a: \(x+\dfrac{3}{9}=\dfrac{7}{6}\cdot\dfrac{2}{3}\)

=>\(x+\dfrac{1}{3}=\dfrac{14}{18}=\dfrac{7}{9}\)

=>\(x=\dfrac{7}{9}-\dfrac{1}{3}=\dfrac{7}{9}-\dfrac{3}{9}=\dfrac{4}{9}\)

b: \(x-\dfrac{2}{3}=\dfrac{1}{8}:\dfrac{5}{4}\)

=>\(x-\dfrac{2}{3}=\dfrac{1}{8}\cdot\dfrac{4}{5}=\dfrac{1}{10}\)

=>\(x=\dfrac{1}{10}+\dfrac{2}{3}=\dfrac{3+20}{30}=\dfrac{23}{30}\)

TThế giới oi oi oi 

\(\dfrac{1}{2}-\dfrac{5}{12}x=\dfrac{2}{3}\)

\(\dfrac{5}{12}x=\dfrac{1}{2}-\dfrac{2}{3}=\dfrac{3}{6}-\dfrac{4}{6}\)

\(\dfrac{5}{12}x=\dfrac{-1}{6}\)

\(x=\dfrac{-1}{6}:\dfrac{5}{12}=\dfrac{-1}{6}.\dfrac{12}{5}\)

\(x=\dfrac{-2}{5}\)

Giúp mik ik 

\(\dfrac{1}{2}-\dfrac{5}{12}x=\dfrac{2}{3}\)

\(\dfrac{5}{12}x=\dfrac{-1}{6}\)

\(x=\dfrac{-2}{5}\)

ta có \(\left(x+2\right)^2-2\left(x+2\right)\left(x+3\right)+\left(x+5\right)^2=7\)

\(\Leftrightarrow x^2+4x+4-2\left(x^2+5x+6\right)+x^2+10x+25=7\)

\(\Leftrightarrow4x+10=0\Leftrightarrow x=-\frac{5}{2}\)

Bạn áp dụng hằng đẳng thức số 1, nhân phá ngoặc là Ok nhé

\(\left(x+2\right)^2-2\left(x+2\right)\left(x+3\right)+\left(x+5\right)^2=7\)

\(\Leftrightarrow x^2+4x+4-2\left(x^2+3x+2x+6\right)+x^2+10x+25-7=0\)

\(\Leftrightarrow2x^2+14x+22-2x^2-6x-4x-12=0\)

\(\Leftrightarrow4x+10=0\)

\(\Leftrightarrow4x=-10\)

\(\Leftrightarrow x=\frac{-5}{2}\)

Trả lời:

( x + 2 )² - 2 ( x + 2 ) ( x + 3 ) + ( x + 5 )² = 7

<=> x² + 4x + 4 - 2 ( x² + 5x + 6 ) + x² + 10x + 25 = 7

<=> x² + 4x + 4 - 2x² - 10x - 12 + x² + 10x + 25 =  7

<=> 4x + 17 = 7

<=> 4x = - 10

<=> x = - 5/2

Vậy x = - 5/2 là nghiệm của pt.

a)  \(\left(\frac{2x}{5}-1\right):\left(-5\right)=\frac{1}{7}\)

\(\frac{2x}{5}-1=\frac{1}{7}.\left(-5\right)\)

\(\frac{2x}{5}-1=\frac{-5}{7}\)

\(\frac{2x}{5}=\frac{-5}{7}+\frac{7}{7}\)

\(\frac{2x}{5}=\frac{2}{7}\)

\(=>2x.7=2.5\)

\(=>14x=10\)

\(=>x=\frac{5}{7}\)

c) \(\left|3,5+2,5x\right|-2,5=3,5\)

\(\left|3,5+2,5x\right|=3,5+2,5\)

\(\left|3,5+2,5x\right|=6\)

\(TH1\)  \(3,5+2,5x=6\)                                        \(TH2\)   \(3,5+2,5x=-6\)

            \(2,5x=6-3,5\)                                                            \(2,5x=-6-3,5\)

             \(2,5x=2,5\)                                                                    \(2,5x=-9.5\)

                    \(x=1\)                                                                    \(x=-3,8\)

     vậy \(x=1\) hoặc  \(x=-3,8\)

câu d) làm tương tự như câu c) 

a/\(x-\dfrac{5}{7}-\dfrac{13}{14}=1\)
\(x=1+\dfrac{5}{7}+\dfrac{13}{14}\)
\(x=\dfrac{14}{14}+\dfrac{10}{14}+\dfrac{13}{14}\)
\(x=\dfrac{37}{14}\)
Vậy \(x=\dfrac{37}{14}\)
b/\(\dfrac{3}{5}+x+1\dfrac{1}{5}=\dfrac{11}{3}\)
\(x+\dfrac{3}{5}+\dfrac{6}{5}=\dfrac{11}{3}\)
\(x+\dfrac{9}{5}=\dfrac{11}{3}\)
\(x=\dfrac{11}{3}-\dfrac{9}{5}\)
\(x=\dfrac{55}{15}-\dfrac{27}{15}\)
\(x=\dfrac{28}{15}\)
Vậy \(x=\dfrac{28}{15}\)
#kễnh

a) \(x-\dfrac{5}{7}-\dfrac{13}{14}=1\)

\(x-\dfrac{23}{14}=1\)

\(x=1+\dfrac{23}{14}\)

