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Kaijo
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Uchiha
15 tháng 3 2020 lúc 21:39

1,\(\frac{3}{2x+6}-\frac{x-6}{x\left(2x+6\right)}\)

=\(\frac{3x}{x\left(2x+6\right)}+\frac{x-6}{x\left(2x+6\right)}\)

=\(\frac{3x+x-6}{x\left(2x+6\right)}\)=\(\frac{4x-6}{x\left(2x+6\right)}=\frac{2\left(2x-3\right)}{x\left(2x+6\right)}\)

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Uchiha
15 tháng 3 2020 lúc 21:42

2, \(\frac{1}{1-x}-\frac{2x}{1-x^2}\)=\(\frac{1+x}{\left(1-x\right)\left(1+x\right)}+\frac{2x}{\left(1-x\right)\left(1+x\right)}\)=\(\frac{1+x+2x}{\left(1-x\right)\left(1+x\right)}=\frac{3x+1}{\left(1-x\right)\left(1+x\right)}\)

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Uchiha
15 tháng 3 2020 lúc 21:46

3,1/x(y-x)-1/y(y-x)=y/xy(y-x)-x/xy(y-x)=(y-x)/xy(x-y)=1/xy

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Vô Danh
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l҉o҉n҉g҉ d҉z҉
16 tháng 12 2020 lúc 15:12

Bài 1.

a)\(\frac{4x-4}{x^2-4x+4}\div\frac{x^2-1}{\left(2-x\right)^2}=\frac{4\left(x-1\right)}{\left(x-2\right)^2}\div\frac{\left(x-1\right)\left(x+1\right)}{\left(x-2\right)^2}=\frac{4\left(x-1\right)}{\left(x-2\right)^2}\times\frac{\left(x-2\right)^2}{\left(x-1\right)\left(x+1\right)}=\frac{4}{x+1}\)

b) \(\frac{2x+1}{2x^2-x}+\frac{32x^2}{1-4x^2}+\frac{1-2x}{2x^2+x}=\frac{2x+1}{x\left(2x-1\right)}+\frac{-32x^2}{4x^2-1}+\frac{1-2x}{x\left(2x+1\right)}\)

\(=\frac{\left(2x+1\right)\left(2x+1\right)}{x\left(2x-1\right)\left(2x+1\right)}+\frac{-32x^3}{x\left(2x-1\right)\left(2x+1\right)}+\frac{\left(1-2x\right)\left(2x-1\right)}{x\left(2x-1\right)\left(2x+1\right)}\)

\(=\frac{4x^2+4x+1}{x\left(2x-1\right)\left(2x+1\right)}+\frac{-32x^3}{x\left(2x-1\right)\left(2x+1\right)}+\frac{-4x^2+4x-1}{x\left(2x-1\right)\left(2x+1\right)}\)

\(=\frac{4x^2+4x+1-32x^3-4x^2+4x-1}{x\left(2x-1\right)\left(2x+1\right)}=\frac{-32x^3+8x}{x\left(2x-1\right)\left(2x+1\right)}\)

\(=\frac{-8x\left(4x^2-1\right)}{x\left(2x-1\right)\left(2x+1\right)}=\frac{-8x\left(2x-1\right)\left(2x+1\right)}{x\left(2x-1\right)\left(2x+1\right)}=-8\)

c) \(\left(\frac{1}{x+1}+\frac{1}{x-1}-\frac{2x}{1-x^2}\right)\times\frac{x-1}{4x}\)

\(=\left(\frac{1}{x+1}+\frac{1}{x-1}+\frac{2x}{x^2-1}\right)\times\frac{x-1}{4x}\)

\(=\left(\frac{x-1}{\left(x-1\right)\left(x+1\right)}+\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{2x}{\left(x-1\right)\left(x+1\right)}\right)\times\frac{x-1}{4x}\)

\(=\left(\frac{x-1+x+1+2x}{\left(x-1\right)\left(x+1\right)}\right)\times\frac{x-1}{4x}\)

\(=\frac{4x}{\left(x-1\right)\left(x+1\right)}\times\frac{x-1}{4x}=\frac{1}{x+1}\)

Bài 3.

N = ( 4x + 3 )2 - 2x( x + 6 ) - 5( x - 2 )( x + 2 )

= 16x2 + 24x + 9 - 2x2 - 12x - 5( x2 - 4 )

= 14x2 + 12x + 9 - 5x2 + 20

= 9x2 + 12x + 29

= 9( x2 + 4/3x + 4/9 ) + 25

= 9( x + 2/3 )2 + 25 ≥ 25 > 0 ∀ x 

=> đpcm

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Nguyễn Diệu Linh
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shitbo
14 tháng 12 2018 lúc 19:39

