tìm x a)
\(\dfrac{7}{2}\)-\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-5}{4}\)
\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-5}{4}\) + \(\dfrac{7}{2}\)
\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-5}{12}+\dfrac{7}{12}\)
\(\left(x+\dfrac{7}{10}\right)\): \(\dfrac{6}{5}\) = \(\dfrac{-12}{12}=1\)
\(x+\dfrac{7}{10}\)= 1 . \(\dfrac{6}{5}\)
*Rồi tự làm phần tt đi

Mình ghi kết quả luôn nha bạn

Các bạn ơi,mình ghi thiếu,còn 3 câu nữa nha!!!~~nya

e)| \(\dfrac{5}{2}\)x-\(\dfrac{1}{2}\) |-(-2²).\(\dfrac{1}{3}\)(0,75-\(\dfrac{1}{7}\))=\(\dfrac{-5}{13}\):2\(\dfrac{9}{13}\)-0,5.(\(\dfrac{-2}{3}\))

f)| 5x+21 | = | 2x -63 |

g) -45 - |-3x-96 | - 54=-207

Làm ơn giúp mình với ạ!Mình đang cần gấp lắm trong ngày hôm nay ạ!!!Mình xin cảm ơn các bạn nhiều nhiều lắm luôn đó!!!Thank you very much!!!(^-^)

a, (\(\dfrac{2}{9}\)(6x - \(\dfrac{3}{4}\)) - 3(\(\dfrac{1}{4}x-\dfrac{1}{5}\)) = \(\dfrac{-8}{15}\)

<=> (\(\dfrac{4}{3}x-\dfrac{1}{6}\)) - (\(\dfrac{3}{4}x-\dfrac{3}{5}\)) = \(\dfrac{-8}{15}\)

<=> \(\dfrac{4}{3}x-\dfrac{1}{6}-\dfrac{3}{4}x+\dfrac{3}{5}=\dfrac{-8}{15}\)

<=> \(\dfrac{7}{12}x+\dfrac{13}{30}=\dfrac{-8}{15}\)

<=> \(\dfrac{7}{12}x=\dfrac{-8}{15}-\dfrac{13}{30}\)

<=> \(\dfrac{7}{12}x=-\dfrac{29}{30}\)

<=> x = \(-\dfrac{58}{35}\)
@Nguyễn Gia Hân

bài 1:

a) \(4\dfrac{1}{2}x:\dfrac{5}{12}=0,5\) ; b)\(1,5+1\dfrac{1}{4}x=\dfrac{2}{3}\)

\(\dfrac{9}{2}x:\dfrac{5}{12}=\dfrac{1}{2}\) \(\dfrac{3}{2}+\dfrac{5}{4}x=\dfrac{2}{3}\)

\(\dfrac{9}{2}x\) \(=\dfrac{1}{2}.\dfrac{5}{12}\) \(\dfrac{5}{4}x=\dfrac{2}{3}-\dfrac{3}{2}\)

\(\dfrac{9}{2}x\) \(=\dfrac{5}{24}\) \(\dfrac{5}{4}x=\dfrac{-5}{6}\)

\(x\) \(=\dfrac{5}{24}:\dfrac{9}{2}\) \(x=\dfrac{-5}{6}:\dfrac{5}{4}\)

\(x\) \(=\dfrac{5}{108}\) \(x=\dfrac{-2}{3}\)

c) Cho mình hỏi x ở đâu vậy ???

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

\(\left(x-5\right)\) \(=\dfrac{2}{5}.\dfrac{1}{3}\) \(\left(\dfrac{9}{2}-2x\right):\dfrac{3}{4}=\dfrac{4}{3}\)

\(x-5\) \(=\dfrac{2}{15}\) \(\dfrac{9}{2}-2x\) =\(\dfrac{4}{3}.\dfrac{3}{4}\)

\(x\) \(=\dfrac{2}{15}+5\) \(\dfrac{9}{2}-2x=1\)

\(x\) \(=\dfrac{77}{15}\) \(2x=\dfrac{9}{2}-1\)

f) \(\left(2,7x-1\dfrac{1}{2}x\right):\dfrac{2}{7}=\dfrac{-21}{7}\) \(2x=\dfrac{7}{2}\)

\(\left(\dfrac{27}{10}x-\dfrac{3}{2}x\right):\dfrac{2}{7}=-3\) \(x=\dfrac{7}{2}:2\)

\(\left[x\left(\dfrac{27}{10}-\dfrac{3}{2}\right)\right]=-3.\dfrac{2}{7}\) \(x=\dfrac{7}{4}\)

