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Kiều Duy Hiếu
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tthnew
17 tháng 1 2021 lúc 19:10

7,3, -6

ĐKXĐ: \(x\ne7;x\ne2\)

BPT \(\Leftrightarrow f\left(x\right)=\dfrac{\left(6-2x\right)^3\left(x+6\right)}{\left(x-7\right)^3}\le0\)

Lập bảng xét dấu ta có:

Từ đây ta thấy \(-6\le x\le3\) hoặc \(x>7\) thỏa mãn bất phương trình ban đầu.

Vậy...

 

Hoài An
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Yeutoanhoc
24 tháng 2 2021 lúc 19:49

`a,(x+3)(x^2+2021)=0`

`x^2+2021>=2021>0`

`=>x+3=0`

`=>x=-3`

`2,x(x-3)+3(x-3)=0`

`=>(x-3)(x+3)=0`

`=>x=+-3`

`b,x^2-9+(x+3)(3-2x)=0`

`=>(x-3)(x+3)+(x+3)(3-2x)=0`

`=>(x+3)(-x)=0`

`=>` $\left[ \begin{array}{l}x=0\\x=-3\end{array} \right.$

`d,3x^2+3x=0`

`=>3x(x+1)=0`

`=>` $\left[ \begin{array}{l}x=0\\x=-1\end{array} \right.$

`e,x^2-4x+4=4`

`=>x^2-4x=0`

`=>x(x-4)=0`

`=>` $\left[ \begin{array}{l}x=0\\x=4\end{array} \right.$

Nguyễn Trần Thành Đạt
24 tháng 2 2021 lúc 19:13

1) a) \(\left(x+3\right).\left(x^2+2021\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2021=0\end{matrix}\right.\\\left[{}\begin{matrix}x=-3\left(nhận\right)\\x^2=-2021\left(loại\right)\end{matrix}\right. \)

=> S={-3}

 

Nguyễn Lê Phước Thịnh
24 tháng 2 2021 lúc 20:07

Bài 1: 

a) Ta có: \(\left(x+3\right)\left(x^2+2021\right)=0\)

mà \(x^2+2021>0\forall x\)

nên x+3=0

hay x=-3

Vậy: S={-3}

Bài 2: 

b) Ta có: \(x\left(x-3\right)+3\left(x-3\right)=0\)

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

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

Vậy: S={3;-3}

tl:)
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Thanh Hoàng Thanh
14 tháng 1 2022 lúc 20:23

\(1.\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}.\Leftrightarrow\dfrac{x-1-3x}{3}=\dfrac{x-2}{2}.\Leftrightarrow\dfrac{-2x-1}{3}-\dfrac{x-2}{2}=0.\)

\(\Leftrightarrow\dfrac{-4x-2-3x+6}{6}=0.\Rightarrow-7x+4=0.\Leftrightarrow x=\dfrac{4}{7}.\)

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

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

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

\(3.3x^2-4x+1=0.\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{3}\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=\dfrac{1}{3}.\end{matrix}\right.\)

\(4.\left|2x-4\right|=0.\Leftrightarrow2x-4=0.\Leftrightarrow x=2.\)

\(5.\left|3x+2\right|=4.\Leftrightarrow\left[{}\begin{matrix}3x+2=4.\\3x+2=-4.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}.\\x=-2.\end{matrix}\right.\)

ILoveMath
14 tháng 1 2022 lúc 20:26

\(1,\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\\ \Leftrightarrow\dfrac{x-1}{3}-x=\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{2\left(x-1\right)-6x}{6}=\dfrac{3\left(x-2\right)}{6}\\ \Leftrightarrow2\left(x-1\right)-6x=3\left(x-2\right)\\ \Leftrightarrow2x-2-6x=3x-6\\ \Leftrightarrow-4x-2=3x-6\)

\(\Leftrightarrow3x-6+4x+2=0\\ \Leftrightarrow7x-4=0\\ \Leftrightarrow x=\dfrac{4}{7}\)

\(2,\left(x-2\right)\left(2x-1\right)=x^2-2x\\ \Leftrightarrow2x^2-4x-x+2=x^2-2x\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\\ \Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

