c.\(\dfrac{3}{7}+\dfrac{5}{7}:x=\dfrac{1}{3}\)

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

\(\dfrac{5}{7}:x=-\dfrac{2}{21}\)

\(x=\dfrac{5}{7}:-\dfrac{2}{21}\)

\(x=-\dfrac{15}{2}\)

d.\(3\dfrac{1}{4}:\left|2x-\dfrac{5}{12}\right|=\dfrac{39}{16}\)

\(\left|2x-\dfrac{5}{12}\right|=3\dfrac{1}{4}:\dfrac{39}{16}\)

\(\left|2x-\dfrac{5}{12}\right|=\dfrac{4}{3}\)

\(\rightarrow\left[{}\begin{matrix}2x-\dfrac{5}{12}=\dfrac{4}{3}\\2x-\dfrac{4}{12}=-\dfrac{4}{3}\end{matrix}\right.\) \(\rightarrow\left[{}\begin{matrix}2x=\dfrac{7}{4}\\2x=-\dfrac{11}{12}\end{matrix}\right.\) \(\rightarrow\left[{}\begin{matrix}x=\dfrac{7}{8}\\x=-\dfrac{11}{24}\end{matrix}\right.\)

A, \(\dfrac{4}{9}+x=\dfrac{5}{3}\)

\(x\)\(=\dfrac{5}{3}-\dfrac{4}{9}\)

\(x\)\(=\dfrac{11}{9}\)

B,\(\dfrac{3}{4}.x=\dfrac{-1}{2}\)

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

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

a)

\(\frac{4}{9} + x = \frac{5}{3}\)

=> \(x = \frac{5}{3}-\frac{4}{9}\)

=> \(x = \) \(\frac{11}{9}\)

Vậy \(x = \dfrac{11}{9}\)

b) 

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

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

=> \(x = \dfrac{-2}{3}\)

Vậy \(x = \dfrac{-2}{3}\)

c)

\( \dfrac{3}{7}+ \dfrac{5}{7}:x = \dfrac{1}{3}\)

=> \(\dfrac{5}{7}:x = \dfrac{1}{3}-\) \( \dfrac{3}{7}\)

=> \(\dfrac{5}{7}:x = \dfrac{-2}{21}\)

=> \(x = \dfrac{5}{7}:\dfrac{-2}{21}\)

=> \(x = \dfrac{-15}{2}\)

Vậy \(x = \dfrac{-15}{2}\)

d) 

\(3\dfrac{1}{4} : |2x - \dfrac{5}{12} | = \dfrac{39}{16}\)

=> \(\dfrac{13}{4} : |2x - \dfrac{5}{12} | = \dfrac{39}{16}\)

=> \( |2x - \dfrac{5}{12} | =\dfrac{13}{4} : \dfrac{39}{16}\)

=> \(|2x-\dfrac{5}{12} |= \dfrac{4}{3}\)

=> \(\left[\begin{matrix} 2x - \dfrac{5}{12} = \dfrac{4}{3}\\ 2x - \dfrac{5}{12} = \dfrac{4}{3}\end{matrix}\right.\)

=> \(\left[\begin{matrix} 2x = \dfrac{-4}{3}+\dfrac{5}{12}\\ 2x = \dfrac{-4}{3}+\dfrac{5}{12} \end{matrix}\right.\)

=> \(\left[\begin{matrix} 2x = \dfrac{7}{4}\\ 2x = \dfrac{-11}{12} \end{matrix}\right.\)

=> \(\left[\begin{matrix} x = \dfrac{7}{8}\\ x = \dfrac{-11}{24} \end{matrix}\right.\)

Vậy \(x \in \) { \(\dfrac{7}{8} ; \dfrac{-11}{24}\) }

\(a.\dfrac{-4}{7}-\dfrac{5}{13}\times\dfrac{-39}{25}+\dfrac{-1}{42}:\dfrac{-5}{6}\) 

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

\(b.\dfrac{2}{9}\times\left[\dfrac{4}{5}:\left(\dfrac{1}{5}-\dfrac{2}{15}\right)+1\dfrac{2}{3}\right]-\dfrac{-5}{27}\) 

\(=\dfrac{2}{9}\times\left[\dfrac{4}{5}:\dfrac{1}{15}+\dfrac{5}{3}\right]-\dfrac{-5}{27}\) 

\(=\dfrac{2}{9}\times\left(12+\dfrac{5}{3}\right)-\dfrac{-5}{27}\) 

\(=\dfrac{2}{9}\times\dfrac{41}{3}-\dfrac{-5}{27}=\dfrac{82}{27}-\dfrac{-5}{27}=\dfrac{29}{9}\)

