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

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

\(\dfrac{2}{3}x=\dfrac{5}{4}+\dfrac{5}{6}\)

\(\dfrac{2}{3}x=\dfrac{25}{12}\)

\(x=\dfrac{25}{12}:\dfrac{2}{3}\)

=>\(x=\dfrac{25}{8}\)

a) \(\dfrac{2}{3}x-\dfrac{5}{6}=1\dfrac{1}{4}\) b) \(2\dfrac{1}{3}-\dfrac{4}{5}:x=0,2\)

\(\dfrac{2}{3}x-\dfrac{5}{6}=\dfrac{5}{4}\) \(\dfrac{7}{3}-\dfrac{4}{5}:x=\dfrac{1}{5}\)

\(\dfrac{2}{3}x=\dfrac{5}{4}-\dfrac{5}{6}\) \(\dfrac{4}{5}:x=\dfrac{7}{3}-\dfrac{1}{5}\)

\(\dfrac{2}{3}x=\dfrac{30}{24}-\dfrac{20}{24}\) \(\dfrac{4}{5}:x=\dfrac{35}{15}-\dfrac{3}{15}\)

\(\dfrac{2}{3}x=\dfrac{5}{12}\) \(\dfrac{4}{5}:x=\dfrac{32}{15}\)

\(x=\dfrac{5}{12}:\dfrac{2}{3}\) \(x=\dfrac{4}{5}:\dfrac{32}{15}\)

\(x=\dfrac{5}{12}:\dfrac{8}{12}\) \(x=\dfrac{4}{5}.\dfrac{15}{32}\)

\(x=\dfrac{5}{12}.\dfrac{12}{8}=\dfrac{5}{8}\) \(x=\dfrac{4.15}{5.32}\)

\(x=\dfrac{1.3}{1.8}=\dfrac{3}{8}\)

d)\(\left(\dfrac{4}{3}-\dfrac{1}{4}x\right)^3=\dfrac{-8}{27}\)

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

\(\Rightarrow\dfrac{4}{3}-\dfrac{1}{4}x=\dfrac{-2}{3}\)

\(\Rightarrow\dfrac{1}{4}x=\dfrac{4}{3}-\dfrac{-2}{3}\)

\(\Rightarrow\dfrac{1}{4}x=2\)

\(\Rightarrow x=2:\dfrac{1}{4}\)

\(\Rightarrow x=2.4=8\)

b)\(2\dfrac{1}{3}-\dfrac{4}{5}:x=0,2\)

\(\dfrac{4}{5}:x=2\dfrac{1}{3}-0,2\)

\(\dfrac{4}{5}:x=\dfrac{32}{15}\)

\(x=\dfrac{4}{5}:\dfrac{32}{15}\)

\(x=0.375\)

d)\(\left(\dfrac{4}{3}-\dfrac{1}{4}x\right)^3=\dfrac{-8}{27}\)

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

=>\(\dfrac{4}{3}-\dfrac{1}{4}x=\dfrac{-2}{3}\)

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

\(\dfrac{1}{4}x=2\)

=>\(x=2:\dfrac{1}{4}\)

=>x=8

a) x.(\(\dfrac{6}{7}\)+\(\dfrac{5}{6}\))=\(\dfrac{3}{4}\)

x.\(\dfrac{71}{42}\)=\(\dfrac{3}{4}\)

x=\(\dfrac{3}{4}\):\(\dfrac{71}{42}\)

x=\(\dfrac{63}{142}\)

a.\(\dfrac{6}{7}x+\dfrac{5}{6}x=\dfrac{3}{4}\)

\(x.\left(\dfrac{6}{7}+\dfrac{5}{6}\right)=\dfrac{3}{4}\)

\(x.\dfrac{71}{42}=\dfrac{3}{4}\)

\(x=\dfrac{3}{4}:\dfrac{71}{42}\)

\(x=\dfrac{63}{142}\)

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

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

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

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

\(x=\dfrac{-36}{5}\)

c. \(\left(\dfrac{4}{7}x-\dfrac{1}{3}\right):3\dfrac{1}{2}=0,5\)

\(\left(\dfrac{4}{7}x-\dfrac{1}{3}\right)=0,5:3\dfrac{1}{2}\)

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

\(\dfrac{4}{7}x=\dfrac{1}{7}+\dfrac{1}{3}\)

\(\dfrac{4}{7}x=\dfrac{10}{21}\)