\(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x^2+x}\)

b, \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{y^2-xy-xy+x^2}{\left(xy-x^2\right)\left(y^2-xy\right)}=\frac{x^2+y^2}{xy^3-xyxy-xyxy+x^3y}\)Tu rut gon tiep

c, tt

d, cx r

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Trần Thanh Phương
14 tháng 12 2018 lúc 19:40

a) \(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1}{x\left(x+1\right)}-\frac{x}{x\left(x+1\right)}\)

\(=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x\left(x+1\right)}\)

b) \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{1}{x\left(y-x\right)}-\frac{1}{y\left(y-x\right)}\)

\(=\frac{y}{xy\left(y-x\right)}-\frac{x}{xy\left(y-x\right)}=\frac{y-x}{xy\left(y-x\right)}=\frac{1}{xy}\)

c) \(\frac{9x-3}{4x-1}-\frac{3x}{1-4x}=\frac{9x-3}{4x-1}+\frac{3x}{4x-1}\)

\(=\frac{9x-3+3x}{4x-1}=\frac{6x-3}{4x-1}\)

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❤  Hoa ❤
14 tháng 12 2018 lúc 19:41

\(a,\frac{1}{x}-\frac{1}{x+1}\)

\(=\frac{x+1}{x\left(x+1\right)}-\frac{x}{x\left(x+1\right)}=\frac{x+1-x}{x\left(x+1\right)}\)

\(=\frac{1}{x\left(x+1\right)}\)

\(b,\frac{1}{xy-x^2}-\frac{1}{y^2-xy}\)

\(=\frac{1}{x\left(y-x\right)}-\frac{1}{y\left(y-x\right)}\)

\(=\frac{y}{xy\left(y-x\right)}-\frac{x}{xy\left(y-x\right)}=\frac{x-y}{xy\left(x-y\right)}=\frac{1}{xy}\)

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Nguyễn Châu Mỹ Linh
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Thanh Thanh
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❤ ~~ Yến ~~ ❤
22 tháng 4 2020 lúc 19:36

a) \(\frac{5x-2}{2-2x}+\frac{2x-1}{2}+\frac{x^2+x-3}{1-x}=1\)

ĐK: x≠1

<=>\(\frac{5x-2}{2\left(1-x\right)}+\frac{2x-1}{2}\frac{x^2+x-3}{1-x}=1\)

<=>\(\frac{5x-2+\left(1-x\right).\left(2x-1\right)+2\left(x^2+x-3\right)}{2\left(1-x\right)}=1\)

<=>\(\frac{5x-2+2x-1-2x^2+x+2x^2+2x-6}{2\left(1-x\right)}=1\)

<=>\(\frac{10x-9}{2\left(1-x\right)}=1\)

<=> 10x-9=2(1-x)

<=>10x-9=2-2x

<=> 10x+2x= 2+9

<=> 12x=11

<=> x= \(\frac{11}{12}\left(tm\right)\)

b) \(\frac{6x-1}{2-x}+\frac{9x+4}{x+2}=\frac{3x^2-2x+1}{x^2-4}\)

ĐK: x≠2, x≠-2

<=>\(\frac{6x-1}{-\left(x-2\right)}+\frac{9x+4}{x+2}-\frac{3x^2-2x+1}{\left(x-2\right)\left(x+2\right)}=0\)

<=> -(x+2).(6x-1)+(x-2).(9x+4)-(3x2-2x+1)=0

<=> -(6x2-x+12x-2)+9x2+4x-18x-8-3x2+2x-1 = 0

<=> -6x2-11x+2+9x2+4x-18x-8-3x2+2x-1=0

<=> -23x-7=0

<=> -23x=7

<=> x= \(\frac{-7}{23}\left(tm\right)\)

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Absolute
22 tháng 4 2020 lúc 20:22

tham khảo câu d trong

https://hoc24.vn/hoi-dap/question/919967.html

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Absolute
22 tháng 4 2020 lúc 20:32

c) \(\frac{1}{x-1}\)+\(\frac{2x^2-5}{x^3-1}\)=\(\frac{4}{x^2+x+1}\) (ĐKXĐ:x≠1)

\(\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)+\(\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}\)=\(\frac{4\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

⇒x2+x+1+2x2-5=4x-4

⇔3x2-3x=0

⇔3x(x-1)=0

⇔x=0 (TMĐK) hoặc x=1 (loại)

Vậy tập nghiệm của phương trình đã cho là:S={0}

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Kaijo
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Ánh Tuyết
16 tháng 3 2020 lúc 8:42

a, \(\frac{4x+1}{2}-\frac{3x+2}{3}=\frac{12x+3}{6}-\frac{6x+4}{6}=\frac{12x+3-6x-4}{6}=\frac{6x-1}{6}\)

b, \(\frac{x+3}{x^2-1}-\frac{1}{x^2+x}=\frac{x+3}{\left(x-1\right)\left(x+2\right)}-\frac{1}{x\left(x+1\right)}\)

\(=\frac{x\left(x+3\right)}{x\left(x-1\right)\left(x+1\right)}-\frac{x-1}{x\left(x-1\right)\left(x+1\right)}\)

