a)

\(\frac{864\cdot48-432\cdot96}{864\cdot48\cdot432}=\frac{432\cdot2\cdot48-432\cdot96}{864\cdot48\cdot432}=\frac{432\cdot96-432\cdot96}{864\cdot48\cdot432}=\frac{0}{864\cdot48\cdot432}=0\)

Làm từng phần một nha =))

b)

\(\frac{16\cdot17-5}{16\cdot16+11}=\frac{16\cdot\left(16+1\right)-5}{16\cdot16+11}=\frac{16^2+16-5}{16^2+11}=\frac{16^2+11}{16^2+11}=1\)

@_@

câu khó quá mmk ko làm đc

Theo bài ta có: a \(⋮\)48; a\(⋮\)60; a\(⋮\)30            ( a\(\in\)N, a nhỏ nhất khac 0)

=> x \(\in\)BC(48,60,30)

Mà a nhỏ nhất

=> a= BCNN( 48,60,30)

+ 48=2⁴.3       60=2².3.5        30=2.3.5

=> BCNN( 48,60,30)=2⁴.3.5=240

=> x=240

x\(⋮\)120

=> X\(\in\){0;120;240;....}

Mà x nhỏ nhất khác 0

=> x=120

Chúc bn học tốt

Ta có \(\dfrac{2}{1\cdot5}+\dfrac{2}{5\cdot9}+\dfrac{2}{9\cdot13}+...+\dfrac{2}{x\left(x+4\right)}=\dfrac{56}{113}\)

\(\dfrac{1}{2}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{x\left(x+4\right)}\right)=\dfrac{56}{113}\)

\(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{x}-\dfrac{1}{x+4}=\dfrac{56}{113}:\dfrac{1}{2}\)

\(1-\dfrac{1}{x+4}=\dfrac{112}{113}\)

\(\dfrac{1}{x+4}=1-\dfrac{112}{113}=\dfrac{1}{113}\)

x + 4 = 113 ⇒ x = 109

\(\dfrac{2}{1.5}+\dfrac{2}{5.9}+...+\dfrac{2}{x\left(x+4\right)}=\dfrac{56}{113}\)

Xét: \(A=\dfrac{2}{1.5}+\dfrac{2}{5.9}+...+\dfrac{2}{x\left(x+4\right)}\)

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

\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{x+4}\right)\)

Với \(A=\dfrac{56}{113}\) thì

 \(\dfrac{1}{2}.\left(1-\dfrac{1}{x+4}\right)=\dfrac{56}{113}\)

\(\left(1-\dfrac{1}{x+4}\right)=\dfrac{112}{113}\)

\(\dfrac{1}{x+4}=\dfrac{1}{113}\)

\(x=109\)

\(\dfrac{2}{1.5}\) + \(\dfrac{2}{5.9}\)  + \(\dfrac{2}{9.13}\)+...+  \(\dfrac{2}{x\left(x+4\right)}\)   = \(\dfrac{56}{113}\)

(\(\dfrac{2}{1.5}+\dfrac{2}{5.9}+\dfrac{2}{9.13}+....+\dfrac{2}{x\left(x+4\right)}\) ) \(\times\) 2  = \(\dfrac{56}{113}\) \(\times\) 2

\(\dfrac{4}{1.5}\) + \(\dfrac{4}{5.9}\) +\(\dfrac{4}{9.13}\)+...+ \(\dfrac{4}{x\left(x+4\right)}\)                = \(\dfrac{112}{113}\)

\(\dfrac{1}{1}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{9}\) + \(\dfrac{1}{9}\) - \(\dfrac{1}{13}\) +...+ \(\dfrac{1}{x}\) - \(\dfrac{1}{x+4}\) = \(\dfrac{112}{113}\)

1  - \(\dfrac{1}{x+4}\) = \(\dfrac{112}{113}\)

        \(\dfrac{1}{x+4}\) = 1 - \(\dfrac{112}{113}\)