\(x=\dfrac{10}{21}:\dfrac{4}{7}\)

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

d.\(\dfrac{4}{5}-\dfrac{2}{3}x=1\dfrac{1}{4}+2,5x\)

\(\dfrac{4}{5}-\left(\dfrac{2}{3}x-2,5x\right)=1\dfrac{1}{4}\)

\(\dfrac{4}{5}-\dfrac{-11}{6}x=1\dfrac{1}{4}\)

\(\dfrac{-11}{6}x=\dfrac{4}{5}-1\dfrac{1}{4}\)

\(\dfrac{-11}{6}x=\dfrac{-9}{20}\)

\(x=\dfrac{-9}{20}:\dfrac{-11}{6}\)

\(x=\dfrac{27}{110}\)

có sai sót j xin bn thông cảm !

`a, 2/3 +3/4 = (8+9)/12=17/12.`

`1 1/3+4/5 = 4/3 + 4/5 = (20+12)/15=32/15`.

`=> x=2.`

`b, 5/6-1/4=(20-6)/24=7/12`.

`2 1/3-2/5= 7/3-2/5 = (35-6)/15=29/15`.

`=> x=1`.

a) \(\dfrac{2}{3}+\dfrac{3}{4}=\dfrac{8+9}{12}=\dfrac{17}{12}\)

-> 1 1/3 + 4/5 = 4/3 + 4/5 =  20+12/15 = 32/15

vậy x có thể = 14/14 = 1 (x thuộc N)

ét ô ét

a.25/27                                                                                                                 b.0                          c.0

a: \(=4+\sqrt{11}+\dfrac{3}{2}-\dfrac{1}{2}\sqrt{7}-4-2\sqrt{7}-\dfrac{1}{2}\sqrt{7}+\dfrac{5}{2}\)

\(=4+\sqrt{11}-3\sqrt{7}\)

b: \(VT=\dfrac{x+2\sqrt{xy}+y-x+2\sqrt{xy}-y+2x+2y}{2\left(x-y\right)}\)

\(=\dfrac{2x+4\sqrt{xy}+2y}{2\left(x-y\right)}=\dfrac{x+2\sqrt{xy}+y}{x-y}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)

a.\(\dfrac{-4}{5}-\left(\dfrac{2}{3}x+1\dfrac{1}{4}\right)=\dfrac{2}{7}\)

\(\left(\dfrac{2}{3}x+1\dfrac{1}{4}\right)=\dfrac{-4}{5}-\dfrac{2}{7}=\dfrac{-38}{35}\)

\(\dfrac{2}{3}x=\dfrac{-38}{35}-1\dfrac{1}{4}\)

\(\dfrac{2}{3}x=\dfrac{-327}{140}\Rightarrow x=\dfrac{-327}{140}:\dfrac{2}{3}=\dfrac{-981}{280}\)

Vậy \(x=\dfrac{-981}{280}\)

b. \(\dfrac{5}{6}+\left(\dfrac{3}{4}-\dfrac{1}{2}:x\right)=\dfrac{-2}{3}\)

\(\left(\dfrac{3}{4}-\dfrac{1}{2}:x\right)=\dfrac{-2}{3}-\dfrac{5}{6}=\dfrac{-3}{2}\)

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

\(\dfrac{1}{2}:x=\dfrac{9}{4}\Rightarrow x=\dfrac{1}{2}:\dfrac{9}{4}=\dfrac{2}{9}\)

Vậy \(x=\dfrac{2}{9}\)

c. \(\left(\dfrac{4}{5}x-1\dfrac{1}{3}\right):\dfrac{3}{4}=0,7\)

\(\left(\dfrac{4}{5}x-1\dfrac{1}{3}\right)=0,7.\dfrac{3}{4}=\dfrac{21}{40}\)

\(\dfrac{4}{5}x=\dfrac{21}{40}+1\dfrac{1}{3}=\dfrac{223}{120}\)

\(\Rightarrow x=\dfrac{223}{120}:\dfrac{4}{5}=\dfrac{223}{96}\)

Vậy \(x=\dfrac{223}{96}\)

d. \(\dfrac{5}{6}-\dfrac{3}{4}x=1\dfrac{1}{3}+0,5x\)

\(0,5x+\dfrac{3}{4}x=\dfrac{5}{6}-1\dfrac{1}{3}\)

