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

\(\dfrac{-13}{9}\le x\le\dfrac{-11}{12}\)

\(\dfrac{-468}{36}\le\dfrac{36.x}{36}\le\dfrac{-396}{36}\)

\(=>36.x\in\left\{-467;-466;-465;-464;...;-398;-397\right\}\)

\(=>x=-12\)

a)\(\left(\frac{1}{2}-\frac{1}{3}\right).6^x+6^{x+2}=6^{15}+6^{18}\)

\(\frac{1}{6}.6^x+6^{x+2}=6^{15}\left(1+6^3\right)\)

\(\frac{1}{6}.6^x\left(1+6^3\right)=6^{15}.217\)

\(6^{x-1}.217=6^{15}.217\)

\(6^{x-1}=6^{15}\)

\(x-1=15\)

\(x=16\)

b) \(\left(\frac{1}{2}-\frac{1}{6}\right).3^{x+4}-4.3^x=3^{16}-4.3^{13}\)

\(\frac{1}{3}.3^x.4\left(3^4-1\right)=3^{13}.4\left(3^3-1\right)\)

\(3^x.4.\left(3^3-1\right)=3^{13}.4.\left(3^3-1\right)\)

\(3^x=3^{13}\)

\(x=13\)

\(\left(\frac{1}{2}-\frac{1}{6}\right).\left(3^x.3^4\right)-4.3^x=3^{16}-4.3^{13}\)

=> \(\frac{1}{3}.3^x.3^4-4.3^x=3^{16}-4.3^{13}\)

=> \(3^x.3^4-4.3^x=\left(3^{16}-4.3^{13}\right):\frac{1}{3}\)

=> \(3^x.3^4-4.3^x=-386339074,3\)

=> \(3^x.\left(3^4-4\right)=-386339074,3\)

=> \(3^x.77=-386339074,3\)

=> \(3^x=-386339074,3:77\)

=> \(3^x=-5017390,575\)

=> x = ... chắc tự ngồi tính đc

a.\(\frac{1}{6}.6^x+6^x.36=6^{15}\left(1+6^3\right)\)

\(6^x.\frac{217}{6}=6^{15}.217\)

\(6^x=6^{16}\)

\(x=16\)

sao bạn ko làm câu b) lun

Mình đang cần gấp.Các bạn giúp nha

Mình chỉ làm được bài một thôi:

BÀI 1:                                                                                Giải

Gọi ƯCLN(a;b)=d (d thuộc N*)

=> a chia hết cho d ; b chia hết cho d

=> a=dx ; b=dy  (x;y thuộc N , ƯCLN(x,y)=1)

Ta có : BCNN(a;b) . ƯCLN(a;b)=a.b

=> BCNN(a;b) . d=dx.dy

=> BCNN(a;b)=\(\frac{dx.dy}{d}\)

=> BCNN(a;b)=dxy

mà BCNN(a;b) + ƯCLN(a;b)=15

=> dxy + d=15

=> d(xy+1)=15=1.15=15.1=3.5=5.3(vì x; y ; d là số tự nhiên)

TH 1: d=1;xy+1=15

=> xy=14 mà ƯCLN(a;b)=1

Ta có bảng sau:

TH2: d=15; xy+1=1

=> xy=0(vô lý vì ƯCLN(x;y)=1)

TH3: d=3;xy+1=5

=>xy=4

mà ƯCLN(x;y)=1

TA có bảng sau:

TH4:d=5;xy+1=3

=> xy = 2

Ta có bảng sau:

.Vậy (a;b) thuộc {(1;14);(14;1);(2;7);(7;2);(3;12);(12;3);(5;10);(10;5)}

\(3.\)

\(\frac{x-1}{2011}+\frac{x-2}{2010}+\frac{x-3}{2009}=\frac{x-4}{2008}\)

\(\Rightarrow\)\(\frac{x-1}{2011}-1+\frac{x-2}{2010}-1+\frac{x-3}{2009}-1-\frac{x-4}{2008}+1+2=0\)

\(\Rightarrow\)\(\frac{x-1}{2011}-\frac{2011}{2011}+\frac{x-2}{2010}-\frac{2010}{2010}+\frac{x-3}{2009}-\frac{2009}{2009}-\frac{x-4}{2008}+\frac{2008}{2008}=0\)

\(\Rightarrow\)\(\frac{x-2012}{2011}+\frac{x-2012}{2010}+\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\)

\(\Rightarrow\)\(x-2012\left(\frac{1}{2011}+\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2008}\right)=0\)

\(\Rightarrow\)\(x=2012\)

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

\(-4\le x\le\frac{11}{18}\)

\(-4\le x\le0,6\left(1\right)\)

\(\Rightarrow x\in\left\{-4;-3;-2;-1;0\right\}\)

\(\left(\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right):\frac{-5}{6}< x< \frac{4}{21}.\frac{4}{7}\)

\(\Rightarrow\left(\frac{6}{12}+\frac{9}{12}-\frac{4}{12}\right):\frac{-10}{12}< x< \frac{16}{147}\)

\(\Rightarrow\frac{11}{12}.\frac{-12}{10}< x< \frac{16}{147}\)

\(\Rightarrow\frac{-11}{10}< x< \frac{16}{147}\)

\(\Rightarrow\frac{-1617}{1470}< x< \frac{16}{1470}\)

\(x=\left\{-1;0\right\}\)

Câu a:

\(\frac{-8}{3x-1}\) = \(\frac{4}{-7}\)

-8.(-7) = 4.(3\(x\) - 1)

56 = 12\(x\) - 4

12\(x\) = 56+ 4

12\(x\) = 60

\(x\) = 60 : 12

\(x\) = 5

Vậy \(x\) = 5

Câu b:

\(\frac{x}{-3}\) = \(\frac{-3}{x}\)

\(x^2\) = (-3)\(^2\)

\(\left[\begin{array}{l}x=-3\\ x=3\end{array}\right.\)

Vậy \(x\in\left\lbrace-3;3\right\rbrace\)

Câu c:

\(-\frac{4}{y}=\frac{x}{2}\)

-4.2 = \(x.y\)

\(xy=-8\)

Ư(8) = (-8; -4; -2; -1; 1; 2; 4; 8}

Vậy (\(x;y\)) = (-8; 1); (-4; 2); (-2; 4); (-1; 8); (1; -8); (2; -4); (4; -2); (8; -1)

Câu 2:

(\(x-1)\)(y + 2) = 7

Ư(7) = {-7; -1; 1; 7}

Lập bảng ta có:

Theo bảng trên ta có:

(\(x;y\)) = (-6; -3); (0; -9); (2; 5); (8; - 1)

Vậy (\(x;y\)) = (-6; -3); (0; -9); (2; 5); (8; -1)