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a) ĐKXĐ: \(x\notin\left\{-1;2;3\right\}\)
Ta có: \(E=\left(1-\frac{x}{x+1}\right):\left(\frac{x+3}{x-2}+\frac{x+2}{3-x}+\frac{x+2}{x^2-5x+6}\right)\)
\(=\left(\frac{x+1}{x+1}-\frac{x}{x+1}\right):\left(\frac{x^2-9}{\left(x-2\right)\left(x-3\right)}-\frac{x^2-4}{\left(x-3\right)\left(x-2\right)}+\frac{x+2}{\left(x-3\right)\left(x-2\right)}\right)\)
\(=\frac{1}{x+1}:\left(\frac{x^2-9-x^2+4+x+2}{\left(x-2\right)\left(x-3\right)}\right)\)
\(=\frac{1}{x+1}:\frac{x-3}{\left(x-2\right)\left(x-3\right)}\)
\(=\frac{1}{x+1}:\frac{1}{x-2}\)
\(=\frac{1}{x+1}\cdot\frac{x-2}{1}=\frac{x-2}{x+1}\)
b) Thay \(x=-1\frac{1}{2}=-\frac{3}{2}\) vào biểu thức \(E=\frac{x-2}{x+1}\), ta được:
\(\left(-\frac{3}{2}-2\right):\left(\frac{-3}{2}+1\right)\)
\(=\left(-\frac{3}{2}-\frac{4}{2}\right):\left(-\frac{3}{2}+\frac{2}{2}\right)\)
\(=\frac{-7}{2}:\frac{-1}{2}\)
\(=\frac{-7}{2}\cdot\frac{2}{-1}=\frac{-7}{-1}=7\)
Vậy: 7 là giá trị của biểu thức \(E=\frac{x-2}{x+1}\) tại \(x=-1\frac{1}{2}=-\frac{3}{2}\)
\(E=\left(\frac{x-2}{x^2-1}-\frac{x+2}{x^2+2x+1}\right).\left(\frac{1-x^2}{2}\right)^2\)
\(E=\left(\frac{x-2}{\left(x-1\right)\left(x+1\right)}-\frac{x-2}{\left(x+1\right)^2}\right).\left(\frac{\left(1-x\right)\left(1+x\right)}{2}\right)^2\)
\(E=\left(\frac{\left(x+1\right)\left(x-2\right)}{\left(x-1\right)\left(x+1\right)^2}-\frac{\left(x-2\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)^2}\right).\frac{\left(1-x\right)^2\left(x+1\right)^2}{4}\)
\(E=\frac{\left(x-2\right)\left(x+1-x+1\right)}{\left(x-1\right)\left(x+1\right)^2}.\frac{\left(x-1\right)^2\left(x+1\right)^2}{4}\)
\(E=\frac{2\left(x-2\right)\left(x-1\right)}{4}\)
\(E=\frac{\left(x-2\right)\left(x-1\right)}{2}\)
a) \(E=\left(\frac{x-2}{x^2-1}-\frac{x+2}{x^2+2x+1}\right).\left(\frac{1-x^2}{2}\right)^2\)
\(=\left(\frac{x-2}{\left(x-1\right)\left(x+1\right)}-\frac{x+2}{\left(x+1\right)^2}\right).\frac{\left(x^2-1\right)^2}{4}\)
\(=\left(\frac{\left(x-2\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}-\frac{\left(x+2\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}\right).\frac{\left(x^2-1\right)^2}{4}\)
\(=\left(\frac{x^2-3x+2-x^2-3x-2}{\left(x-1\right)^2\left(x+1\right)}\right).\frac{\left(x^2-1\right)^2}{4}\)
\(=\frac{-6x.\left(x^2-1\right)^2}{\left(x-1\right)^2\left(x+1\right).4}=\frac{-3x\left(x^2-1\right)^2}{\left(x^2-1\right)\left(x-1\right).4}=\frac{-3x\left(x-1\right)\left(x+1\right)}{\left(x-1\right).4}\)\(=\frac{-3x\left(x+1\right)}{4}\)
b) Muốn \(\frac{E-4}{5}=x\) thì \(\frac{\frac{-3x\left(x+1\right)}{4}-4}{5}=x\)
\(\Rightarrow\frac{\frac{-3x^2\left(x+1\right)}{4}-\frac{16}{4}}{5}=x\)
\(\Rightarrow\frac{-3x^3-3x^2-16}{4}=5x\)
\(\Rightarrow-3x^3-3x^2-16=20x\)
\(\Rightarrow-3x^3-3x^2-16=20x\).....................................................................
