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Nguyễn Linh
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Đào Tùng Dương
5 tháng 2 2022 lúc 11:06

e) ĐK : \(\left\{{}\begin{matrix}1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x\ne-1\\3x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}=\dfrac{\left(1-3x\right)^2-\left(1+3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}\)

\(\Leftrightarrow12\left(1+3x\right)\left(1-3x\right)=\left(1-3x\right)\left(1+3x\right)\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)\)

\(\Leftrightarrow12=\left(-6x\right).2\Leftrightarrow6=-6x\)

\(\Leftrightarrow x=-1\left(TM\right)\)

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camcon
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Yeutoanhoc
24 tháng 6 2021 lúc 16:01

`a)P=(x/(x+2)-(x^3-8)/(x^3+8)*(x^2-2x+4)/(x^2-4)):4/(x+2)`

`đk:x ne 0,x ne -2`

`P=(x/(x+2)-((x-2)(x^2+2x+4))/((x+2)(x^2-2x+4))*(x^2-2x+4)/((x-2)(x+2)))*(x+2)/4`

`=(x/(x+2)-(x^2+2x+4)/(x+2)^2)*(x+2)/4`

`=(x^2+2x-x^2-2x-4)/(x+2)^2*(x+2)/4`

`=-4/(x+2)^2*(x+2)/4`

`=-1/(x+2)`

`b)P<0`

`<=>-1/(x+2)<0`

Vì `-1<0`

`<=>x+2>0`

`<=>x> -2`

`c)P=1/x+1(x ne 0)`

`<=>-1/(x+2)=1/x+1`

`<=>1/x+1+1/(x+2)=0``

`<=>x+2+x(x+2)+x=0`

`<=>x^2+4x+2=0`

`<=>` \(\left[ \begin{array}{l}x=\sqrt2-2\\x=-\sqrt2-2\end{array} \right.\) 

`d)|2x-1|=3`

`<=>` \(\left[ \begin{array}{l}2x=4\\2x=-2\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=2(l)\\x=-1(tm)\end{array} \right.\) 

`x=-1=>P=-1/(-1+2)=-1`

`e)P=-1/(x+2)` thì nhỏ nhất cái gì nhỉ?

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๖ۣۜDũ๖ۣۜN๖ۣۜG
24 tháng 6 2021 lúc 16:04

a) đk: \(x\ne-2;2\)

 \(P=\left[\dfrac{x}{x+2}-\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}.\dfrac{x^2-2x+4}{\left(x-2\right)\left(x+2\right)}\right]:\dfrac{4}{x+2}\)

\(\left[\dfrac{x}{x+2}-\dfrac{x^2+2x+4}{\left(x+2\right)^2}\right].\dfrac{x+2}{4}\)

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

b) Để P < 0

<=> \(\dfrac{-1}{x+2}< 0\)

<=> x +2 > 0

<=> x > -2 ( x khác 2)

c) Để P= \(\dfrac{1}{x}+1\)

<=> \(\dfrac{-1}{x+2}=\dfrac{1}{x}+1\)

<=> \(\dfrac{1}{x}+\dfrac{1}{x+2}+1=0\)

<=> \(\dfrac{x+2+x+x\left(x+2\right)}{x\left(x+2\right)}=0\)

<=> x2 + 4x + 2 = 0

<=> (x+2)2 = 2

<=> \(\left[{}\begin{matrix}x=\sqrt{2}-2\left(c\right)\\x=-\sqrt{2}-2\left(c\right)\end{matrix}\right.\)

d) Để \(\left|2x-1\right|=3\)

<=> \(\left[{}\begin{matrix}2x-1=3< =>x=2\left(l\right)\\2x-1=-3< =>x=-1\left(c\right)\end{matrix}\right.\)

Thay x = -1, ta có:

P = \(\dfrac{-1}{-1+2}=-1\)

 

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An Thy
24 tháng 6 2021 lúc 16:06

a) ĐKXĐ: \(x\ne2;-2\)

\(P=\left(\dfrac{x}{x+2}-\dfrac{x^3-8}{x^3+8}.\dfrac{x^2-2x+4}{x^2-4}\right):\dfrac{4}{x+2}\)

\(=\left(\dfrac{x}{x+2}-\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}.\dfrac{x^2-2x+4}{\left(x-2\right)\left(x+2\right)}\right):\dfrac{4}{x+2}\)

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

\(=\left(\dfrac{x}{x+2}-\dfrac{x^2+2x+4}{\left(x+2\right)^2}\right):\dfrac{4}{x+2}\)