\(x=\dfrac{37}{14}\)

b) \(\dfrac{3}{5}+x+1\dfrac{1}{5}=\dfrac{11}{3}\)

\(x+1+\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{11}{3}\)

\(x+\dfrac{9}{5}=\dfrac{11}{3}\)

\(x=\dfrac{11}{3}-\dfrac{9}{5}\)

\(x=\dfrac{28}{15}\)

`a)|2x+1|=5`

`<=>` \(\left[ \begin{array}{l}2x+1=5\\2x+1=-5\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}2x=4\\2x=-6\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=2\\x=-3\end{array} \right.\) 

`b)|2x+1|=0`

`<=>2x+1=0`

`<=>2x=-1`

`<=>x=-1/2`

`c)|2x+1|=7`

`<=>` \(\left[ \begin{array}{l}2x+1=7\\2x+1=-7\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}2x=6\\2x=-8\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=4\\x=-4\end{array} \right.\) 

`d)|2x+5|=|3x-7|`

`<=>` \(\left[ \begin{array}{l}2x+5=3x-7\\2x+5=7-3x\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=12\\5x=2\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=12\\x=\dfrac25\end{array} \right.\) 

`e)|2x+7|=1`

`<=>` \(\left[ \begin{array}{l}2x+7=1\\2x+7=-1\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}2x=-6\\2x=-8\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=3\\x=-4\end{array} \right.\) 

`g)|x-2|+|2x-3|=2`

Nếu `x>=2=>|x-2|=x-2,|2x-3|=2x-3`

`pt<=>x-2+2x-3=2`

`<=>3x-5=2`

`<=>3x=7`

`<=>x=7/3(tm)`

Nếu `x<=3/2=>|x-2|=2-x,|2x-3|=3-2x`

`pt<=>2-x+3-2x=2`

`<=>5-3x=2`

`<=>3x=3`

`<=>x=1(tm)`

Nếu `3/2<=x<=2=>|x-2|=2-x,|2x-3|=2x-3`

`pt<=>2-x+2x-3=2`

`<=>x-1=2`

`<=>x=3(l)`

`h)|x+2|+|1-x|=3x+2`

Vì `VT>=0=>3x+2>=0=>x>=-2/3`

`=>|x+2|=x+2`

`pt<=>x+2+|1-x|=3x+2`

`<=>|1-x|=2x(x>=0)`

`<=>` \(\left[ \begin{array}{l}2x=1-x\\2x=x-1\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}3x=1\\x=-1\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=\dfrac13(TM)\\x=-1(KTM)\end{array} \right.\) 

a.

$|2x+1|=5$
\(\Leftrightarrow \left[\begin{matrix} 2x+1=5\\ 2x+1=-5\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=2\\ x=-3\end{matrix}\right.\)

b.

$|2x+1|=0$

$\Leftrightarrow 2x+1=0$

$\Leftrightarrow x=-\frac{1}{2}$
c.

$|2x+1|=7$

\(\Leftrightarrow \left[\begin{matrix} 2x+1=7\\ 2x+1=-7\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=3\\ x=-4\end{matrix}\right.\)

d.

$|2x+5|=|3x-7|$

\(\Leftrightarrow \left[\begin{matrix} 2x+5=3x-7\\ 2x+5=7-3x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=12\\ x=0,4\end{matrix}\right.\)

e.

$|2x+7|=x-1\Rightarrow x-1\geq 0\Leftrightarrow x\geq 1$
Với $x\geq 1$ thì $|2x+7|=2x+7$

Khi đó pt trở thành:
$2x+7=x-1$

$\Leftrightarrow x=-8< 1$ (vô lý)

Vậy pt vô nghiệm.

g.

$|x-2|+|2x-3|=2$

Nếu $x\geq 2$ thì pt trở thành:

$x-2+2x-3=2$

$\Leftrightarrow 3x-5=2$

$\Leftrightarrow x=\frac{7}{3}$ (thỏa mãn)

Nếu $\frac{3}{2}\leq x< 2$ thì pt trở thành:

$2-x+2x-3=2$

$\Leftrightarrow x=3$ (không thỏa mãn)

Nếu $x< \frac{3}{2}$ thì pt trở thành:

$2-x+3-2x=2$

$\Leftrightarrow 5-3x=2$

$\Leftrightarrow x=1$ (thỏa mãn)

Vậy..........

h.

Từ đề suy ra $x\geq \frac{-2}{3}$

$\Rightarrow |x+2|=x+2$

Nếu  $x\geq 1$ thì $|1-x|=x-1$. PT trở thành:

$x+2+x-1=3x+2$

$\Leftrightarrow 2x+1=3x+2$

$\Leftrightarrow x=-1$ (vô lý)

Nếu $\frac{-2}{3}\leq x< 1$ thì $|1-x|=1-x$. PT trở thành:
$x+2+1-x=3x+2$

$\Leftrightarrow 3=3x+2$

$\Leftrightarrow x=\frac{1}{3}$ (thỏa mãn)

a, 2x-1 thuộc ước của 2,rồi giải ra  

b,c tương tự

d\(\frac{x^2-64-123}{x+8}=\frac{\left(x+8\right)\left(x-8\right)-123}{x+8}=x-8-\frac{123}{X+8}\) .........rồi làm tương tự như câu a,,,,,,,,,,,,còn câu e cũng gần giống câu d

mik cảm ơn nhiều nhé mik cx vừa lam ra ạ