\(\dfrac{1}{5}+\dfrac{2}{11}< \dfrac{x}{55}< \dfrac{2}{5}+\dfrac{1}{5}\)

\(\dfrac{11+10}{55}< \dfrac{x}{55}< \dfrac{3}{5}\)

\(\dfrac{21}{55}< \dfrac{x}{55}< \dfrac{33}{55}\)

Vậy \(x\in\left\{22;23;24;...\right\}\)

\(\dfrac{????????}{????????????}\)

a) \(\dfrac{1}{5}+\dfrac{2}{11}< \dfrac{x}{55}< \dfrac{2}{5}+\dfrac{1}{5}\)

   \(\dfrac{11}{55}+\dfrac{10}{55}< \dfrac{x}{55}< \dfrac{22}{55}+\dfrac{1}{55}\)

   \(\dfrac{21}{55}< \dfrac{x}{55}< \dfrac{23}{55}\)

\(\Rightarrow\) \(x=22\)

b) \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}< x\le\dfrac{13}{4}+\dfrac{14}{8}\)

  \(\dfrac{3}{6}+\dfrac{2}{6}+\dfrac{1}{6}< x\le\dfrac{26}{8}+\dfrac{14}{8}\)

  \(1< x\le5\)

  \(\Rightarrow\) \(x\in\) {\(2;3;4;5\)}

c) \(\dfrac{1}{3}+\dfrac{5}{12}+\dfrac{-1}{13}< x< \dfrac{7}{5}+\dfrac{2}{10}+\dfrac{1}{2}\)

 Ko biết làm

d) \(\dfrac{79}{15}+\dfrac{7}{5}+\dfrac{-8}{3}\le x\le\dfrac{10}{3}+\dfrac{15}{4}+\dfrac{23}{12}\)

   \(\dfrac{79}{15}+\dfrac{21}{15}+\dfrac{-40}{15}\le x\le\dfrac{40}{12}+\dfrac{45}{12}+\dfrac{23}{12}\)

   \(4\le x\le9\)

   \(\Rightarrow\) \(x\in\) {\(4;5;6;7;8;9\)}

Giải:

a) \(\dfrac{12}{16}=\dfrac{-x}{4}=\dfrac{21}{y}=\dfrac{z}{80}\)  

\(\Rightarrow x=\dfrac{12.-4}{16}=-3\) 

\(\Rightarrow y=\dfrac{16.21}{12}=28\) 

\(\Rightarrow z=\dfrac{12.80}{16}=60\) 

b) \(\dfrac{1}{3}x+\dfrac{2}{5}\left(x-1\right)\)  =0

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

     \(x.\left(\dfrac{1}{3}+\dfrac{2}{5}\right)\)   \(=0+\dfrac{2}{5}\) 

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

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

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

c) (2x-3)(6-2x)=0

⇒2x-3=0 hoặc 6-2x=0

        x=3/2 hoặc x=3

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

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

               \(\dfrac{1}{3}\left(2x-5\right)=\dfrac{-13}{6}\)  

                   \(2x-5=\dfrac{-13}{6}:\dfrac{1}{3}\) 

                   \(2x-5=\dfrac{-13}{2}\) 

                         \(2x=\dfrac{-13}{2}+5\)

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

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

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

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

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

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

\(\Rightarrow\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{1}{8}\)  hoặc \(\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-1}{8}\) 

                \(x=\dfrac{11}{12}\) hoặc \(x=\dfrac{5}{12}\)

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

a) Ta có: \(\dfrac{2}{3}x-1=\dfrac{3}{2}\)

\(\Leftrightarrow x\cdot\dfrac{2}{3}=\dfrac{5}{2}\)

hay \(x=\dfrac{5}{2}:\dfrac{2}{3}=\dfrac{5}{2}\cdot\dfrac{3}{2}=\dfrac{15}{4}\)

b) Ta có: \(\left|5x-\dfrac{1}{2}\right|-\dfrac{2}{7}=25\%\)

\(\Leftrightarrow\left|5x-\dfrac{1}{2}\right|=\dfrac{1}{4}+\dfrac{2}{7}=\dfrac{15}{28}\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-\dfrac{1}{2}=\dfrac{15}{28}\\5x-\dfrac{1}{2}=\dfrac{-15}{28}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{29}{28}\\5x=\dfrac{-1}{28}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{140}\\x=\dfrac{-1}{140}\end{matrix}\right.\)

c) Ta có: \(\dfrac{x-3}{4}=\dfrac{16}{x-3}\)

