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20 tháng 11 2018

24 tháng 4 2022

a) \(\dfrac{3}{x-7}+\dfrac{2}{x+7}=\dfrac{5}{x^2-49}\)

(ĐKXĐ: x khác 7; x khác -7)

<=>\(\dfrac{3.\left(x+7\right)}{\left(x-7\right).\left(x+7\right)}+\dfrac{2.\left(x-7\right)}{\left(x+7\right).\left(x-7\right)}=\dfrac{5}{\left(x+7\right).\left(x-7\right)}\)

=> 3x + 21 + 2x - 14 = 5

<=> 3x + 2x = 5 + 14 - 21

<=> 5x = -2

<=> x = \(\dfrac{-2}{5}\)

Vậy S = { \(\dfrac{-2}{5}\) }

24 tháng 4 2022

b) \(\dfrac{2x-1}{3}-\dfrac{x+3}{2}>1+\dfrac{5x}{6}\)

<=> \(\dfrac{2.\left(2x-1\right)}{3.2}-\dfrac{3.\left(x+3\right)}{3.2}>\dfrac{1.6}{6}+\dfrac{5x}{6}\)

=> 4x - 2 - 3x - 9 > 6 + 5x

<=> 4x - 3x - 5x > 6 + 9 + 2

<=> -4x > 17

<=> \(\dfrac{-17}{4}\)

Vậy S = { \(\dfrac{-17}{4}\) }

9 tháng 1 2023

a. 3(x-2)-10=5(2x + 1)

<=> 3x - 6 - 10 = 10x + 5

<=> 3x - 10x = 5 + 6 + 10

<=> -7x = 21

<=> x = -3

b. 3x + 2=8 -2(x-7)

<=> 3x + 2 = 8 - 2x + 14

<=> 3x + 2x = 8 + 14 - 2

<=> 5x = 20

<=> x = 4

c. 2x-(2+5x)= 4(x + 3)

<=> 2x - 2 - 5x = 4x + 12

<=> 2x - 5x - 4x = 12 + 2

<=> -7x = 14

<=> x = -2

d. 5-(x +8)=3x + 3(x-9)

<=> 5 - x - 8 = 3x + 3x - 27

<=> -x - 3x - 3x = -27 + 8 - 5

<=> -7x = -24

<=> x = 24/7

e. 3x - 18 + x= 12-(5x + 3)

<=> 3x - 18 + x = 12 - 5x - 3

<=> 3x + x - 5x = 12 - 3 + 18

<=> -x = 27

<=> x = - 27

a. 3(x-2)-10=5(2x + 1)

<=> 3x - 6 - 10 = 10x + 5

<=> 3x - 10x = 5 + 6 + 10

<=> -7x = 21

<=> x = -3

b. 3x + 2=8 -2(x-7)

<=> 3x + 2 = 8 - 2x + 14

<=> 3x + 2x = 8 + 14 - 2

<=> 5x = 20

<=> x = 4

c. 2x-(2+5x)= 4(x + 3)

<=> 2x - 2 - 5x = 4x + 12

<=> 2x - 5x - 4x = 12 + 2

<=> -7x = 14

<=> x = -2

d. 5-(x +8)=3x + 3(x-9)

<=> 5 - x - 8 = 3x + 3x - 27

<=> -x - 3x - 3x = -27 + 8 - 5

<=> -7x = -24

<=> x = 24/7

e. 3x - 18 + x= 12-(5x + 3)

<=> 3x - 18 + x = 12 - 5x - 3

<=> 3x + x - 5x = 12 - 3 + 18

<=> -x = 27

<=> x = - 27

a) Ta có: \(\left|x^2-x+2\right|-3x-7=0\)

\(\Leftrightarrow\left|x^2-x+2\right|=3x+7\)

\(\Leftrightarrow x^2-x+2=3x+7\)(Vì \(x^2-x+2>0\forall x\))

\(\Leftrightarrow x^2-x+2-3x-7=0\)

\(\Leftrightarrow x^2-4x-5=0\)

\(\Leftrightarrow x^2-5x+x-5=0\)

\(\Leftrightarrow x\left(x-5\right)+\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

