1. \(\frac{-7}{12}\)< \(\frac{x-1}{4}\)< \(\frac{2}{3}\)

=> \(\frac{-7}{12}\)< \(\frac{3.\left(x-1\right)}{12}\)< \(\frac{8}{12}\)

=> 3 . ( x - 1 ) thuộc { - 6 ; - 5 ; - 4 ; - 3 ; - 2 ; - 1 ; 0 ; 1 ; 2 ; 3 ; 4 ; 5 ; 6 ; 7}

Lập bảng tính giá trị x , cái này dễ lên bạn tự làm nha

1/ \(-\frac{7}{12}< \frac{x-1}{4}< \frac{2}{3}\)

hay \(\frac{-7}{12}< \frac{3.\left(x-1\right)}{12}< \frac{8}{12}\)

Vậy \(-7< 3.\left(x-1\right)< 8\)

Vậy \(3.\left(x-1\right)\in\left\{-6;-5;-4;...;7\right\}\)

mà \(x\in Z\)nên \(3.\left(x-1\right)⋮3\)

Vậy \(3.\left(x-1\right)\in\left\{-6;-3;0;3;6\right\}\)

hay \(x-1\in\left\{-2;-1;0;1;2\right\}\)

tới đây dễ rồi thì làm nốt nhé, để thời gian làm mấy câu sau!

post từng câu một thôi bn nhìn mệt quá

\(\frac{x-2}{27}+\frac{x-3}{26}+\frac{x-4}{25}+\frac{x-5}{24}+\frac{x-44}{5}=1\)

\(\Leftrightarrow\left(\frac{x-2}{27}-1\right)+\left(\frac{x-3}{26}-1\right)+\left(\frac{x-4}{25}-1\right)+\left(\frac{x-5}{24}-1\right)\)\(+\left(\frac{x-44}{5}+3\right)=1-1\)

\(\Leftrightarrow\frac{x-29}{27}+\frac{x-29}{26}+\frac{x-29}{25}+\frac{x-29}{24}\)\(+\frac{x-29}{5}=0\)

\(\Leftrightarrow\left(x-29\right)\left(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{5}\right)=0\)

Mà \(\frac{1}{27}+\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{5}\ne0\)

=> x - 29 = 0

=> x = 29.

x(x+1)+y(y+1)+z(z+1) \(\le18\)

<=> \(x^2+y^2+z^2+\left(x+y+z\right)\le18\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\)

\(\Rightarrow3\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)

\(\Rightarrow54\ge\left(x+y+z\right)^2+3\left(x+y+z\right)\)

\(\Leftrightarrow-9\le x+y+z\le6\)

\(\Rightarrow0\le x+y+z\le6\)

\(\hept{\begin{cases}\frac{1}{x+y+1}+\frac{x+y+1}{25}\ge\frac{2}{5}\\\frac{1}{y+z+1}+\frac{y+z+1}{25}\ge\frac{2}{5}\\\frac{1}{z+x+1}+\frac{z+x+1}{25}\ge\frac{2}{5}\end{cases}}\Rightarrow B+\frac{2\left(x+y+z\right)+3}{25}\ge\frac{6}{5}\)

\(\Rightarrow B\ge\frac{27}{25}-\frac{2}{25}\left(x+y+z\right)\ge\frac{15}{25}=\frac{3}{5}\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x=y=z>0;x+y+z=6\\\left(x+y+1\right)^2=\left(y+z+1\right)^2=\left(z+x+1\right)^2=25\end{cases}\Leftrightarrow x=y=z=2}\)

vậy giá trị nhỏ nhất cho B=3_/5 khi x=y=z=2

Hai Ngox  Xem laị  từ dòng thứ 2  và dòng thứ 3 xuống dưới. Nhiều lỗi quá!

Cô Chi giúp em với!!!

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

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

\(\left(x-3\right)\times y=4=\left(-1\right)\times\left(-4\right)=\left(-4\right)\times\left(-1\right)=4\times1=1\times4=2\times2=\left(-2\right)\times\left(-2\right)\)

Vậy \(\left(x;y\right)\in\left\{\left(2;-4\right);\left(-1;-1\right);\left(7;1\right);\left(4;4\right);\left(5;2\right);\left(1;-2\right)\right\}\)

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

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

\(\Rightarrow\)y.(x-3)=4 hay y và x-3 \(\in\)Ư(4)

Ta có bảng sau:

Vậy (x;y)\(\in\){(5;1);(-3;-1);(3;2);(-1;-2);(2;4);(0;-4)}

Đặt: \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}=k\)

\(\Rightarrow x=k\)

     \(y=2k\)

     \(z=3k\)

Thay x = k , y = 2k , z = 3k vào biểu thức cần cm ,ta đc:

\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\right)=\left(k+2k+3k\right)\left(\frac{1}{k}+\frac{4}{2k}+\frac{9}{3k}\right)\)

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

\(=6k.\frac{6}{k}\)

\(=\frac{36k}{k}=36\)

=.= hok tốt!!

