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MA
2
23 tháng 12 2021
g: \(=\dfrac{x^2+2x-x^2-4x-2x+4}{x\left(x-2\right)\left(x+2\right)}=\dfrac{-4x+4}{x\left(x-2\right)\left(x+2\right)}\)
h: \(=\dfrac{2x^2+1-x^2+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{1}{x^2-x+1}\)
23 tháng 12 2021
\(e,=\dfrac{1}{x-1}-\dfrac{2x}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{x^2-2x+1}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{\left(x-1\right)^2}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{x-1}{x^2+1}\\ f,=\dfrac{3x-1}{2\left(3x+1\right)}+\dfrac{3x+1}{2\left(3x-1\right)}-\dfrac{6x}{\left(3x-1\right)\left(3x+1\right)}\\ =\dfrac{9x^2-6x+1+9x^2+6x+1-12x}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{2\left(3x-1\right)^2}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{3x-1}{3x+1}\)
\(g,=\dfrac{x}{x\left(x-2\right)}-\dfrac{x^2+4x}{x\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x\left(x+2\right)}\\ =\dfrac{x^2+2x-x^2-4x-2x+4}{x\left(x-2\right)\left(x+2\right)}=\dfrac{-4x+4}{x\left(x-2\right)\left(x+2\right)}\\ h,=\dfrac{2x^2+1-x^2+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{1}{x^2-x+1}\)
DM
1
8 tháng 11 2021
Bài 1:
\(a,\dfrac{25}{14x^2y}=\dfrac{75y^4}{42x^2y^5};\dfrac{14}{21xy^5}=\dfrac{28x}{42x^2y^5}\\ b,\dfrac{3x+1}{12xy^4}=\dfrac{3x\left(3x+1\right)}{36x^2y^4};\dfrac{y-2}{9x^2y^3}=\dfrac{4y\left(y-2\right)}{36x^2y^4}\\ c,\dfrac{1}{6x^3y^2}=\dfrac{6y^2}{36x^3y^4};\dfrac{x+1}{9x^2y^4}=\dfrac{4x\left(x+1\right)}{36x^3y^4};\dfrac{x-1}{4xy^3}=\dfrac{9x^2y\left(x-1\right)}{36x^3y^4}\\ d,\dfrac{3+2x}{10x^4y}=\dfrac{12y^4\left(3+2x\right)}{120x^4y^5};\dfrac{5}{8x^2y^2}=\dfrac{75x^2y^3}{120x^4y^5};\dfrac{2}{3xy^5}=\dfrac{80x^3}{120x^4y^5}\)
HN
1
giúp e vs
22 tháng 10 2021
1: \(\left(x+2y\right)^2=x^2+4xy+4y^2\)
2: \(\left(2x+3y\right)^2=4x^2+12xy+9y^2\)
3: \(\left(3x-2y\right)^2=9x^2-12xy+4y^2\)
VD
29 tháng 5 2022
Hướng dẫn: A đạt GTLN khi \(\dfrac{1}{A}\) đạt GTNN
Ta có: \(x^2+2\ge0\forall x\)
\(\Rightarrow A=\dfrac{1}{x^2+2}\le\dfrac{1}{2}\forall x\)
Vậy GTLN của A là 1/2
=> A
20 tháng 8 2021
1) \(x^3-8x+7=\left(x-1\right)\left(x^2+x-7\right)\)
2) \(x^3+8x^2-9=\left(x-1\right)\left(x^2+9x+9\right)\)
3) \(3x^3-4x+1=\left(x-1\right)\left(3x^2+3x-1\right)\)
4) \(x^4-3x^2+3x-1=\left(x-1\right)\left(x^3+x^2-2x+1\right)\)
5) \(x^4-5x^2+4=\left(x-1\right)\left(x-2\right)\left(x+1\right)\left(x+2\right)\)
20 tháng 8 2021
1: Ta có: \(x^3-8x+7\)
\(=x^3-x-7x+7\)
\(=x\left(x-1\right)\left(x+1\right)-7\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x-7\right)\)
2: Ta có: \(x^3+8x^2-9\)
\(=x^3-x^2+9x^2-9\)
\(=x^2\left(x-1\right)+9\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x^2+9x+9\right)\)
3: Ta có: \(3x^3-4x+1\)