Vậy: S={5;-1}

6 tháng 3 2021

bạn giải giúp mình câu b nữa với

mai mình phải nộp bài rồi!!!khocroi

a) ĐKXĐ: \(x\notin\left\{5;-5\right\}\)

Ta có: \(\dfrac{-\left(x^2+5\right)}{x^2-25}=\dfrac{3}{x+5}+\dfrac{x}{x-5}\)

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

Suy ra: \(3x-15+x^2+5x+x^2+5=0\)

\(\Leftrightarrow2x^2+8x-10=0\)

\(\Leftrightarrow2x^2+10x-2x-10=0\)

\(\Leftrightarrow2x\left(x+5\right)-2\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(2x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\2x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)

Vậy: S={1}

28 tháng 2 2021

`a,(-(x^2+5))/(x^2-25)=3/(x+5)+x/(x-5)`

`ĐK:x ne +-5`

`pt<=>-x^2+5=3(x-5)+x(x+5)`

`<=>-x^2+5=3x-15+x^2+5x`

`<=>-x^2+5=x^2+8x-15`

`<=>2x^2+8x-20=0`

`<=>x^2+4x-5=0`

`<=>x^2-x+5x-5=0`

`<=>x(x-1)+5(x-1)=0`

`<=>` $\left[ \begin{array}{l}x=1\\x=-5\end{array} \right.$

Vậy `S={1,-5}`

18 tháng 7 2021

\(a,=>x^3-2x^2+4x+2x^2-4x+8-x^3+2x-15=0\)

\(< =>2x-7=0< =>x=\dfrac{7}{2}\)

b,\(=>x\left(x^2-25\right)-\left(x+2\right)\left(x^2-2x+4\right)-3=0\)

\(< =>x^3-25x-x^3+2x^2-4x-2x^2+4x-8-3=0\)

\(< =>-25x-11=0\)

\(< =>x=-0,44\)

18 tháng 7 2021

cảm ơn bạn nhiều nha!

 

22 tháng 4 2022

a.\(x^2-25=8\left(5-x\right)\)

\(\Leftrightarrow\left(x-5\right)\left(x+5\right)-8\left(5-x\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+5\right)+8\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+13\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-13\end{matrix}\right.\)

b.\(\dfrac{x-2}{x+2}-\dfrac{2\left(x-11\right)}{x^2-4}=\dfrac{3}{x-2}\) ; \(ĐK:x\ne\pm2\)

\(\Leftrightarrow\dfrac{\left(x-2\right)\left(x-2\right)-2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow\left(x-2\right)^2-2\left(x-11\right)=3\left(x+2\right)\)

\(\Leftrightarrow x^2-4x+4-2x+22=3x+6\)

\(\Leftrightarrow x^2-9x+20=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=4\left(tm\right)\end{matrix}\right.\)

 

13 tháng 9 2021

\(a,\Leftrightarrow x^2+2x+1+2x+3-2\sqrt{2x+3}+1=0\\ \Leftrightarrow\left(x+1\right)^2+\left(\sqrt{2x+3}-1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-1\\2x+3=1\end{matrix}\right.\Leftrightarrow x=-1\left(N\right)\)

13 tháng 9 2021

\(b,\Leftrightarrow3x^2+3x-2\sqrt{x^2+x}=0\left(x\le-1;x\ge0\right)\\ \Leftrightarrow3x\left(x-1\right)-2\sqrt{x\left(x+1\right)}=0\\ \Leftrightarrow\sqrt{x\left(x+1\right)}\left(3\sqrt{x\left(x-1\right)}-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x\left(x-1\right)=0\\\sqrt{x\left(x-1\right)}=\dfrac{2}{3}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x^2-x-\dfrac{4}{9}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\9x^2-9x-4=0\left(1\right)\end{matrix}\right.\)

\(\Delta\left(1\right)=81-4\left(-4\right)\cdot9=225\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\dfrac{9-15}{18}\\x=\dfrac{9+15}{18}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(N\right)\\x=1\left(N\right)\\x=-\dfrac{1}{3}\left(L\right)\\x=\dfrac{4}{3}\left(N\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\dfrac{4}{3}\end{matrix}\right.\)