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10 tháng 10 2016

a) (x + 1/2)^2 = 1/16

=> (x + 1/2)^2 = (1/4)^2 hoặc (x + 1/2)^2 = (-1/4)^2

=> x + 1/2 = 1/4 hoặc x + 1/2 = -1/4

* x + 1/2 = 1/4

   x           = 1/4 - 1/2

   x           = -1/4

* x + 1/2 = -1/4

   x           = -1/4 - 1/2

   x           = -3/4

Vậy x = -1/4 hoặc x = -3/4

b) 2^x+2 - 2^x      = 9^6

=> 2^x . 2^2 - 2^x = 9^6

=> 2^x . (2^2 - 1) = 9^6

=> 2^x . (4 - 1)     = 9^6

=> 2^x . 3             = (3^2)^6

=> 2^x . 3             = 3^12

=> 2^x                  = 3^12 : 3

=> 2^x                  = 3^11

Vì 3^11 không chia hết cho 2

=> Không có giá trị nào của x thõa mãn đề bài

c)  (3^x)^2 : 3^3  = 1/243

=>   3^2x = 1/243 . 3^3

=>   3^2x = 1/243 . 27

=>   3^2x = 1/9

=>   3^2x . 9 = 1

=>  3^2x . 3^2 = 1

=>  3^2x+2 = 1

=>  3^2x+2 = 3^0

=> 2x + 2 = 0

=> 2x   = 0 - 2

=> 2x   = -2

=> x = -2 : 2

=> x = -1

1 tháng 10 2016

a\(\left(x+\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2 \)
\(\Rightarrow x+\frac{1}{2}=\frac{1}{4}\)
\(\Leftrightarrow x=\frac{1}{4}-\frac{1}{2}\)
\(\Rightarrow x=\frac{-1}{4}\)
 

13 tháng 6 2019

\(\left(\frac{5}{x+3}-2\right).4=7-\left(\frac{9}{x+3}+\frac{1}{2}\right).2\)

\(\Leftrightarrow\frac{20}{x+3}-8=7-\frac{18}{x+3}+1\)

\(\Leftrightarrow\frac{20}{x+3}-8=8-\frac{18}{x+3}\)

\(\Leftrightarrow\frac{20}{x+3}+\frac{18}{x+3}=8+8\)

\(\Leftrightarrow\frac{38}{x+3}=16\)

\(\Leftrightarrow x+3=2,375\)

\(\Leftrightarrow x=-0,625\)

13 tháng 6 2019

\(\left(\frac{5}{x+3}-2\right).4=7-\left(\frac{9}{x+3}+\frac{1}{2}\right).2\)

\(\Leftrightarrow\frac{20}{x+3}-8=7-\left(\frac{18}{x+3}+1\right)\)

\(\Leftrightarrow\frac{20}{x+3}-8=7-\frac{18}{x+3}-1\)

\(\Leftrightarrow\frac{20}{x+3}+\frac{18}{x+3}=7-1+8\)

\(\Leftrightarrow\frac{38}{x+3}=14\)

\(\Leftrightarrow\left(x+3\right)14=38\)

\(\Leftrightarrow14x+42=38\)

\(\Leftrightarrow14x=-4\Leftrightarrow x=-\frac{4}{14}=-\frac{2}{7}\)

Vậy \(x=-\frac{2}{7}\)

1 tháng 8 2018

a) Ta có : (x - 5)2 - 16

= (x - 5)2 - 42

= (x - 5 - 4)(x - 5 + 4)

= (x - 1)(x - 9)

b) 25 - (3 - x)2

= 52 - (3 - x)2

= (5 - 3 + x)(5 + 3 - x)

= (x + 2)(8 - x)

1 tháng 8 2018

c) (7x - 4)2 - (2x + 1)2

= (7x - 4 - 2x - 1)(7x - 4 + 2x + 1)

= (5x - 5)(9x - 3)

= 5(x - 1)3(3x - 1)

= 15(x - 1)(3x - 1)

20 tháng 6 2018

\(\Leftrightarrow2.\left(\frac{-1}{2}\right).\left(\frac{2}{3}\right)^2-3\left(-\frac{1}{3}\right)^2.\frac{2}{9}:x=3.\left(-\frac{1}{2}\right)-\frac{2}{3}\)

