\(\Rightarrow x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{4}+x+\frac{1}{5}-x+\frac{1}{6}=0\)  

\(\Rightarrow3x+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\)

k cho minh

\(x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{4}+x+\frac{1}{5}=x+\frac{1}{6}\)

\(\Leftrightarrow x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{4}+x+\frac{1}{5}-x-\frac{1}{6}=0\)

\(\Leftrightarrow3x+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}=0\)

Tính ra nhé !

\(x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{4}+x+\frac{1}{5}=x+\frac{1}{6}\)

\(\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)=x+\frac{1}{6}\)

\(\Rightarrow4x+\frac{77}{60}=x+\frac{1}{6}\)

\(\Rightarrow3x=\frac{1}{6}-\frac{77}{60}\)

\(\Rightarrow3x=-\frac{67}{60}\)

\(\Rightarrow x=-\frac{67}{60}\div3=\frac{-67}{60.3}=-\frac{67}{180}\)

Vậy x = .........

Ta có x/2 = 1/6 + 3/y ⇒ x/2 - 1/6 = 3/y ⇒ 3x - 1/ 6 = 3/y

Vậy y( 3x - 1 ) = 18

Mà x; y nguyên nên 3x - 1 nguyên và y; 3x - 1 ϵ Ư( 18 ) = { -1; 1; 2; -2; -3; 3; -6; 6; 18; -18 }

Vì 3x - 1 chia 3 dư 2 nên ( 3x - 1 ) ϵ { 2; -1 }

Nếu 3x - 1 = 2 ⇒ x = 1; y = 9

Nếu 3x - 1 = -1 ⇒ x = 0; y = -18

Vậy các cặp số nguyên ( x; y ) cần tìm là ( 1; 9 ) ; ( 0; -18 )

X + 1+2+3+4+5-6-7-8-9=1-2-3-4-5+6+7+8+9

X+  (-15)                     =  17

X                                = 17-(-15)

X                                = 32

vậy x = 32 

tk nha

x *(1/2+1/3+1/6)=425

x *         1        =  425

x                     =425:1

x                   =425

=> x*(5/6+1/6)=425

=> x*1=425

=> x=425:1

=> x=425

x . (1/2+1/3+1/6)=425

x .1=425

x=425:1=425

\(x.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right)=425\)

\(x.\left(\frac{3}{6}+\frac{2}{6}+\frac{1}{6}\right)=425\)

\(x.1=425\)

\(x=425\)

Ai giải giùm mình với

Mình đang cần gấp

\(x+y+z+8=2\sqrt[]{x-1}+4\sqrt[]{y-2}+6\sqrt[]{z-3}\left(1\right)\)

Áp dụng Bđt Bunhiacopxki :

\(\left(2\sqrt[]{x-1}+4\sqrt[]{y-2}+6\sqrt[]{z-3}\right)^2\le\left(2^2+4^2+6^2\right)\left(x-1+y-2+z-3\right)\)

\(\Leftrightarrow\left(2\sqrt[]{x-1}+4\sqrt[]{y-2}+6\sqrt[]{z-3}\right)^2\le56^{ }\left(x+y+z-6\right)\)

\(\Leftrightarrow\left(2\sqrt[]{x-1}+4\sqrt[]{y-2}+6\sqrt[]{z-3}\right)^2\le56^{ }\left(x+y+z+8\right)-784\)

Dấu "=" xảy ra khi và chỉ khi

\(\dfrac{x-1}{2}=\dfrac{y-2}{4}=\dfrac{z-3}{8}=\dfrac{x+y+z-6}{14}\left(2\right)\)

Đặt \(t=x+y+z+8\)

\(\left(1\right)\Leftrightarrow t^2=56t-784\)

\(\Leftrightarrow t^2-56t+784=0\)

\(\Leftrightarrow\left(t-28\right)^2=0\)

\(\Leftrightarrow t=28\)

\(\Leftrightarrow x+y+z+8=28\)

\(\Leftrightarrow x+y+z-6=14\)

\(\left(2\right)\Leftrightarrow\dfrac{x-1}{2}=\dfrac{y-2}{4}=\dfrac{z-3}{8}=1\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=1.2=2\\y-2=1.4=4\\z-2=1.8=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=6\\z=10\end{matrix}\right.\) thỏa mãn đề bài

a x * 4 + [1/2 + 1/4 + 1/8 + 1/6] = 23/16

x * 4 +             25/24             = 23/16

x * 4                                     = 23/16 - 25/24

x * 4                                     =   38/96 = 19/48

x                                          = 19/48 / 4

x                                          = 19/172

b x *         1              = 425

x                              = 425 / 1

x                              =    425

số số hạng: ( 2500 - 2 ) : 2 +1  = 1250 số

tổng: ( 2500 +2  ) . 1250 : 2 = 1563750

Vì: 1250 . 1251 = 1563750

=> x = 1250