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

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

\(\Leftrightarrow xy-3=x\)

\(\Leftrightarrow xy-x=3\)

\(\Leftrightarrow x\left(y-1\right)=3=\left(-1\right).\left(-3\right)=3.1\)( vì x, y là các số nguyên )

\(TH1:\)

\(\orbr{\begin{cases}x=1\\y-1=3\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\y=4\end{cases}}\)

\(\orbr{\begin{cases}x=3\\y-1=1\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\y=2\end{cases}}}\)

\(TH2:\)

\(\orbr{\begin{cases}x=-1\\y-1=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\y=-2\end{cases}}\)

\(\orbr{\begin{cases}x=-3\\y-1=-1\end{cases}\Rightarrow}\orbr{\begin{cases}x=-3\\y=0\end{cases}}\)

Vậy .......

Giải:     Có y/3-1/x=1/3

y/3-1/3=1/x

Suy ra y-1/3=1/x

Suy ra (y-1).x=3

Suy ra y-1 và x thuộc Ư(3)

Vì x,y thuộc Z

Do đó ta có bảng giá trị:

Vậy (x,y)= {...........}

nha
 

có \(\frac{28}{-36}=\frac{7}{-9}\)

gọi phân số cần tìm là\(\frac{a}{b}\)

\(\Rightarrow\hept{\begin{cases}a=7k\\b=-9k\end{cases}}\)

theo đầu bài b-a=-36

\(\Rightarrow-9k-7k=36\)

\(\Rightarrow-16k=36\)

\(\Rightarrow k=-\frac{9}{4}\)

\(\Rightarrow\hept{\begin{cases}a=-\frac{63}{4}\\b=\frac{81}{4}\end{cases}}\)

\(\Rightarrow\frac{a}{b}=\frac{-63}{4}\div\frac{81}{4}\)

\(\Rightarrow\frac{a}{b}=\frac{-63}{81}\)

\(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-...-\frac{100}{3^{100}}\)
\(\Leftrightarrow3A=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...-\frac{100}{3^{99}}\)
\(\Rightarrow4A=A+3A=1-\frac{1}{3}+\frac{1}{3^2}-...-\frac{100}{3^{100}}\)
\(\Rightarrow12A=3.4A=3-1+\frac{1}{3}-\frac{1}{3^2}+...-\frac{100}{3^{99}}\)
\(\Rightarrow16A=12A+4A=3-\frac{1}{3^{99}}-\frac{100}{3^{99}}-\frac{100}{3^{101}}\)
\(\Leftrightarrow16A=3-\frac{101}{3^{99}}-\frac{100}{3^{100}}\)
\(\Leftrightarrow A=\frac{3}{16}-\frac{\left(\frac{101}{3^{99}}+\frac{100}{3^{100}}\right)}{16}< \frac{3}{16}\)

Suy ra A<3/16

7/-78+9/65=-35/390+54/390=-19/390

-3/16+5/18=-27/144+40/144=13/144

23/-18+19/-57=-437/342+(-114/342)=-29/18

\(\left\{x^2-\left[6^2-\left(8^2-9.7\right)^3-7.5\right]^3-5.3\right\}^3=1\)

\(\Rightarrow\left\{x^2-\left(36-1^3-35\right)^3-15\right\}^3=1\)

\(\Rightarrow x^2-\left(0^3-15\right)^3=1\)

\(\Rightarrow x^2-\left(-3375\right)=1\)

\(\Rightarrow x^2=-3374\)

\(\Rightarrow x\in\varnothing\)

{ x² - [ 6² - ( 8² - 9.7)³ - 7.5]³ - 5.3 }³ = 1

{ x² + [ 36 - (64 - 63)³ - 35]³ - 15}³ = 1

[ x² - ( 36 - 1³ - 35 ) - 15 ]³ = 1

[ x² - ( 36 - 1 - 35 ) - 15]³ = 1

[ x² - ( 35 - 35 ) - 15]³ = 1

[ x² - 0 - 15]³ = 1

( x² - 15 )³ = 1

<=> ( x² - 15)³ = 1³

=> x² - 15 = 1

<=> x² = 16

=> x = 4

\(\frac{-7}{-16}+\frac{-6}{9}\)

\(=\frac{-7}{-16}+\frac{-2}{3}\)

\(=\frac{-53}{48}\)

AI GIÚP NHANH TỚ K

2^22 lớn nhất

Ta có: \(A=\frac{1}{5}+\frac{1}{14}+\frac{1}{27}+\frac{1}{43}+\frac{1}{61}+\frac{1}{89}+\frac{1}{111}\)

               \(=\frac{1}{5}+\left(\frac{1}{14}+\frac{1}{27}+\frac{1}{43}\right)+\left(\frac{1}{61}+\frac{1}{89}+\frac{1}{111}\right)\)

                 \(< \frac{1}{5}+\left(\frac{1}{12}+\frac{1}{12}+\frac{1}{12}\right)+\left(\frac{1}{60}+\frac{1}{60}+\frac{1}{60}\right)\)

                    \(=\frac{1}{5}+\frac{1}{4}+\frac{1}{20}=\frac{1}{2}\)

Vậy \(A< \frac{1}{2}\)