\(a,x-\dfrac{3}{4}=\dfrac{1}{7}\\ x=\dfrac{1}{7}+\dfrac{3}{4}\\ x=\dfrac{25}{28}\\ b,x+\dfrac{7}{5}=\dfrac{9}{8}.\dfrac{4}{27}\\ x+\dfrac{7}{5}=\dfrac{1}{6}\\ x=\dfrac{1}{6}-\dfrac{7}{5}\\ x=\dfrac{-37}{30}\\ c,\dfrac{2}{5}-\dfrac{3}{7}=\dfrac{x}{70}\\ \dfrac{-1}{35}=\dfrac{x}{70}\\ \dfrac{-2}{70}=\dfrac{x}{70}\\ x=-2\\ d,\dfrac{2}{9}-\dfrac{7}{8}.x=1\\ \dfrac{7}{8}.x=\dfrac{2}{9}-1\\ \dfrac{7}{8}.x=\dfrac{-7}{9}\\ x=\dfrac{-7}{9}:\dfrac{7}{8}\\ x=\dfrac{-8}{9}\)

\(a,x-\dfrac{3}{4}=\dfrac{1}{7}\)

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

\(\Rightarrow x=\dfrac{25}{28}\)

\(b,x+\dfrac{7}{5}=\dfrac{9}{8}.\dfrac{4}{27}\)

\(\Rightarrow x+\dfrac{7}{5}=\dfrac{1}{6}\)

\(\Rightarrow x=\dfrac{1}{6}-\dfrac{7}{5}\)

\(\Rightarrow x=-\dfrac{37}{30}\)

\(c,\dfrac{2}{5}-\dfrac{3}{7}=\dfrac{x}{70}\)

\(\Rightarrow\dfrac{-1}{35}=\dfrac{x}{70}\)

\(\Rightarrow35x=-70\)

\(\Rightarrow x=-2\)

\(d,\dfrac{2}{9}-\dfrac{7}{8}.x=1\)

\(\Rightarrow\dfrac{7}{8}x=\dfrac{2}{9}-1\)

\(\Rightarrow\dfrac{7}{8}x=-\dfrac{7}{9}\)

\(\Rightarrow x=-\dfrac{8}{9}\)

`a)x-3/4=1/7`

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

`=>x=4/28+21/28`

`=>x=25/28`

`b)x+7/5=9/8 . 4/27`

`=>x+7/5=(9.4)/(8.27)`

`=>x+7/5=1/6`

`=>x=1/6+7/5`

`=>x=5/30+42/30`

`=>x=47/30`

`c)2/5-3/7=x/70`

`=>14/35-15/35=x/70`

`=>-1/35=x/70`

`=>-2/70=x/70`

`=>x=-2`

`d)2/9-7/8x=1`

`=>7/8x=2/9-1`

`=>7/8x=-7/9`

`=>x=-7/9:7/8`

`=>x=-7/9 . 8/7`

`=>x=-8/9`

`#lv`

a)4/3:x=-4/7                         b)5/3.x=7/12

⇒x=-16/21                             ⇒x=7/20

a. \(\dfrac{5}{7}+\dfrac{4}{3}:x=\dfrac{1}{7}\)

<=> \(\dfrac{5}{7}+\dfrac{4}{3}.\dfrac{1}{x}=\dfrac{1}{7}\)

<=> \(\dfrac{5}{7}+\dfrac{4}{3x}=\dfrac{1}{7}\)         ĐKXĐ: x \(\ne\) 0

<=> \(\dfrac{15x}{21x}+\dfrac{28}{21x}=\dfrac{3x}{21x}\)

<=> 15x + 28 = 3x

<=> 15x - 3x = -28

<=> 12x = -28

<=> x = \(\dfrac{-28}{12}=-\dfrac{7}{3}\)

b. \(\dfrac{5}{3}x.\dfrac{-1}{4}=\dfrac{2}{6}\)

<=> \(\dfrac{-5x}{12}=\dfrac{2}{6}\)