\(3,3x^2-4x+1=0\\ \Leftrightarrow\left(3x^2-3x\right)-\left(x-1\right)=0\\ \Leftrightarrow3x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(4,\left|2x-4\right|=0\\ \Leftrightarrow2x-4=0\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\)

\(5,\left|3x+2\right|=4\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)

\(6,\left|2x-5\right|=\left|-x+2\right|\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=-x+2\\2x-5=x-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=7\\x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=3\end{matrix}\right.\)

Nguyễn Linh
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Trần Đức Huy
3 tháng 2 2022 lúc 9:01

f. 5 – (x – 6) = 4(3 – 2x)

<=>5-x+6=12-8x

<=>7x=1

<=>x=\(\dfrac{1}{7}\)

g. 7 – (2x + 4) = – (x + 4)

<=>7-2x-4=-x-4

<=>x=7

h. 2x(x+2)\(^2\)−8x\(^2\)=2(x−2)(x\(^2\)+2x+4)

<=>\(2x\left(x^2+4x+4\right)-8x^2=2\left(x^3-8\right)\)

<=>\(2x^3+8x^2+8x-8x^2=2\left(x^3-8\right)\)

<=>\(2x^3+8x=2x^3-16\)

<=>\(8x=-16\)

<=>\(x=-2\)

i. (x−2\(^3\))+(3x−1)(3x+1)=(x+1)\(^3\)

<=>\(x-8+9x^2-1=x^3+3x^2+3x+1\)

<=>\(6x^2-2x-10=0\)

<=>\(3x^2-x-5=0\)

<=>\(\left[{}\begin{matrix}x=\dfrac{1+\sqrt{61}}{6}\\x=\dfrac{1-\sqrt{61}}{6}\end{matrix}\right.\)

k. (x + 1)(2x – 3) = (2x – 1)(x + 5)

<=>\(2x^2-x-3=2x^2+9x-5\)

<=>10x=2

<=>\(x=\dfrac{1}{5}\)

Trần Đức Huy
3 tháng 2 2022 lúc 9:16

f. 5 – (x – 6) = 4(3 – 2x)

<=>5-x+6=12-8x

<=>7x=1

<=>x=\(\dfrac{1}{7}\)

g. 7 – (2x + 4) = – (x + 4)

<=>7-2x-4=-x-4

<=>x=7

h. \(2x\left(x+2\right)^2-8x^2=2\left(x-2\right)\left(x^2+2x+4\right)\)

<=>\(2x\left(x^2+4x+4\right)-8x^2=2\left(x^3-8\right)\)

<=>\(2x^3+8x^2+8x-8x^2=2x^3-16\)

<=>\(8x=-16\)

<=>x=-2

i.\(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)=\left(x+1\right)^3\)

<=>\(x^3-6x^2+12x+8+9x^2-1=x^3+3x^2+3x+1\)

<=>\(9x+6=0\)

<=>x=\(\dfrac{-2}{3}\)

k. (x + 1)(2x – 3) = (2x – 1)(x + 5)

<=>\(2x^2-x-3=2x^2+9x-5\)

<=>10x=2

<=>

Mot So
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Nguyễn Lê Phước Thịnh
28 tháng 1 2022 lúc 21:47

4: =>2x-3>5 hoặc 2x-3<-5

=>x>4 hoặc x<-1

5: =>-4<=2x-1<=4

=>-3/2<=x<=5/2

Kinder
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tl:)
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ILoveMath
28 tháng 1 2022 lúc 21:04

\(1,\) thiếu đề

\(2,\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)

\(\Leftrightarrow\dfrac{5\left(5x+2\right)}{30}-\dfrac{10\left(8x-1\right)}{30}=\dfrac{6\left(4x+2\right)}{30}-\dfrac{150}{30}\)

\(\Leftrightarrow5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-150\)

\(\Leftrightarrow25x+10-80x+10=24x+12-150\)

\(\Leftrightarrow-55x+20=24x-138\)

\(\Leftrightarrow24x-138+55x-20=0\)

\(\Leftrightarrow79x-158=0\)

\(\Leftrightarrow x=2\)

