\(3\dfrac{1}{2}+4\dfrac{5}{7}-5\dfrac{5}{14}\)

= \(\dfrac{7}{2}+\dfrac{33}{7}-\dfrac{75}{14}\)

= \(\dfrac{49}{14}+\dfrac{66}{14}-\dfrac{75}{14}\)

= \(\dfrac{40}{14}=\dfrac{20}{7}\)

\(4\dfrac{1}{2}+\dfrac{1}{2}\div5\dfrac{1}{2}\)

=\(\dfrac{9}{2}+\dfrac{1}{2}\div\dfrac{11}{2}\)

=\(\dfrac{9}{2}+\dfrac{1}{2}\times\dfrac{2}{11}\)

=\(\dfrac{9}{2}+\dfrac{1}{11}\)

=\(\dfrac{101}{22}\)

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

\(x\times\dfrac{10}{3}=\dfrac{10}{3}\div\dfrac{17}{4}\)

\(x\times\dfrac{10}{3}=\dfrac{10}{3}\times\dfrac{4}{17}\)

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

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

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

\(x=\dfrac{120}{510}=\dfrac{12}{51}=\dfrac{4}{7}\)

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

\(\dfrac{17}{3}\div x=\dfrac{11}{3}-\dfrac{5}{2}\)

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

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

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

\(x=\dfrac{102}{21}=\dfrac{34}{7}\)

Bài 4: 

a) \(\dfrac{4}{3}+\left(1,25-x\right)=2,25\)

\(1,25-x=2,25-\dfrac{4}{3}=\dfrac{9}{4}-\dfrac{4}{3}\)

\(1,25-x=\dfrac{11}{12}\)

\(x=1,25-\dfrac{11}{12}=\dfrac{5}{4}-\dfrac{11}{12}\)

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

b) \(\dfrac{17}{6}-\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)

\(x-\dfrac{7}{6}=\dfrac{17}{6}-\dfrac{7}{4}=\dfrac{34}{12}-\dfrac{21}{12}\)

\(x-\dfrac{7}{6}=\dfrac{13}{12}\)

\(x=\dfrac{13}{12}+\dfrac{7}{6}=\dfrac{13}{12}+\dfrac{14}{12}\)

\(x=\dfrac{27}{12}=\dfrac{9}{4}\)

c) \(4-\left(2x+1\right)=3-\dfrac{1}{3}=\dfrac{9}{3}-\dfrac{1}{3}\)

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

\(2x+1=\dfrac{8}{3}+4=\dfrac{8}{3}+\dfrac{12}{3}\)

\(2x+1=\dfrac{20}{3}\)

\(2x=\dfrac{20}{3}-1=\dfrac{20}{3}-\dfrac{3}{3}\)

\(2x=\dfrac{17}{3}\)

\(x=\dfrac{17}{3}.\dfrac{1}{2}=\dfrac{17}{6}\)

Bài 15:

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

\(x=\left(\dfrac{-2}{3}\right)^9:\dfrac{-2}{3}=\left(\dfrac{-2}{3}\right)^{9-1}\)

\(=>x=\left(\dfrac{-2}{3}\right)^8\)

b) \(x:\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^4\)

\(x=\left(\dfrac{4}{9}\right)^4.\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^{4+5}\)

\(=>x=\left(\dfrac{4}{9}\right)^9\)

c) \(\left(x+4\right)^3=-125\)

\(\left(x+4\right)^3=\left(-5\right)^3\)

\(=>x+4=-5\)

\(x=-5-4\)

\(=>x=-9\)

d) \(\left(10-5x\right)^3=64\)

\(\left(10-5x\right)^3=4^3\)

\(=>10-5x=4\)

\(5x=10-4\)

\(5x=6\)

\(=>x=\dfrac{6}{5}\)

e) \(\left(4x+5\right)^2=81\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(4x+5\right)^2=\left(-9\right)^2\\\left(4x+5\right)^2=9^2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+5=-9\\4x+5=9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=-14\\4x=4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-14}{4}\\x=1\end{matrix}\right.\)

Bài 16:

a) \(4-1\dfrac{2}{5}-\dfrac{8}{3}\)

\(=4-\dfrac{7}{5}-\dfrac{8}{3}\)

\(=\dfrac{60-21-40}{15}=\dfrac{-1}{15}\)

b) \(-0,6-\dfrac{-4}{9}-\dfrac{16}{15}\)

\(=\dfrac{-3}{5}+\dfrac{4}{9}-\dfrac{16}{15}\)

\(=\dfrac{\left(-27\right)+20-48}{45}=\dfrac{-55}{45}=\dfrac{-11}{9}\)

c) \(-\dfrac{15}{4}.\left(\dfrac{-7}{15}\right).\left(-2\dfrac{2}{5}\right)\)

\(=\dfrac{7}{4}.\dfrac{-12}{5}\)

\(=\dfrac{-21}{5}\)

\(#Wendy.Dang\)

