Cho biểu thức E=\(1+\left(\frac{2x^3+x^2-x}{x^3-1}-\frac{2x-1}{x-1}\right)\frac{x^2-x}{2x-1}\)
a)rút gọn E
b) Chứng minh E > \(\frac{2}{3}\)
Những câu hỏi liên quan
1.Cho biểu thức Cfrac{x+2}{x^2+x+1}-frac{2}{x-1}-frac{2x^2+4}{1-x^3}
a, Rút gọn C
b,So sánh C và frac{1}{3}
2.Cho biểu thức Dfrac{3x}{x-3}-frac{8x+3}{2x+1}-frac{70}{2x^2-5x-3}
a, Rút gọn D
b, Tính giá trị của D biết |x+3|7
c, Tìm x nguyên để D nguyên
3.Cho biểu thức E frac{x-1}{2}:left(frac{x^2+2}{x^3-1}-frac{x}{x^2+x+1}+frac{1}{1-x}right)
a, Rút gọn E
b, Chứng minh rằng E 0 với mọi x khác 1
C, Tìm giá trị nhỏ nhất của E
Đọc tiếp
1.Cho biểu thức C=\(\frac{x+2}{x^2+x+1}-\frac{2}{x-1}-\frac{2x^2+4}{1-x^3}\)
a, Rút gọn C
b,So sánh C và \(\frac{1}{3}\)
2.Cho biểu thức D=\(\frac{3x}{x-3}-\frac{8x+3}{2x+1}-\frac{70}{2x^2-5x-3}\)
a, Rút gọn D
b, Tính giá trị của D biết |x+3|=7
c, Tìm x nguyên để D nguyên
3.Cho biểu thức E =\(\frac{x-1}{2}:\left(\frac{x^2+2}{x^3-1}-\frac{x}{x^2+x+1}+\frac{1}{1-x}\right)\)
a, Rút gọn E
b, Chứng minh rằng E> 0 với mọi x khác 1
C, Tìm giá trị nhỏ nhất của E
Cho biểu thức :
\(E=\left(\frac{x-2}{x^2-1}-\frac{x+2}{x^2+2x+1}\right).\left(\frac{1-x^2}{2}\right)^2.\)
a) Rút gọn E
b) Tìm giá trị của x để \(\frac{E-4}{5}=x.\)
\(E=\left(\frac{x-2}{x^2-1}-\frac{x+2}{x^2+2x+1}\right).\left(\frac{1-x^2}{2}\right)^2\)
\(E=\left(\frac{x-2}{\left(x-1\right)\left(x+1\right)}-\frac{x-2}{\left(x+1\right)^2}\right).\left(\frac{\left(1-x\right)\left(1+x\right)}{2}\right)^2\)
\(E=\left(\frac{\left(x+1\right)\left(x-2\right)}{\left(x-1\right)\left(x+1\right)^2}-\frac{\left(x-2\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)^2}\right).\frac{\left(1-x\right)^2\left(x+1\right)^2}{4}\)
\(E=\frac{\left(x-2\right)\left(x+1-x+1\right)}{\left(x-1\right)\left(x+1\right)^2}.\frac{\left(x-1\right)^2\left(x+1\right)^2}{4}\)
\(E=\frac{2\left(x-2\right)\left(x-1\right)}{4}\)
\(E=\frac{\left(x-2\right)\left(x-1\right)}{2}\)
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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 \(\frac{\frac{-3x\left(x+1\right)}{4}-4}{5}=x\)
\(\Rightarrow\frac{\frac{-3x^2\left(x+1\right)}{4}-\frac{16}{4}}{5}=x\)
\(\Rightarrow\frac{-3x^3-3x^2-16}{4}=5x\)
\(\Rightarrow-3x^3-3x^2-16=20x\)
\(\Rightarrow-3x^3-3x^2-16=20x\).....................................................................
