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\(\frac{A}{n}=\frac{4n+4}{n}=4+\frac{4}{n}\)
\(\Rightarrow n\in U\left(4\right)\)
Lập bảng tiếp nhé!
\(\frac{B}{n}=\frac{5n+6}{n}=5+\frac{6}{n}\)
Lập bảng
\(2.\)
a)\(\left(\frac{3}{29}-\frac{1}{5}\right)\cdot\frac{29}{3}=\frac{3}{29}\cdot\frac{29}{3}-\frac{1}{5}\cdot\frac{29}{3}=1-\left(1+\frac{14}{15}\right)=1-1-\frac{14}{15}=\frac{14}{15}\)
b)\(\frac{1}{7}\cdot\frac{5}{9}+\frac{5}{9}\cdot\frac{1}{7}+\frac{5}{9}\cdot\frac{3}{7}=\frac{5}{9}\cdot\left(\frac{1}{7}+\frac{1}{7}+\frac{3}{7}\right)=\frac{5}{9}\cdot\frac{5}{7}=\frac{25}{63}\)
\(A=\frac{2n+8}{5}+\frac{-n-7}{5}\)
\(\Leftrightarrow A=\frac{2n+8-n-7}{5}\)
\(\Leftrightarrow A=\frac{n+1}{5}\)
Để A nguyên thì \(\frac{n+1}{5}\)nguyên
\(\Rightarrow\left(n+1\right)⋮5\)
\(\Leftrightarrow n+1\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
Ta có bảng sau :
\(n+1\) | \(-5\) | \(-1\) | \(1\) | \(5\) |
\(m\) | \(-6\) | \(-2\) | \(0\) | \(4\) |
\(ĐểA\in Z\)thì:
\(n+2⋮n-5\)
=> \(\left[n-5\right]+7⋮n-5\)
=> 7 chia hết cho n - 5
=> n -5 E Ư[7] E {-7;-1;1;7}
=> n E {-2;4;6;12}
Vậy: n = -2; n = 4 n = 6; n = 12
\(A=\frac{n+2}{n-5}=\frac{n-5+7}{n-5}=1+\frac{7}{n-5}\)
Để \(A\in Z\)thì n-5 là ước nguyên của 7
\(n-5=1\Rightarrow n=6\)
\(n-5=7\Rightarrow n=12\)
\(n-5=-1\Rightarrow n=4\)
\(n-5=-7\Rightarrow n=-2\)
Ai thấy đúng k cho mink nha !!!
a, \(A=\frac{7}{n-3}\)
Để \(\frac{7}{n-3}\in Z\)thì \(7⋮n-3\Leftrightarrow n-3\inƯ\left(7\right)=\left\{\text{±}1;\text{±}7\right\}\)
Ta có bảng sau:
n - 3 | -1 | -7 | 1 | 7 |
n | 2 | -4 | 4 | 10 |
Vậy \(n\in\left\{-4;2;4;10\right\}\)để\(\frac{7}{n-3}\in Z\)
b,\(B=\frac{13}{2n-5}\)
Để \(\frac{13}{2n-5}\in Z\)thì \(13⋮2n-5\Leftrightarrow2n-5\inƯ\left(13\right)=\left\{\text{±}1;\text{±}13\right\}\)
Ta có bảng sau:
2n - 5 | -1 | -13 | 1 | 13 |
2n | 4 | -8 | 6 | 18 |
n | 2 | -4 | 3 | 9 |
Vậy \(n\in\left\{-4;2;3;9\right\}\)để\(\frac{13}{2n-5}\in Z\)
c, \(C=\frac{-6}{3n+2}\)
Để \(\frac{-6}{3n+2}\in Z\)thì \(-6⋮3n+2\Leftrightarrow3n+2\inƯ\left(-6\right)=\left\{\text{±}1;\text{±}2;\text{±}3;\text{±}6\right\}\)
Ta có bảng sau:
3n + 2 | -1 | -2 | -3 | -6 | 1 | 2 | 3 | 6 |
3n | -3 | -4 | -5 | -8 | -1 | 0 | 1 | 4 |
n | -1 | \(\frac{-4}{3}\) | \(\frac{-5}{3}\) | \(\frac{-8}{3}\) | \(\frac{-1}{3}\) | 0 | \(\frac{1}{3}\) | \(\frac{4}{3}\) |
Vậy \(n\in\left\{\frac{-8}{3};\frac{-5}{3};\frac{-4}{3};\frac{-1}{3};-1;0;\frac{1}{3};\frac{4}{3}\right\}\)để \(\frac{-6}{3n+2}\in Z\)
mà \(n\in Z\)
Vậy \(n\in\left\{-1;0\right\}\)để\(\frac{-6}{3n+2}\in Z\)
a,Để \(A\in Z\)
\(\Rightarrow\)\(\frac{7}{n-3}\in Z\)
\(\Rightarrow\)n-3\(\in\)Ư(7)
