Tìm x:
b) 1/3.x+2/5.(x-1)=0

\(<=> \dfrac{1}{3} .x +\dfrac{2}{5}x - \dfrac{2}{5} =0\)

\(<=> \dfrac{11}{15}x = \dfrac{2}{5}\)

\(<=> x= \dfrac{6}{11}\)

Vậy \( x= \dfrac{6}{11}\)
c) (2x-3).(6-2x)=0
\(<=> \begin{cases} 2x-3=0 \\ 6-2x=0 \end{cases}\) \(<=> \begin{cases} 2x=3 \\ -2x=-6 \end{cases}\) \(<=>\begin{cases} x=\dfrac{3}{2} \\ x=3 \end{cases}\)

Vậy \(x=( \dfrac{3}{2} ; 3)\)
d) -2/3-1/3.(2x-5)= 3/2

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

\(<=> 2x= \dfrac{15}{2}\)

\(<=> x= \dfrac{15}{4}\)

Vậy \(x= \dfrac{15}{4}\)
f) 1/3.x-1/2=4 và 1/2 (Hỗn số ý '^')

\(<=> \dfrac{1}{3} x -\dfrac{1}{2} = \dfrac{9}{2}\)

\(<=> \dfrac{1}{3}x =5\)

\(<=> x= 15\)

Vậy \(x= 15\)

Cảm ơn cậu nhiều nhé!

e) \(\left(x-3\right)\left(x^2+1\right)=0\)

\(\Rightarrow\left(x-3\right)=0\)  ( \(x^2+1>0\forall x\))

\(\Rightarrow x=3\)

đ) \(4.8^2=2^x\)

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

\(2^2.2^6=2^x\)

\(2^8=2^x\)

\(\Rightarrow x=8\)

d) \(\left|x+3\right|=8\)

\(\Rightarrow\orbr{\begin{cases}x+3=8\\x+3=-8\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-11\end{cases}}\)

mấy câu trên dễ rồi tự làm em nhé 

a, 2x + 5 = 7

=> 2x = 2

=> x = 1

vậy____

a) \(\left(x-2\right).\left(2x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\2x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\2x=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{1}{2}\end{cases}}\)

b) \(\left(3x+9\right).\left(1-3x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x+9=0\\1-3x=0\end{cases}}\Rightarrow\orbr{\begin{cases}3x=-9\\3x=1\end{cases}\Rightarrow}\orbr{\begin{cases}x=-3\\x=\frac{1}{3}\end{cases}}\)

c) (31 - 2x)³ =27

    (31 - 2x)³ = 3³

=> 31 - 2x = 3

            2x = 31 - 3 

           2x = 28

             x = 14

a. \(\left(x-2\right).\left(2x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x-2=0\\2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{1}{2}\end{cases}}}\)

Vậy \(x=2\)hoặc \(x=\frac{1}{2}\)

b.\(\left(3x+9\right).\left(1-3x\right)=0\Leftrightarrow\orbr{\begin{cases}3x+9=0\\1-3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=\frac{1}{3}\end{cases}}}\)

Vậy \(x=-3\)hoặc \(x=\frac{1}{3}\)

c.\(\left(31-2x\right)^3=-27\)

\(\Leftrightarrow\left(31-2x\right)^3=\left(-3\right)^3\)

\(\Leftrightarrow31-2x=-3\)

\(2x=34\)

\(x=17\)

d.\(\left(x-2\right).\left(7-x\right)=0\Leftrightarrow\orbr{\begin{cases}x-2=0\\7-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=7\end{cases}}}\)

Vậy \(x=2\)hoặc \(x=7\)

e.\(\left(x-5\right)^5=32\)

\(\Leftrightarrow\left(x-5\right)^5=2^5\)

\(\Leftrightarrow x-5=2\Leftrightarrow x=7\)

f.\(\left(2-x\right)^4=81\)

\(\Leftrightarrow\left(2-x\right)^4=3^4\)

\(2-x=3\Leftrightarrow x=-1\)

g.\(\left|x-7\right|< 3\Leftrightarrow-3< x-7< 3\Leftrightarrow4< x< 10\)

Câu b:

2x + |x| = 3x

|x| = 3x - 2x

|x| = x

|x| = x khi và chỉ khi x ≥ 0

Vậy x ≥ 0

Câu a chưa rõ đề bài, em nhé.