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

b) \(P< 0\Rightarrow-\dfrac{1}{x+2}< 0\Rightarrow x+2>0\Rightarrow x>-2\)

\(\Rightarrow x>-2;x\ne2\)

c) \(P=\dfrac{1}{x}+1\Rightarrow\dfrac{-1}{x+2}=\dfrac{x+1}{x}\Rightarrow-x=\left(x+2\right)\left(x+1\right)\)

\(\Rightarrow-x=x^2+3x+2\Rightarrow x^2+4x+2=0\)

\(\Delta=4^2-2.4=8\Rightarrow\left[{}\begin{matrix}x=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-4-2\sqrt{2}}{2}=-2-\sqrt{2}\\x=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-4+2\sqrt{2}}{2}=-2+\sqrt{2}\end{matrix}\right.\)

d) \(\left|2x-1\right|=3\Rightarrow\left[{}\begin{matrix}2x-1=3\\1-2x=3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}P=-\dfrac{1}{2+2}=-\dfrac{1}{4}\\P=-\dfrac{1}{-1+2}=-1\end{matrix}\right.\)

 

 

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Khánh Linh Đỗ
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Nguyễn Lê Phước Thịnh
26 tháng 10 2023 lúc 21:46

6:

a: ĐKXĐ: x<>0

\(\dfrac{x^3+3x^2+3x+1}{x^2+x}\)

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

b: ĐKXĐ: x<>1

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

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

c: ĐKXĐ: x<>-2

\(\dfrac{x^2+4x+4}{2x+4}\)

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

\(=\dfrac{x+2}{2}\)

d: ĐKXĐ: x<>-2

\(\dfrac{\left(x-1\right)\left(-x-2\right)}{x+2}\)

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

e: ĐKXĐ: x<>-y

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

g: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

\(\dfrac{-3x^2-6x}{4-x^2}=\dfrac{3x^2+6x}{x^2-4}\)

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

7:

a: \(\dfrac{2}{5x^3y^2}=\dfrac{2\cdot4}{20x^3y^2}=\dfrac{8}{20x^3y^2}\)

\(\dfrac{3}{4xy}=\dfrac{3\cdot5\cdot x^2y}{20x^3y^2}=\dfrac{15x^2y}{20x^3y^2}\)

b: \(\dfrac{x}{x^2-2xy+y^2}=\dfrac{x}{\left(x-y\right)^2}\)

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

c: \(\dfrac{1}{x+2}=\dfrac{6}{6\left(x+2\right)}\)

\(\dfrac{2}{2x+4}=\dfrac{2}{2\left(x+2\right)}=\dfrac{1}{x+2}=\dfrac{6}{6\left(x+2\right)}\)

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

d:

\(\dfrac{2}{2x-6}=\dfrac{2}{2\left(x-3\right)}=\dfrac{1}{x-3};\dfrac{3}{3x-9}=\dfrac{3}{3\left(x-3\right)}=\dfrac{1}{x-3}\)

\(\dfrac{2}{2x-6}=\dfrac{1}{x-3}=\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}\)

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

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

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Hoàng Huy
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Huỳnh Thị Thanh Ngân
29 tháng 7 2021 lúc 9:23

\(\dfrac{4x^2-3x+5}{x^3-1}-\dfrac{1+2x}{x^2+x+1}-\dfrac{6}{x-1}\)

\(\Leftrightarrow\dfrac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{1+2x}{x^2+x+1}-\dfrac{6}{x-1}\)

\(ĐKXĐ:x\ne1\)

\(\dfrac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{(1+2x)\left(x-1\right)}{(x^2+x+1)\left(x-1\right)}-\dfrac{6\left(x^2+x+1\right)}{(x-1)\left(x^2+x+1\right)}\)

\(\Rightarrow4x^2-3x+5-\left(1+2x\right)\left(x-1\right)-6\left(x^2+x+1\right)\)

\(\Rightarrow4x^2-3x+5-\left(x-1+2x^2-2x\right)-6x^2-6x-6\)

\(\Rightarrow4x^2-3x+5-x+1-2x^2+2x-6x^2-6x-6\)

\(\Rightarrow-4x^2-8x\)

⇒-4x(x-4)

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le thi yen chi
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Duy Đỗ Ngọc Tuấn
30 tháng 4 2018 lúc 0:10

4)a)\(\dfrac{x+5}{x-5}-\dfrac{x-5}{x+5}=\dfrac{20}{x^2-25}\)(1)

ĐKXĐ:\(\left\{{}\begin{matrix}x-5\ne0\\x+5\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne5\\x\ne-5\end{matrix}\right.\)