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) \(ĐKXĐ:\hept{\begin{cases}x^3+1\ne0\\x^3-2x^2\ne0\\x+1\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-1\\x\ne2\end{cases}}\)(chỗ chữ và là do OLM thiếu ngoặc 4 cái nên mk để thế nha! trình bày thì kẻ thêm 1 ngoặc nưax)
\(Q=1+\left(\frac{x+1}{x^3+1}-\frac{1}{x-x^2-1}-\frac{2}{x+1}\right):\frac{x^3-2x^2}{x^3-x^2+x}\)
\(=1+\left[\frac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{1}{x^2-x+1}-\frac{2}{x+1}\right]:\frac{x^2\left(x-2\right)}{x\left(x^2-x+1\right)}\)
\(=1+\frac{\left(x+1\right)+\left(x+1\right)-2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{x^2-x+1}{x\left(x-2\right)}\)
\(=1+\frac{4x-2x^2}{x+1}.\frac{1}{x\left(x-2\right)}\)
\(=1-\frac{2x\left(x-2\right)}{x\left(x+1\right)\left(x-2\right)}=1-\frac{2}{x+1}=\frac{x-1}{x+1}\)
b, Với \(x\ne0;x\ne-1;x\ne2\)Ta có:
\(|x-\frac{3}{4}|=\frac{5}{4}\)
*TH1:
\(x-\frac{3}{4}=\frac{5}{4}\Rightarrow x=2\)(ko thảo mãn)
*TH2:
\(x-\frac{3}{4}=-\frac{5}{4}\Rightarrow x=-\frac{1}{2}\)
\(\Rightarrow Q=\frac{-\frac{1}{2}-1}{-\frac{1}{2}+1}=-3\)
c,
\(Q=\frac{x-1}{x+1}=1-\frac{2}{x+1}\)
Để Q nguyên thì x+1 phải thuộc ước của 2!! tự làm tiếp dễ rồi!!
1. Ta có:
\(\frac{1}{x}+\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+2013\right)\left(x+2014\right)}\)
\(=\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+2013}-\frac{1}{x+2014}\)
\(=\frac{2}{x}-\frac{1}{x+2014}\)
\(=\frac{2\left(x+2014\right)}{x\left(x+2014\right)}-\frac{x}{x\left(x+2014\right)}\)
\(=\frac{2x+4028-x}{x\left(x+2014\right)}=\frac{x+4028}{x\left(x+2014\right)}\)
2a) ĐKXĐ: x \(\ne\)1 và x \(\ne\)-1
b) Ta có: A = \(\frac{x^2-2x+1}{x-1}+\frac{x^2+2x+1}{x+1}-3\)
A = \(\frac{\left(x-1\right)^2}{x-1}+\frac{\left(x+1\right)^2}{x+1}-3\)
A = \(x-1+x+1-3\)
A = \(2x-3\)
c) Với x = 3 => A = 2.3 - 3 = 3
c) Ta có: A = -2
=> 2x - 3 = -2
=> 2x = -2 + 3 = 1
=> x= 1/2
b/ \(3-100x+8x^2=8x^2+x-300\)
\(\Leftrightarrow-101x=-303\)
\(\Rightarrow x=3\)
c/ \(5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-150\)
\(\Leftrightarrow25x+10-80x+10=24x+12-150\)
\(\Leftrightarrow-79x=-158\)
\(\Rightarrow x=2\)
d/ \(3\left(3x+2\right)-\left(3x+1\right)=12x+10\)
\(\Leftrightarrow9x+6-3x-1=12x+10\)
\(\Leftrightarrow-6x=5\)
\(\Rightarrow x=-\frac{5}{6}\)
e/ \(30x-6\left(2x-5\right)+5\left(x+8\right)=210+10\left(x-1\right)\)
\(\Leftrightarrow30x-12x+30+5x+40=210+10x-10\)
\(\Leftrightarrow13x=130\)
\(\Rightarrow x=10\)
\(A=x^2-4x+1=\left(x-2\right)^2-3\ge-3\)
\(\Rightarrow A_{min}=-3\) khi \(x=2\)
\(B=4x^2+4x+11=\left(2x+1\right)^2+10\ge10\)
\(\Rightarrow B_{min}=10\) khi \(x=-\frac{1}{2}\)
\(C=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(=\left(x^2+5x\right)^2-36\ge-36\)
\(\Rightarrow C_{min}=-36\) khi \(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
\(D=-x^2-8x-16+21=21-\left(x+4\right)^2\le21\)
\(\Rightarrow C_{max}=21\) khi \(x=-4\)
\(E=-x^2+4x-4+5=5-\left(x-2\right)^2\le5\)
\(\Rightarrow E_{max}=5\) khi \(x=2\)
có gì sai sai ý bạn
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