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Rút gọn biểu thức sau:\(\left(\frac{1}{x}+1-\frac{3}{x^3+1}+\frac{3}{x^2-x+1}\right).\frac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2x-2}{x^2+2x}\)
\(\left(\frac{1}{x}+1-\frac{3}{x^3+1}-\frac{3}{x^2-x+1}\right)\cdot\frac{3x^2-3x+3}{\left(x+1\right).\left(x+2\right)}-\frac{2x-2}{x^2+2x}\)
\(=\left(\frac{x+1}{x}-\frac{3}{\left(x+1\right).\left(x^2-x+1\right)}+\frac{3.\left(x+1\right)}{\left(x+1\right).\left(x^2-x+1\right)}\right)\cdot\frac{3.\left(x^2-x+1\right)}{\left(x+1\right).\left(x+2\right)}-\frac{2.\left(x-1\right)}{x.\left(x+2\right)}\)
\(=\left[\frac{\left(x+1\right)^2.\left(x^2-x+1\right)-3x+3x^2+3x}{x.\left(x+1\right).\left(x^2-x+1\right)}\right]\cdot\frac{3.\left(x^2-x+1\right)}{\left(x+1\right).\left(x+2\right)}-\frac{2.\left(x-1\right)}{x.\left(x+2\right)}\)
\(=\left[\frac{x^4+x^3+x+1+3x^2}{x.\left(x+1\right).\left(x^2-x+1\right)}\right]\cdot\frac{3.\left(x^2-x+1\right)}{\left(x+1\right).\left(x+2\right)}-\frac{2.\left(x-1\right)}{x.\left(x+2\right)}\)
\(=\frac{3x^4+3x^3+3x+3+9x^2}{x.\left(x+1\right)^2.\left(x+2\right)}-\frac{2.\left(x-1\right)}{x.\left(x+2\right)}=\frac{3x^4+3x^3+3x+3+9x^2}{x.\left(x+1\right)^2.\left(x+2\right)}-\frac{2x^3+2x^2-2x-2}{x.\left(x+1\right)^2.\left(x+2\right)}\)
\(=\frac{3x^4+x^3+7x^2+5x+5}{x.\left(x+1\right)^2.\left(x+2\right)}\)
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rút gọn biểu thức
\(\left(\frac{1}{x+1}-\frac{3}{x^3+1}+\frac{3}{x^2-x+1}\right):\frac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2x-2}{x^2+2x}\)
\(\left(\frac{1}{x+1}-\frac{3}{x^3+1}+\frac{3}{x^2-x+1}\right):\frac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2x-2}{x^2+2x}\left(x\ne-1;x\ne0;x\ne-2\right)\)
\(=\left(\frac{1}{x+1}-\frac{3}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{3}{x^2-x+1}\right):\frac{3x^3-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\left(\frac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}-\frac{3}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{3x+3}{\left(x+1\right)\left(x^2-x+1\right)}\right)\)\(:\frac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{x^2-x+1-3+3x+3}{\left(x+1\right)\left(x^2-x+1\right)}:\frac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{x^2+2x+1}{\left(x+1\right)\left(x^2-x+1\right)}:\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+1\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x^2-x+1\right)}\cdot\frac{\left(x+1\right)\left(x+2\right)}{3\left(x^2-x+1\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{\left(x+2\right)^2\left(x+1\right)}{3\left(x^2-x+1\right)^2}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
Cho biểu thức: A=\(\left[\frac{3\left(x+2\right)}{2x^3+2x+2x^2+2}+\frac{2x^2-x-10}{2x^3-2-2x^2+2x}\right]:\left[\frac{5}{x^2+1}+\frac{3}{2x+2}-\frac{3}{2x-2}\right]\)
Rút gọn A.
Bài làm
Như đã nhắn là mình sẽ làm theo quan điểm của mình là 5/(x^2 - 1) nha
\(A=\left[\frac{3\left(x+2\right)}{2x^3+2x+2x^2+2}+\frac{2x^2-x-10}{2x^3-2-2x^2+2x}\right]:\left[\frac{5}{x^2-1}+\frac{3}{2x+2}-\frac{3}{2x-2}\right]\)
\(A=\left[\frac{3\left(x+2\right)}{2x^2\left(x+1\right)+2\left(x+1\right)}+\frac{2x^2+4x-5x-10}{\left(2x^3-2x^2\right)+\left(2x-2\right)}\right]:\left[\frac{5}{x^2-1}+\frac{3}{2\left(x+1\right)}-\frac{3}{2\left(x-1\right)}\right]\)
