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20 tháng 1

\(\dfrac{x^2}{2x^2+8x+8}\cdot\dfrac{3x+6}{-x}\left(dkxd:x\ne0;x\ne-2\right)\)

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

\(=\dfrac{x\cdot3\left(x+2\right)}{-2\left(x+2\right)^2}\)

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

10 tháng 11 2023

a: \(\lim\limits_{x\rightarrow3}\dfrac{\sqrt{2x+10}-4}{3x-9}\)

\(=\lim\limits_{x\rightarrow3}\dfrac{2x+10-16}{3x-9}\cdot\dfrac{1}{\sqrt{2x+10}+4}\)

\(=\lim\limits_{x\rightarrow3}\dfrac{2\left(x-3\right)}{3\left(x-3\right)\cdot\left(\sqrt{2x+10}+4\right)}\)

\(=\lim\limits_{x\rightarrow3}\dfrac{2}{3\left(\sqrt{2x+10}+4\right)}\)

\(=\dfrac{2}{3\cdot\sqrt{6+10}+3\cdot4}=\dfrac{2}{3\cdot4+3\cdot4}=\dfrac{2}{24}=\dfrac{1}{12}\)

b: \(\lim\limits_{x\rightarrow7}\dfrac{\sqrt{4x+8}-6}{x^2-9x+14}\)

\(=\lim\limits_{x\rightarrow7}\dfrac{4x+8-36}{\sqrt{4x+8}+6}\cdot\dfrac{1}{\left(x-2\right)\left(x-7\right)}\)

\(=\lim\limits_{x\rightarrow7}\dfrac{4x-28}{\left(\sqrt{4x+8}+6\right)\cdot\left(x-2\right)\left(x-7\right)}\)

\(=\lim\limits_{x\rightarrow7}\dfrac{4}{\left(\sqrt{4x+8}+6\right)\left(x-2\right)}\)

\(=\dfrac{4}{\left(\sqrt{4\cdot7+8}+6\right)\left(7-2\right)}\)

\(=\dfrac{4}{5\cdot12}=\dfrac{4}{60}=\dfrac{1}{15}\)

c: \(\lim\limits_{x\rightarrow5}\dfrac{x^2-8x+15}{2x^2-9x-5}\)

\(=\lim\limits_{x\rightarrow5}\dfrac{\left(x-3\right)\left(x-5\right)}{2x^2-10x+x-5}\)

\(=\lim\limits_{x\rightarrow5}\dfrac{\left(x-3\right)\left(x-5\right)}{\left(x-5\right)\left(2x+1\right)}\)

\(=\lim\limits_{x\rightarrow5}\dfrac{x-3}{2x+1}=\dfrac{5-3}{2\cdot5+1}=\dfrac{2}{11}\)

11 tháng 7 2017

\(D=\dfrac{x-2}{x+2}.\left(\dfrac{5x+10}{7x-14}+\dfrac{x-2}{3x-6}\right)+\dfrac{3x^2-12}{2x^2-8x+8}\)

\(D=\dfrac{x-2}{x+2}.\left(\dfrac{5\left(x+2\right)}{7\left(x-2\right)}+\dfrac{x-2}{3\left(x-2\right)}\right)+\dfrac{3\left(x^2-4\right)}{2\left(x^2-4x+4\right)}\)

\(D=\dfrac{x-2}{x+2}.\dfrac{5\left(x+2\right)}{7\left(x-2\right)}+\dfrac{x-2}{3\left(x-2\right)}.\dfrac{x-2}{x+2}+\dfrac{3\left(x^2-4\right)}{2\left(x^2-4x+4\right)}\)

\(D=\dfrac{5}{7}+\dfrac{x-2}{2\left(x+2\right)}+\dfrac{3\left(x-2\right)\left(x+2\right)}{2\left(x-2\right)^2}\)

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

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

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

\(D=\dfrac{5}{7}-\dfrac{2x+4}{2\left(x-2\right)}\)

\(D=\dfrac{5}{7}+\dfrac{2\left(x-2\right)}{2\left(x-2\right)}=\dfrac{5}{7}+\dfrac{x-2}{x-2}\)

\(D=\dfrac{5}{7}+1=\dfrac{12}{7}\)

Vậy \(D=\dfrac{12}{7}\)

29 tháng 5 2020

5) 3x - 1 < 8

⇔ 3x < 9

⇔ x < 3

29 tháng 5 2020

4) -8x > 24

<=> x > 32

5 tháng 5 2023

Nhớ tick cho mình nha

\(\dfrac{1}{3}\)x\(\dfrac{1}{x^2}\) - 8x + 32 = \(\dfrac{1}{x^2}\) - 2x + 8  ĐK: x ≠ 0

\(\dfrac{1}{3}\)x\(\dfrac{1}{x^2}\) - \(\dfrac{1}{x^2}\) - 8x + 2x + 32 - 8 = 0

\(\dfrac{1}{3}\)x2 - 6x +24 = 0

\(\left(x-12\right)\) \(\left(x-6\right)\) = 0

\(\left[{}\begin{matrix}x-12=0\\x-6=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=12\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)

⇒ S = \(\left\{12;6\right\}\)

 

 

a: =>5-x+6=12-8x

=>-x+11=12-8x

=>7x=1

hay x=1/7

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

\(\Leftrightarrow9x+6-3x-1=12x+10\)

=>12x+10=6x+5

=>6x=-5

hay x=-5/6

d: =>(x-2)(x-3)=0

=>x=2 hoặc x=3

29 tháng 1 2022

1.

