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) \(x-\dfrac{3}{4}=6\times\dfrac{3}{8}\)

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

=> \(x=\dfrac{9}{4}+\dfrac{3}{4}=3\)

b) \(\dfrac{7}{8}:x=3-\dfrac{1}{2}\)

\(\dfrac{7}{8}:x=\dfrac{5}{2}\)

=> \(x=\dfrac{7}{8}:\dfrac{5}{2}=\dfrac{7}{20}\)

c) \(x+\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{3}{4}\)

\(x+\dfrac{1}{6}=\dfrac{3}{4}\)

=> \(x=\dfrac{3}{4}-\dfrac{1}{6}=\dfrac{7}{12}\)

d) \(\dfrac{3}{2}\times\dfrac{4}{5}-x=\dfrac{2}{3}\)

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

=> \(x=\dfrac{6}{5}-\dfrac{2}{3}=\dfrac{8}{15}\)

e) \(x\times3\dfrac{1}{3}=3\dfrac{1}{3}:4\dfrac{1}{4}\)(?)

\(x\times\dfrac{10}{3}=\dfrac{40}{51}\)

=> \(x=\dfrac{40}{51}:\dfrac{10}{3}=\dfrac{4}{17}\)

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

\(\dfrac{17}{3}:x=\dfrac{5}{3}\)

=> \(x=\dfrac{17}{3}:\dfrac{5}{3}=\dfrac{17}{5}\)

a: =>x-3/4=18/8=9/4

=>x=9/4+3/4=12/4=3

b: =>7/8:x=5/2

=>x=7/8:5/2=7/8*2/5=14/40=7/20

c: x+1/2*1/3=3/4

=>x+1/6=3/4

=>x=3/4-1/6=9/12-2/12=7/12

d: =>12/10-x=2/3

=>6/5-x=2/3

=>x=6/5-2/3=18/15-10/15=8/15

e: =>x*10/3=10/3:17/4=10/3*4/17

=>x=4/17

f: =>17/3:x=13/3-5/2=26/6-15/6=11/6

=>x=17/3:11/6=17/3*6/11=34/11

`#040911`

a,

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

\(\Rightarrow\dfrac{1}{2}x-2-\dfrac{1}{4}x+\dfrac{1}{3}=2x-1\\\Rightarrow\left(\dfrac{1}{2}x-\dfrac{1}{4}x-2x\right)=2-\dfrac{1}{3}-1\\ \Rightarrow-\dfrac{7}{4}x=\dfrac{2}{3}\\ \Rightarrow x=\dfrac{2}{3}\div\left(-\dfrac{7}{4}\right)\\ \Rightarrow x=-\dfrac{8}{21}\)

Vậy, \(x=-\dfrac{8}{21}\)

b,

\(\dfrac{3}{4}-\left(x-\dfrac{1}{2}\right)^2=-\dfrac{11}{2}\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{3}{4}-\left(-\dfrac{11}{2}\right)\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{25}{4}\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\left(\pm\dfrac{5}{2}\right)^2\)

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

Vậy, \(x\in\left\{-2;3\right\}\)

c,

\(\dfrac{3}{16}+1\dfrac{1}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}\)

\(\Rightarrow\dfrac{17}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}-\dfrac{3}{16}\\ \Rightarrow\dfrac{17}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\\ \Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\div\dfrac{17}{16}\\ \Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{17}\)

Bạn xem lại đề có sai kh nhỉ?

c) \(\dfrac{3}{16}+\dfrac{1}{\dfrac{1}{16}}\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}\)

\(\Rightarrow16\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}-\dfrac{3}{16}\)

\(\Rightarrow16\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\)

\(\Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}:16\)

\(\Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{256}=\left(\dfrac{3}{16}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{2}{3}=\dfrac{3}{16}\\x-\dfrac{2}{3}=-\dfrac{3}{16}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{16}+\dfrac{2}{3}\\x=-\dfrac{3}{16}+\dfrac{2}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{41}{48}\\x=\dfrac{23}{48}\end{matrix}\right.\)

a) 5-(x-6)=4(3-2x)

