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Bài 10:
a: Để A là phân số thì n+2<>0
hay n<>-2
b: Khi n=0 thì A=3/2
Khi n=2 thì A=3/(2+2)=3/4
Khi n=-7 thì A=3/(-7+2)=-3/5
Bài 9:
1)9/x = -35/105 2) 12/5 = 32/x 3)x/2 = 32/x x = 9. (-35)/105 x.12/5 = x.32/x 2x.x/2 = 2x.32/x
x = -3 x.12/5=32 xx = 2.32
x= 32:12/5 x^2 = 2.32
x = 40/3 x^2 = 64
x = 8
4) x-2/4 = x-1/5
5(x-2) = 4(x-1)
5x - 10 = 4x - 4
5x - 4x = 10 - 4
x = 6
Bài 10:Cho biểu thức A=3/n+2
a) Để A là phân số thì mẫu số phải khác 0
Do đó: n + 2 ≉ 0. Suy ra: n ≉ -2
b) Khi n = 0 thì A = 3/0+2 = 3/2
Khi n = 2 thì A = 3/2+2 = 3/4
Khi n = -7 thì A = 3/-7+2 = 3/-5
a) Ta có: \(-3\dfrac{1}{4}\cdot x-75\%+\dfrac{3x}{2}=-1.2:\dfrac{-9}{10}-1\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{-13x}{4}-\dfrac{3}{4}+\dfrac{3x}{2}=\dfrac{-6}{5}\cdot\dfrac{10}{-9}-\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{-13x-3+6x}{4}=\dfrac{4}{3}-\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{-7x-3}{4}=\dfrac{1}{12}\)
\(\Leftrightarrow-7x-3=\dfrac{1}{3}\)
\(\Leftrightarrow-7x=\dfrac{10}{3}\)
hay \(x=-\dfrac{10}{21}\)
b) Ta có: \(\dfrac{5}{3}+\dfrac{5}{15}+\dfrac{5}{35}+...+\dfrac{5}{x\left(x+2\right)}=2\dfrac{8}{17}\)
\(\Leftrightarrow\dfrac{5}{2}\left(\dfrac{2}{3}+\dfrac{2}{15}+\dfrac{2}{35}+...+\dfrac{2}{x\left(x+2\right)}\right)=2\dfrac{8}{17}\)
\(\Leftrightarrow\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=2+\dfrac{8}{17}\)
\(\Leftrightarrow\left(1-\dfrac{1}{x+2}\right)=\dfrac{42}{17}:\dfrac{5}{2}\)
\(\Leftrightarrow\dfrac{x+1}{x+2}=\dfrac{42}{17}\cdot\dfrac{2}{5}=\dfrac{84}{85}\)
\(\Leftrightarrow85x+85=84x+168\)
\(\Leftrightarrow x=83\)
1. \(\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
\(=\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right).0\)
\(=0\)
\(\Leftrightarrow-\dfrac{16}{279}< \dfrac{x}{9}< =\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{x}{9}=0\)
hay x=0
a) Ta có \(A=\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot\dfrac{24}{25}\cdot...\cdot\dfrac{2499}{2500}\)
\(=\dfrac{2\cdot4}{3\cdot3}\cdot\dfrac{3\cdot5}{4\cdot4}\cdot\dfrac{4\cdot6}{5\cdot5}\cdot...\cdot\dfrac{49\cdot51}{50\cdot50}\)
\(=\dfrac{2\cdot4\cdot3\cdot5\cdot4\cdot6\cdot...\cdot49\cdot51}{3\cdot3\cdot4\cdot4\cdot5\cdot5\cdot...\cdot50\cdot50}\)
\(=\dfrac{2\cdot3\cdot4\cdot...\cdot49}{3\cdot4\cdot5\cdot...\cdot50}\cdot\dfrac{4\cdot5\cdot6\cdot...\cdot51}{3\cdot4\cdot5\cdot...\cdot50}\)
= \(\dfrac{2}{50}\cdot17=\dfrac{17}{25}\)
b) Vì n nguyên nên 3n - 1 nguyên
Để phân số \(\dfrac{12}{3n-1}\) có giá trị nguyên thì 12 ⋮ ( 3n - 1 ) hay ( 3n - 1 ) ϵ Ư( 12 )
Ư( 12 ) = { \(\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\) }
Lập bảng giá trị
| 3n - 1 | 1 | -1 | 2 | -2 | 3 | -3 | 4 | -4 | 6 | -6 | 12 | -12 |
| n | \(\dfrac{2}{3}\) | 0 | 1 | \(\dfrac{-1}{3}\) | \(\dfrac{3}{4}\) | \(\dfrac{-2}{3}\) | \(\dfrac{5}{3}\) | -1 | \(\dfrac{7}{3}\) | \(\dfrac{-5}{3}\) | \(\dfrac{13}{3}\) | \(\dfrac{-11}{3}\) |
Vì n nguyên nên n ϵ { 0; 1; -1 }
Vậy n ϵ { 0; 1; -1 } để phân số \(\dfrac{12}{3n-1}\) có giá trị nguyên
a,2/5 = 2/5 ; 3/8=6/16 ; 1/9=3/27
b, 4/3=8/6 ; -1=-1 ; -4/-2=-8/4
tick cho mik nhé
a) \(\dfrac{5}{x}=\dfrac{-10}{12}.\Rightarrow x=-6.\)
b) \(\dfrac{4}{-6}=\dfrac{x+3}{9}.\Rightarrow x+3=-6.\Leftrightarrow x=-9.\)
c) \(\dfrac{x-1}{25}=\dfrac{4}{x-1}.\left(đk:x\ne1\right).\Leftrightarrow\dfrac{x-1}{25}-\dfrac{4}{x-1}=0.\)
\(\Leftrightarrow\dfrac{x^2-2x+1-100}{25\left(x-1\right)}=0.\Leftrightarrow x^2-2x-99=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=11.\\x=-9.\end{matrix}\right.\) \(\left(TM\right).\)
Cho \(A=\dfrac{\dfrac{-5}{8}.\dfrac{3}{7}+\dfrac{3}{7}.\dfrac{3}{4}+\dfrac{1}{7}.\dfrac{1}{2}+\dfrac{15}{8}}{a+\dfrac{5}{6}-\left(\dfrac{-1}{3}\right)}\)
a) Rút gọn A?
