a, \(\dfrac{x^3+27}{x^2-3x+9}=\dfrac{x+3}{M}\Leftrightarrow\dfrac{\left(x+3\right)\left(x^2-3x+9\right)}{x^2-3x+9}=\dfrac{x+3}{M}\)

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

b, \(\dfrac{M}{x+4}=\dfrac{x^2-8x+16}{16-x^2}=\dfrac{\left(x-4\right)^2}{\left(4-x\right)\left(x+4\right)}=\dfrac{4-x}{x+4}\)

\(\Rightarrow M=\dfrac{\left(4-x\right)\left(x+4\right)}{x+4}=4-x\)

c, tương tự 

a:

ĐKXĐ: x<>-1/2

Để \(\dfrac{2x^3+x^2+2x+2}{2x+1}\in Z\) thì

\(2x^3+x^2+2x+1+1⋮2x+1\)

=>\(2x+1\inƯ\left(1\right)\)

=>2x+1 thuộc {1;-1}

=>x thuộc {0;-1}

b:

ĐKXĐ: x<>1/3

 \(\dfrac{3x^3-7x^2+11x-1}{3x-1}\in Z\)

=>3x^3-x^2-6x^2+2x+9x-3+2 chia hết cho 3x-1

=>2 chia hết cho 3x-1

=>3x-1 thuộc {1;-1;2;-2}

=>x thuộc {2/3;0;1;-1/3}

mà x nguyên

nên x thuộc {0;1}

c: 

ĐKXĐ: x<>2

\(\dfrac{x^4-16}{x^4-4x^3+8x^2-16x+16}\in Z\)

=>\(\left(x^2-4\right)\left(x^2+4\right)⋮\left(x-2\right)^2\left(x^2+4\right)\)

=>\(x+2⋮x-2\)

=>x-2+4 chia hết cho x-2

=>4 chia hết cho x-2

=>x-2 thuộc {1;-1;2;-2;4;-4}

=>x thuộc {3;1;4;0;6;-2}

Lời giải:

a. $\frac{2-x}{4}=\frac{3x-1}{3}$

$\Rightarrow 3(2-x)=4(3x-1)$

$\Rightarrow 6-3x=12x-4$

$\Rightarrow 6+4=12x+3x$

$\Rightarrow 10=15x$

$\Rightarrow x=\frac{10}{15}=\frac{2}{3}$

b.

$\frac{x}{7}=\frac{x+16}{35}$

$\Rightarrow \frac{5x}{35}=\frac{x+16}{35}$

$\Rightarrow 5x=x+16$

$\Rightarrow 4x=16$

$\Rightarrow x=4$

c.

$\sqrt{x^2+1}=3$

$\Rightarrow x^2+1=9$

$\Rightarrow x^2=8\Rightarrow x=\pm \sqrt{8}=\pm 2\sqrt{2}$

\(\dfrac{1}{2}-\left|2-3x\right|=\sqrt{\dfrac{19}{16}}-\sqrt{\left(-0,75\right)^2}\\ \Rightarrow\dfrac{1}{2}-\left|2-3x\right|=\dfrac{\sqrt{19}}{4}-\dfrac{3}{4}\\ \Rightarrow\left|2-3x\right|=\dfrac{1}{2}-\dfrac{\sqrt{19}-3}{4}\)

\(\Rightarrow\left|2-3x\right|=\dfrac{5-\sqrt{19}}{4}\)

\(TH_1:x\le\dfrac{2}{3}\\ 2-3x=\dfrac{5-\sqrt{19}}{4}\\ \Rightarrow3x=\dfrac{3+\sqrt{19}}{4}\\ \Rightarrow x=\dfrac{3+\sqrt{19}}{12}\left(tm\right)\)

\(TH_2:x>\dfrac{2}{3}\\ 3x-2=\dfrac{5-\sqrt{19}}{4}\\ \Rightarrow3x=\dfrac{13-\sqrt{19}}{4}\\ \Rightarrow x=\dfrac{13-\sqrt{19}}{12}\left(tm\right)\)

