`@` `\text {Ans}`

`\downarrow`

`a,`

`(x-1954) \times 5=50`

`x-1954 = 50 \div 5`

`x-1954 = 10`

`x=10 + 1954`

`x= 1964`

`B.`

`48 - 40 \times x=8`

`40 \times x=48-8`

`40 \times x = 40`

` x=40 \div 40`

`x=1`

`C.`

`2+70 \div x = 3`

`70 \div x = 3-2`

`70 \div x = 1`

`x=70 \div 1`

`x=70`

`D. [3 \times (x+2) \div 7]  \times  4=120`

`3 \times (x+2) \div 7 = 120 \div 4`

`3 \times (x+2) \div 7 =30`

`3 \times (x+2) = 30 \times 7`

`3 \times (x+2)=210`

`x+2=210 \div 3`

`x+2=70`

`x=70-2`

`x=68`

`E.`

`7,2 \div [ (0,6 \times x+8) \div 20 + 59] = 0,12?` Thiếu ngoặc bạn nhé._.

\(a,\left(x-1954\right)\times5=50\\ x-1954=50:5\\ x-1954=10\\ x=1954+10\\ x=1964\\b.48-40\times x=8\\40\times x=48-8\\40\times x=40\\ x=40:40\\ x=1 \\ c.2+70:X=3\\ 70:x=3-2\\ 70:x=1\\ x=70:1\\ x=70\\ d,\left[3\times\left(x+2\right):7\right]\times4=120\\ 3\times\left(x+2\right):7-120:4\\ 3\times\left(x+2\right):7=30\\ \left(x+2\right):7=30:3\\ \left(x+2\right):7=10\\ x+2=10\times7\\ x+2=70\\ x=70-2\\ x=68\)

Mình ghi nhầm đó, lẽ ra phải thế này cơ:\(\left(30:7\frac{1}{2}+0,5x3-1,5\right)x\left(4\frac{1}{2}-\frac{9}{2}\right):\left(14,5x100\right)\)

\((30:7\frac{1}{2}\)\(+0,5x3\)\(-1,5)x(4\frac{1}{2}\)\(-\frac{9}{2}\)\():(14,5x100)\)
 

\(=\)\((30:7\frac{1}{2}\)\(+0,5x3\) \(-1,5)x(\frac{9}{2}\)\(-\frac{9}{2}\)\()\)\(:(14,5x100)\)
\(=30:7\frac{1}{2}\)\(+0,5x3-1,5)x0x(14,5x100)\)
\(=0\)
Chúc bạn học tốt!

Tìm x :

2,8 : x - 1,5 = 2,3 : 5,75

=> x - 1,5 = 5,75

     x         = 5,75 + 1,5

     x         = 7,25

Tính nhanh :

\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{13\times14}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{13}-\frac{1}{14}\)

\(=1-\frac{1}{14}+0+...+0\)

\(=\frac{13}{14}\)

1. 2,8 : x - 1,5 = 2,3 : 5,75

    2,8 : x - 1,5 = 0,4

    2,8 : x = 0,4 + 1,5

    2,8 : x = 1,9

            x = 2,8 : 1,9

            x = 1,473

2. \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{13.14}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{13}-\frac{1}{14}\)

= \(1-\frac{1}{14}\)

= \(\frac{13}{14}\)

(30:15/2+0,5.3-1,5).(9/2-7/2):(14,5-x)=10

<=>(4+1,5-1,5).1:(14,5-x)=10

<=>4:(14,5-x)=10

<=>14,5-x=2/5

<=>x=14,1

a) 187 - {[497 - ( 8 x X + 11) : X] : 3 - 78} = 150

=> {[497 - ( 8 x X + 11) : X] : 3 - 78} = 187 - 150

=> {[497 - (8 x X + 11) : X] : 3 - 78} = 37

=> [497 - (8 x X +11): X ] : 3 - 78 = 37

=> [497 - (8 x X + 11) : X] : 3 = 115

=> 497 - ( 8 x X + 11) : X = 345

=> (8 x X  + 11) : X = 497 - 345 = 152

=> 8X + 11 = 152X

=> 152X - 8X = 11

=> 144X = 11

=> X = 11/144

b) 19,96 + 4,19 - 24,15 : \(\left(x:\frac{1}{4}-\frac{1}{4}\right)=23,15\)

=> 19,96 + 4,19 - 24,15 : \(\left(x\cdot4-\frac{1}{4}\right)=23,15\)

=> 24,15 - 24,15 : \(\left(x\cdot4-\frac{1}{4}\right)\)= 23,15

=> 24,15 : \(\left(x\cdot4-\frac{1}{4}\right)\)= 1

=> \(x\cdot4-\frac{1}{4}=24,15\)

=> \(x\cdot4=24,15+\frac{1}{4}=24,4\)

=> x = 24,4 : 4 = 6,1

Còn câu c tương tự

Để \(x\left(2+x^2\right)\left(7-x\right)=0\hept{\begin{cases}x=0\\2+x^2=0\\7-x=0\end{cases}}\)

Mà \(2+x^2>0\) => \(\orbr{\begin{cases}x=0\\7-x=0\end{cases}}\)=> \(\orbr{\begin{cases}x=0\\x=7\end{cases}}\)

Vậy x = { 0; 7 }

bn có chắc ko,dù sao cũng cảm ơn bn

a.\(y:\dfrac{4}{5}=\dfrac{11}{8}\)

\(y=\dfrac{11}{8}\times\dfrac{4}{5}\)

\(y=\dfrac{11}{10}\)

b.\(\dfrac{11}{3}-y=\dfrac{1}{9}\)

\(y=\dfrac{11}{3}-\dfrac{1}{9}\)

\(y=\dfrac{32}{9}\)

c.\(\dfrac{1}{7}\times x=\dfrac{8}{5}\)

\(x=\dfrac{8}{5}:\dfrac{1}{7}\)

\(x=\dfrac{56}{5}\)

\(y=\dfrac{11}{8}.\dfrac{4}{5}\)

\(y=\dfrac{44}{40}\)

\(x=\dfrac{11}{3}-\dfrac{1}{9}\)

\(x=\dfrac{99-3}{27}=\dfrac{96}{27}\)

\(x=\dfrac{8}{5}:\dfrac{1}{7}\)

\(x=\dfrac{56}{5}\)

a)\(y=\dfrac{4}{5}\times\dfrac{11}{8}=\dfrac{44}{40}=1\dfrac{4}{40}\)

b)\(y=\dfrac{11}{3}-\dfrac{1}{9}=\dfrac{33}{9}-\dfrac{1}{9}=\dfrac{32}{9}\)

c)\(x=\dfrac{8}{5}\times7=\dfrac{56}{5}\)