a, \(x\) : \(\dfrac{13}{3}\) = -2,5

    \(x\)         = -2,5 . \(\dfrac{13}{3}\)

    \(x\)         = \(\dfrac{65}{6}\)

b,\(\dfrac{3}{5}\)\(x\)         = \(\dfrac{1}{10}-\)\(\dfrac{1}{4}\)

   \(\dfrac{3}{5}x\)         = \(\dfrac{-3}{20}\)

      \(x\)          = \(\dfrac{-3}{20}\) :  \(\dfrac{3}{5}\)

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

c, \(\dfrac{25}{9}-\dfrac{12}{13}x=\dfrac{7}{9}\)

              \(\dfrac{12}{13}x\)\(=\dfrac{25}{9}-\dfrac{7}{9}\)

               \(\dfrac{12}{13}x=2\)

                    \(x=2:\dfrac{12}{13}\)

                    \(x=\dfrac{13}{6}\)

Câu 1: D

Câu 2: A

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)

1.

x : \(\dfrac{2}{9}=18\)

x = 18 x \(\dfrac{2}{9}\)

x = 4

\(\Rightarrow\) Chọn C

2. 

\(\dfrac{3}{5}\) của 4 tấn:

Đổi: 4 tấn = 4000 kg.

\(\dfrac{3}{5}\) của 4000 kg là:

  4000 x \(\dfrac{3}{5}\) = 2400 kg

\(\Rightarrow\) Chọn A

3. 25ha = ... km²

Ta có: 1ha = 0,01 km²

\(\Rightarrow\) 25ha = 0,01 x 25 = 0,25 km²

\(\Rightarrow\) Chọn C

`#3107.101107`

1.

`a,`

`2 - 2/3 + 2,5 + 1/3 + 3 + 1/2`

`= (2 + 2,5 + 3) - (2/3 - 1/3 - 1/2)`

`= 7,5 - (-1/6)`

`= 7,5 + 1/6`

`= 23/3`

`b,`

`9/10 - (6/5 * 3/2 + 7/4)`

`= 9/10 - (18/10 + 7/4)`

`= 9/10 - 18/10 - 7/4`

`= -9/10 - 7/4`

`= -53/20`

\(\text{a) }3x+\dfrac{4}{9}=2x+\dfrac{11}{18}\\ \Leftrightarrow3x-2x=\dfrac{11}{18}-\dfrac{4}{9}\\ \Leftrightarrow x=\dfrac{1}{6}\\ \text{Vậy }x=\dfrac{1}{6}\\ \)

\(\text{b) }\dfrac{7}{12}+\dfrac{2}{3}:x=\dfrac{5}{8}\\ \Leftrightarrow\dfrac{2}{3}:x=\dfrac{1}{24}\\ \Leftrightarrow x=16\\ \text{Vậy }x=16\\ \)

\(\text{c) }\left|2.5-x\right|-\dfrac{1}{5}=1.2\\ \Leftrightarrow\left|2.5-x\right|=1.4\\ \Leftrightarrow\left[{}\begin{matrix}2.5-x=-1.4\\2.5-x=1.4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3.9\\x=1.1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{39}{10}\\x=\dfrac{11}{10}\end{matrix}\right.\\ \text{Vậy }x=\dfrac{39}{10}\text{ hoặc }x=\dfrac{11}{10}\\ \)

\(\text{d) }2^{x+1}+2^{x+2}=192\\ \Leftrightarrow2^x\cdot2+2^x\cdot4=192\\ \Leftrightarrow2^x\left(2+4\right)=192\\ \Leftrightarrow2^x\cdot6=192\\ \Leftrightarrow2^x=32\\ \Leftrightarrow2^x=2^5\\ \Leftrightarrow x=5\\ \text{Vậy }x=5\\ \)

a) Ta có: |x- 1,5| + |2,5-x| =0

=>x-1,5=0 và 2,5 - x =0

=> x = 1,5 và x = 2,5

Vậy x thuộc {1,5; 2,5}

Còn câu b mik hk bt lm

phần b cậu vào câu hỏi của tớ mà xem

\(\left|x-1,5\right|+\left|2,5-x\right|\ge\left|x-1,5+2,5-x\right|=1>0\)

Không có giá trị \(x\) thỏa mãn

\(\dfrac{x+1}{9}+\dfrac{x+1}{8}=\dfrac{x+1}{11}+\dfrac{x+1}{12}\)

\(\Rightarrow\dfrac{x+1}{9}+\dfrac{x+1}{8}-\dfrac{x+1}{11}-\dfrac{x+1}{12}=0\)

\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{9}+\dfrac{1}{8}-\dfrac{1}{11}-\dfrac{1}{12}\right)=0\)

Vì \(\dfrac{1}{9}+\dfrac{1}{8}-\dfrac{1}{11}-\dfrac{1}{12}\ne0\Leftrightarrow x=-1\)

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

b)\(=-\dfrac{35}{9}-\dfrac{10}{9}=-\dfrac{45}{9}=-5\)

c)\(=\left[-10+9\right].\left[1,3.\left(2,5+7,5\right)\right]=-1.\left[1,3.10\right]=-1.13=-13\)

a) Ta có: \(\dfrac{x}{-15}=\dfrac{-60}{x}\)

\(\Leftrightarrow x^2=\left(-15\right)\cdot\left(-60\right)=900\)

hay \(x\in\left\{30;-30\right\}\)

Vậy: \(x\in\left\{30;-30\right\}\)

b) Ta có: \(\left|x\right|+0.573=2\)

\(\Leftrightarrow\left|x\right|=1.427\)

hay \(x\in\left\{1.427;-1.427\right\}\)

Vậy: \(x\in\left\{1.427;-1.427\right\}\)

c) Ta có: \(\left|x+\dfrac{1}{3}\right|-4=-1\)

\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=3\\x+\dfrac{1}{3}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{10}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{8}{3};-\dfrac{10}{3}\right\}\)

d) Ta có: \(0.01:2.5=\left(0.75x\right):0.75\)

\(\Leftrightarrow\dfrac{0.75\cdot x}{0.75}=\dfrac{0.01}{2.5}\)

\(\Leftrightarrow x=\dfrac{1}{250}\)

Vậy: \(x=\dfrac{1}{250}\)