\(\dfrac{5}{4}x=\dfrac{-1}{2}\Rightarrow x=\dfrac{-1}{2}:\dfrac{5}{4}=\dfrac{-2}{5}\)

Vậy \(x=\dfrac{-2}{5}\)

bài 1

a)\(=\dfrac{16}{40}+\dfrac{15}{40}=\dfrac{31}{40}\)

b)\(=\dfrac{7}{6}-\dfrac{4}{6}=\dfrac{3}{6}=\dfrac{1}{2}\)

c)\(=\dfrac{30}{9}=\dfrac{10}{3}\)

d)\(=\dfrac{8}{5}\times\dfrac{7}{4}=\dfrac{56}{20}=\dfrac{14}{5}\)

bài 2

a)\(x=\dfrac{5}{6}-\dfrac{4}{5}=\dfrac{25}{30}-\dfrac{24}{30}=\dfrac{1}{30}\)

b)\(x=5\times\dfrac{10}{7}=\dfrac{50}{7}\)

bài 4 :

145  69 + 22 x  145 +145 x 8 + 145 

\(=145\times\left(69+22+8+1\right)=145\times100=14500\)

bài 1:

a, \(\dfrac{2}{5}+\dfrac{3}{8}=\dfrac{16}{40}+\dfrac{15}{40}=\dfrac{31}{40}\)

b,\(\dfrac{7}{6}-\dfrac{2}{3}=\dfrac{7}{6}-\dfrac{4}{6}=\dfrac{3}{6}=\dfrac{1}{2}\)

c,\(\dfrac{5}{9}x6=\dfrac{5}{9}x\dfrac{6}{1}=\dfrac{30}{9}\)

d,\(\dfrac{8}{5}:\dfrac{4}{7}=\dfrac{8}{5}x\dfrac{7}{4}=\dfrac{14}{5}\)

bài 2 :

\(a,\dfrac{4}{5}+x=\dfrac{5}{6}\)

            \(x=\dfrac{5}{6}-\dfrac{4}{5}\)

            \(x=\dfrac{1}{30}\)

b, \(x:\dfrac{7}{10}=5\)

    \(x\)         \(=5x\dfrac{7}{10}\)

    \(x\)         \(=\dfrac{35}{10}\)    

bài 3 :

đổi :16 tấn 8 tạ = 168 tạ

        2 tấn 6 tạ = 26 tạ 

xe ô tô thứ nhất chở số tạ hàng là:

      ( 168 + 26 ) : 2= 97 ( tạ)

xe ô tô thứ hai chở số tạ hàng là:

         97 - 26 = 71 ( tạ)

               đáp số :xe ô tô thứ nhất : 97 tạ thóc

                            xe ô tô thứ hai  : 71 tạ thóc 

1, \(x\left(x+\dfrac{2}{3}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-2}{3}\end{matrix}\right.\)

2, a, \(\left|x+\dfrac{4}{6}\right|\ge0\)

Để \(\left|x+\dfrac{4}{6}\right|\) đạt GTNN thì \(\left|x+\dfrac{4}{6}\right|=0\)

\(\Leftrightarrow x+\dfrac{4}{6}=0\Rightarrow x=\dfrac{-2}{3}\)

Vậy, ...

b, \(\left|x-\dfrac{1}{3}\right|\ge0\)

Để \(\left|x-\dfrac{1}{3}\right|\) đạt GTLN thì \(\left|x-\dfrac{1}{3}\right|=0\)

\(\Leftrightarrow x-\dfrac{1}{3}=0\Rightarrow x=\dfrac{1}{3}\)

Vậy, ...

1)

a)

\(x\cdot\left(x+\dfrac{2}{3}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{2}{3}\end{matrix}\right.\)

2)

a)

\(\left|x+\dfrac{4}{6}\right|\ge0\)

Dấu \("="\) xảy ra khi \(x+\dfrac{4}{6}=0\Leftrightarrow x=\dfrac{-4}{6}\Leftrightarrow x=\dfrac{-2}{3}\)

Vậy \(Min_{\left|x+\dfrac{4}{6}\right|}=0\text{ khi }x=\dfrac{-2}{3}\)

b)

\(\left|x-\dfrac{1}{3}\right|\ge0\)

Dấu \("="\) xảy ra khi \(x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\)

Vậy \(Min_{\left|x-\dfrac{1}{3}\right|}=0\text{ khi }x=\dfrac{1}{3}\)