\(\frac{1}{2}\)|+|x+\(\frac{1}{3}\)|+|x+
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10 tháng 10 2020

Ta có: \(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{4}\right|\ge0\)

\(\Rightarrow4x\ge0\Rightarrow x\ge0\)

\(\Rightarrow x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{4}=4x\)

\(\Leftrightarrow3x+\frac{13}{12}=4x\)

\(\Leftrightarrow x=\frac{13}{12}\left(tm\right)\)

25 tháng 8 2020

a) \(\frac{4}{9}x+\frac{2}{5}-\frac{1}{3}x=\frac{2}{9}-\frac{1}{4}x\)

\(\Leftrightarrow\frac{13}{36}x=-\frac{8}{45}\)

\(\Rightarrow x=-\frac{32}{65}\)

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

\(\Leftrightarrow-\frac{4}{9}x+\frac{1}{3}+\frac{1}{5}=-\frac{3}{4}\)

\(\Leftrightarrow\frac{4}{9}x=\frac{77}{60}\)

\(\Rightarrow x=\frac{231}{80}\)

25 tháng 8 2020

a) \(\frac{4}{9}x+\frac{2}{5}-\frac{1}{3}x=\frac{2}{9}-\frac{1}{4}x\)

=> \(\frac{4}{9}x-\frac{1}{3}x+\frac{2}{5}-\frac{2}{9}+\frac{1}{4}x=0\)

=> \(\left(\frac{4}{9}x-\frac{1}{3}x+\frac{1}{4}x\right)+\left(\frac{2}{5}-\frac{2}{9}\right)=0\)

=> \(\frac{13}{36}x+\frac{8}{45}=0\)

=> \(\frac{13}{36}x=-\frac{8}{45}\)

=> \(x=-\frac{32}{65}\)

b) \(\left(\frac{2}{3}x-\frac{1}{2}\right)\cdot\frac{-2}{3}+\frac{1}{5}=\frac{-3}{4}\)

=> \(\left(\frac{2}{3}x-\frac{1}{2}\right)\cdot\frac{-2}{3}=-\frac{19}{20}\)

=> \(\frac{2}{3}x-\frac{1}{2}=\left(-\frac{19}{20}\right):\left(-\frac{2}{3}\right)=\left(-\frac{19}{20}\right)\cdot\left(-\frac{3}{2}\right)=\frac{57}{40}\)

=> \(\frac{2}{3}x=\frac{57}{40}+\frac{1}{2}=\frac{77}{40}\)

=> \(x=\frac{77}{40}:\frac{2}{3}=\frac{77}{40}\cdot\frac{3}{2}=\frac{231}{80}\)

27 tháng 9 2019

a)\(\left(\frac{4}{5}\right)^{2x+7}=\left(\frac{4}{5}\right)^4\)

=> 2x + 7 = 4 

     2x        = 4 - 7 

     2x        = -3

       x        = -3 : 2

       x         = -1,5

   Vậy x = -1,5

30 tháng 5 2018

\(c)\)

\(2x-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-...-\frac{1}{49.50}=\left(7-\frac{1}{50}+x\right)\)

\(\Rightarrow2x-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{49.50}\right)=\left(\frac{350}{50}-\frac{1}{50}+x\right)\)

\(\Rightarrow2x-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\right)=\frac{349}{50}+x\)

\(\Rightarrow2x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)-x=\frac{349}{50}\)

\(\Rightarrow x-\left(1-\frac{1}{50}\right)=\frac{349}{50}\)

\(\Rightarrow x-\frac{49}{50}=\frac{349}{50}\)

\(\Rightarrow x=\frac{349}{50}+\frac{49}{50}\)

\(\Rightarrow x=\frac{199}{25}\)

Vậy \(x=\frac{199}{25}\)

~ Ủng hộ nhé 

30 tháng 5 2018

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

\(\Rightarrow2x-3-x=\frac{1}{2}\)

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

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

\(\Rightarrow x=\frac{1}{2}+\frac{6}{2}\)

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

Vậy \(x=\frac{7}{2}\)

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

\(\Rightarrow4.x-2.x-1=\frac{9}{3}-\frac{1}{3}+x\)

\(\Rightarrow2.x-1=\frac{8}{3}+x\)

\(\Rightarrow2x-1-x=\frac{8}{3}\)

\(\Rightarrow x-1=\frac{8}{3}\)

\(\Rightarrow x=\frac{8}{3}+1\)

\(\Rightarrow x=\frac{8}{3}+\frac{3}{3}\)

\(\Rightarrow x=\frac{11}{3}\)

Vậy \(x=\frac{11}{3}\)

~ Ủng hộ nhé 

11 tháng 10 2020

a) Đặt \(\frac{x}{-2}=\frac{y}{-3}=k\Rightarrow\hept{\begin{cases}x=-2k\\y=-3k\end{cases}}\)

Khi đó 4x - 3y = 9

<=> -8k + 9k = 9

=> k = 9

=> x = -18 ; y = -27

b) Ta có : \(2x=3y\Rightarrow\frac{2x}{6}=\frac{3y}{6}\Rightarrow\frac{x}{2}=\frac{y}{3}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có : 

