X2x—7=15

2x=15+7

2x=22

x=22:2

x=11

4x +5=205

4x=205–5

4x=200

x=200:4

x=50

80/(x+1)=40

==> 80: x+1=40

80:x=40-1

80:x=39

x=80:39

x=80/39

72–(2x—4)=2

2x—4=72-2

2x—4=70

2x=70+4

2x=74

x=74:2

x=37

75-5(x+3)=5

5(x+3)=75–5

5(x+3)=70

x+3=70:5

x+3=14

x=14–3

x=11

7x:7=4

7x=4.7

7x=28

x=28:7

x=4

20:(x—5)=20

x—5=20:20

x—5=1

x=1+5

x=6

9(2x—8)=0

2x=0:9

2x=0

x=0:2

x=0

2080:4x=40

4x=2080:40

4x=52

x=52:4

x=13

2*X-7=15

2*X   =15+7

2*X   =22

X      =22/2

X      =11

mik chỉ giải một nửa thôi nha,mệt lắm rồi

4*x+5=205

4*x    = 205-5

4*x    = 200

x       = 200/4

x       = 50

 80/(x+1)=40

x+1       =80/40

x+1       =2

x           =2-1

x           =1

72-(2*x-4) =2

2*X-4       =72-2

2*X-4       =70

2*X          =70+4

2*X          =74

X             =74/2

X             =37

3: 

\(=\dfrac{1}{7}\cdot\dfrac{3}{5}\cdot\dfrac{5}{6}\cdot\dfrac{5}{8}=\dfrac{1}{7}\cdot\dfrac{1}{2}\cdot\dfrac{5}{8}=\dfrac{5}{112}\)

4:

=>2/3:x=-2-1/3=-7/3

=>x=-2/3:7/3=-2/7

5:

AC=CB=12/2=6cm

IB=6/2=3cm

(7.x + 1) : 2 - 23 = 2

=> (7.x + 1) : 2 = 2 + 23

=> (7.x + 1) : 2 = 25

=> 7.x + 1 = 25 . 2

=> 7.x + 1 = 50

=> 7.x = 50 - 1

=> 7.x = 49

=> x = 49 : 7 = 7

(7.x + 1) : 2 - 23 = 2

=> (7.x + 1) : 2 = 2 + 23

=> 7.x + 1 = 25 . 2

=> 7.x = 50 - 1

=> 7.x = 49

 x = 49 : 7 = 7

Vậy.............

hok tốt

80 - 3 ( x - 7 ) = 20

<=> 3 ( x - 7 ) = 60

<=> x - 7 = 20

<=> x = 27

4 ( x - 1 ) + 5 = 25

<=> 4 ( x - 1 ) = 20

<=> x - 1 = 5

<=> x = 6

\(\text{Giải:}\)

\(80-3\left(x-7\right)=20\)

\(3\left(x-7\right)=80-20\)

\(3\left(x-7\right)=60\)

\(x-7=60:3\)

\(x-7=20\)

\(x=20+7\)

\(x=27\)

\(\text{Vậy x = 27}\)

\(4\left(x-1\right)+5=25\)

\(4\left(x-1\right)=25-5\)

\(4\left(x-1\right)=20\)

\(x-1=20:4\)

\(x-1=5\)

\(x=5+1\)

\(x=6\)

\(\text{Vậy x = 6}\)

\(\text{Học tốt!!!}\)

1) \(80-3\left(x-7\right)=20\)

\(\Leftrightarrow80-3x+21=20\)

\(\Leftrightarrow3x=81\)

\(\Rightarrow x=27\)

2) \(4\left(x-1\right)+5=25\)

\(\Leftrightarrow4x-4+5=25\)

\(\Leftrightarrow4x=24\)

\(\Rightarrow x=6\)

1) \(5^{x+1}-5^x=20\Leftrightarrow5^x\left(5-1\right)=20\Leftrightarrow5^x=5\Leftrightarrow x=1\)

2) \(2^x+2^{x+4}=544\Leftrightarrow2^x\left(1+2^4\right)=544\Leftrightarrow2^x=32\Leftrightarrow x=5\)

3) \(4^{2x+1}+4^{2x}=80\Leftrightarrow4^{2x}\left(4+1\right)=80\Leftrightarrow16^x=16\Leftrightarrow x=1\)

4) \(3^{2x+2}+3^{2x+1}=108\Leftrightarrow3^{2x}\left(3^2+3\right)=108\Leftrightarrow9^x=9\Leftrightarrow x=1\)

