a) \(\left(2x-3\right)\left(6-2x\right)=0\)

\(\circledast\)TH₁: \(2x-3=0\\ 2x=0+3\\ 2x=3\\ x=\dfrac{3}{2}\)

\(\circledast\)TH₂: \(6-2x=0\\ 2x=6-0\\ 2x=6\\ x=\dfrac{6}{2}=3\)

Vậy \(x\in\left\{\dfrac{3}{2};3\right\}\).

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

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

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

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

\(-\dfrac{11}{15}=-x\left(x-1\right)\)

\(\Rightarrow x=1.491631652\)

Vậy \(x=1.491631652\)

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

\(\circledast\)TH₁: \(3x-1=0\\ 3x=0+1\\ 3x=1\\ x=\dfrac{1}{3}\)

\(\circledast\)TH₂: \(-\dfrac{1}{2}x+5=0\\ -\dfrac{1}{2}x=0-5\\ -\dfrac{1}{2}x=-5\\ x=-5:-\dfrac{1}{2}\\ x=10\)

Vậy \(x\in\left\{\dfrac{1}{3};10\right\}\).

d) \(\dfrac{x}{5}=\dfrac{2}{3}\\ x=\dfrac{5\cdot2}{3}\\ x=\dfrac{10}{3}\)

Vậy \(x=\dfrac{10}{3}\).

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

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

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

\(x=\dfrac{3\cdot7}{10}\)

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

Vậy \(x=\dfrac{21}{10}\).

f) \(\dfrac{x}{5}-\dfrac{1}{2}=\dfrac{6}{10}\)

\(\dfrac{x}{5}=\dfrac{6}{10}+\dfrac{1}{2}\)

\(\dfrac{x}{5}=\dfrac{11}{10}\)

\(x=\dfrac{5\cdot11}{10}\)

\(x=\dfrac{55}{10}=\dfrac{11}{2}\)

Vậy \(x=\dfrac{11}{2}\).

g) \(\dfrac{x+3}{15}=\dfrac{1}{3}\\ x+3=\dfrac{15}{3}=5\\ x=5-3\\ x=2\)

Vậy \(x=2\).

h) \(\dfrac{x-12}{4}=\dfrac{1}{2}\\ x-12=\dfrac{4}{2}=2\\ x=2+12\\ x=14\)

Vậy \(x=14\).

a: (x+1/2)(2/3-2x)=0

=>x+1/2=0 hoặc 2/3-2x=0

=>x=-1/2 hoặc x=1/3

b: 

c: \(\Leftrightarrow x\cdot\left(\dfrac{13}{4}-\dfrac{7}{6}\right)=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{5}{12}+\dfrac{20}{12}=\dfrac{25}{12}\)

\(\Leftrightarrow x=\dfrac{25}{12}:\dfrac{39-14}{12}=\dfrac{25}{25}=1\)

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)

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

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: Tính ( hợp lý nếu có thể )

\(A=\dfrac{-3}{8}+\dfrac{12}{25}+\dfrac{5}{-8}+\dfrac{2}{-5}+\dfrac{13}{25}\)

\(=\left(\dfrac{-3}{8}+\dfrac{5}{-8}\right)+\left(\dfrac{12}{25}+\dfrac{13}{25}\right)+\dfrac{2}{-5}\)

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

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

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

\(B=\dfrac{-3}{15}+\left(\dfrac{2}{3}+\dfrac{3}{15}\right)\)

\(=\left(\dfrac{-3}{15}+\dfrac{3}{15}\right)+\dfrac{2}{3}\)

\(=0+\dfrac{2}{3}\)

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

\(C=\dfrac{-5}{21}+\left(\dfrac{-16}{21}+1\right)\)

\(=\left(\dfrac{-5}{21}+\dfrac{-16}{21}\right)+1\)

\(=-1+1\)

\(=0\)

\(D=\left(\dfrac{-1}{6}+\dfrac{5}{-12}\right)+\dfrac{7}{12}\)

\(=\left(\dfrac{5}{-12}+\dfrac{7}{12}\right)+\dfrac{-1}{6}\)

\(=\dfrac{1}{6}+\dfrac{-1}{6}\)

\(=0\)

Bài 2: Tìm x,biết:

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

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

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

Vậy \(x=\dfrac{2}{15}\)

b) \(x-\dfrac{2}{3}=\dfrac{7}{21}\)

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

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

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

Vậy \(x=1\)

c) sai đề hay sao ấy bạn.bỏ dấu - ở x thì đúng đề.mk giải luôn nha!

