d) \(x.\left(y+2\right)-y=15\)

\(\Rightarrow x.\left(y+2\right)=15+y\)

\(\Rightarrow x=\frac{y+15}{y+2}=\frac{y+2+13}{y+2}=1+\frac{13}{y+2}\)

y + 2 là ước nguyên của 13

\(y+2=1\Rightarrow y=-1\Rightarrow x=14\)

\(y+2=-1\Rightarrow y=-3\Rightarrow x=-12\)

\(y+2=13\Rightarrow y=11\Rightarrow x=2\)

\(y+2=-13\Rightarrow y=-15\Rightarrow x=0\)

Ai thấy đúng thì ủng hộ, mink chỉ làm được vậy thuu

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

=> \(\frac{2}{3}x=\frac{4}{5}-\frac{1}{2}=\frac{3}{10}\)

=> \(x=\frac{3}{10}:\frac{2}{3}=\frac{9}{20}\)

Vậy \(x\in\left\{\frac{9}{20}\right\}\)

\(b,x+\frac{1}{4}=\frac{4}{3}\)

=> \(x=\frac{4}{3}-\frac{1}{4}=\frac{13}{12}\)

Vậy \(x\in\left\{\frac{13}{12}\right\}\)

\(c,\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)

=> \(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}=\frac{5}{14}\)

=> \(x=\frac{5}{14}:\frac{3}{5}=\frac{25}{42}\)

Vậy \(x\in\left\{\frac{25}{42}\right\}\)

\(d,\left|x+5\right|-6=9\)

=> \(\left|x+5\right|=9+6=15\)

=> \(\left[{}\begin{matrix}x+5=15\\x+5=-15\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=15-5=10\\x=-15-5=-20\end{matrix}\right.\)

Vậy \(x\in\left\{10;-20\right\}\)

\(e,\left|x-\frac{4}{5}\right|=\frac{3}{4}\)

=> \(\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{4}\\x-\frac{4}{5}=-\frac{3}{4}\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{3}{4}+\frac{4}{5}=\frac{31}{20}\\x=-\frac{3}{4}+\frac{4}{5}=\frac{1}{20}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{31}{20};\frac{1}{20}\right\}\)

\(f,\frac{1}{2}-\left|x\right|=\frac{1}{3}\)

=> \(\left|x\right|=\frac{1}{2}-\frac{1}{3}\)

=> \(\left|x\right|=\frac{1}{6}\)

=> \(\left[{}\begin{matrix}x=\frac{1}{6}\\x=-\frac{1}{6}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{1}{6};-\frac{1}{6}\right\}\)

\(g,x^2=16\)

=> \(\left|x\right|=\sqrt{16}=4\)

=> \(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

vậy \(x\in\left\{4;-4\right\}\)

\(h,\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)

=> \(x-\frac{1}{2}=\sqrt[3]{\frac{1}{27}}=\frac{1}{3}\)

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

Vậy \(x\in\left\{\frac{5}{6}\right\}\)

\(i,3^3.x=3^6\)

\(x=3^6:3^3=3^3=27\)

Vậy \(x\in\left\{27\right\}\)

\(J,\frac{1,35}{0,2}=\frac{1,25}{x}\)

=> \(x=\frac{1,25.0,2}{1,35}=\frac{5}{27}\)

Vậy \(x\in\left\{\frac{5}{27}\right\}\)

\(k,1\frac{2}{3}:x=6:0,3\)

=> \(\frac{5}{3}:x=20\)

=> \(x=\frac{5}{3}:20=\frac{1}{12}\)

Vậy \(x\in\left\{\frac{1}{12}\right\}\)

Cậu có chắc của lớp 6 không ???

