Rút gọn biểu thức:
\(\left(\frac{x-1}{x^3}-\frac{x-1}{x^3-x^2}+\frac{3}{x^3-2x^2+x}\right):\frac{16x^2+16x-8}{x^5-2x^4+x^3}\)
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Cho biểu thức
A=\(\left(\frac{2x+1}{1-2x}-\frac{1-2x}{1+2x}-\frac{16x^2-1}{x^2-1}\right)\div\frac{16x^3-4x}{4x^2-1}\)
a)Rút gọn A
b)Tìm x để A có giá trị dương
Cho biểu thức:
\(A=x-\left(\frac{16x-x^2}{x^2-4}+\frac{3+2x}{2-x}-\frac{2-3x}{x+2}\right):\frac{x-1}{x^3+4x^2+4x}\)
1) Rút gọn biểu thức A.
2) Tính giá trị của biểu thức A với các giá trị x thỏa mãn:\(|x^2-3|=3-x\)
1.CHO BIỂU THỨC Aleft(frac{x^2-2x}{2x^2+8}-frac{2x^2}{8-4x+2x^2-x^3}right)left(1-frac{1}{x}-frac{2}{x^2}right)a. Tìm x để giá trị của A được xác định. Rút gọn biểu thức Ab. Tìm giá trị nguyến của x để A nhận giá trị nguyên2. Giaỉ các phương trình sau:a. xleft(x+2right)left(x^2+2x+2right)+10b. y^2+4^x+2y-2^{x+1}+20c. frac{x^2+4x+6}{x+2}+frac{x^2+16x+72}{x+8}frac{x^2+8x+20}{x+4}+frac{x^2+12x+42}{x+6}
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1.CHO BIỂU THỨC A=\(\left(\frac{x^2-2x}{2x^2+8}-\frac{2x^2}{8-4x+2x^2-x^3}\right)\left(1-\frac{1}{x}-\frac{2}{x^2}\right)\)
a. Tìm x để giá trị của A được xác định. Rút gọn biểu thức A
b. Tìm giá trị nguyến của x để A nhận giá trị nguyên
2. Giaỉ các phương trình sau:
a. \(x\left(x+2\right)\left(x^2+2x+2\right)+1=0\)
b. \(y^2+4^x+2y-2^{x+1}+2=0\)
c. \(\frac{x^2+4x+6}{x+2}+\frac{x^2+16x+72}{x+8}=\frac{x^2+8x+20}{x+4}+\frac{x^2+12x+42}{x+6}\)
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 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=\(\left(\frac{2x+1}{1-2x}-\frac{1-2x}{1+2x}-\frac{16x^2}{4x^2-1}\right):\frac{16x^3-4x}{4x^2-4x+1}\)
a) Tìm ĐKXĐ
b) Rút gọn
c) Tìm x để A có giá trị dương
2) Giải phương trìnha) frac{x+1}{x-2}+frac{x-1}{x+2}frac{2left(x^2+2right)}{x^2-4}b) left(2x+3right).left(frac{3x+8}{2-7x}+1right)left(x-5right).left(frac{3x+8}{2-7x}+1right)3) Rút gọna) frac{2x-1}{x^3+1}+frac{2x}{x^2-x+1}+frac{-x}{x+1}+2b) frac{x+1}{2x-2}+frac{x^2+3}{2-2x^2}+frac{1}{1-x}-1,5c) left(frac{x^2}{x^3-4x}-frac{6}{3x-6}+frac{1}{x+2}right).frac{x+2}{6}d) left(frac{x}{xy-y^2}+frac{2x-y}{xy-x^2}right):frac{x^2-2xy+y^2}{x^2y-xy^2}e) [frac{1}{left(2x-yright)^2}+frac{2}{4x^2-y^2}-frac{1}{le...
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2) Giải phương trình
a) \(\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)
b) \(\left(2x+3\right).\left(\frac{3x+8}{2-7x}+1\right)=\left(x-5\right).\left(\frac{3x+8}{2-7x}+1\right)\)
3) Rút gọn
a) \(\frac{2x-1}{x^3+1}+\frac{2x}{x^2-x+1}+\frac{-x}{x+1}+2\)
b) \(\frac{x+1}{2x-2}+\frac{x^2+3}{2-2x^2}+\frac{1}{1-x}-1,5\)
c) \(\left(\frac{x^2}{x^3-4x}-\frac{6}{3x-6}+\frac{1}{x+2}\right).\frac{x+2}{6}\)
d) \(\left(\frac{x}{xy-y^2}+\frac{2x-y}{xy-x^2}\right):\frac{x^2-2xy+y^2}{x^2y-xy^2}\)
e) \([\frac{1}{\left(2x-y\right)^2}+\frac{2}{4x^2-y^2}-\frac{1}{\left(2x+y\right)^2}].\frac{x^2+4xy+y^2}{16x}\)
Mn giúp mik vs mik đang cần gấp
\(a,\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)\(\Leftrightarrow\frac{x^2+3x+2+x^2-3x+2}{x^2-4}=\frac{2\left(x^2+2\right)}{x^2-4}\)
\(\Leftrightarrow2\left(x^2+2\right)=2\left(x^2+2\right)\)(luôn đúng)
Vậy pt có vô số nghiệm
\(b,\Leftrightarrow\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\)
\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left(2x+3-x+5\right)=0\)\(\Leftrightarrow\left(\frac{-4x+10}{2-7x}\right)\left(x+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-4x+10=0\\x+8=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-8\end{cases}}\)
Mấy câu rút gọn bạn quy đồng nha
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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 :>>