\(\frac{x}{2}=\frac{y}{3}=\frac{x+y}{2+3}=\frac{10}{5}=2\)

=> x = 4 ; y = 6 

c) Đặt \(\frac{x}{3}=\frac{y}{4}=k\Rightarrow\hept{\begin{cases}x=3k\\y=4k\end{cases}}\)

Khi đó (3k)2 + (4k)2 = 100

<=> 9k2 + 16k2 = 100

=> 25k2 = 100

=> k2 = 4

=> k = \(\pm\)2

Khi k = 2 => x = 6 ; y = 8

Khi k = -2 =>  x = -6 ; y = -8

Vậy các cặp (x;y) thỏa mãn cần tìm là (6;8);(-6;-8)

d) Đặt \(\frac{x}{3}=\frac{y}{4}=k\Rightarrow\hept{\begin{cases}x=3k\\y=4k\end{cases}}\)

Khi đó x3 + y3 = 91 

<=> (3k)3 + (4k)3 = 91

=> 27k3 + 64k3 = 91

=> 91k3 = 91

=> k3 = 1

=> k = 1

=> x = 3 ; y = 4

e) Đặt \(\frac{x}{5}=\frac{y}{4}=k\Rightarrow\hept{\begin{cases}x=5k\\y=4k\end{cases}}\) 

Khi đó x2y = 100

<=> (5k)2.4k = 100

=> 25k2.4k = 100

=> 100k3 = 100

=> k = 1

=> x = 5 ; y = 4

Bài 1 : Thực hiện phép tính(1) D = \(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{16}\left(1+2+...+16\right)\)(2) M =\(\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)Bài 2 : Tìm x biết(1) \(x-\left\{x-\left[x-\left(-x+1\right)\right]\right\}=1\)(2) \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right]\cdot...
Đọc tiếp

Bài 1 : Thực hiện phép tính

(1) D = \(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{16}\left(1+2+...+16\right)\)

(2) M =\(\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

Bài 2 : Tìm x biết

(1) \(x-\left\{x-\left[x-\left(-x+1\right)\right]\right\}=1\)

(2) \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right]\cdot x=\frac{2015}{1}+\frac{2014}{2}+...+\frac{1}{2015}\)

(3) \(\frac{x}{\left(a+5\right)\left(4-a\right)}=\frac{1}{a+5}+\frac{1}{4-a}\)

(4) \(\frac{x+2}{11}+\frac{x+2}{12}+\frac{x+2}{13}=\frac{x+2}{14}+\frac{x+2}{15}\)

(5) \(\frac{x+1}{2015}+\frac{x+2}{2014}+\frac{x+3}{2013}+\frac{x+4}{2012}+4=0\)

Bài 3 : 

(1) Cho : A =\(\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+...+\frac{1}{9}\); B =\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\)

CMR : \(\frac{A}{B}\)Là 1 số nguyên

(2) Cho : D =\(\frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+...+\frac{1}{2000}\)CMR : \(D< \frac{3}{4}\)

Bài 4 : Ký hiệu [x] là số nguyên lớn nhất không vượt quá x , gọi là phần nguyên của x.

VD : [1.5] =1 ; [3] =3 ; [-3.5] = -4

(1) Tính :\(\left[\frac{100}{3}\right]+\left[\frac{100}{3^2}\right]+\left[\frac{100}{3^3}\right]+\left[\frac{100}{3^4}\right]\)

(2) So sánh : A =\(\left[X\right]+\left[X+\frac{1}{5}\right]+\left[X+\frac{2}{5}\right]+\left[X+\frac{3}{5}\right]+\left[X+\frac{4}{5}\right]\)và B = [5x]. Biết x=3.7

0
24 tháng 6 2018

\(a)\) \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)

\(\Leftrightarrow\)\(\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+4}{96}+1\right)=-4+4\)

\(\Leftrightarrow\)\(\frac{x+1+99}{99}+\frac{x+2+98}{98}+\frac{x+3+97}{97}+\frac{x+4+96}{96}=0\)

\(\Leftrightarrow\)\(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)

\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)

Vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\)

Nên \(x+100=0\)

\(\Rightarrow\)\(x=-100\)

Vậy \(x=-100\)

Chúc bạn học tốt ~ 

24 tháng 6 2018

\(b)\) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{2008}{2009}\)

\(\Leftrightarrow\)\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2008}{2009}\)

\(\Leftrightarrow\)\(1-\frac{1}{x+1}=\frac{2008}{2009}\)

\(\Leftrightarrow\)\(\frac{1}{x+1}=1-\frac{2008}{2009}\)

\(\Leftrightarrow\)\(\frac{1}{x+1}=\frac{1}{2009}\)

\(\Leftrightarrow\)\(x+1=2009\)

\(\Leftrightarrow\)\(x=2009-1\)

\(\Leftrightarrow\)\(x=2008\)

Vậy \(x=2008\)

Chúc bạn học tốt ~