\(\frac{1}{2}+\frac{5}{14}+\frac{2}{63}+\frac{3}{108}+\frac{1}{156}\)\(=\frac{12}{13}\)

TL:

\(\frac{1}{2}+\frac{5}{14}+\frac{2}{63}+\frac{3}{108}+\frac{1}{156}\)

\(=\frac{12}{13}\)

TL :

\(\frac{1}{2}+\frac{5}{14}+\frac{2}{63}+\frac{3}{108}+\frac{1}{156}\)

\(=\frac{12}{13}\)

Hok tốt

ê bạn ơi tìm y sao lại có x

\(\frac{2}{3}\cdot y-\frac{12}{3}:\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\right)=\frac{1}{3}\)\(\frac{1}{3}\)

\(\frac{2}{3}\cdot y-4:\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}\right)=\frac{1}{3}\)

\(\frac{2}{3}\cdot y-4:\left(\frac{3-1}{1\cdot3}+\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+\frac{9-7}{7\cdot9}+\frac{11-9}{9\cdot11}+\frac{13-11}{11\cdot13}\right)=\frac{1}{3}\)

\(\frac{2}{3}\cdot y-4:\left(1+\frac{1}{3}-\frac{1}{3}+\frac{1}{5}-\frac{1}{5}+\frac{1}{7}-\frac{1}{7}+\frac{1}{9}-\frac{1}{9}+\frac{1}{11}-\frac{1}{11}+\frac{1}{13}\right)\)\(=\frac{1}{3}\)

\(\frac{2}{3}\cdot y-4:\left(\frac{1}{1}+\frac{1}{3}\right)=\frac{1}{3}\)

\(\frac{2}{3}\cdot y-4:\frac{4}{3}\)\(=\frac{1}{3}\)

\(\frac{2}{3}\cdot y-4\cdot\frac{3}{4}=\frac{1}{3}\)

\(\frac{2}{3}\cdot y-3=\frac{1}{3}\)

\(\frac{2}{3}\cdot y=\frac{1}{3}+3\)

\(\frac{2}{3}\cdot y=\frac{10}{3}\)

\(y=\frac{10}{3}:\frac{2}{3}\)

y=5

dấu nhân bạn ạ

7/2 nhé bạn

    \(\frac{29}{12}:\frac{1}{2}-\frac{5}{12}:\frac{1}{2}-\frac{1}{2}\)

=  \(\left(\frac{29}{12}-\frac{5}{12}\right):\frac{1}{2}-\frac{1}{2}\)

=   \(2:\frac{1}{2}-\frac{1}{2}\)

=  \(4-\frac{1}{2}\)

=\(\frac{8}{2}-\frac{1}{2}\)

=\(\frac{7}{2}\)

Cảm ơn các bạn nhiều nhé

S=1/5=(1/13+1/14+1/15)+(1/61+1/62+1/63)

suy ra S<1/5+1/12x3+1/60x3

S<1/5+1/4+1/20

=>S<1/2

1) a. ( 2x + 1 ) = 125

=> 2x +1 = 125

=> 2x = 125 - 1 

=> 2x = 124

=> x = 124 : 2

=> x = 64

Vậy x = 64

2) 3^ 2x +1 = 243

=> 3^ 2x +1 = 3^5

=> 2x +1 = 5

=> 2x= 5-1

=> 2x = 4

=> x= 4:2 

=> x = 2

Vậy x = 2

a) x⁴+x³+2x²+x+1=(x⁴+x³+x²)+(x²+x+1)=x²(x²+x+1)+(x²+x+1)=(x²+x+1)(x²+1)

b)a³+b³+c³-3abc=a³+3ab(a+b)+b³+c³ -(3ab(a+b)+3abc)=(a+b)³+c³-3ab(a+b+c)

=(a+b+c)((a+b)²-(a+b)c+c²)-3ab(a+b+c)=(a+b+c)(a²+2ab+b²-ac-ab+c²-3ab)=(a+b+c)(a²+b²+c²-ab-ac-bc)

c)Đặt x-y=a;y-z=b;z-x=c

a+b+c=x-y-z+z-x=o

đưa về như bài b

d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung

e)x²(y-z)+y²(z-x)+z²(x-y)=x²(y-z)-y²((y-z)+(x-y))+z²(x-y)

=x²(y-z)-y²(y-z)-y²(x-y)+z²(x-y)=(y-z)(x²-y²)-(x-y)(y²-z²)=(y-z)(x²-2y²+xy+xz+yz)

0,2x-2/3(x+1)=1/3

<=>0,2x-2/3x-2/3=1/3

<=>-7/15x=1

<=>x=-15/7

Vậy..............

Hok tốt

\(0,2x-\frac{2}{3}\left(x+1\right)=\frac{1}{3}\)

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

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

\(-\frac{7}{15}x=-\frac{1}{3}\)

\(x=\frac{5}{7}\)

   0,2x - 2/3 .(x+1)=1/3

=> 1/5 .x - 2(x+1)/3 =1/3 

=> x/5 -(2x+2)/3 -1/3 =0

=> x/5 - (2x+2-1)/3=0

=> x/5 -(2x+1)/3=0

=> 3x/15 -5(2x+1)/15=0

=> (3x-5(2x+1))=0

=> 3x-10x-5=0

=>-7x-5=0

=> -7x=5

=> x=-7/5=-1/4.

Vậy x= -7/5