<=>5-x+6-12+8x=0

<=>7x-1=0

=>x=1/7

a: =>x-2/5=3/4:1/3=3/4*3=9/4

=>x=9/4+2/5=45/20+8/20=53/20

b: =>x-2/3=7/3:4/5=7/3*5/4=35/12

=>x=35/12+2/3=43/12

c: 1/3(x-2/5)=4/5

=>x-2/5=4/5*3=12/5

=>x=12/5+2/5=14/5

d: =>2/3x-1/3-1/4x+1/10=7/3

=>5/12x-7/30=7/3

=>5/12x=7/3+7/30=77/30

=>x=77/30:5/12=154/25

e: \(\Leftrightarrow x\cdot\dfrac{3}{7}-\dfrac{2}{7}+\dfrac{1}{2}-\dfrac{5}{4}x+\dfrac{5}{2}=0\)

=>\(x\cdot\dfrac{-23}{28}=\dfrac{2}{7}-3=\dfrac{-19}{7}\)

=>x=19/7:23/28=76/23

f: =>1/2x-3/2+1/3x-4/3+1/4x-5/4=1/5

=>13/12x=1/5+3/2+4/3+5/4=257/60

=>x=257/65

i: =>x^2-2/5x-x^2-2x+11/4=4/3

=>-12/5x=4/3-11/4=-17/12

=>x=17/12:12/5=85/144

`x + 2 3/4 = 5 2/3`

`=> x + 11/4 = 17/3`

`=> x= 17/3 -11/4`

`=>x=35/12`

__

`x - 1 4/5 = 3 2/7`

`=> x- 9/5 = 23/7`

`=> x= 23/7 +9/5`

`=>x=178/35`

__

`x xx 3 1/2 =4 3/4`

`=> x xx 7/2 =19/4`

`=> x= 19/4 : 7/2`

`=> x= 19/4 xx 2/7`

`=> x= 19/14`

__

`x : 2 2/3 = 4 1/3`

`=> x : 8/3 = 13/3`

`=> x= 13/3 xx 8/3`

`=>x=104/9`

a: =>x+11/4=17/3

=>x=17/3-11/4=68/12-33/12=35/12

b: =>x-9/5=23/7

=>x=23/7+9/5=178/35

c: =>x*7/2=4,75

=>x=19/4:7/2=19/14

d: =>x:8/3=13/3

=>x=13/3*8/3=104/9

a) \(x+2\dfrac{3}{4}=5\dfrac{2}{3}\) 

    \(x+\dfrac{11}{4}=\dfrac{17}{3}\) 

              \(x=\dfrac{17}{3}-\dfrac{11}{4}\) 

               \(x=\dfrac{35}{12}\) 

b) \(x-1\dfrac{4}{5}=3\dfrac{2}{7}\) 

      \(x-\dfrac{9}{5}=\dfrac{23}{7}\) 

              \(x=\dfrac{23}{7}+\dfrac{9}{5}\) 

              \(x=\dfrac{178}{35}\) 

c) \(x\times3\dfrac{1}{2}=4\dfrac{3}{4}\) 

    \(x\times\dfrac{7}{2}=\dfrac{19}{4}\) 

            \(x=\dfrac{19}{4}\div\dfrac{7}{2}\) 

            \(x=\dfrac{19}{14}\) 

d) \(x\div2\dfrac{2}{3}=4\dfrac{1}{3}\) 

    \(x\div\dfrac{8}{3}=\dfrac{13}{3}\) 

            \(x=\dfrac{13}{3}\times\dfrac{8}{3}\) 

            \(x=\dfrac{104}{9}\)

a) \(...=x+\dfrac{11}{3}=\dfrac{17}{3}\Rightarrow x=\dfrac{17}{3}-\dfrac{11}{3}=\dfrac{6}{3}=2\)

b) \(...\Rightarrow x-\dfrac{9}{5}=\dfrac{23}{7}\Rightarrow x=\dfrac{23}{7}+\dfrac{9}{5}=\dfrac{115}{35}+\dfrac{36}{35}=\dfrac{151}{35}\)

c) \(...\Rightarrow x.\dfrac{7}{2}=\dfrac{19}{4}\Rightarrow x=\dfrac{19}{4}:\dfrac{7}{2}\Rightarrow x=\dfrac{19}{4}.\dfrac{2}{7}=\dfrac{19}{14}\)

d) \(...\Rightarrow x:\dfrac{8}{3}=\dfrac{13}{3}\Rightarrow x=\dfrac{13}{3}.\dfrac{8}{3}=\dfrac{124}{9}\)

a) 35/12

b) 178/35

c) 19/14

d) 104/9

a) Ta có: \(A=\left(x+3\right)^2+\left(x-3\right)\left(x+3\right)-2\left(x+2\right)\left(x-4\right)\)

\(=x^2+6x+9+x^2-9-2\left(x^2-2x-8\right)\)

