x.5+x.6-x.7=78x12

x.(5+6-7)=78x12

x.4=936

x   =936:4

x   =234

X x ( 5 + 6 - 7 ) = 936

X x 4 = 936

X = 936 : 4

X = 234

\(x\cdot5+x\cdot6-x\cdot7=78\cdot12\)

\(\Leftrightarrow x\left(5+6-7\right)=936\)

\(\Leftrightarrow4x=936\)

\(\Leftrightarrow x=234\)

dau . la dau x

a/ 1.3.2.4.3.5.4.6.5.7/2.2.3.3.4.4.5.5.6.6=1.7/2.6=7/12

b/ ab.aba=abab

        aba=abab:ab

        aba=101

=>a=1     b=0

aabb : ab = 99 hay ab x 99 = aabb hay  ab x100 – ab = aabb

Ta có phép tính

                  __  ab00

                      ___ab___

                        aabb

 b=0 hoặc b=5

Nếu b=0 thì    a000 – a0 = aa00  (sai)

Nếu b=5 thì  

                   __  a500

                        __a5___

                        aa55

            a=4

c) thay a=7/6 b=6/5 thi 3 x a + 4 : b - 5/12=3.7/6+4.6/5-5/12=7/2+24/5-5/12=210/60+288/60-25/60=473/60

**** nha

\(\frac{1.3.2.4.3.5.4.6.5.7}{2.2.3.3.4.4.5.5.6.6}=\frac{\left(2.3.4.5.6\right).\left(3.4.5.7\right)}{\left(2.3.4.5.6\right).\left(2.3.4.5.6\right)}=\frac{7}{12}\)

 chỉ trả lời được câu a thôi: phần a bằng 7/12

Ta có: 

\(\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+.....+\frac{1}{14\times15}=\frac{4-3}{3\times4}+\frac{5-4}{4\times5}+\frac{6-5}{6\times5}+...+\frac{15-14}{14\times15}\)

\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{14}-\frac{1}{15}=\frac{1}{3}-\frac{1}{15}=\frac{4}{15}\)

Do đó: \(\frac{4}{15}=\frac{x}{30}\)

               \(x=\frac{4}{15}\times30=8\)

\(\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{14\times15}=\frac{x}{30}\)

\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{5}+...+\frac{1}{14}-\frac{1}{15}=\frac{x}{30}\)

\(\frac{1}{3}-\frac{1}{15}=\frac{x}{30}\)

\(\frac{4}{15}=\frac{x}{30}\)

\(\Rightarrow4\times30=15\times x\)

\(120=15\times x\)

\(x=120\div15\)

\(x=8.\)

\(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{14.15}=\frac{x}{30}\)

\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{14}-\frac{1}{15}=\frac{x}{30}\)

\(\frac{1}{3}-\frac{1}{15}=\frac{x}{30}\)

\(\frac{5}{15}-\frac{1}{15}=\frac{x}{30}\)

\(\frac{4}{15}=\frac{x}{30}\)

\(\frac{8}{30}=\frac{x}{30}\)

\(x=8\)

hok tốt!!

\(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{14.15}=\frac{x}{30}\)

\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{14}-\frac{1}{15}=\frac{x}{30}\)

\(\frac{1}{3}-\frac{1}{15}=\frac{x}{30}\)

\(\frac{4}{15}=\frac{x}{30}\)

\(\Leftrightarrow120=15x\Leftrightarrow x=8\)

34 - x = 12

       x = 34 - 12

      x = 22

x X 4 = 8 

      x = 8 : 4 

      x = 2

x : 9 = 5

     x = 5 x 9 

     x =45

tk mk nha

~ chúc bạn học giỏi ~

\(34-x=12\)

\(x=34-12\)

\(x=22\)

\(x\times4=8\)

\(x=8\div4\)

\(x=2\)

\(x\div9=5\)

\(x=5\times9\)

\(x=45\)

34 - x = 12                                            

       x = 34 - 12 

       x = 22

 vậy x = 22

x . 4 = 8

     x = 8 : 2 

     x = 4

x : 9 = 5

     x = 5 . 9 

     x = 45

rút gọn là được mà ! 

\(\frac{1}{2}\)

\(\frac{7}{14}\)

=546 k mik nha Aikatsu Starts Hime

chúc bạn học tốt ^^

12 + 89 x 6 = 546

= 546 bạn nhé !

6x4=24

Số cần tìm là :

6 x 4 = 24

Đáp số : 24

6 x 4 = 24 nha bạn

\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)

\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)

\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)

\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

\(\Rightarrow x+2004=0\Rightarrow x=-2004\)

\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\Rightarrow\dfrac{x+4}{2000}+\dfrac{x+3}{2001}-\dfrac{x+2}{2002}-\dfrac{x+1}{2003}=0\)

\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1-\dfrac{x+2}{2002}-1-\dfrac{x+1}{2003}-1=0\)

\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)

\(\Rightarrow x+2004\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

\(\Rightarrow x+2004=0\)

\(\Rightarrow x=-2004\)

Vậy \(x=-2004\)

1/ Ta có :

\(\dfrac{1}{1\times2}+\dfrac{1}{3\times4}+\dfrac{1}{5\times6}+....+\dfrac{1}{49\times50}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+.....+\dfrac{1}{49}-\dfrac{1}{50}\)

\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+....+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+.....+\dfrac{1}{50}\right)\)

\(\Rightarrow\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{50}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+....+\dfrac{1}{50}\right)\times2\)

\(\Rightarrow\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{25}\right)\)

\(\Rightarrow\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+.....+\dfrac{1}{50}=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+.....+\dfrac{1}{50}\)

Hay \(\dfrac{1}{1\times2}+\dfrac{1}{3\times4}+\dfrac{1}{5\times6}+...+\dfrac{1}{49\times50}=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}\)

~ Học tốt nha ~