\(x.\dfrac{6}{5}=\dfrac{-6}{7}\)

\(x=\dfrac{-6}{7}:\dfrac{6}{5}\)

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

bài 2:

Theo bài ra ta có :\(\dfrac{a}{27}=\dfrac{-5}{9}=\dfrac{-45}{b}\)

\(\Rightarrow9a=27.\left(-5\right)\Rightarrow a=\dfrac{27.\left(-5\right)}{9}=-15\)

\(\Rightarrow\left(-5\right)b=\left(-45\right).9\Rightarrow b=\dfrac{\left(-45\right).9}{-5}=81\)

Vậy \(a=-15;b=81\)

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

Theo định nghĩa về hai phân số bằng nhau, ta có:

\(4\cdot9=x^2\\ 36=x^2\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)

8)

\(x:\dfrac{5}{3}+\dfrac{1}{3}=-\dfrac{2}{5}\\ x:\dfrac{5}{3}=-\dfrac{2}{5}+\dfrac{1}{3}\\ x:\dfrac{5}{3}=-\dfrac{1}{15}\\ x=\dfrac{1}{15}\cdot\dfrac{5}{3}\\ x=\dfrac{1}{9}\)

7)

\(2x-16=40+x\\ 2x-x=40+16\\ x\left(2-1\right)=56\\ x=56\)

6)

\(1\dfrac{1}{2}+x=\dfrac{3}{2}-7\\ \dfrac{3}{2}+x=\dfrac{3}{2}-7\\ \dfrac{3}{2}-\dfrac{3}{2}=-7-x\\ -7-x=0\\ x=-7-0\\ x=-7\)

5)

\(3\dfrac{1}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\\ \dfrac{7}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\\ \dfrac{1}{2}x=\dfrac{7}{2}-\dfrac{2}{3}\\ \dfrac{1}{2}x=\dfrac{17}{6}\\ x=\dfrac{17}{6}:\dfrac{1}{2}\\ x=\dfrac{17}{3}\)

4)

\(x\cdot\left(x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

3)

\(\left(\dfrac{2x}{5}+2\right):\left(-4\right)=-1\dfrac{1}{2}\\ \left(\dfrac{2x}{5}+2\right):\left(-4\right)=-\dfrac{3}{2}\\ \dfrac{2x}{5}+2=-\dfrac{3}{2}\cdot\left(-4\right)\\ \dfrac{2x}{5}+2=6\\ \dfrac{2x}{5}=6-2\\ \dfrac{2x}{5}=4\\ 2x=4\cdot5\\ 2x=20\\ x=20:2\\ x=10\)

2)

\(\dfrac{1}{3}+\dfrac{1}{2}:x=-0,25\\ \dfrac{1}{3}+\dfrac{1}{2}:x=-\dfrac{1}{4}\\ \dfrac{1}{2}:x=-\dfrac{1}{4}-\dfrac{1}{3}\\ \dfrac{1}{2}:x=-\dfrac{7}{12}\\ x=\dfrac{1}{2}:-\dfrac{7}{12}\\ x=-\dfrac{6}{7}\)

1)

\(\dfrac{4}{3}+x=\dfrac{2}{15}\\ x=\dfrac{2}{15}-\dfrac{4}{3}x=-\dfrac{6}{5}\)

a: \(\Leftrightarrow x\cdot\dfrac{62}{7}=\dfrac{29}{9}\cdot\dfrac{56}{3}=\dfrac{1624}{27}\)

hay \(x=\dfrac{1624}{27}:\dfrac{62}{7}=\dfrac{5684}{837}\)

b: \(\Leftrightarrow\dfrac{1}{5}:x=\dfrac{12}{35}\)

nên \(x=\dfrac{1}{5}:\dfrac{12}{35}=\dfrac{1}{5}\cdot\dfrac{35}{12}=\dfrac{7}{12}\)

c: \(\Leftrightarrow\left|2x+\dfrac{1}{3}\right|=\dfrac{30-7}{42}=\dfrac{23}{42}\)

=>2x+1/3=23/42 hoặc 2x+1/3=-23/42

=>2x=3/14 hoặc 2x=-37/42

=>x=3/28 hoặc x=-37/84

\(4)\)

\(\dfrac{-\left(-x\right)}{5}-\dfrac{2}{10}=\dfrac{1}{-5}-\dfrac{7}{50}\)

\(\Leftrightarrow\dfrac{x}{5}-\dfrac{2}{10}=\dfrac{1}{-5}-\dfrac{7}{50}\)