(1)\(\Rightarrow\left(x+5\right)\left(x+5\right)-\left(x-5\right)\left(x-5\right)=20\)

\(\Leftrightarrow x^2+10x+25-\left(x^2-10x+25\right)=20\)

\(\Leftrightarrow x^2+10x+25-x^2+10x-25=20\)

\(\Leftrightarrow x^2-x^2+10x+10x=-25+25=20\)

\(\Leftrightarrow20x=20\)

\(\Leftrightarrow x=1\left(nh\text{ậ}n\right)\)

S=\(\left\{1\right\}\)

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Duy Đỗ Ngọc Tuấn
30 tháng 4 2018 lúc 0:11

mấy bài còn lại dễ ẹt cứ bình tĩnh làm là ok

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Phan van thach
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Lê Bùi
28 tháng 4 2018 lúc 20:56

câu nào cũng ghi lại đề nha

a) \(x\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

b)\(x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

c) \(\left(x+1\right)\left(x+2\right)+\left(x+2\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x+1+x-2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{2}\end{matrix}\right.\)

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Lê Bùi
28 tháng 4 2018 lúc 21:03

d) \(\dfrac{1}{x-2}+3-\dfrac{3-x}{x-2}=0\)

\(\Leftrightarrow\dfrac{1+3\left(x-2\right)-\left(3-x\right)}{x-2}=0\)

\(\Leftrightarrow\dfrac{1+3x-6-3+x}{x-2}=0\) ( đk \(x\ne2\) )

\(\Leftrightarrow4x-8=0\Rightarrow x=2\)

đ) \(\dfrac{8-x}{x-7}-8-\dfrac{1}{x-7}=0\)

\(\Leftrightarrow\dfrac{8-x-8\left(x-7\right)-1}{x-7}=0\) (đk \(x\ne7\))

\(\Leftrightarrow8-x-8x+56-1=0\)

\(\Leftrightarrow-9x+63=0\)

\(\Leftrightarrow x=7\)

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Lê Bùi
28 tháng 4 2018 lúc 21:09

e) \(\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x+1\right)}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\) (MTC 2(x+1)(x-3) )

\(\Leftrightarrow\dfrac{x\left(x+1\right)+x\left(x-3\right)-2x.2}{2\left(x-3\right)\left(x+1\right)}=0\) (đk \(x\ne3vsx\ne-1\) )

\(\Leftrightarrow x^2+x+x^2-3x-4x=0\)

\(\Leftrightarrow2x^2-6x=0\)

\(\Leftrightarrow2x\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

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My Nguyễn
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Nguyễn Lê Phước Thịnh
11 tháng 7 2023 lúc 22:00

1: Sửa đề: 2/x+2

\(\dfrac{2x+1}{x^2-4}+\dfrac{2}{x+2}=\dfrac{3}{2-x}\)

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

=>4x-3=-3x-6

=>7x=-3

=>x=-3/7(nhận)

2: \(\Leftrightarrow\dfrac{\left(3x+1\right)\left(3-x\right)+\left(3+x\right)\left(1-3x\right)}{\left(1-3x\right)\left(3-x\right)}=2\)

=>9x-3x^2+3-x+3-9x+x-3x^2=2(3x-1)(x-3)

=>-6x^2+6=2(3x^2-10x+3)

=>-6x^2+6=6x^2-20x+6

=>-12x^2+20x=0

=>-4x(3x-5)=0

=>x=5/3(nhận) hoặc x=0(nhận)

3: \(\Leftrightarrow x\cdot\dfrac{8}{3}-\dfrac{2}{3}=1+\dfrac{5}{4}-\dfrac{1}{2}x\)

=>x*19/6=35/12

=>x=35/38

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Kyun Diệp
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Nguyễn Thành Trương
25 tháng 2 2019 lúc 19:12

Hỏi đáp Toán

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Nguyễn Thành Trương
25 tháng 2 2019 lúc 19:06

Hỏi đáp Toán

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Kyun Diệp
25 tháng 2 2019 lúc 2:04

Nguyễn TrươngNguyễn Việt LâmNguyenTruong Viet TruongKhôi BùiAkai HarumaÁnh LêDƯƠNG PHAN KHÁNH DƯƠNGPhùng Tuệ Minhsaint suppapong udomkaewkanjana

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Trần An
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Nguyễn Ngọc Lộc
7 tháng 2 2021 lúc 22:50

Chia câu hỏi ra cho thành nhiều phần cho dễ trả lời á bạn

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Trần An
8 tháng 2 2021 lúc 14:00

Giúp dới

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