n-3 \(\in\){1;-1;7;-7}
n\(\in\){4;2;10;-4}
Vậy n\(\in\){4;2;10;-4}
b,Để \(B\in Z\)
\(\Rightarrow\frac{13}{2n-5}\in Z\)
\(\Rightarrow\)2n-5\(\in\)Ư(13)
2n-5\(\in\){1;-1;13;-13}
2n\(\in\){6;4;18;-8}
n\(\in\){3;2;9;-4}
Vậy n\(\in\){3;2;9;-4}
c,Để \(C\in Z\)
\(\Rightarrow\frac{-6}{3n+2}\in Z\)
\(\Rightarrow\)3n+2\(\in\)Ư(-6)
3n+2\(\in\){1;-1;2;-2;3;-3;6;-6}
n\(\in\){-1;0}
Vậy n \(\in\){-1;0}
Bài 1:
\(B=\frac{\frac{1}{2}+\frac{3}{4}-\frac{5}{6}}{\frac{1}{4}+\frac{3}{8}-\frac{5}{12}}+\frac{\frac{3}{4}+\frac{3}{5}-\frac{3}{8}}{\frac{1}{4}+\frac{1}{5}-\frac{1}{8}}\)\(=\frac{\frac{1}{2}+\frac{3}{4}-\frac{5}{6}}{\frac{1}{2}\left(\frac{1}{2}+\frac{3}{4}-\frac{5}{6}\right)}+\frac{3\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{8}\right)}{\frac{1}{4}+\frac{1}{5}-\frac{1}{8}}\)
\(=\frac{1}{\frac{1}{2}}+3\) \(=2+3\) \(=5\)
Vậy B=5
Bài 2:
a) x3 - 36x = 0
=> x(x2-36)=0
=> x(x2+6x-6x-36)=0
=> x[x(x+6)-6(x+6) ]=0
=> x(x+6)(x-6)=0
\(\Rightarrow\orbr{\begin{cases}^{x=0}x+6=0\\x-6=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}^{x=0}x=-6\\x=6\end{cases}}\)
Vậy x=0; x=-6; x=6
b) (x - y = 4 => x=4+y)
x−3y−2 =32
=>2(x-3) = 3(y-2)
=>2x-6= 3y-6
=>2x-3y=0
=>2(4+y)-3y=0
=>8+2y-3y=0
=>8-y=0
=>y=8 (thỏa mãn)
Do đó x=4+y=4+8=12 (thỏa mãn)
Vậy x=12 và y =8
B= 1/2 + 3/4 - 5/6/1/2(1.2 + 3/4 - 5/6) + 3(1/4+ 1/5 - 1/8)/ 1/4 1/5 - 1/8
B= 1/ 1/2 + 3
B= 2+3
B=5
B2:
a) x^3 - 36x = 0
x(x^2 - 36) = 0
=> x=0 hoặc x^2-36=0
=> x= 0 hoặc x^2=36
=> x=0 hoặc x= +- 6
a) Ta có \(\frac{x}{x+5}=\frac{x}{x+y}\)
\(\Rightarrow x.\left(x+y\right)=x.\left(x+5\right)\Rightarrow x^2+xy=x^2+5y\Rightarrow xy=5x\)
\(\Rightarrow xy-5x=0\Rightarrow x.\left(y-5\right)=0\Rightarrow\orbr{\begin{cases}x=0\\y=5\end{cases}}\)
Vậy x = 0 hoặc y = 5
b) \(\frac{x-7}{y-8}=\frac{7}{8}\)
\(\Rightarrow\left(x-7\right).8=7.\left(y-8\right)\Rightarrow8x-56=7y-56\Rightarrow8x=7y\)(1)
Từ x - y = 4 nên x = y + 4 . Thay x = y + 4 vào (1) ta có :
\(8.\left(y+4\right)=7y\Rightarrow8y+32=7y\Rightarrow y=-32\)
Do đó x = - 28
Vậy x = -28 ; y = -32
a) Ta có: \(\frac{x}{x+5}=\frac{x}{x+y}\)
\(\Leftrightarrow\frac{x}{x+5}=\frac{x}{x+5}\Rightarrow y=5\)
\(\Leftrightarrow\frac{x}{x+5}=\frac{x}{x}.\frac{x}{5}\)
Đặt mẫu số chung là 5. Ta quy đồng phân số:\(\frac{x}{x}=\frac{x.5}{x.5}=\frac{5}{5}\)
\(\Rightarrow\orbr{\begin{cases}x=5\\y=5\end{cases}}\)
b) Ta có: \(\frac{x-7}{y-8}=\frac{7}{8}\Rightarrow\frac{x}{y}=\frac{7}{8}+\frac{7}{8}=\frac{14}{8}\)
Mà \(\frac{x}{y}=\frac{14}{8}\Rightarrow\orbr{\begin{cases}x=14\\y=8\end{cases}}\)
\(\Rightarrow40+8a=40a\)
\(\Leftrightarrow40=32a\)
\(\Leftrightarrow a=\frac{40}{32}=\frac{5}{4}\)