\(\dfrac{2x}{10}-\dfrac{2}{10}=\dfrac{-10}{50}-\dfrac{7}{50}\)

\(\Leftrightarrow\dfrac{2x-2}{10}=\dfrac{-10-7}{50}\)

\(\dfrac{2x-2}{10}=\dfrac{-17}{50}\)

\(\Leftrightarrow50\left(2x-2\right)=-17.10\)

\(100x-100=-170\)

\(100x=-170+100=-70\)

\(x=-70:100=\dfrac{-7}{10}\)

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

\(\left(x+1\right)\left(x-1\right)5.7\)

\(x\left(x-1\right)+1\left(x-1\right)=35\)

\(x^2-x+x-1=35\)

\(x^2-1=35\)

\(x^2=36\)

\(\Leftrightarrow x=\left\{\pm6\right\}\)

bạn có thể giải đc các bài còn lại k ? K phải mk ép bạn đâu nhưng nếu bạn lm đc thì giúp mk nha

a) Để phân số \(\dfrac{3}{x-1}\) có giá trị là số nguyên thì:

3 \(⋮\) x - 1

Vì x \(\in\) Z => x - 1 \(\in\) Z

=> x \(\in\) Ư₍₃₎ = {1;3;-1;-3}

Ta lập bảng sau

Vậy x \(\in\) {2;4;0;-2} là giá trị cần tìm

b)Để phân số \(\dfrac{x-2}{x+3}\) có giá trị là số nguyên thì:

x - 2 \(⋮\) x + 3

=> \(\left\{{}\begin{matrix}x-2⋮x+3\\x+3⋮x+3\end{matrix}\right.\)

=> 5\(⋮\)x+3

Vì x \(\in\) Z => x + 3 \(\in\) Z

=> x + 3 \(\in\) Ư₍₅₎ = {1;5;-1;-5}

Ta lập bảng sau

Vậy x \(\in\) {-2;2;-4;-8} là giá trị cần tìm

a) Để phân số \(A=\dfrac{3}{x-1}\) có giá trị là 1 số nguyên thì :

\(3⋮x-1\)

\(\Rightarrow x-1\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)

Ta có bảng :

Vậy \(x\in\left\{2;0;4;-2\right\}\) là giá trị cần tìm

b) Để phân số \(A=\dfrac{x-2}{x+3}\) có giá trị là 1 số nguyên thì :

\(x-2⋮x+3\)

\(\Rightarrow x+3-5⋮x+3\)

\(\Rightarrow-5⋮x+3\)

\(\Rightarrow x+3\inƯ\left(-5\right)=\left\{1;-1;5;-5\right\}\)

Ta có bảng :

Vậy \(x\in\left\{-2;-4;2;-8\right\}\) là giá trị cần tìm

Bài 1:

\(\left(-\dfrac{72}{40}-\dfrac{144}{60}-2\dfrac{1}{3}\right):\left(\dfrac{45}{100}-\dfrac{25}{60}+-\dfrac{75}{25}\right)\)

\(=\left(-\dfrac{9}{5}-\dfrac{12}{5}-\dfrac{7}{3}\right):\left(\dfrac{9}{20}-\dfrac{5}{12}+-3\right)\)

\(=\left(-\dfrac{27}{15}-\dfrac{36}{15}-\dfrac{21}{15}\right):\left(\dfrac{27}{60}-\dfrac{25}{60}+-3\right)\)

\(=\left(-\dfrac{28}{5}\right):\left(-\dfrac{89}{30}\right)\)

\(=\left(-\dfrac{28}{5}\right).\left(-\dfrac{30}{89}\right)\)

\(=\dfrac{168}{89}\)

a)\(\dfrac{3}{10}\)-x=\(\dfrac{25}{30}\)-\(\dfrac{4}{30}\)

\(\dfrac{3}{10}-x=\dfrac{7}{10}\)

x = \(\dfrac{3}{10}-\dfrac{7}{10}\)

x=\(\dfrac{-4}{10}\)

b)\(\dfrac{-5}{8}+x=\dfrac{4}{9}-\dfrac{63}{9}\)

\(\dfrac{-5}{9}+x=\dfrac{-59}{9}\)

\(x=\dfrac{-59}{9}-\dfrac{-5}{9}\)

\(x=\dfrac{-64}{9}\)

c)=>2.18=(x-3).(x-3)

=>36=(x-3)\(^2\)

=>6\(^2\)=(x-3)\(^2\)

6= x-3

x=6+3=9