<=> -5x . 6 = 12 . 2

<=> -30x = 24

<=> x = \(-\dfrac{4}{5}\)

b: =>x/23=1+3/4+4/7=65/28

=>x=23*65/28=1495/28

c: =>3/5:x=3/5-1/4-1/2=9/40

=>x=3/5:9/40=8/3

a) \(P=2\left|2x-5\right|+7\)

Ta có \(\left|2x-5\right|\ge0\) với \(\forall x\in R\). \(\left|2x-5\right|=0\Leftrightarrow x=\dfrac{5}{2}\)

\(\Rightarrow2\left|2x-5\right|\ge0\) với \(\forall x\in R\)

\(\Rightarrow2\left|2x-5\right|+7\ge7\) với \(\forall x\in R\)

\(\Rightarrow P\ge7\) với \(\forall x\in R\)

Vậy GTNN của P là 7 tại \(x=\dfrac{5}{2}\)

b) \(Q=-4+3\left|x-7\right|\)

Ta có: \(\left|x-7\right|\ge0\) với \(\forall x\in R\). \(\left|x-7\right|=0\Leftrightarrow x=7\)

\(\Rightarrow3\left|x-7\right|\ge0\) với \(\forall x\in R\)

\(\Rightarrow-4+3\left|x-7\right|\ge-4\) với \(\forall x\in R\)

\(\Rightarrow Q\ge-4\) với \(\forall x\in R\)

Vậy GTNN của Q là -4 tại x = 7

~~ Chúc bạn học tốt ~~

a) `|2x-5|>=0`

`-> 2|2x+5|>=0`

`->2|2x+5|+7>=7`

`->P>=7`

`=> P_(min)=7<=>x=-5/2`

b) `|x-7|>=0`

`3|x-7|>=0`

`-4+3|x-7|>=-4`

`=> Q_(min)=-4 <=>x=7`.

a)Ta có:\(P=2\left|2x-5\right|+7\)

Do \(\left|2x-5\right|\ge0\Rightarrow2\left|2x-5\right|\ge0\Rightarrow P\ge7\)

Dấu "=" xảy ra <=> 2x-5=0<=>x=5/2

b)Do \(\left|x-7\right|\ge0\Rightarrow3\left|x-7\right|\ge0\Rightarrow Q\ge-4\)

Dấu '=' xảy ra <=> x-7=0<=>x=7

1) \(\Leftrightarrow\left(x-4\right)\left(x+4\right)-x\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+4-x\right)=0\)

\(\Leftrightarrow\left(x-4\right)4=0\)

\(\Leftrightarrow x=4\)

2) \(\left(x+3\right)^2-\left(x-3\right)\left(x+5\right)=x^2+6x+9-x^2-2x+15=4x+24\)

3) \(2x^3+3x^2-2x+a=2x^2\left(x-2\right)+7x\left(x-2\right)+16\left(x-2\right)+32+a\)

Để \(2x^3+3x^2-2x+a⋮x-2\) thì \(32+a=0\Leftrightarrow a=-32\)

1. 

x² - 16 - x(x - 4) = 0

<=> (x² - 4²) - x(x - 4) = 0

<=> (x - 4)(x + 4) - x(x - 4) = 0

<=> (x + 4 - x)(x + 4) = 0

<=> 4(x + 4) = 0

<=> x + 4 = 0

<=> x = -4

2.

(x + 3)² - (x - 3)(x + 5)

= x² + 6x + 9 - (x² + 5x - 3x - 15)

= x² + 6x + 9 - x² + 5x - 3x - 15

= x² - x² + 6x + 5x - 3x + 9 - 15

= 8x - 6

⇒x=4

⇒⇒ 2x = -4 ⇒ x = - 2

Vậy  hoặc .