Vậy \(x\in\left\{\dfrac{3+\sqrt{19}}{12};\dfrac{13-\sqrt{19}}{12}\right\}\)

\(\dfrac{1}{2}-\left|2-3x\right|=\sqrt[]{\dfrac{19}{16}}-\sqrt[]{\left(-0,75\right)^2}\)

\(\Rightarrow\dfrac{1}{2}-\left|2-3x\right|=\dfrac{\sqrt[]{19}}{4}-0,75\)

\(\Rightarrow\dfrac{1}{2}-\left|2-3x\right|=\dfrac{\sqrt[]{19}}{4}-\dfrac{3}{4}\)

\(\Rightarrow\left|2-3x\right|=\dfrac{1}{2}-\dfrac{\sqrt[]{19}}{4}+\dfrac{3}{4}\)

\(\Rightarrow\left|2-3x\right|=\dfrac{5-\sqrt[]{19}}{4}\)

\(\Rightarrow\left[{}\begin{matrix}2-3x=\dfrac{5-\sqrt[]{19}}{4}\\2-3x=\dfrac{-5+\sqrt[]{19}}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=2-\dfrac{5-\sqrt[]{19}}{4}\\3x=2-\dfrac{\sqrt[]{19}-5}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=\dfrac{3+\sqrt[]{19}}{4}\\3x=\dfrac{13-\sqrt[]{19}}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt[]{19}}{12}\\x=\dfrac{13-\sqrt[]{19}}{12}\end{matrix}\right.\)

a) \(\left(2\dfrac{3}{4}-1\dfrac{4}{5}\right)\cdot x=1\)

\(\left(\dfrac{11}{4}-\dfrac{9}{5}\right)\cdot x=1\)

\(\dfrac{19}{20}x=1\)

\(x=\dfrac{20}{19}\)

Vậy \(x=\dfrac{20}{19}\)

b) \(\left(x^2-9\right)\left(3-5x\right)=0\)

TH1:

\(x^2-9=0\)

\(x^2=9\)

\(x^2=3^2=\left(-3\right)^2\)

=>\(x\in\left\{3;-3\right\}\)

TH2:

\(3-5x=0\)

\(5x=3\)

\(x=\dfrac{3}{5}\)

Vậy \(x\in\left\{3;-3;\dfrac{3}{5}\right\}\)

bạn viết lại câu b được không?

a: ĐKXĐ: x>=2/3

\(\dfrac{x-2}{\sqrt{3x-2}+2}=9\)

=>\(x-2=9\sqrt{3x-2}+18\)

=>\(9\sqrt{3x-2}=x-2-18=x-20\)

=>\(\Leftrightarrow\left\{{}\begin{matrix}x>=20\\81\left(3x-2\right)=x^2-40x+400\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=20\\x^2-40x+400-243x+162=0\end{matrix}\right.\)

=>x>=20 và x^2-283x+562=0

=>x=281(nhận) hoặc x=2(loại)

b: ĐKXĐ: x>=2/5

\(\sqrt{5x-2}=9\)

=>5x-2=81

=>5x=83

=>x=83/5

c: ĐKXĐ: x>=-1; x<>8

\(\dfrac{2x-16}{\sqrt{x+1}-3}=5\)

=>\(2x-16=5\sqrt{x+1}-15\)

=>\(\sqrt{25x+25}=2x-16+15=2x-1\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{1}{2}\\4x^2-4x+1=25x+25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{1}{2}\\4x^2-29x-24=0\end{matrix}\right.\)

=>x=8(nhận) hoặc x=-3/4(loại)