Uh, chừa sau k dám học muộn nx

Đăng từng bài `1` thôi cậu ơii

a: =>x-3=9

=>x=12

b: =>10-x=-26

=>x=36

c: =>x:4-1=2

=>x:4=3

=>x=12

d: =>x^2=4

=>x=2 hoặc x=-2

e: =>(x-2)^2=100

=>x-2=10 hoặc x-2=-10

=>x=12 hoặc x=-8

Bài 1:

a: \(\sqrt{27}+\dfrac{1}{2}\sqrt{48}-\sqrt{108}\)

\(=3\sqrt{3}+\dfrac{1}{2}\cdot4\sqrt{3}-6\sqrt{3}\)

\(=-3\sqrt{3}+2\sqrt{3}=-\sqrt{3}\)

b: \(\left(\sqrt{14}-\sqrt{10}\right)\cdot\sqrt{6+\sqrt{35}}\)

\(=\left(\sqrt{7}-\sqrt{5}\right)\cdot\sqrt{2}\cdot\sqrt{6+\sqrt{35}}\)

\(=\left(\sqrt{7}-\sqrt{5}\right)\cdot\sqrt{12+2\sqrt{35}}\)

\(=\left(\sqrt{7}-\sqrt{5}\right)\cdot\sqrt{\left(\sqrt{7}+\sqrt{5}\right)^2}\)

\(=\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)=7-5=2\)

c: \(\dfrac{\sqrt{15}+\sqrt{3}}{1+\sqrt{5}}-\dfrac{2}{\sqrt{3}-1}\)

\(=\dfrac{\sqrt{3}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}-\dfrac{2\left(\sqrt{3}+1\right)}{3-1}\)

\(=\sqrt{3}-\sqrt{3}-1=-1\)

Bài 2:

a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >1\end{matrix}\right.\)

\(A=\dfrac{x-5}{x+2\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}+\dfrac{2}{\sqrt{x}-1}\)

\(=\dfrac{x-5}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}+\dfrac{1}{\sqrt{x}+3}+\dfrac{2}{\sqrt{x}-1}\)

\(=\dfrac{x-5+\sqrt{x}-1+2\sqrt{x}+6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)

b: A=2

=>\(\sqrt{x}=2\left(\sqrt{x}-1\right)\)

=>\(2\sqrt{x}-2=\sqrt{x}\)

=>\(\sqrt{x}=2\)

=>x=4(nhận)

c: Để A là số nguyên thì \(\sqrt{x}⋮\sqrt{x}-1\)

=>\(\sqrt{x}-1+1⋮\sqrt{x}-1\)

=>\(\sqrt{x}-1\inƯ\left(1\right)\)

=>\(\sqrt{x}-1\in\left\{1;-1\right\}\)

=>\(\sqrt{x}\in\left\{2;0\right\}\)

=>\(x\in\left\{4;0\right\}\)

`( 1 + 1/12 ) xx ( 1 + 1/13 ) xx ( 1 + 1/14 ) xx ( 1 + 1/15 )`

`= 13/12 xx14/13 xx 15/14 xx 16/15`

`= ( 13 xx 14 xx 15 xx 16 )/( 12xx13xx14xx15 )`

`= 16/12`

`=4/3` 

`x : 4/7 xx 2/3 = 2`

` x  : 4/7 = 2 : 2/3`

` x : 4/7 = 3`

`x = 3 xx 4/7`

`x=12/7` 

Đổi `: 17` tấn `5` tạ `= 175` tạ 

Ô tô thứ nhất chở là `:`

`175 : ( 3 + 4 ) xx 3 = 75` `(` tạ `)`

Ô tô thứ hai chở là `:`

`175-75=100` `(` tạ `)`

Đ/s : Ô tô thứ nhất `: 75` tạ 

Ô tô thứ hai `: 100` tạ 

a)4/5+x=2/3

x=2/3-4/5

x=-2/15

b)-5/6-x=2/3

x=-5/6-2/3

x=-3/2

c)1/2x+3/4=-3/10

1/2x=-3/10-3/4

1/2x=-21/20

x=-21/20:1/2

x=-21/10

d)x/3-1/2=1/5

x/3=1/5+1/2

x/3=7/10

10x/30=21/30

10x=21

x=21:10

x=21/10

a) Ta có \(-4\dfrac{3}{5}\cdot2\dfrac{4}{3}=-\dfrac{23}{5}\cdot\dfrac{10}{3}=-\dfrac{46}{3}\) và \(-2\dfrac{3}{5}\div1\dfrac{6}{15}=-\dfrac{13}{5}\div\dfrac{7}{5}=-\dfrac{13}{7}\)

Do đó \(-\dfrac{46}{3}< x< -\dfrac{13}{7}\)

Lại có \(-\dfrac{46}{3}\le-15\) và \(-\dfrac{13}{7}\ge-2\)