        \(\dfrac{1}{x+4}\) = \(\dfrac{1}{113}\)

         \(x\) + 4 = 113

         \(x\)        = 113 - 4

         \(x\)        = 109

a, => x=0 hoặc x+5=0

=> x=0 hoặc x=-5

b, => 2x-6=0 hoặc -4x-8=0

=> x=3 hoặc x=-2

Tk mk nha

a.x.(x+5)=0 <=>\(\orbr{\begin{cases}x=0\\x+5=0\end{cases}}\)^{<=> \(\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)}

b.(2x-6)(-4x-8)=0 <=> \(\orbr{\begin{cases}2x-6=0\\-4x-8=0\end{cases}}\) <=>\(\orbr{\begin{cases}2x=6\\-4x=8\end{cases}}\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)

a) \(x\left(x+5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x+5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)

Vậy x=0 ; x=-5

b) (2x-6)(-4x-8)=0

\(\Rightarrow\orbr{\begin{cases}2x-6=0\\-4x-8=0\end{cases}}\Rightarrow\orbr{\begin{cases}2x=6\\-4x=8\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)

Vậy x=-2 hoặc x=3

X : 100 + X x 3,99 = 5,2

X x 0,01 + X x 3,99 = 5,2

X x (0,01 + 3,99) = 5,2

X x 4 = 5,2

X = 5,2 : 4

X = 1,3

Nhớ k cho mik nha. Chúc bạn học tốt

cho mk hỏi bn lấy o,o1 ở đâu đó

a) 128 - 3(x + 4) = 23

             3(x + 4) = 128 -23

             3(x + 4) = 105

                x + 4  = 105 : 3

                x + 4  = 35

                x        = 35 - 4

                x        = 31

\(a.128-3\left(x+4\right)=23.\)

\(3\left(x+4\right)=128-23\)

\(3\left(x+4\right)=105\)

\(x+4=105:3=35\)

\(x=35-4=31\)

\(b.\left[\left(4x+28\right)\cdot3+55\right]:5=35\)

\(\left(4x+28\right)\cdot3+55=35\cdot5=175\)

\(\left(4x+28\right)\cdot3=175-55=120\)

\(4x+28=120:3=40\)

\(4x=40-28=12\)

\(x=12:4=3\)

a, \(128-3.\left(x+4\right)=23\)

\(\Rightarrow\) \(3.\left(x+4\right)=105\)

\(\Rightarrow\) \(x+4=35\)

\(\Rightarrow\) \(x=31\)

b, \(\left[\left(4x+28\right).3+55\right]:5=35\)

\(\Rightarrow\) \(\left(4x+28\right).3+55=175\)

\(\Rightarrow\) \(\left(4x+28\right).3=120\)

\(\Rightarrow\) \(4x+28=40\)

\(\Rightarrow\) \(4x=12\)

\(\Rightarrow\) \(x=3\)

4.\(\left(2-x\right)\left(5-x\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}2-x=0\\5-x=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\x=5\end{cases}}}\)

5.\(\left(4x-16\right)\left(8-2x\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}4x-16=0\\8-2x=0\end{cases}\Leftrightarrow\hept{\begin{cases}4x=16\\-2x=-8\end{cases}\Leftrightarrow}\hept{\begin{cases}x=4\\x=4\end{cases}}}\)

Câu 4:

(  2- x ) . ( 5 - x ) = 0

=> 2 - x = 0 hoặc 5 - x = 0

=> x = 2 hoặc x = 5

Vậy x \(\in\){ 2 ; 5 }

Cậu 5:

( 4x - 16 ) . ( 8 - 2x ) = 0

=> 4x - 16 = 0 hoặc 8 - 2x = 0

=> 4x = 16 hoặc 2x = 8

=> x = 4 hoặc x = 4

Vậy x = 4

câu 4 

a)x=2;x=5

câu5

x=4;

~~~~~~~~~ai đi ngang qua nhớ để lại ti k ~~~~~~~~~~~