3. Ta có : 2x³ + 3x² - 2x + a = (x - 2)(2x² + 7x + 12) + (a - 24)
Để phép chia trên là phép chia hết thì a - 24 = 0 => a = 24

Còn bài 2 mình khum biéc làm 😢😢😢

a) x=3/7

b)x=8/7

c)x=6/7

a)\(x=\dfrac{2}{7}:\dfrac{2}{3}=\dfrac{3}{7}\)

b)\(x=\dfrac{13}{7}\times\dfrac{8}{13}=\dfrac{8}{7}\)

c)\(x=\dfrac{3}{2}:\dfrac{7}{4}=\dfrac{6}{7}\)

=3/7

=8/7

=6/7

\(a,\left(x-\dfrac{1}{2}\right):\dfrac{1}{3}+\dfrac{5}{7}=9\dfrac{5}{7}\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right).3=\dfrac{68}{7}-\dfrac{5}{7}\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right).3=9\)

\(\Leftrightarrow x-\dfrac{1}{3}=3\)

\(\Leftrightarrow x=3+\dfrac{1}{3}\)

\(\Leftrightarrow x=\dfrac{9}{3}+\dfrac{1}{3}\)

\(\Leftrightarrow x=\dfrac{10}{3}\)

\(b,x+30\%x=-1,31\)

\(\Leftrightarrow x+\dfrac{3}{10}.x=-\dfrac{131}{100}\)

\(\Leftrightarrow x.\left(1+\dfrac{3}{10}\right)=-\dfrac{131}{100}\)

\(\Leftrightarrow x.\dfrac{13}{10}=-\dfrac{131}{100}\)

\(\Leftrightarrow x=-\dfrac{131}{100}.\dfrac{10}{13}\)

\(\Leftrightarrow x=-\dfrac{131}{130}\)

\(c,-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{1}{10}\)

\(\Leftrightarrow\dfrac{-2}{3}x=\dfrac{1}{10}-\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{-2}{3}x=\dfrac{1}{10}-\dfrac{2}{10}\)

\(\Leftrightarrow-\dfrac{2}{3}x=-\dfrac{1}{10}\)

\(\Leftrightarrow x=-\dfrac{1}{10}.\left(-\dfrac{3}{2}\right)\)

\(\Leftrightarrow x=\dfrac{3}{20}\)

b: =>x-3=12

hay x=15

a. 4.(x+41) = 7

x + 41 = 7 : 4 = 1,75

x = 1,75 - 41 = -39,25

b. 4.(x-3) = 7² - 1¹⁰ = 49 - 1 = 48

x - 3 = 48 : 4 = 12

x = 12 + 3 = 15

a) 3/35 - (3/5 + x) = 2/7

=> 3/5 + x= 3/35- 2/7

=> 3/5 +x = -1/5

=> x = -1/5 -3/5

=> x = -4/5

b) 3/7 +1/7 : x = 3/14

=> 1/7 : x= 3/14 -3/7

=> 1/7 : x = -3/14

=> x = 1/7 : -3/14 

=> x = -2/3

c) (5x-1).(2x-1/3)=0

=> \(\left[{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}5x=0+1=1\\2x=0+\dfrac{1}{3}=\dfrac{1}{3}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{3}:2=\dfrac{1}{6}\end{matrix}\right.\)

Học tốt :D

a)x=-4/5

b)x=-2/3

c)\(\left\{{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x=1\\2x=\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{6}\end{matrix}\right.\)

Vậy.........

mik lười mong bn thông cảm

a) \(\dfrac{3}{35}-\left(\dfrac{3}{5}+x\right)=\dfrac{2}{7}\\ \Rightarrow\dfrac{3}{5}+x=-\dfrac{1}{5}\\ \Rightarrow x=-\dfrac{4}{5}\)

b) \(\dfrac{3}{7}+\dfrac{1}{7}:x=\dfrac{3}{14}\\ \Rightarrow\dfrac{1}{7}:x=-\dfrac{3}{14}\\ \Rightarrow x=-\dfrac{2}{3}\)

c) \(\left(5x-1\right)\left(2x-\dfrac{1}{3}\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}5x=1\\2x=\dfrac{1}{3}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{6}\end{matrix}\right.\)