`#040911`

`a)`

`3 1/3 x + 16 3/4 = -13,25`

`=> 3 1/3 x = -13,25 - 16 3/4`

`=> 3 1/3 x = -30`

`=> x = -30 \div 3 1/3`

`=> x =-9`

Vậy, `x = -9`

`b)`

`3 2/7*x - 1/8 = 2 3/4`

`=> 3 2/7x = 2 3/4 + 1/8`

`=> 3 2/7x = 23/8`

`=> x = 23/8 \div 3 2/7`

`=> x = 7/8`

Vậy, `x = 7/8`

`c)`

`x \div 4 1/3 = -2,5`

`=> x = -2,5 * 4 1/3`

`=> x = -65/6`

Vậy, `x = -65/6`

`d)`

`( (3x)/7 + 1) \div (-4) = (-1)/28`

`=> (3x)/7 +1 = (-1)/28 * (-4)`

`=> (3x)/7 + 1 = 1/7`

`=> (3x)/7 = 1/7 - 1`

`=> (3x)/7 = -6/7`

`=> 3x = -6`

`=> x = -6 \div 3`

`=> x = -2`

Vậy, `x = -2.`

a

=>10/3 . x + 16 + 3/4 = -13,25

=>10/3 x + 3/4 = -29,25

=>10/3 x = -30

=>x=-30 : 10/3

=>x=-30 . 3/10

=>x=-9

b.

=>23/7 x - 1/8 = = 11/4

=>23/7 x = 11/4 + 1/8

=>23/7 x= 22/8 + 1/8

=>23/7 x= 23/8

=>x=23/8 : 23/7

=>x=23/8 . 7/23

=>x=7/8

c.

=>x : 13/3 =-5/2

=>x=-5/2 . 13/3

=>x=-65/6

d.

=>3x/7 +1 = (-1/28) . (-4)

=>3x/7 + 1 = 1/7

=>3x/7 = -6/7

=>3x=-6

=>x=-2

a) \(3\dfrac{1}{3}x+16\dfrac{3}{4}=-13,25\)

\(\Rightarrow\dfrac{10}{3}x+\dfrac{67}{4}=-\dfrac{53}{4}\)

\(\Rightarrow\dfrac{10}{3}x=-30\)

\(\Rightarrow x=-30:\dfrac{10}{3}\)

\(\Rightarrow x=-9\)

b) \(3\dfrac{2}{7}x-\dfrac{1}{8}=2\dfrac{3}{4}\)

\(\Rightarrow\dfrac{23}{7}x-\dfrac{1}{8}=\dfrac{11}{4}\)

\(\Rightarrow\dfrac{23}{7}x=\dfrac{11}{4}+\dfrac{1}{8}\)

\(\Rightarrow\dfrac{23}{7}x=\dfrac{23}{8}\)

\(\Rightarrow x=\dfrac{23}{8}:\dfrac{23}{7}\)

\(\Rightarrow x=\dfrac{7}{8}\)

c) \(x:4\dfrac{1}{3}=-2,5\)

\(\Rightarrow x:\dfrac{13}{3}=-\dfrac{5}{2}\)

\(\Rightarrow x=-\dfrac{5}{2}\cdot\dfrac{13}{3}\)

\(\Rightarrow x=-\dfrac{65}{6}\)

d) \(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{28}\)

\(\Rightarrow\dfrac{3x}{7}+1=\dfrac{-1}{28}\cdot-4\)

\(\Rightarrow\dfrac{3x}{7}+1=\dfrac{1}{7}\)

\(\Rightarrow\dfrac{3x}{7}=-\dfrac{6}{7}\)

\(\Rightarrow x=-\dfrac{6}{7}:\dfrac{3}{7}\)

\(\Rightarrow x=-2\)

làm ơn giúp 🙏🙏🙏

a: =>1/3x+2/5x-2/5=0

=>11/15x-2/5=0

=>11/15x=2/5

=>x=2/5:11/15=2/5*15/11=30/55=6/11

b: =>-5x-1-1/2x+1/3=x

=>-11/2x-2/3-x=0

=>-13/2x=2/3

=>x=-2/3:13/2=-2/3*2/13=-4/39

c: (x+1/2)(2/3-2x)=0

=>x+1/2=0 hoặc 2/3-2x=0

=>x=1/3 hoặc x=-1/2

d: 9(3x+1)^2=16

=>(3x+1)^2=16/9

=>3x+1=4/3 hoặc 3x+1=-4/3

=>3x=1/3 hoặc 3x=-7/3

=>x=1/9 hoặc x=-7/9