Đặt \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}=k\)

Do đó  \(x=k;y=2k;z=3k\)

Thay \(x=k;y=2k;z=3k\)vào \(\left(x+y+z\right).\left(\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\right)\)ta có 

\(\left(k+2k+3k\right).\left(\frac{1}{k}+\frac{4}{2k}+\frac{9}{3k}\right)\)

\(=6k.\left(\frac{6}{6k}+\frac{12}{6k}+\frac{18}{6k}\right)\)

\(=6k.\frac{6+12+18}{6k}\)

\(=\frac{6k.\left(6+12+18\right)}{6k}\)

\(=36\)

Do đó \(\left(x+y+z\right).\left(\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\right)=36\)

Ta có:

\(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}=\frac{x+y+z}{1+2+3}=\frac{x+y+z}{6}\)(Tính chất dãy tỉ số bằng nhau

=>  \(\hept{\begin{cases}\frac{x}{1}=\frac{x+y+z}{6}\\\frac{y}{2}=\frac{x+y+z}{6}\\\frac{z}{3}=\frac{x+y+z}{6}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{x+y+z}{6}\\y=\frac{x+y+z}{3}\\z=\frac{x+y+z}{2}\end{cases}}\)

Đặt biểu thức cần chứng minh là A và x + y + z = k

=> \(\hept{\begin{cases}x=\frac{k}{6}\\y=\frac{k}{3}\\z=\frac{k}{2}\end{cases}}\)

=> A = \(k\left(\frac{1}{\frac{k}{6}}+\frac{4}{\frac{k}{3}}+\frac{9}{\frac{k}{2}}\right)\)

    A = \(k.\left(\frac{6}{k}+\frac{12}{k}+\frac{18}{k}\right)=k.\frac{36}{k}=36\)(đpcm)

Với mọi số thực ta luôn có:

`(x-y)^2>=0`

`<=>x^2-2xy+y^2>=0`

`<=>x^2+y^2>=2xy`

`<=>(x+y)^2>=4xy`

`<=>(x+y)^2>=16`

`<=>x+y>=4(đpcm)`

\(\dfrac{1}{x+3}+\dfrac{1}{y+3}=\dfrac{x+3+y+3}{\left(x+3\right)\left(y+3\right)}\)

\(=\dfrac{x+y+6}{3x+3y+13}\)(vì \(xy=4\))

=> \(\dfrac{x+y+6}{3x+3y+13}\)≤\(\dfrac{2}{5}\)

<=> \(5\left(x+y+6\right)\)≤\(2\left(3x+3y+13\right)\)

<=>\(6x+6y+26-5x-5y-30\)≥\(0\)

<=> \(x+y-4\)≥\(0\)

Áp dụng BĐT AM-GM \(\dfrac{a+b}{2}\)≥\(\sqrt{ab}\)

Ta có \(\dfrac{x+y}{2}\)≥\(\sqrt{xy}\)

<=>\(x+y\) ≥ 2\(\sqrt{xy}\)

=>2\(\sqrt{xy}-4\)≥\(0\)

<=> \(4-4\)≥0

<=>0≥0 ( Luôn đúng )

Vậy \(\dfrac{1}{x+3}+\dfrac{1}{y+3}\)≤\(\dfrac{2}{5}\)

\(\frac{-5}{x}=\frac{-y}{8}=\frac{18}{72}\)

\(\Leftrightarrow\frac{-5}{x}=\frac{-y}{8}=\frac{1}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}-\frac{5}{x}=\frac{1}{4}\\-\frac{y}{8}=\frac{1}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5.4:1\\-y=8.1:4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-20\\-y=2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-20\\y=-2\end{cases}}}\)

vậy x=-20 và y=-2

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

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

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

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

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

\(x=-\frac{4}{12}-\frac{9}{12}\)

\(x=-\frac{13}{12}\)

\(x=-\frac{18}{24}+\frac{15}{21}\)

\(x=-\frac{3}{4}+\frac{5}{7}\)

\(x=-\frac{21}{28}+\frac{20}{28}\)

\(x=-\frac{1}{28}\)