\(=3x^3-3x-x+1\)
\(=3x\left(x-1\right)\left(x+1\right)-\left(x-1\right)\)
\(=\left(x-1\right)\left(3x^2+3x-1\right)\)
4: Ta có: \(x^4-3x^2+3x-1\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)-3x\cdot\left(x-1\right)\)
\(=\left(x-1\right)\cdot\left(x^3+x+x^2+1-3x\right)\)
\(=\left(x-1\right)\left(x^3+x^2-2x+1\right)\)
NT
3
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
8 tháng 12 2023
câu a, \(\dfrac{x}{x+1}\); \(\dfrac{x^2}{1-x}\); \(\dfrac{1}{x^2-1}\) (đk \(x\)≠ -1; 1)
\(x^2\) - 1 = ( \(x\) - 1).(\(x\) + 1)
\(\dfrac{x}{x+1}\) = \(\dfrac{x.\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}\);
\(\dfrac{x^2}{1-x}\) = \(\dfrac{-x^2}{x-1}\)= \(\dfrac{-x^2.\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(\dfrac{1}{x^2-1}\) = \(\dfrac{1}{\left(x-1\right)\left(x+1\right)}\)
b, \(\dfrac{10}{x+2}\); \(\dfrac{5}{2x-4}\); \(\dfrac{1}{6-3x}\) (đk \(x\) ≠ -2; 2)
2\(x-4\) = 2.(\(x\) - 2); 6 - 3\(x\) = - 3.(\(x\) - 2)
\(\dfrac{10}{x+2}\) = \(\dfrac{10.2.3\left(x-2\right)}{2.3\left(x+2\right)\left(x-2\right)}\) = \(\dfrac{60\left(x-2\right)}{6\left(x-2\right)\left(x+2\right)}\)
\(\dfrac{5}{2x-4}\) = \(\dfrac{5.3\left(x+2\right)}{2.3\left(x-2\right).\left(x+2\right)}\) = \(\dfrac{15.\left(x+2\right)}{6.\left(x-2\right)\left(x+2\right)}\)
\(\dfrac{1}{6-3x}\) = \(\dfrac{-1}{3.\left(x-2\right)}\) = \(\dfrac{-1.\left(x+2\right)}{3.2.\left(x-2\right)\left(x+2\right)}\) = \(\dfrac{-2.\left(x+2\right)}{6.\left(x-2\right).\left(x+2\right)}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
8 tháng 12 2023
c, \(\dfrac{x}{2x-4}\); \(\dfrac{1}{2x+4}\) và \(\dfrac{3}{4-x^2}\) đk \(x\) ≠ 2; -2
\(\dfrac{x}{2x-4}\) = \(\dfrac{x}{2.\left(x-2\right)}\) = \(\dfrac{x.\left(x+2\right)}{2.\left(x-2\right).\left(x+2\right)}\)
\(\dfrac{1}{2x+4}\) = \(\dfrac{1}{2.\left(x+2\right)}\) = \(\dfrac{\left(x-2\right)}{2.\left(x+2\right).\left(x-2\right)}\)
\(\dfrac{3}{4-x^2}\) = \(\dfrac{-3}{\left(x-2\right)\left(x+2\right)}\) = \(\dfrac{-6}{2.\left(x-2\right)\left(x+2\right)}\)
TK
2
9 tháng 12 2021
\(a,=\dfrac{4xy-1-2xy+1}{5x^2y}=\dfrac{6xy}{5x^2y}=\dfrac{6}{5x}\\ b,=\dfrac{x^2+8x-2x+8}{x\left(x-4\right)\left(x+4\right)}=\dfrac{\left(x+2\right)\left(x+4\right)}{x\left(x-4\right)\left(x+4\right)}=\dfrac{x+2}{x\left(x-4\right)}\\ c,=\dfrac{x^2+3x-x+1}{x\left(x+1\right)\left(x-1\right)}=\dfrac{\left(x+1\right)^2}{x\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{x\left(x-1\right)}\\ d,=\dfrac{x-3-x-3-2x}{\left(x-3\right)\left(x+3\right)}=\dfrac{-2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{3-x}\\ e,=\dfrac{x+1-1}{x+1}=\dfrac{x}{x+1}\\ f,=\dfrac{3x+5-5+9x}{6x^2y}=\dfrac{12x}{6xy}=\dfrac{2}{y}\)
9 tháng 12 2021
