a) ( 1/2-1/3-1/6).(1/2+2/3+3/4+...+2017/2018) + 3/4.x = 9/10

0.(1/2+2/3+3/4+...+2017/2018) + 3/4.x = 9/10

0+3/4.x = 9/10

3/4.x = 9/10

x = 9/10: 3/4

x = 6/5

b) x + ( 3/1.3+3/3.5+...+3/13.15) = 11/5

x + 3/2. ( 1-1/3 + 1/3 - 1/5 + ...+ 1/13 - 1/15) = 11/5

x + 3/2. ( 1-1/15) = 11/5

x + 3/2.14/15 = 11/5

x + 7/5 = 11/5

x = 11/5 - 7/5

x = 4/5

..... là gì?

a) ( 1/2-1/3-1/6).(1/2+2/3+3/4+...+2017/2018) + 3/4.x = 9/10

0.(1/2+2/3+3/4+...+2017/2018) + 3/4.x = 9/10

0+3/4.x = 9/10

3/4.x = 9/10

x = 9/10: 3/4

x = 6/5

b) x + ( 3/1.3+3/3.5+...+3/13.15) = 11/5

x + 3/2. ( 1-1/3 + 1/3 - 1/5 + ...+ 1/13 - 1/15) = 11/5

x + 3/2. ( 1-1/15) = 11/5

x + 3/2.14/15 = 11/5

x + 7/5 = 11/5

x = 11/5 - 7/5

x = 4/5

đơn giản 

......dễ.....

áp dụng công thức 1/x.(x+1)=1/x-1/(x+1) là ra mà bạn

Giải chi tiết ra cho mình với

a) (x-2)² -(x-3)(x-3)=6

=>x² -4x+4-x²+3=6

=>7-4x=6

=>4x=1 =>x=\(\frac{1}{4}\)

b)4(x-3)² -(2x-1)(2x+1)=10

=>4(x²-6x+9)-4x²+1=10

=>4x²-24x+36-4x²+1=10

=>37-24x=10 =>24x=27 =>x=\(\frac{9}{8}\)

c)x²-16-3(x+4)=0

=>(x-4)(x+4)-3(x+4)=0

=>(x-7)(x+4)=0

=>\(\orbr{\begin{cases}x-7=0\\x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-4\end{cases}}}\)

=>x\(\in\left\{-4;7\right\}\)

d)(x-4)²-(x-2)(x+2)=6

=>x²-8x+16-x²+4=6

=>20-8x=6

=>8x=14 =>x=\(\frac{4}{7}\)

e) 9(x+1)²-(3x-2)(3x+2)=10

=>9(x² +2x+1)-9x²+4=10

=>9x²+18x+9-9x²+4=10

=>18x+13=10

=>18x=-3

=>x=\(\frac{-1}{6}\)

mình chỉ làm bài 1 nha

nhớ chon mk đúng nha

Cảm ơn bạn nha . Ai giúp mình làm bài 2 với TT

2,

B=-2x²+6x-11=-2(x²-3x+\(\frac{4}{9}\)) - \(\frac{91}{9}\)

=-2(x+\(\frac{2}{3}\))²-\(\frac{91}{9}\)

vì (x+\(\frac{2}{3}\))² \(\ge\)0  với mọi x\(\in\)R

=> -2(x+\(\frac{2}{3}\))²\(\le\)0 với mọi x\(\in\)R

=> -2(x+\(\frac{2}{3}\))² -\(\frac{91}{9}\)\(\le-\frac{91}{9}\)với mọi x\(\in\)R

dấu bằng xảy ra khi -2(x+\(\frac{2}{3}\))² =0 hay x+\(\frac{2}{3}\)=0 =>x=-\(\frac{2}{3}\)

vậy GTLN của B=-\(\frac{91}{9}\)khi x=-\(\frac{2}{3}\)

\(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right):2}=\frac{2009}{2011}\)

\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2009}{2011}\)

\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)

\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{2011}:2\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)

\(\frac{1}{x+1}=\frac{1}{2}-\frac{2009}{4022}\)

\(\frac{1}{x+1}=\frac{1}{2011}\)

=>x+1=2011

=>x=2010

419 bn nha 

= 419^^

         HT

lớp 6 ai chơi được