giải phương trình
a)\(\frac{7x+10}{x+1}\left(x^2-x-2\right)=\frac{7x+10}{x+1}\left(2x^2-3x-5\right)\)
b)\(\frac{1}{x^2-5x+6}+\frac{1}{x^2-7x+12}+\frac{1}{x^2-9x+20}+\frac{1}{x^2-11x+30}=\frac{1}{8}\)
c)\(x^2+\frac{1}{x^2}+\frac{9x}{2}-\frac{9}{2x}+7=0\)
3) \(\frac{x-2}{x-5}-\frac{5}{x^2-5x}=\frac{1}{x}\)
\(\Leftrightarrow\frac{x-2}{x-5}-\frac{5}{x.\left(x-5\right)}=\frac{1}{x}\)
\(\Leftrightarrow\frac{x.\left(x-2\right)}{x.\left(x-5\right)}-\frac{5}{x.\left(x-5\right)}=\frac{1.\left(x-5\right)}{x.\left(x-5\right)}\)
Mc: \(x.\left(x-5\right)\)
\(\Leftrightarrow\) \(x^2\) - 2\(x\) - 5 = \(x\) - 5
\(\Leftrightarrow\) \(x^2\) - 2\(x\) - \(x\) - 5 + 5 = 0
\(\Leftrightarrow\) \(x^2\) - 3\(x\) = 0
\(\Leftrightarrow\) \(x\) . (\(x\) - 3) = 0
\(\Leftrightarrow\) \(x\) = 0 hoặc \(x\) - 3 = 0
\(\Leftrightarrow\) \(x\) = 0 hoặc \(x\) = 3
Vậy \(x\) = 0 hoặc \(x\) = 3
\(x-5\ne0\Rightarrow x\ne5\)
\(x^2-5\ne0\Rightarrow x\ne5\) và \(x\ne0\) \(\Rightarrow\left\{{}\begin{matrix}x\ne0\\x\ne5\end{matrix}\right.\)
\(x\ne0\)
Vậy S = {3}
4) \(\frac{x-4}{x+7}-\frac{1}{x}=\frac{-7}{x^2+7x}\)
\(\Leftrightarrow\frac{x-4}{x+7}-\frac{1}{x}=\frac{-7}{x.\left(x+7\right)}\)
\(\Leftrightarrow\frac{x.\left(x-4\right)}{x.\left(x+7\right)}-\frac{1.\left(x+7\right)}{x.\left(x+7\right)}=\frac{-7}{x.\left(x+7\right)}\)
Mc: \(x.\left(x+7\right)\)
\(\Leftrightarrow x^2-4x-x-7=-7\)
\(\Leftrightarrow x^2-4x-x=-7+7\)
\(\Leftrightarrow\) \(x^2-5x=0\)
\(\Leftrightarrow x.\left(x-5\right)=0\)
\(\Leftrightarrow x=0\) hoặc \(x-5=0\)
\(\Leftrightarrow x=0\) hoặc \(x=5\)
Vậy \(x=0\) hoặc \(x=5\)
\(x+7\ne0\Rightarrow x\ne-7\)
\(x^2+7\ne0\Rightarrow x\ne-7\) và \(x\ne0\) \(\Rightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-7\end{matrix}\right.\)
\(x\ne0\)
Vậy S = {5}
5) \(\frac{x+2}{x-2}+\frac{x-2}{x+2}=\frac{8x}{x^2-4}\)
\(\left\{{}\begin{matrix}x-2\ne0\\x+2\ne0\\x^2-4\ne0\end{matrix}\right.\Rightarrow TXĐ\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)
Mc : \(\left(x-2\right).\left(x+2\right)\)
\(\Leftrightarrow\frac{\left(x+2\right).\left(x+2\right)}{\left(x-2\right).\left(x+2\right)}+\frac{\left(x-2\right).\left(x-2\right)}{\left(x+2\right).\left(x-2\right)}=\frac{8x}{\left(x-2\right).\left(x+2\right)}\)
\(\Leftrightarrow x^2+2x+2x+4+x^2-2x-2x+4=8x\)
\(\Leftrightarrow x^2+x^2+2x+2x-2x-2x-8x+4+4=0\)
\(\Leftrightarrow2x^2-8x+8=0\)
\(\Leftrightarrow\) \(2x^2-4x-4x+8=0\)
\(\Leftrightarrow\) \(2x.\left(x-2\right)-4.\left(x-2\right)=0\)
\(\Leftrightarrow\left(2x-4\right).\left(x-2\right)=0\)
\(\Leftrightarrow2x-4=0\) hoặc \(x-2=0\)
\(\Leftrightarrow x=2\) hoặc \(x=2\)