\(g,=\dfrac{x^2+6x-2x+4}{x\left(x+2\right)\left(x-2\right)}=\dfrac{\left(x+2\right)^2}{x\left(x+2\right)\left(x-2\right)}=\dfrac{x+2}{x\left(x-2\right)}\\ h,=\dfrac{3x+1-3x+1+2x-3}{\left(3x-1\right)\left(3x+1\right)}=\dfrac{2x-1}{\left(3x-1\right)\left(3x+1\right)}\\ j,=\dfrac{5x+30+x^2-30}{x\left(x+6\right)}=\dfrac{x^2+5x}{x^2+6x}\\ k,=\dfrac{\left(x-7\right)\left(x+7\right)}{2x+1}\cdot\dfrac{-3}{x-7}=\dfrac{-3\left(x+7\right)}{2x+1}\\ l,=\dfrac{x\left(3x-2\right)}{x^2-1}\cdot\dfrac{\left(x^2+1\right)\left(x^2-1\right)}{\left(3x-2\right)^3}=\dfrac{x\left(x^2+1\right)}{\left(3x-2\right)^2}\)
HN
2
29 tháng 4 2022
a.\(A=\dfrac{1}{x-1}-\dfrac{x^2+x}{x^2+1}.\left(\dfrac{1}{x-1}-\dfrac{1}{x+1}\right)\);\(ĐK:x\ne\pm1\)
\(A=\dfrac{1}{x-1}-\dfrac{x\left(x+1\right)}{x^2+1}.\left(\dfrac{x+1-x+1}{\left(x-1\right)\left(x+1\right)}\right)\)
\(A=\dfrac{1}{\left(x-1\right)}-\dfrac{2x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}\)
\(A=\dfrac{1}{x-1}-\dfrac{2x}{\left(x-1\right)\left(x^2+1\right)}\)
\(A=\dfrac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}\)
\(A=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x^2+1\right)}\)
\(A=\dfrac{x-1}{x^2+1}\)
b.\(A=0,2=\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{x-1}{x^2+1}=\dfrac{1}{5}\)
\(\Leftrightarrow x^2+1=5x-5\)
\(\Leftrightarrow x^2-5x+6=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
c.\(A< 0\) mà \(x^2+1\ge1>0\)
--> A<0 khi \(x-1< 0\)
\(\Leftrightarrow x< 1\)
29 tháng 4 2022
a. -ĐKXĐ:\(x\ne\pm1\)
\(A=\dfrac{1}{x-1}-\dfrac{x^2+x}{x^2+1}.\left(\dfrac{1}{x-1}-\dfrac{1}{x+1}\right)\)
\(=\dfrac{1}{x-1}-\dfrac{x\left(x+1\right)}{x^2+1}.\left(\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{x-1}{\left(x-1\right)\left(x+1\right)}\right)\)
\(=\dfrac{1}{x-1}-\dfrac{x\left(x+1\right)}{x^2+1}.\dfrac{x+1-x+1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{1}{x-1}-\dfrac{x\left(x+1\right)}{x^2+1}.\dfrac{2}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{1}{x-1}-\dfrac{2x}{\left(x^2+1\right)\left(x-1\right)}\)
\(=\dfrac{x^2+1}{\left(x^2+1\right)\left(x-1\right)}-\dfrac{2x}{\left(x^2+1\right)\left(x-1\right)}\)
\(=\dfrac{\left(x-1\right)^2}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{x-1}{x^2+1}\)
b. \(A=\dfrac{x-1}{x^2+1}=0,2\)
\(\Leftrightarrow\dfrac{x-1}{x^2+1}=\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{5\left(x-1\right)}{5\left(x^2+1\right)}=\dfrac{x^2+1}{5\left(x^2+1\right)}\)
\(\Rightarrow5x-5=x^2+1\)
\(\Leftrightarrow x^2-5x+1+5=0\)
\(\Leftrightarrow x^2-5x+6=0\)
\(\Leftrightarrow x^2-2x-3x+6=0\)
\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=3\left(nhận\right)\end{matrix}\right.\)
c. \(A=\dfrac{x-1}{x^2+1}< 0\)
\(\Leftrightarrow x-1< 0\) (vì \(x^2+1>0\forall x\))
\(\Leftrightarrow x< 1\)