\(\Leftrightarrow-\frac{4}{9}-\frac{1}{3}.\frac{2}{9}:x=-\frac{3}{2}-\frac{2}{3}\)

\(\Leftrightarrow-\frac{4}{6}-\frac{2}{27}:x=-\frac{13}{6}\)

\(\Leftrightarrow\frac{2}{27}:x=-\frac{4}{9}:\frac{-13}{6}\)

\(\Leftrightarrow\frac{2}{27}:x=\frac{31}{18}\)

\(\Leftrightarrow x=\frac{2}{27}:\frac{31}{18}\)

\(\Rightarrow x=\frac{4}{93}\)

Vậy \(x=\frac{4}{93}\)

\(-4\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)

\(\Rightarrow-\frac{13}{3}.\left(\frac{3}{6}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{4}{12}-\frac{6}{12}-\frac{9}{12}\right)\)

\(\Rightarrow-\frac{13}{3}.\frac{2}{6}\le x\le-\frac{2}{3}.\frac{-11}{12}\)

\(\Rightarrow\frac{-13}{9}\le x\le\frac{11}{18}\)

\(\Rightarrow\frac{-26}{18}\le x\le\frac{11}{18}\)

=> -1,44444444444........... ≤ x ≤ 0,6111111111...........

Mà x ∈ Z

=> x ∈ { -1 ; 0 }

14 tháng 7 2021

\(x\in\varnothing\) 

21 tháng 7 2019

a) \(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)

=> \(\left(\frac{x-6}{7}+1\right)+\left(\frac{x-7}{8}+1\right)+\left(\frac{x-8}{9}+1\right)=\left(\frac{x-9}{10}+1\right)+\left(\frac{x-10}{11}+1\right)+\left(\frac{x-11}{12}+1\right)\)

=> \(\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}+\frac{x+1}{12}=0\)

=> \(\left(x+1\right)\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)=0\)

=>  x + 1 = 0

=> x = -1

21 tháng 7 2019

b) \(\frac{x-1}{2020}+\frac{x-2}{2019}-\frac{x-3}{2018}=\frac{x-4}{2017}\)

=> \(\left(\frac{x-1}{2020}-1\right)+\left(\frac{x-2}{2019}-1\right)-\left(\frac{x-3}{2018}-1\right)=\left(\frac{x-4}{2017}-1\right)\)

=> \(\frac{x-2021}{2020}+\frac{x-2021}{2019}-\frac{x-2021}{2018}=\frac{x-2021}{2017}\)

=> \(\left(x-2021\right)\left(\frac{1}{2020}+\frac{1}{2019}-\frac{1}{2018}-\frac{1}{2017}\right)=0\)

=> x - 2021 = 0

=> x = 2021

c) \(\left(\frac{3}{4}x+3\right)-\left(\frac{2}{3}x-4\right)-\left(\frac{1}{6}x+1\right)=\left(\frac{1}{3}x+4\right)-\left(\frac{1}{3}x-3\right)\)

=> \(\frac{3}{4}x+3-\frac{2}{3}x+4-\frac{1}{6}x-1=\frac{1}{3}x+4-\frac{1}{3}x+3\)

=> \(-\frac{1}{12}x+6=7\)

=> \(-\frac{1}{12}x=1\)

=> x = -12

cho 3 k 

\(\left(1-\frac{1}{2^2}\right)\cdot\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{10^2}\right)\)

=> \(\left(1-\frac{1}{2}\right)\left(1+\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1+\frac{1}{3}\right)\)\(...\left(1-\frac{1}{10}\right)\cdot\left(1+\frac{1}{10}\right)\)

=> \(\left(1-\frac{1}{2}\right)\cdot\frac{3}{2}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\cdot\cdot\frac{9}{10}\cdot\frac{10}{11}\)

=> \(\frac{1}{2}\cdot\frac{3\cdot2\cdot4\cdot\cdot\cdot9\cdot10}{2\cdot3\cdot3\cdot\cdot\cdot10\cdot11}=\frac{1}{2}\cdot\frac{11}{10}=\frac{11}{20}\)

Chúc bn học tốt !

cho mk 3 k nha bn

thanks nhìu

bài này mk ko copy, ko chép mạng, tự nghĩ mất 6 phút . 

có công thức rùi nha !

chúc bn học tốt