Áp dụng Bất đẳng thức Cauchy-Schwarz dạng Engel , có :

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{\left(1+1+1\right)^2}{x+y+z}=\frac{9}{6}=\frac{3}{2}\) 

Đẳng thức xảy ra : \(\Leftrightarrow\frac{1}{x}=\frac{1}{y}=\frac{1}{z}=\frac{1}{2}\)

Xét \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\left(x+y+z\right)=3+\frac{x}{y}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{x}{z}+\frac{z}{x}\)

Với \(x,y,z\inℕ^∗\)áp dụng bất đẳng thức Cô si  \(\frac{x}{y}+\frac{y}{x}\ge2\sqrt{\frac{x}{y}.\frac{y}{x}}=2\),\(\frac{y}{z}+\frac{z}{y}\ge2\sqrt{\frac{y}{z}.\frac{z}{y}}=2\),\(\frac{x}{z}+\frac{z}{x}\ge2\sqrt{\frac{x}{z}.\frac{z}{x}}=2\)

\(\Rightarrow\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\left(x+y+z\right)\ge3+2+2+2=9\)

\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}=\frac{9}{6}=\frac{3}{2}\left(x+y+z=6theogt\right)\)

a) x = 1

a.X=\(\frac{25}{7}\)

b./ \(\Leftrightarrow\frac{x+1}{2009}+1+\frac{x+2}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1.\)

\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)

\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)(b)

Mà \(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}< 0\)

(b) \(\Leftrightarrow x+2010=0\Leftrightarrow x=-2010\)

a./

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

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

Mà \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}>0\)

(a) \(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

Vì -24:-6=4

mà -24:-6=x:3=4:y^2=z^3:-2

Suy ra x=4x3=12

             y^2=4:4=1; y=1

             z^3=4x-2=-8;z=-2

\(b,\frac{z}{7}=-\frac{11}{-28}\)

\(\Leftrightarrow z.\left(-28\right)=-11.7\)

\(\Leftrightarrow z.\left(-28\right)=-77\)

\(\Leftrightarrow z=\frac{11}{4}\)

\(a,-\frac{2}{3}=\frac{x-3}{-6}=\frac{10}{5-y}=\frac{4-2z}{9}\)

Xét :

\(-\frac{2}{3}=\frac{x-3}{-6}\)

\(\Leftrightarrow-2.\left(-6\right)=\left(x-3\right).3\)

\(\Leftrightarrow12=\left(x-3\right).3\)

\(\Leftrightarrow4=x-3\Leftrightarrow x=7\)

Xét 

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

\(\Leftrightarrow-2.\left(5-y\right)=10.3\)

\(\Leftrightarrow-10+2y=30\)

\(\Leftrightarrow2y=40\Leftrightarrow y=20\)

Xét : 

\(-\frac{2}{3}=\frac{4-2z}{9}\)

\(\Leftrightarrow-2.9=\left(4-2z\right).3\)

\(\Leftrightarrow-18=\left(4-2z\right).3\)

\(\Leftrightarrow-6=4-2z\)

\(\Leftrightarrow10=2z\Leftrightarrow z=5\)

Vậy \(\left(x;y;z\right)=\left(7;20;5\right)\)

Theo đề bài \(\Rightarrow\frac{x}{-3}=\frac{2}{2}\Rightarrow2x=2.\left(-3\right)\Rightarrow2x=-6\Rightarrow x=-3\)

\(\frac{y}{6}=\frac{2}{2}\Rightarrow2y=2.6\Rightarrow2y=12\Rightarrow y=6\)

\(\frac{t+1}{3}=\frac{8}{3}\Rightarrow3\left(t+1\right)=8.3\Rightarrow3t+3=24\Rightarrow3t=21\Rightarrow t=7\)

Vậy x = -3 ; y = 6 và t = 7

\(\frac{x}{-3}=\frac{y.1}{6}=\frac{2}{2}=\frac{t+1}{3}=\frac{8}{3}\)

=> \(\frac{x}{-3}=\frac{y}{6}=\frac{2}{2}=\frac{t+1}{3}=\frac{8}{3}\)

=> 2x = - 3 . 2

2x = -6

x = - 6 : 2

x = - 3 

2y = 6 . 2

y = \(\frac{6.2}{2}\)

y = 6

Vì \(\frac{t+1}{3}=\frac{8}{3}\)=> t + 1 = 8 => t = 7

Vậy x = 2 ; y = 6 ; t = 7