-\frac{1}{3}\)
Hay là \(x=1\) hoặc \(x=\frac{1}{3}\)
Do để P có nghĩa thì \(x\ne\pm1\) nên \(x=\frac{1}{3}\), khi đó:
\(P=\frac{\frac{1}{3}}{\frac{1}{3}+1}=\frac{1}{4}\)
c) P > 1 khi \(\frac{x}{x+1}>1\)
\(\Leftrightarrow1-\frac{1}{x+1}>1\)
\(\Leftrightarrow\frac{1}{x+1}< 0\)
\(\Leftrightarrow x< -1\)
e) Đề không rõ ràng
I don't now
...............
.................
) \(\dfrac{x^3+8y^3}{2y+x}\)
\(=\dfrac{x^3+\left(2y\right)^3}{x+2y}\)
\(=\dfrac{\left(x+2y\right)\left[x^2+x.2y+\left(2y\right)^2\right]}{x+2y}\)
\(=x^2+2xy+4y^2\)
b) \(\dfrac{a-1}{2\left(a-4\right)}+\dfrac{a}{a-4}\) MTC: \(2\left(a-4\right)\)
\(=\dfrac{a-1}{2\left(a-4\right)}+\dfrac{2a}{2\left(a-4\right)}\)
\(=\dfrac{a-1+2a}{2\left(a-4\right)}\)
\(=\dfrac{3a-1}{2\left(a-4\right)}\)
c) \(\dfrac{x^3+3x^2y+3xy^2+y^3}{2x+2y}\)
\(=\dfrac{\left(x+y\right)^3}{2\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2}\)
d) \(\left(x-5\right)^2+\left(7-x\right)\left(x+2\right)\)
\(=\left(x^2-2.x.5+5^2\right)+\left(7x+14-x^2-2x\right)\)
\(=x^2-10x+25+7x+14-x^2-2x\)
\(=39-5x\)
e) \(\dfrac{3x}{x-2}-\dfrac{2x+1}{2-x}\)
\(=\dfrac{3x}{x-2}+\dfrac{2x+1}{x-2}\)
\(=\dfrac{3x+2x+1}{x-2}\)
\(=\dfrac{5x+1}{x-2}\)
h) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x+6}{4-9x^2}\)
\(=\dfrac{1}{3x-2}-\dfrac{1}{3x+2}+\dfrac{3x+6}{9x^2-4}\)
\(=\dfrac{1}{3x-2}-\dfrac{1}{3x+2}+\dfrac{3x+6}{\left(3x-2\right)\left(3x+2\right)}\) MTC: \(\left(3x-2\right)\left(3x+2\right)\)
\(=\dfrac{3x+2}{\left(3x-2\right)\left(3x+2\right)}-\dfrac{3x-2}{\left(3x-2\right)\left(3x+2\right)}+\dfrac{3x+6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{\left(3x+2\right)-\left(3x-2\right)+\left(3x+6\right)}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{3x+2-3x+2+3x+6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{3x+10}{\left(3x-2\right)\left(3x+2\right)}\)
a) A có nghĩa \(\Leftrightarrow\left(x+1\right)^2-3x\ne0\), \(x^3+1\ne0\),\(x+1\ne0\),\(3x^2+6x\ne0\) và \(x^2-4\ne0\)
+) \(\left(x+1\right)^2-3x\ne0\Leftrightarrow x^2+2x+1-3x\ne0\)
\(\Leftrightarrow x^2-x+1\ne0\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ne0\)(luôn đúng)
+) \(x^3+1\ne0\Leftrightarrow x^3\ne-1\Leftrightarrow x\ne-1\)
+) \(x+1\ne0\Leftrightarrow x\ne-1\)
+) \(3x^2+6x\ne0\Leftrightarrow3x\left(x+2\right)\ne0\)
\(\Leftrightarrow x\ne0;x\ne-2\)
+) \(x^2-4\ne0\Leftrightarrow x^2\ne4\Leftrightarrow x\ne\pm2\)
Vậy ĐKXĐ của A là \(x\ne-1;x\ne0;x\ne\pm2\)
a, \(Đkxđ:\hept{\begin{cases}x\ne-1\\x\ne0\\x\ne-2\end{cases}}\)
\(A=\left[\frac{\left(x+1\right)^2}{\left(x+1\right)^2-3x}-\frac{2x^2+4x-1}{x^3+1}-\frac{1}{x+1}\right]:\frac{x^2-4}{3x^2+6x}\)
\(=\left[\frac{x^2+2x+1}{x^2-x+1}-\frac{2x^2+4x-1}{\left(x+1\right)\left(x^2-x+1\right)}-\frac{1}{x+1}\right].\frac{3x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{\left(x^2+2x+1\right)\left(x+1\right)-2x^2-4x+1-\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{3x}{x-2}\)
\(=\frac{x^3+1}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{3x}{x-2}\)
\(=\frac{3x}{x-2}=3+\frac{6}{x-2}\)
b, Để A nguyên thì \(\Leftrightarrow6\)chia hết cho \(x-2\)
Hay \(\left(x-2\right)\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
| x-2 | -6 | -3 | -2 | -1 | 1 | 2 | 3 | 6 |
| x | -4 | -1 | 0 | 1 | 3 | 4 | 5 | 8 |
Vậy ............................
\(N=\left(\frac{1}{x^2+x}-\frac{2-x}{x+1}\right).\frac{3x}{x^2-2x+1}\)
\(N=\left[\frac{1-2x+x^2}{x\left(x+1\right)}\right].\frac{3x}{x^2-2x+1}\)
Để \(N=\frac{3}{x+1}\)là số nguyên
\(\Rightarrow3⋮x+1\Rightarrow x+1\in U\left(3\right)=\left\{\pm1;\pm3\right\}\)
bn tự lập bảng ha ~