1) 2. I2x-3l = 1/2

         |2x-3| =1/2:2

          |2x-3| =1/4

        =>2x-3 =1/4      hoặc    2x-3 =-1/4

            2x      =1/4+3              2x     =-1/4+3

            2x      =13/4                2x     =11/4

               x      =13/4:2              x     =11/4:2

               x       =13/8                x     =11/8

vậy x=13/8 hoặc 11/8

tich dung cho minh nhe

(3\(\frac12\) + 2\(x\)).2\(\frac23\) = 5\(\frac13\)

(\(\frac72\) + 2\(x\)).\(\frac83\) = \(\frac{16}{3}\)

\(\frac72+2x\) = \(\frac{16}{3}\) : \(\frac83\)

\(\frac72\) + 2\(x\) = \(\frac{16}{3}\) x \(\frac38\)

\(\frac72\) + 2\(x\) = 2

2\(x\) = 2 - \(\frac72\)

2\(x\) = \(\frac42-\frac72\)

2\(x\) = - \(\frac32\)

\(x\) = - \(\frac32\) : 2

\(x\) = - \(\frac32\times\frac12\)

\(x\) = - \(\frac34\)

Vậy \(x=-\frac34\)

Câu a:

(x - 5)^2 - 2x - 2^4 = 2x

(x - 5)(x - 5) - 2x - 16 - 2x = 0

x^2 - 5x - 5x + 25 - 2x - 16 - 2x = 0

x^2 - (5x + 5x +2x + 2x) + (25 - 16) = 0

x^2 - 14x + 9 = 0

(x^2 - 7x) - (7x - 49) - 40 = 0

x(x - 7) - 7(x - 7) - 40 = 0

(x - 7)(x - 7) - 40 = 0

(x - 7)^2 = 40

x - 7 = \(\sqrt{40}\) hoặc x - 7 = -\(\sqrt{40}\)

x - 7 = - \(\sqrt{40}\)

x = 7 - \(\sqrt{40}\)

x - 7 = \(\sqrt{40}\)

x = 7+ \(\sqrt{40}\)

Vậy x ∈ {7 - \(\sqrt{40}\) ; 7+ \(\sqrt{40}\))

Câu b:

3^(x -1) - 7^2 = 2^5 + 0^3

3^(x -1) - 49 = 32 + 0

3^(x - 1) - 49 = 32

3^(x -1) = 32 + 49

3^(x -1) = 81

3^(x-1) = 3^4

x - 1 = 4

x = 4 + 1

x = 5

Vậy x = 5

\(\left(x+1\right)\left(x+7\right)< 0\)

thì \(x+1;x+7\)khác dấu

 th1\(\hept{\begin{cases}x+1< 0\\x+7>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< -1\\x>-7\end{cases}\Rightarrow}-7< x< -1\left(tm\right)}\)

th2\(\hept{\begin{cases}x+1>0\\x+7< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>-1\\x< -7\end{cases}\Rightarrow}-1< x< -7\left(vl\right)}\)

vậy với\(-7< x< -1\)thì \(\left(x+1\right)\left(x+7\right)< 0\)

a) (2x - 3) = 5

<=> 2x - 3 = 5

<=> 2x = 5 + 3

<=> 2x = 8

<=> x = 4

=> x = 4

b) (5x - 3) = 1/2

<=> 5x - 3 = 1/2

<=> 5x = 1/2 + 3

<=> 5x = 7/2

<=> x = 7/10

=> x = 7/10

c) (x + 1)(x + 7) < 0

<=> x = -1; -7

<=> x < -7 <=> x = -8 <=> (-8 + 1)(-8 + 7) < 0 <=> 7 < 0 (loại)

<=> -7 < x < -1 <=> x = -6 <=> (-6 + 1)(-6 + 7) < 0 <=> -5 < 0 (nhận)

<=> x > -1 <=> x = 0 <=> (x + 1)(x + 7) < 0 <=> 7 < 0 (loại)

Vậy: -7 < x < -1

a) \(x +(x + 1) + (x + 2) + ... + (x +30) = 620\)

\(=\left(x+x+...+x+x\right)+\left(1+2+...+30\right)\)

\(=31x+465=620\)

\(=31x=620-465\)

\(=31x=155\)

\(=x=155\div31\)

\(x=5\)

b) \(2+4+6+8+....+2x = 210\)

\(\Rightarrow2.1+2.2+2.3+2.4+...+2.x\)

\(\Rightarrow2.\left(2+4+6+8+...+x\right)=210\)

\(\Rightarrow2+4+6+8+x=210\div2\)

\(\Rightarrow2+4+6+8+...+x=105\)

\(\Rightarrow x=14\)

sao đến 2+4+6+8+...+x = 105 lại => x=14 được bạn

a)/6x-3/=15                                    

Th1:                                       th2

6x-3=15                                 6x-3=-15

6x    =15+3                            6x   =-15+3

6x    =18                               6x    =-12

x      =18:6                             x    =-12:6

x     =3                                   x    =-2

b)(x+7)(8-x)=0

=>x+7=0 hoặc 8-x=0

x=-7       hoặc     x=8-0=8