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Long Nguyễn
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Phương An
1 tháng 12 2016 lúc 18:06

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

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

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

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

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

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

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

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

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

\(=\frac{1}{x}\)

Huyền
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Huỳnh Thành Lợi
6 tháng 11 2018 lúc 22:15

quy đồng nha bạn

Thái Bùi Ngọc
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Mtrangg
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HT.Phong (9A5)
2 tháng 8 2023 lúc 9:12

a) \(\dfrac{3}{4xy}+\dfrac{5x}{2x^2z}+\dfrac{7}{6yz^2}\) (MSC: \(12x^2yz^2\))

\(=\dfrac{3\cdot3xz^2}{4xy\cdot3xz^2}+\dfrac{5x\cdot6yz}{2x^2z\cdot6yz}+\dfrac{7\cdot2x^2}{6yz^2\cdot2x^2}\)

\(=\dfrac{9xz^2}{12x^2yz^2}+\dfrac{30xyz}{12x^2yz^2}+\dfrac{14x^2}{12x^2yz^2}\)

\(=\dfrac{9xz^2+30xyz+14x^2}{12x^2yz^2}\)

\(=\dfrac{x\left(9z^2+30yz+14x\right)}{12x^2yz^2}\)

\(=\dfrac{9z^2+30yz+14x}{12x^2yz^2}\)

b) \(\dfrac{x^2}{x^2+3x}+\dfrac{3}{x+3}+\dfrac{3}{x}\)

\(=\dfrac{x^2}{x\left(x+3\right)}+\dfrac{3}{x+3}+\dfrac{3}{x}\)

\(=\dfrac{x}{x+3}+\dfrac{3}{x+3}+\dfrac{3}{x}\)

\(=\dfrac{x+3}{x+3}+\dfrac{3}{x}\)

\(=1+\dfrac{3}{x}\)

\(=\dfrac{x}{x}+\dfrac{3}{x}\)

\(=\dfrac{x+3}{x}\)

Nguyễn Lê Phước Thịnh
2 tháng 8 2023 lúc 9:06

a: \(=\dfrac{3\cdot3\cdot xz^2+5x\cdot6\cdot y+7\cdot x^2\cdot2}{12x^2yz^2}=\dfrac{9xz^2+30xy+14x^2}{12x^2yz^2}\)

\(=\dfrac{9z^2+30y+14x}{12xyz^2}\)

b: \(=\dfrac{x}{x+3}+\dfrac{3}{x+3}+\dfrac{3}{x}=1+\dfrac{3}{x}=\dfrac{x+3}{x}\)

nhóc naruto
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Ngô Chi Lan
14 tháng 8 2020 lúc 20:11

Xin phép sửa đề:

Ta có: \(\frac{3x+1}{\left(x-1\right)^2}-\frac{1}{x+1}=\frac{x+3}{1-x^2}\) \(\left(x\ne\pm1\right)\)

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

\(\Rightarrow3x^2+4x+1-1+2x-x^2=-x^2-2x+3\)

\(\Leftrightarrow3x^2+8x-3=0\)

\(\Leftrightarrow\left(3x^2+9x\right)-\left(x+3\right)=0\)

\(\Leftrightarrow3x\left(x+3\right)-\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\3x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{1}{3}\end{cases}}\)

Vậy tập nghiệm PT \(S=\left(-3;\frac{1}{3}\right)\)

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Phương Kuro
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Diệu Vy
20 tháng 12 2016 lúc 10:35

a. \(=\frac{x+1}{2.\left(x+3\right)}+\frac{2x+3}{x.\left(x+3\right)}=\frac{x^2+x+4x+6}{2x.\left(x+3\right)}=\frac{x^2+5x+6}{2x.\left(x+3\right)}=\frac{\left(x+2\right).\left(x+3\right)}{2x.\left(x+3\right)}=\frac{x+2}{2x}\)

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

Chắc chắn đúng, mik nhaaaaaa

Nguyễn Hà Phương
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Phạm Tuán Quang
2 tháng 9 2021 lúc 10:21

\(\frac{3\left(x-2\right)}{4}\div\frac{2-x}{2}=\frac{3\left(x-2\right)}{4}\times\frac{-2}{x-2}=\frac{-3}{2}\)

học tốt

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Member lỗi thời :>>...
2 tháng 9 2021 lúc 10:18

Rút gọn nhé !

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

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

\(=\frac{\left(12x-24\right)-\left(6x^2+12x\right)}{4}=\frac{-24-6x^2}{4}\)

\(=\frac{-12-3x^2}{2}=\frac{-3.\left(4+x^2\right)}{2}\)

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Hàn Uyển Nghi
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Minh Nguyen
3 tháng 3 2020 lúc 12:38

\(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)

\(\left(\frac{9}{x^3-9x}+\frac{1}{x+3}\right):\left(\frac{x-3}{x^2+3x}-\frac{x}{3x+9}\right)\)

\(=\left(\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)

\(=\frac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\frac{3x-9-x^2}{3x\left(x+3\right)}\)

\(=\frac{\left(9+x^2-3x\right)\left(x+3\right)3x}{x\left(x-3\right)\left(x+3\right)\left(3x-9-x^2\right)}\)

\(=\frac{-3}{x-3}\)

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Hạ Nhi
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Kiyotaka Ayanokoji
6 tháng 6 2020 lúc 17:07

a, \(\frac{2}{5}.\frac{1}{3}-\frac{2}{15}:\frac{1}{5}+\frac{3}{5}.\frac{1}{3}\)

\(=\frac{1}{3}.\left(\frac{2}{5}+\frac{3}{5}\right)-\frac{2}{15}.5\)

\(=\frac{1}{3}.1-\frac{2}{3}\)

\(=\frac{1}{3}-\frac{2}{3}\)

\(=\frac{-1}{3}\)

b, \(\left(6-2\frac{4}{5}\right).3\frac{1}{8}+1\frac{3}{8}:\frac{1}{4}\)

\(=\left(6-\frac{14}{5}\right).\frac{25}{8}+\frac{11}{8}.4\)

\(=\frac{16}{5}.\frac{25}{8}+\frac{11}{2}\)

\(=10+\frac{11}{2}\)

\(=\frac{31}{2}\)

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ミ★ngũッhoàngッluffy★...
6 tháng 6 2020 lúc 17:56

1/3×(3/5+2/5)-2/15×1/5

1/3×1-2/15×1/5

1/3-2/15×1/5

1/3-2/75

25/75-2/75

23/75

(6-14/5)×25/8-11/8:4/1

16/5×25/8-11/8:4/1

10/1-11/8:4/1

10/1-11/8×1/4

10/1-11/32

320/32-11/32

309/32

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