\(\Leftrightarrow x=2\) (Loại) hoặc x = 2 (Loại)
Vậy S = \(\left\{\varnothing\right\}\)
6) \(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{4}{x^2-1}\)
\(\Leftrightarrow\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}-\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}=\frac{4}{\left(x-1\right).\left(x+1\right)}\)
MC: \(\left(x-1\right).\left(x+1\right)\)
\(\Leftrightarrow x^2+x+x+1-x^2+x+x-1=4\)
\(\Leftrightarrow x^2-x^2+x+x+x+x+1-1-4=0\)
\(\Leftrightarrow4x-4=0\)
\(\Leftrightarrow4.\left(x-1\right)=0\)
\(\Leftrightarrow\) 4 = 0 hoặc \(x-1=0\)
\(\Leftrightarrow\) 4 = 0 hoặc \(x=1\)
\(\Leftrightarrow\) 4 = 0 (Loại) hoặc \(x=1\) (Loại)
Vậy S = \(\left\{\varnothing\right\}\)
7) \(\frac{x+1}{x-1}+\frac{-4x}{x^2-1}=\frac{x-1}{x+1}\)
\(\Leftrightarrow\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}+\frac{-4x}{\left(x-1\right).\left(x+1\right)}=\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}\)
\(Mc:\left(x-1\right).\left(x+1\right)\)
\(\Leftrightarrow\) \(x^2+x+x+1-4x=x^2-x-x+1\)
\(\Leftrightarrow x^2-x^2+x+x-4x+x+x=-1+1\)
\(\Leftrightarrow0=0\) (Nhận)
Vậy S = \(\left\{x\in R;x\ne\pm1\right\}\)
Bài 1: Thực hiện phép tính
a) 3x\(^2\) (5x\(^2\) - 2x + 4)
b) (2x\(^2\) - 4)(x\(^2\) - 3)
c) ( \(\frac{x+1}{x-1}\)- \(\frac{x-1}{x+1}\) ) . (1- \(\frac{1}{x^2}\))
d) \(\frac{3x+1}{\left(x-1\right)^2}\) - \(\frac{1}{x+1}\) + \(\frac{x+3}{1-x^2}\)
Mong m.n giúp ạ!! Mai e nộp gấp!!
1/x^2+9x+20+1/x^2+11x+30+1/x^2+13x+42+1/x^2-15x+56=\(\frac{4}{x^2-11x+2020}\)
tìm GTNN
a)\(\frac{-3}{x^2+1}\)
b)\(\frac{x^2+3x-1}{x^2}\)
c)\(\frac{x^4+3x^2+3}{x^2+1}\)
Rút gọn
P = \(\left(\frac{4x^2-3x+17}{x^3-1}+\frac{2x-1}{x^2+x+1}+\frac{6x}{x-x^2}\right):\frac{x+2}{x^3-1}\)
Bài 1: Thực hiện phép tính
a. \(\frac{11x+10}{3x-3}+\frac{15x+13}{4-4x}\)
b. \(\frac{5x+3}{x^2-3x}+\frac{9-x}{9-3x}\)
c. \(\frac{4xy-1}{5x^2y}-\frac{2xy-1}{5x^2y}\)
d. \(\frac{x+8}{x^2-16}-\frac{2}{x^2+4x}\)
e. \(\frac{x^2-49}{2x+1}.\frac{3}{7-x}\)
f. \(\frac{3x^2-2x}{x^2-1}.\frac{1-x^4}{\left(2-3x\right)^3}\)
g. \(\frac{5xy}{2x-3}:\frac{15xy^3}{12-8x}\)
h. \(\frac{x^2+2x}{3x^2-6x+3}:\frac{2x+4}{5x-5}\)
B=\(\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{4}{x-2\sqrt{x}}\right).\left(\frac{1}{\sqrt{x}+2}+\frac{4}{x-4}\right)\)
Rút gọn
1. \(\frac{1}{2}x^2-\left(\frac{1}{2}x-4\right)\frac{1}{2}x=-14\)
2. \(3\left(1-4x\right)\left(x-1\right)+4\left(3x-2\right)\left(x+3\right)=-27\)
3. \(6x\left(5x+3\right)+3x\left(1-10x\right)=7\)
4. \(\left(3x-3\right)\left(5-21x\right)+\left(7x+4\right)\left(9x-5\right)=44\)
5. \(\left(-2+x^3\right)\left(-2+x^2\right)\left(-2+x^2\right)=1\)