\(A=\left[\frac{3\left(x+2\right)}{\left(2x^2+2\right)\left(x+1\right)}+\frac{2x\left(x+2\right)-5\left(x+2\right)}{2x^2\left(x-1\right)+2\left(x-1\right)}\right]:\left[\frac{5\cdot2}{2\left(x+1\right)\left(x-1\right)}+\frac{3}{2\left(x+1\right)}-\frac{3}{2\left(x-1\right)}\right]\)
\(A=\left[\frac{3\left(x+2\right)}{\left(2x^2+2\right)\left(x+1\right)}+\frac{\left(2x-5\right)\left(x+2\right)}{\left(2x^2+2\right)\left(x-1\right)}\right]:\left[\frac{5\cdot2}{2\left(x+1\right)\left(x-1\right)}+\frac{3}{2\left(x+1\right)}-\frac{3}{2\left(x-1\right)}\right]\)
\(A=\left[\frac{3\left(x+2\right)\left(x-1\right)}{\left(2x^2+2\right)\left(x^2-1\right)}+\frac{\left(2x-5\right)\left(x+2\right)\left(x+1\right)}{\left(2x^2+2\right)\left(x^2-1\right)}\right]:\left[\frac{5\cdot2}{2\left(x+1\right)\left(x-1\right)}+\frac{3\left(x-1\right)}{2\left(x^2-1\right)}-\frac{3\left(x+1\right)}{2\left(x^2-1\right)}\right]\)
\(A=\left[\frac{3\left(x+2\right)\left(x-1\right)+\left(2x-5\right)\left(x+2\right)\left(x+1\right)}{\left(2x^2+2\right)\left(x^2-1\right)}\right]:\left[\frac{10}{2\left(x^2-1\right)}+\frac{3x-3}{2\left(x^2-1\right)}-\frac{3x+3}{2\left(x^2-1\right)}\right]\)
\(A=\left[\frac{\left(x+2\right)\left[3x-3+\left(2x-5\right)\left(x+1\right)\right]}{\left(2x^2+2\right)\left(x^2-1\right)}\right]:\left[\frac{10+3x-3-3x-3}{2\left(x^2-1\right)}\right]\)
\(A=\left[\frac{\left(x+2\right)\left(3x-3+2x^2+2x-5x-5\right)}{\left(2x^2+2\right)\left(x^2-1\right)}\right]:\frac{4}{2\left(x^2-1\right)}\)
\(A=\frac{\left(x+2\right)\left(2x^2-8\right)}{\left(2x^2+2\right)\left(x^2-1\right)}\cdot\frac{\left(x^2-1\right)}{2}\)
\(A=\frac{\left(x+2\right)2\left(x^2-4\right)}{2\left(2x^2+2\right)}\)
\(A=\frac{2\left(x+2\right)\left(x-2\right)\left(x+2\right)}{4\left(x^2+1\right)}\)
\(A=\frac{\left(x+2\right)^2\left(x-2\right)}{2\left(x^2+1\right)}\)
:>>> Chả biết đúng không nữa nhưng số to quá :>>
Cho biểu thức: Q= \([\left(x^4-x+\frac{x-3}{x^3+1}\right).\frac{\left(x^3-2x^2+2x-1\right).\left(x+1\right)}{x^9+x^7-3x^2-3}+1-\frac{2\left(x+6\right)}{x^2+1}]\)
a, Tìm điều kiện xác định của biểu thức
b, Rút gọn Q
c, Chứng minh rằng với các giá trị của x thỏa mãn điều kiện xác định thì -5 <= Q <= 0
a, ĐKXĐ: \(\hept{\begin{cases}x^3+1\ne0\\x^9+x^7-3x^2-3\ne0\\x^2+1\ne0\end{cases}}\)
b, \(Q=\left[\left(x^4-x+\frac{x-3}{x^3+1}\right).\frac{\left(x^3-2x^2+2x-1\right)\left(x+1\right)}{x^9+x^7-3x^2-3}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\left[\frac{\left(x^3+1\right)\left(x^4-x\right)+x-3}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\left[\left(x^7-3\right).\frac{\left(x-1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\frac{x-1+x^2+1-2x-12}{x^2+1}\)
\(Q=\frac{\left(x-4\right)\left(x+3\right)}{x^2+1}\)
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Cho \(E=\left(\frac{1}{x^2-x}+\frac{1}{1-x^2}+\frac{2x^2+2}{x^3-x}\right):\frac{x^2}{x^2-4x+4}\)
a, Rút gọn E
b, Tính giá trị của E biết \(x^2-2x=0\)
c, Tìm x thuộc z để giá trị của E là số nguyên
Rút gọn biểu thức sau: A=\(\left[\left(x^4-x+\frac{x-3}{x^3+1}\right).\frac{\left(x^3-2x^2+2x-1\right)\left(x+1\right)}{x^9+x^7-3x^2-3}+1-\frac{2\left(x+6\right)}{x^2+1}\right].\frac{4x^2+4x+1}{\left(x+4\right)\left(3-x\right)}\)
Rút gọn biểu thức :
E=\(\frac{\sqrt{2x+2\sqrt{x^2-4}}}{\sqrt{x^2-4}+x+2}\)
F=\(\frac{\left(\frac{3}{\sqrt{1+a}}+\sqrt{1+a}\right)}{\left(\frac{3}{\sqrt{1-a^2}}+1\right)}\)
HELP ME !