\(=2x^2+6x-2x^2+4x+16\)

\(=10x+16\)

Thay \(x=-\frac{1}{2}\) vào biểu thức \(A=10x+16\), ta được:

\(A=10\cdot\frac{-1}{2}+16=-5+16=11\)

Vậy: 11 là giá trị của biểu thức \(A=\left(x+3\right)^2+\left(x-3\right)\left(x+3\right)-2\left(x+2\right)\left(x-4\right)\) tại \(x=-\frac{1}{2}\)

b) Ta có: \(B=\left(3x+4\right)^2-\left(x-4\right)\left(x+4\right)-10x\)

\(=9x^2+24x+16-\left(x^2-16\right)-10x\)

\(=9x^2+14x+16-x^2+16\)

\(=8x^2+14x+32\)

Thay \(x=-\frac{1}{10}\) vào biểu thức \(B=8x^2+14x+32\), ta được:

\(B=8\cdot\left(-\frac{1}{10}\right)^2+14\cdot\frac{-1}{10}+32\)

\(=8\cdot\frac{1}{100}-\frac{14}{10}+32\)

\(=\frac{2}{25}-\frac{14}{10}+32\)

\(=\frac{4}{50}-\frac{70}{50}+\frac{1600}{50}\)

\(=\frac{1534}{50}\)

Vậy: \(\frac{1534}{50}\) là giá trị của biểu thức \(B=\left(3x+4\right)^2-\left(x-4\right)\left(x+4\right)-10x\) tại \(x=-\frac{1}{10}\)

c) Ta có: \(C=\left(x+1\right)^2-\left(2x-1\right)^2+3\left(x-2\right)\left(x+2\right)\)

\(=x^2+2x+1-\left(4x^2-4x+1\right)+3\left(x^2-4\right)\)

\(=x^2+2x+1-4x^2+4x-1+3x^2-12\)

\(=6x-12\)

Thay x=1 vào biểu thức C=6x-12, ta được:

\(C=6\cdot1-12=6-12=-6\)

Vậy: -6 là giá trị của biểu thức \(C=\left(x+1\right)^2-\left(2x-1\right)^2+3\left(x-2\right)\left(x+2\right)\) tại x=1

d) Ta có: \(D=\left(x-3\right)\left(x+3\right)+\left(x-2\right)^2-2x\left(x-4\right)\)

\(=x^2-9+x^2-4x+4-2x^2+8x\)

\(=4x-5\)

Thay x=-1 vào biểu thức D=4x-5, ta được:

\(D=4\cdot\left(-1\right)-5=-4-5=-9\)

Vậy: -9 là giá trị của biểu thức \(D=\left(x-3\right)\left(x+3\right)+\left(x-2\right)^2-2x\left(x-4\right)\) tại x=-1

`@` `\text {Ans}`

`\downarrow`

`a,`

\(\dfrac{3}{2}\times\dfrac{4}{5}-x=\dfrac{2}{3}\)

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

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

\(x=\dfrac{8}{15}\)

Vậy, `x = \dfrac{8}{15}`

`b,`

\(x\times3\dfrac{1}{3}=3\dfrac{1}{3}\div4\dfrac{1}{4}\)

\(x\times3\dfrac{1}{3}=\dfrac{40}{51}\)

\(x=\dfrac{40}{51}\div3\dfrac{1}{3}\)

\(x=\dfrac{4}{17}\)

Vậy, `x=`\(\dfrac{4}{17}\)

`c,`

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

\(\dfrac{17}{3}\div x=\dfrac{7}{6}\)

\(x=\dfrac{17}{3}\div\dfrac{7}{6}\)

\(x=\dfrac{34}{7}\)

Vậy, `x=`\(\dfrac{34}{7}\)

a,\(\dfrac{3}{2}.\dfrac{4}{5}-x=\dfrac{2}{3}\)

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

\(\dfrac{4}{5}-x=\dfrac{4}{9}\)

\(x=\dfrac{4}{5}-\dfrac{4}{9}\)

\(x=\dfrac{16}{45}\)

b,\(x.3.\dfrac{1}{3}=\dfrac{31}{3}\div\dfrac{4}{14}\)

\(x.3.\dfrac{1}{3}=\dfrac{217}{6}\)

\(x.3=\dfrac{217}{6}\div\dfrac{1}{3}\)

\(x\div3=\dfrac{217}{2}\)

\(x=\dfrac{217}{2}\div3\)

\(x=\dfrac{217}{6}\)