\(3,ĐKXĐ:\left\{{}\begin{matrix}x\ne1\\x\ne-1\\x\ne3\end{matrix}\right.\\ \dfrac{x}{2x-6}+\dfrac{x}{2x-2}=\dfrac{-2x}{\left(x+1\right)\left(3-x\right)}\)

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

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

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

\(\Leftrightarrow x\left(\dfrac{\left(x-1\right)\left(x+1\right)}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}+\dfrac{\left(x-3\right)\left(x+1\right)}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}-\dfrac{4\left(x-1\right)}{2\left(x+1\right)\left(x-3\right)\left(x-1\right)}\right)=0\)

\(\Leftrightarrow x\left(\dfrac{x^2-1}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}+\dfrac{x^2-2x-3}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}-\dfrac{4x-4}{2\left(x+1\right)\left(x-3\right)\left(x-1\right)}\right)=0\)

\(\Leftrightarrow x.\dfrac{x^2-1+x^2-2x-3-4x+4}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}=0\)

\(\Leftrightarrow x.\dfrac{2x^2-6x}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}=0\)

 

 

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

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

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

\(\Leftrightarrow x=0\)

Thanh Hoàng Thanh
28 tháng 1 2022 lúc 21:00

undefinedundefined

FREESHIP Asistant
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Thanh Hoàng Thanh
28 tháng 1 2022 lúc 11:35

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

Mà \(-3< 0.\)

\(\Rightarrow x+1>0.\Leftrightarrow x>-1\left(TMĐK\right).\)

\(\dfrac{x^2-2x+5}{x-2}-x+1\ge0.\left(x\ne2\right).\\ \Leftrightarrow\dfrac{x^2-2x+5-x^2+2x+x-2}{x-2}\ge0.\\ \Leftrightarrow\dfrac{x+3}{x-2}\ge0.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3\ge0.\\x-2\ge0.\end{matrix}\right.\\\left\{{}\begin{matrix}x+3\le0.\\x-2\le0.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-3.\\x\ge2.\end{matrix}\right.\\\left\{{}\begin{matrix}x\le-3.\\x\le2.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ge2.\\x\le-3.\end{matrix}\right.\)

Kết hợp ĐKXĐ.

\(\Rightarrow\left[{}\begin{matrix}x>2.\\x\le-3.\end{matrix}\right.\)

\(\dfrac{\left(1+2x\right)\left(x-2\right)}{\left(2x+3\right)\left(1-x\right)}\le0.\left(x\ne1;x\ne\dfrac{-3}{2}\right).\)

Đặt \(\dfrac{\left(1+2x\right)\left(x-2\right)}{\left(2x+3\right)\left(1-x\right)}=f\left(x\right).\)

Ta có bảng sau:

\(x\)\(-\infty\)              \(-\dfrac{3}{2}\)                       \(-\dfrac{1}{2}\)                       \(1\)                         \(2\)                        \(+\infty\)
\(1+2x\)         -              |            -                 0           +              |           +               |              +           
\(x-2\)         -               |           -                  |             -           |             -             0             +
\(2x+3\)         -              0           +                |             +            |              +           |             +
\(1-x\)         +              |           +                |              +           0             -            |            -                
\(f\left(x\right)\)

          -              ||          +                0               -          ||           +              0            -

Vậy \(f\left(x\right)\ge0.\Leftrightarrow x\in\left(\dfrac{-3}{2};\dfrac{-1}{2}\right)\cup\)(1;2].

 

Mot So
28 tháng 1 2022 lúc 14:18

2) −(x2−2x+5)x−2−x+1≥0                                                        ⇔

tl:)
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Nguyễn Lê Phước Thịnh
7 tháng 3 2022 lúc 0:28

1: \(\Leftrightarrow6\left(3x-1\right)+3\left(6x-2\right)=4\left(1-3x\right)\)

=>18x-6+18x-6=4-12x

=>36x-12=4-12x

=>48x=16

hay x=1/3

2: \(\Leftrightarrow\left(2x-1\right)\left(2x-1+x-3\right)=0\)

=>(2x-1)(3x-4)=0

=>x=1/2 hoặc x=4/3