<=> \(\left[{}\begin{matrix}4-3x=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)

2.

<=>\(\left[{}\begin{matrix}7-2x=0\\4+8x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

3.

<=>\(\left[{}\begin{matrix}9-7x=0\\11-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{7}\\x=\dfrac{11}{3}\end{matrix}\right.\)

4.

<=>\(\left[{}\begin{matrix}7-14x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)

5. 

<=>\(\left[{}\begin{matrix}\dfrac{7}{8}-2x=0\\3x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{16}\\x=-\dfrac{1}{9}\end{matrix}\right.\)

6,7. ko đủ điều kiện tìm

29 tháng 1 2022

Oki pạn cảm ơn

 

20 tháng 6 2017

\(B=\dfrac{x-2}{x+2}\cdot\left(\dfrac{5x+10}{7x-14}+\dfrac{x-2}{3x-6}\right)+\dfrac{3\left(x^2-4\right)}{2x^2-8x+8}\)

\(=\dfrac{x-2}{x+2}\cdot\left(\dfrac{5x+10}{7\left(x-2\right)}+\dfrac{x-2}{3\left(x-2\right)}\right)+\dfrac{3\left(x-2\right)\left(x+2\right)}{2\left(x^2-4x+4\right)}\)

\(=\dfrac{x-2}{x+2}\cdot\left(\dfrac{5x+10}{7\left(x-2\right)}+\dfrac{1}{3}\right)+\dfrac{3\left(x-2\right)\left(x+2\right)}{2\left(x-2\right)^2}\)

\(=\dfrac{x-2}{x+2}\cdot\dfrac{3\left(5x+10\right)+7\left(x-2\right)}{21\left(x-2\right)}+\dfrac{3\left(x+2\right)}{2\left(x-2\right)}\)

\(=\dfrac{1}{x+2}\cdot\dfrac{15x+30+7x-14}{21}+\dfrac{3x+6}{2\left(x-2\right)}\)

\(=\dfrac{22x+16}{21\left(x+2\right)}+\dfrac{3x+6}{2\left(x-2\right)}\)

\(=\dfrac{2\left(x-2\right)\left(22x+16\right)+21\left(x+2\right)\left(3x+6\right)}{42\left(x+2\right)\left(x-2\right)}\)

\(=\dfrac{\left(2x-4\right)\left(22x+16\right)+\left(21x+42\right)\left(3x+6\right)}{42\left(x^2-4\right)}\)

\(=\dfrac{44x^2+32x-88x-64+63x^2+126x+126x+252}{42x^2-168}\)

\(=\dfrac{107x^2+196x+188}{42x^2-168}\)

a: =>10x-4=15-9x

=>19x=19

hay x=1

b: \(\Leftrightarrow3\left(10x+3\right)=36+4\left(8x+6\right)\)

=>30x+9=36+32x+24

=>30x-32x=60-9

=>-2x=51

hay x=-51/2

c: \(\Leftrightarrow2x+\dfrac{6}{5}=5-\dfrac{13}{5}-x\)

=>3x=6/5

hay x=2/5

d: \(\Leftrightarrow\dfrac{7x}{8}-\dfrac{5\left(x-9\right)}{1}=\dfrac{20x+1.5}{6}\)

\(\Leftrightarrow21x-120\left(x-9\right)=4\left(20x+1.5\right)\)

=>21x-120x+1080=80x+60

=>-179x=-1020

hay x=1020/179

e: \(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)

=>35x-5+60x=96-6x

=>95x+6x=96+5

=>x=1

f: \(\Leftrightarrow6\left(x+4\right)+30\left(-x+4\right)=10x-15\left(x-2\right)\)

=>6x+24-30x+120=10x-15x+30

=>-24x+96=-5x+30

=>-19x=-66

hay x=66/19

a: =>A-B=3x^2y-4xy^2+x^2y-2xy^2=4x^2y-6xy^2

b: =>B-A=-7xy^2+8x^2y-5xy^2+6x^2y=-12xy^2+14x^2y

=>A-B=12xy^2-14x^2y

c: =>B-A=8x^2y^3-4x^3y-3x^2y^3+5x^3y^2=5x^2y^3+x^3y^2

=>A-B=-5x^2y^3-x^3y^2

d: =>A-B=2x^2y^3-7x^3y+6x^2y^3+3x^3y^2=8x^2y^3-7x^3y+3x^3y^2