\(\Leftrightarrow\left(x-3\right)^2=64\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=8\\x-3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-5\end{matrix}\right.\)

d) Ta có: \(\dfrac{-8}{13}+\dfrac{7}{17}+\dfrac{21}{31}\le x\le\dfrac{-9}{14}+4-\dfrac{5}{14}\)

\(\Leftrightarrow\dfrac{3246}{6851}\le x\le3\)

\(\Leftrightarrow x\in\left\{1;2;3\right\}\)

ko biết =))

TK

https://lazi.vn/edu/exercise/giai-phuong-trinh-4x-5-x-1-2-x-x-1-7-x-2-3-x-5

a: \(\Leftrightarrow4x-5=2x-2+x\)

=>4x-5=3x-2

=>x=3(nhận)

b: =>7x-35=3x+6

=>4x=41

hay x=41/4(nhận)

c: \(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{-3}{2\left(x-4\right)}-\dfrac{5}{6}\)

\(\Leftrightarrow\dfrac{28}{6\left(x-4\right)}-\dfrac{6\left(x+2\right)}{6\left(x-4\right)}=\dfrac{-9}{6\left(x-4\right)}-\dfrac{5\left(x-4\right)}{6\left(x-4\right)}\)

\(\Leftrightarrow28-6x-12=-9-5x+20\)

=>-6x+16=-5x+11

=>-x=-5

hay x=5(nhận)

d: \(\Leftrightarrow x^2+2x+1-\left(x^2-2x+1\right)=16\)

\(\Leftrightarrow4x=16\)

hay x=4(nhận)

a)

x^2-16/25=0

x^2-4^2/5^2=0

=>x-4/5=0

x=0+4/5

x=0/5

bài 1 :

\(a,\dfrac{2}{7}+\dfrac{1}{3}=\dfrac{6}{21}+\dfrac{7}{21}=\dfrac{13}{21}\)

\(b,\dfrac{3}{5}-\dfrac{1}{3}=\dfrac{9}{15}-\dfrac{5}{15}=\dfrac{4}{15}\)

\(c,\dfrac{13}{4}:5=\dfrac{13}{4}:\dfrac{5}{1}=\dfrac{13}{4}x\dfrac{1}{5}=\dfrac{13}{20}\)

\(d,\dfrac{6}{23}x\dfrac{1}{18}=\dfrac{1}{69}\)

bài 2 :

\(a,x+\dfrac{1}{3}=\dfrac{5}{12}\)

   \(x=\dfrac{5}{12}-\dfrac{1}{3}\)

  \(x=\dfrac{1}{12}\)

\(b,x:\dfrac{7}{4}=\dfrac{2}{5}\)

   \(x=\dfrac{2}{5}x\dfrac{7}{4}\)

   \(x=\dfrac{7}{10}\)

bài 3 :

đổi : 2 dm 1cm = 21cm

chiều cao hình bình hành là;

       21 x\(\dfrac{3}{7}=\)9(cm)

diện tích hình bình hành là;

       21 x 9 =189 (cm²)

                 đáp số : 189 cm²

bài 4 :

\(\dfrac{2}{3}x\dfrac{2}{10}+\dfrac{2}{3}x\dfrac{5}{10}x\dfrac{3}{3}\)

\(\dfrac{2}{3}x\left(\dfrac{2}{10}+\dfrac{5}{10}\right)x\dfrac{2}{3}\)

=\(\dfrac{2}{3}x1x\dfrac{2}{3}\)

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

=\(\dfrac{4}{9}\)

Bài 1)

a) \(\dfrac{6}{21}+\dfrac{7}{21}=\dfrac{13}{21}\)

b) \(\dfrac{9}{15}-\dfrac{5}{15}=\dfrac{4}{15}\)

c) \(\dfrac{13}{4}x\dfrac{1}{5}=\dfrac{13}{20}\)

d) \(\dfrac{6}{414}=\dfrac{1}{69}\)

Bài 2)

a) \(x=\dfrac{5}{12}-\dfrac{1}{3}\)

\(x=\dfrac{1}{12}\)

b) \(x=\dfrac{2}{5}x\dfrac{7}{4}\)

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

Bài 3)

2dm 1cm = 21 cm

Chiều cao tấm bìa la

\(21x\dfrac{3}{7}=9\left(cm\right)\)

Diện tích tấm bìa là

\(21x9=189\left(cm2\right)\)

Bài 4)

\(\dfrac{2}{3}x\dfrac{2}{10}+\dfrac{2}{3}x\dfrac{5}{10}x\dfrac{2}{3}=\dfrac{2}{3}x\left(\dfrac{2}{10}+\dfrac{5}{10}\right)x\dfrac{2}{3}=\dfrac{2}{3}x\dfrac{7}{10}x\dfrac{2}{3}=\dfrac{14}{45}\)