Suy ra \(-15\le x\le-2\), x ϵ Z

b) Ta có \(-4\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)=-\dfrac{13}{3}\cdot\dfrac{1}{3}=-\dfrac{13}{9}\) và \(-\dfrac{2}{3}\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)=-\dfrac{2}{3}\cdot\dfrac{-11}{12}=\dfrac{11}{18}\)

Do đó \(-\dfrac{13}{9}< x< \dfrac{11}{18}\)

Lại có \(-\dfrac{13}{9}\le-1\) và \(\dfrac{11}{18}\ge0\)

Suy ra \(-1\le x\le0\), x ϵ Z

b, -4\(\dfrac{1}{3}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{6}\)) < \(x\) < - \(\dfrac{2}{3}\).(\(\dfrac{1}{3}\) - \(\dfrac{1}{2}\) - \(\dfrac{3}{4}\))

   - \(\dfrac{13}{3}\).\(\dfrac{1}{3}\) < \(x\) < - \(\dfrac{2}{3}\).(-\(\dfrac{11}{12}\))

    - \(\dfrac{13}{9}\) < \(x\) < \(\dfrac{11}{18}\)

     \(x\) \(\in\) { -1; 0; 1}

a, -4\(\dfrac{3}{5}\).2\(\dfrac{4}{3}\) < \(x\) < -2\(\dfrac{3}{5}\): 1\(\dfrac{6}{15}\)

  - \(\dfrac{23}{5}\).\(\dfrac{10}{3}\) <   \(x\)   < - \(\dfrac{13}{5}\): \(\dfrac{21}{15}\)

   -  \(\dfrac{46}{3}\)     <  \(x\) < - \(\dfrac{13}{7}\) 

          \(x\) \(\in\) {-15; -14;-13;..; -2}

`@` `\text {Ans}`

`\downarrow`

`a)`

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

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

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

\(x=\dfrac{18}{15}-\dfrac{10}{15}\)

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

Vậy, `x =`\(\dfrac{8}{15}\)

`b)`

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

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

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

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

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) \(\dfrac{3}{2}x\dfrac{4}{5}-x=\dfrac{2}{3}\Rightarrow\dfrac{6}{5}-x=\dfrac{2}{3}\Rightarrow x=\dfrac{6}{5}-\dfrac{2}{3}=\dfrac{18}{15}-\dfrac{10}{15}=\dfrac{8}{15}\)

b) \(x.3\dfrac{1}{3}=3\dfrac{1}{3}:4\dfrac{1}{4}\Rightarrow\dfrac{10}{3}.x=\dfrac{10}{3}:\dfrac{17}{4}\Rightarrow\dfrac{10}{3}.x=\dfrac{10}{3}.\dfrac{4}{17}\Rightarrow x=\dfrac{10}{3}.\dfrac{4}{17}:\dfrac{10}{3}=\dfrac{10}{3}.\dfrac{4}{17}.\dfrac{3}{10}=\dfrac{4}{17}\)

c) \(5\dfrac{2}{3}:x=3\dfrac{2}{3}-2\dfrac{1}{2}\Rightarrow\dfrac{17}{3}:x=\dfrac{11}{3}-\dfrac{5}{2}\Rightarrow\dfrac{17}{3}:x=\dfrac{22}{6}-\dfrac{15}{6}\Rightarrow\dfrac{17}{3}:x=\dfrac{7}{6}\Rightarrow x=\dfrac{17}{3}:\dfrac{7}{6}=\dfrac{17}{3}.\dfrac{7}{6}=\dfrac{119}{18}\)

a, 2/5 + 3/4 : x = -1/2

3/4 : x = -1/2 - 2/5

3/4 : x = -9/10

x = 3/4 : -9/10

x = -5/6

b, 5/7 - 2/3 . x = 4/5 

2/3 . x = 4/5 + 5/7

2/3 . x = 53/35

x = 53/35 : 2/3

x = 159/70

c và d mình làm dược nhưng ko ghi được cái suy ra

a: \(A=\dfrac{-7}{28}\cdot\dfrac{15}{25}=\dfrac{-1}{4}\cdot\dfrac{3}{5}=\dfrac{-3}{20}\)

b: \(B=\dfrac{-5\cdot7}{14\cdot\left(-3\right)}=\dfrac{35}{42}=\dfrac{5}{6}\)

c: \(C=\dfrac{-1}{5}-\dfrac{1}{5}\cdot\dfrac{3}{5}=\dfrac{-1}{5}-\dfrac{3}{25}=\dfrac{-8}{25}\)

d: \(D=\dfrac{-3}{4}-\dfrac{1}{4}=-1\)

e: \(E=\dfrac{-4}{5}\left(1-\dfrac{15}{16}\right)=\dfrac{-4}{5}\cdot\dfrac{1}{16}=\dfrac{-1}{20}\)

f: \(F=\dfrac{6-7}{4}\cdot\dfrac{4+12}{22}=\dfrac{-1}{4}\cdot\dfrac{8}{11}=\dfrac{-2}{11}\)