b) Tính A khi a=75%
c) Tìm a để A=50%
d) Tìm a thuộc Z để A là số nguyên.
e) Với a = bao nhiêu để A có giá trị bằng với giá trị của biểu thức:
\(B=\dfrac{\dfrac{2}{3}.\dfrac{15}{6}+\left(-0,5\right)^3}{\dfrac{1}{9}.6^2-5\dfrac{1}{3}}\)
Giải
a, Ta có:
\(A=\dfrac{\dfrac{-5}{8}.\dfrac{3}{7}+\dfrac{3}{7}.\dfrac{3}{4}+\dfrac{3}{7}.\dfrac{1}{6}+\dfrac{1}{8}.15}{a+\dfrac{5}{6}+\dfrac{1}{3}}\)
\(A=\dfrac{\dfrac{3}{7}.\left(\dfrac{-5}{8}+\dfrac{3}{4}+\dfrac{1}{6}\right)+\dfrac{1}{8}.15}{a+\dfrac{7}{6}}\)
\(A=\dfrac{\dfrac{3}{7}.\dfrac{7}{24}+\dfrac{1}{8}.15}{a+\dfrac{7}{6}}\)
\(A=\dfrac{\dfrac{1}{8}+\dfrac{1}{8}.15}{a+\dfrac{7}{6}}\)
\(A=\dfrac{\dfrac{1}{8}.\left(15+1\right)}{a+\dfrac{7}{6}}\)
\(A=\dfrac{2}{a+\dfrac{7}{6}}\)
b, Thay \(a=75\%\) vào \(A\), ta được:
\(A=\dfrac{2}{75\%+\dfrac{7}{6}}\)
\(A=\dfrac{2}{\dfrac{3}{4}+\dfrac{7}{6}}\)
\(\Rightarrow A=\dfrac{23}{12}\)
c, Ta có: \(\dfrac{2}{a+\dfrac{7}{6}}=50\%\)
\(\dfrac{2}{a+\dfrac{7}{6}}=\dfrac{1}{2}\)
\(\dfrac{2}{a+\dfrac{7}{6}}=\dfrac{2}{4}\)
\(\Rightarrow a+\dfrac{7}{6}=4\)
\(\Rightarrow a=\dfrac{17}{6}\)
d, Để \(A\in Z\Rightarrow2⋮a+\dfrac{7}{6}\)
\(\Rightarrow a+\dfrac{7}{6}\in\left\{\pm1;\pm2\right\}\)
\(\circledast,a+\dfrac{7}{6}=1\Rightarrow a=\dfrac{-1}{6}\)
\(\circledast,a+\dfrac{7}{6}=-1\Rightarrow a=\dfrac{-13}{6}\)
\(\circledast,a+\dfrac{7}{6}=2+\Rightarrow a=\dfrac{5}{6}\)
\(\circledast,a+\dfrac{7}{6}=-2\Rightarrow a=\dfrac{-19}{6}\)
\(a\in\varnothing\) khi \(A\in Z\)
e, Ta có:
\(B=\dfrac{5}{3}+\dfrac{-1}{8}\Rightarrow B=\dfrac{37}{24}\)
\(\Rightarrow\dfrac{2}{a+\dfrac{7}{6}}=\dfrac{37}{24}\)
\(a+\dfrac{7}{6}=\dfrac{37}{24}.2\)
\(a+\dfrac{7}{6}=\dfrac{37}{12}\)
\(\Rightarrow a=\dfrac{23}{12}\)
Chúc bạn học thiệt giỏi nha!!! 
)
Bài 1 :
Sửa đề :
Tìm \(n\in Z\) để những phân số sau đồng thời có giá trị nguyên
\(\dfrac{-12n}{n};\dfrac{15}{n-2};\dfrac{8}{n+1}\)
Làm
Ta có :
\(\dfrac{-12n}{n}=-12\)
\(\Leftrightarrow\) Với mọi \(n\) thì \(\dfrac{-12n}{n}\) đều có giá trị nguyên \(\left(1\right)\)
Để \(\dfrac{15}{n-2}\in Z\) \(\Leftrightarrow n-2\inƯ\left(15\right)=\left\{\pm1;\pm15;\pm3;\pm5\right\}\)
\(\Leftrightarrow n\in\left\{-13;\pm3;\pm1;5;7;17\right\}\left(1\right)\)
Để \(\dfrac{8}{n+1}\in Z\Leftrightarrow n+1\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
\(\Leftrightarrow n\in\left\{-9;-5;\pm3;-2;0;1;7\right\}\left(3\right)\)
Từ \(\left(1\right)+\left(2\right)+\left(3\right)\Leftrightarrow n\in\left\{\pm3;1;7\right\}\)