5) \(7^{x+3}-7^{x+1}=16464\Leftrightarrow7^x\left(7^3-7\right)=16464\Leftrightarrow7^x=49\Leftrightarrow x=2\)

c:Ta có: \(5^{x+1}-5^x=20\)

\(\Leftrightarrow5^x\cdot5-5^x=20\)

\(\Leftrightarrow5^x\cdot4=20\)

\(\Leftrightarrow5^x=5\)

hay x=1

c: Ta có: \(2^x+2^{x+4}=544\)

\(\Leftrightarrow2^x+2^x\cdot16=544\)

\(\Leftrightarrow2^x\cdot17=544\)

\(\Leftrightarrow2^x=32\)

hay x=5

c: Ta có: \(4^{2x+1}+4^{2x}=80\)

\(\Leftrightarrow16^x\cdot4+16^x=80\)

\(\Leftrightarrow16^x\cdot5=80\)

\(\Leftrightarrow16^x=16\)

hay x=1

c: Ta có: \(3^{2x+2}+3^{2x+1}=108\)

\(\Leftrightarrow9^x\cdot9+9^x\cdot3=108\)

\(\Leftrightarrow9^x\cdot12=108\)

\(\Leftrightarrow9^x=9\)

hay x=1

c: Ta có: \(7^{x+3}-7^{x+1}=16464\)

\(\Leftrightarrow7^x\cdot343-7^x\cdot7=16464\)

\(\Leftrightarrow7^x\cdot336=16464\)

\(\Leftrightarrow7^x=49\)

hay x=2

\(a)ĐK:x\ge-1\\ \Leftrightarrow x+1=2\sqrt{x+1}\\ \Leftrightarrow x^2+2x+1=4x+4\\ \Leftrightarrow x^2+2x-4x+1-4=0\\ \Leftrightarrow x^2-2x-3=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)

Vậy \(S=\left\{3;-1\right\}\)

\(b)ĐK:x\ge2\\ \Leftrightarrow2x-4=\sqrt{x-2}\\ \Leftrightarrow4x^2-16x+16=x-2\\ \Leftrightarrow4x^2-16x-x+16+2=0\\ \Leftrightarrow4x^2-17x+18=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)

Vậy \(S=\left\{\dfrac{9}{4};2\right\}\)

\(c)ĐK:x\ge3\\ \Leftrightarrow2\sqrt{9\left(x-3\right)}-\dfrac{1}{5}\sqrt{25\left(x-3\right)}-\dfrac{1}{7}\sqrt{49\left(x-3\right)}=20\\ \Leftrightarrow2.3\sqrt{x-3}-\dfrac{1}{5}\cdot5\sqrt{x-3}-\dfrac{1}{7}\cdot7\sqrt{x-3}=20\\ \Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\\ \Leftrightarrow4\sqrt{x-3}=20\\ \Leftrightarrow\sqrt{x-3}=5\\ \Leftrightarrow x-3=25\\ \Leftrightarrow x=25+3\\ \Leftrightarrow x=28\left(tm\right)\)

Vậy \(S=\left\{28\right\}\)

Bài 1:

a; (\(\dfrac{1}{4}\)\(x\) - \(\dfrac{1}{8}\)) x \(\dfrac{3}{4}\) = \(\dfrac{1}{4}\)

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

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

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

      \(\dfrac{1}{4}x\) = \(\dfrac{1}{3}\) + \(\dfrac{1}{8}\)

       \(\dfrac{1}{4}\) \(x\)=  \(\dfrac{8}{24}\) + \(\dfrac{11}{24}\)

          \(\dfrac{1}{4}x=\dfrac{11}{24}\)

           \(x=\dfrac{11}{24}:\dfrac{1}{4}\)

           \(x=\dfrac{11}{24}\times4\)

           \(x=\dfrac{11}{6}\) 

b; \(\dfrac{12}{5}:x\) = \(\dfrac{14}{3}\) x \(\dfrac{4}{7}\)

     \(\dfrac{12}{5}\) : \(x\) = \(\dfrac{8}{3}\)

            \(x\) = \(\dfrac{12}{5}\) : \(\dfrac{8}{3}\)

            \(x\) = \(\dfrac{12}{5}\) x \(\dfrac{3}{8}\)

             \(x\) = \(\dfrac{9}{10}\)

c; \(x+\dfrac{2}{3}\) = 8:4 - 1

   \(x+\dfrac{2}{3}\) = 2 - 1

   \(x+\dfrac{2}{3}\) = 1

  \(x=1-\dfrac{2}{3}\)

  \(x\) = \(\dfrac{1}{3}\)

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