\(=\frac{x^2+3x-x+1}{x\left(x-1\right)\left(x+1\right)}=\frac{x^2+2x+1}{x\left(x-1\right)\left(x+1\right)}=\frac{\left(x+1\right)^2}{x\left(x-1\right)\left(x+1\right)}\)

\(=\frac{x+1}{x\left(x-1\right)}\)

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๖²⁴ʱTú❄⁀ᶦᵈᵒᶫ
16 tháng 3 2020 lúc 8:44

\(\frac{4x+1}{2}-\frac{3x+2}{3}\)

\(=\frac{12x+3}{6}-\frac{6x+4}{6}=\frac{6x-1}{6}\)

tương tự đến hết nha a hay cj gì đps ! 

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✰๖ۣۜŠɦαɗøω✰
16 tháng 3 2020 lúc 8:57

a) \(\frac{4.x+1}{2}-\frac{3.x+2}{3}=\frac{3.\left(4.x+1\right)-2.\left(3.x+2\right)}{6}\)

                                                \(=\frac{12.x+3-6.x-4}{6}\)

                                                   \(=\frac{6.x-1}{6}\)

b)\(\frac{x+3}{x^2-1}-\frac{1}{x^2+x}\)

\(=\frac{x+3}{\left(x-1\right).\left(x+1\right)}-\frac{1}{x.\left(x+1\right)}\)

\(=\frac{x.\left(x+3\right)-\left(x-1\right)}{x.\left(x-1\right).\left(x+1\right)}\)

\(=\frac{x^2+3.x-x+1}{x.\left(x-1\right).\left(x+1\right)}\)

\(=\frac{x^2+2.x+1}{x.\left(x-1\right).\left(x+1\right)}\)

\(=\frac{\left(x+1\right)^2}{x.\left(x-1\right).\left(x+1\right)}\)

\(=\frac{x+1}{x.\left(x-1\right)}\)

\(=\frac{x+1}{x^2-x}\)

c)\(\frac{3}{2.x^2+2.x}+\frac{2.x-1}{x^2-1}-\frac{1}{2}\)

\(=\frac{3}{2.x.\left(x+1\right)}+\frac{2.x-1}{\left(x-1\right).\left(x+1\right)}-\frac{1}{2}\)

\(=\frac{3.\left(x-1\right)+2.x.\left(2.x-1\right)-x.\left(x-1\right).\left(x+1\right)}{2.x.\left(x-1\right).\left(x+1\right)}\)

\(=\frac{3.x-3+4.x^2-2.x-x.\left(x^2-1\right)}{2.x.\left(x-1\right).\left(x+1\right)}\)

\(=\frac{3.x-3+4.x^2-2.x-x^3+x}{2.x.\left(x-1\right).\left(x+1\right)}\)

\(=\frac{2.x-3+4.x^2-x^3}{2.x.\left(x-1\right).\left(x+1\right)}\)

\(=\frac{-x^3+4.x^2+2.x-3}{2.x.\left(x-1\right).\left(x+1\right)}\)

\(=\frac{-x^3-x^2+5.x^2+5.x-3.x-3}{2.x.\left(x-1\right).\left(x+1\right)}\)

\(=\frac{-x^2.\left(x+1\right)+5.x.\left(x+1\right)-3.\left(x+1\right)}{2.x.\left(x-1\right).\left(x+1\right)}\)

\(=\frac{-\left(x+1\right).\left(x^2-5.x+3\right)}{2.x.\left(x-1\right).\left(x+1\right)}\)

\(=\frac{-\left(x^2-5.x+3\right)}{2.x.\left(x-1\right)}\)

\(=-\frac{x^2-5.x+3}{2.x^2-2.x}\)

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Nguyễn Châu Mỹ Linh
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viên cổn cổn
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Angel of the eternal lig...
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Tiến_Về_Phía_Trước
2 tháng 12 2019 lúc 21:52

a) \(\frac{3x+5}{2\left(x-1\right)}+\frac{4}{x-2}=\frac{\left(3x+5\right)\left(x-2\right)+4\cdot2\left(x-1\right)}{2\left(x-1\right)\left(x-2\right)}=\frac{3x^2-6x+5x-10+8x-8}{2\left(x-1\right)\left(x-2\right)}\)

\(=\frac{3x^2+7x-18}{2\left(x-1\right)\left(x-2\right)}\)

b) \(\frac{2x^2+1}{4x^2-2x}+\frac{3-3x}{1-2x}+\frac{3}{2x}=\frac{2x^2+1+4x\left(3-3x\right)+2\cdot3\left(1-2x\right)}{4x\left(1-2x\right)}=\frac{2x^2+1+12-12x+6-12x}{4x\left(1-2x\right)}\)\(=\frac{2x^2-24x+19}{4x\left(1-2x\right)}\)

Đề này... bạn xem lại đi. Chứ thế này thì dùng máy tính cũng không làm nổi T-T

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