\(x-\dfrac{3}{4}=\dfrac{-8}{11}\)

\(x=\dfrac{-8}{11}+\dfrac{3}{4}\)

\(x=\dfrac{1}{44}\)

Vậy \(x=\dfrac{1}{44}\)

d) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)

\(\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)

\(\dfrac{2}{5}+x=\dfrac{1}{4}\)

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

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

Vậy \(x=-\dfrac{3}{20}\)

có nhất thiết phải hỏi mấy bài thế này ko ?

có

bài 2:tính hợp lý

1.a) Dễ nhận thấy đề toán chỉ giải được khi đề là tìm x,y. Còn nếu là tìm x ta nhận thấy ngay vô nghiệm. Do đó: Sửa đề: \(\left|x-3\right|+\left|2-y\right|=0\)

\(\Leftrightarrow\left|x-3\right|=\left|2-y\right|=0\)

\(\left|x-3\right|=0\Rightarrow\left\{{}\begin{matrix}x-3=0\\-\left(x-3\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\) (1)

\(\left|2-y\right|=0\Rightarrow\left\{{}\begin{matrix}2-y=0\\-\left(2-y\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\) (2)

Từ (1) và (2) có: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x_1=3\\x_2=-3\end{matrix}\right.\\\left\{{}\begin{matrix}y_1=2\\y_2=-2\end{matrix}\right.\end{matrix}\right.\)

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

c) \(\dfrac{x+1}{35}+\dfrac{x+2}{34}+\dfrac{x+3}{33}=\dfrac{x+4}{32}+\dfrac{x+5}{31}+\dfrac{x+6}{30}\)

\(\Rightarrow\dfrac{x+1}{35}+1+\dfrac{x+2}{34}+1+\dfrac{x+3}{33}+1=\dfrac{x+4}{32}+1+\dfrac{x+5}{31}+1+\dfrac{x+6}{30}+1\)

\(\Rightarrow\dfrac{x+1+35}{35}+\dfrac{x+2+34}{34}+\dfrac{x+3+33}{33}=\dfrac{x+4+32}{32}+\dfrac{x+5+31}{31}+\dfrac{x+6+30}{30}\)

\(\Rightarrow\dfrac{x+36}{35}+\dfrac{x+36}{34}+\dfrac{x+36}{33}=\dfrac{x+36}{32}+\dfrac{x+36}{31}+\dfrac{x+36}{30}\)

\(\Rightarrow\dfrac{x+36}{35}+\dfrac{x+36}{34}+\dfrac{x+36}{33}-\dfrac{x+36}{32}-\dfrac{x+36}{31}-\dfrac{x+36}{30}=0\)

\(\Rightarrow\left(x+36\right)\left(\dfrac{1}{35}+\dfrac{1}{34}+\dfrac{1}{33}+\dfrac{1}{32}+\dfrac{1}{31}+\dfrac{1}{30}\right)=0\)

\(\Rightarrow x+36=0\left(\text{vì }\dfrac{1}{35}+\dfrac{1}{34}+\dfrac{1}{33}+\dfrac{1}{32}+\dfrac{1}{31}+\dfrac{1}{30}\ne0\right)\)

\(\Rightarrow x=-36\)

Vậy ...

a/ Ta có: \(-4\dfrac{3}{5}.2\dfrac{4}{3}\le x\le-2\dfrac{3}{5}:1\dfrac{6}{15}\)

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

\(\Rightarrow\dfrac{-46}{3}\le x\le\dfrac{-13}{5}.\dfrac{15}{21}\)

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

\(\Rightarrow-15,\left(3\right)\le x\le-1,\left(857142\right)\)

Vì x \(\in\) Z nên x \(\in\left\{-1;-2;-3;...;-15\right\}\)

Chúc bạn học tốt!!!