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Bài 3:
\(=\left(\dfrac{3}{5}+\dfrac{2}{7}+\dfrac{4}{35}\right)+\left(-\dfrac{1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right)+\dfrac{1}{131}\)
\(=\dfrac{21+10+4}{35}+\dfrac{-9-2-7}{18}+\dfrac{1}{131}\)
=1/131
Bài 5:
a: Phần nguyên là 0
b: Phần nguyên là -1
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài làm:
c) \(-\frac{2}{5}+\frac{5}{3}\left(\frac{3}{2}-\frac{4}{15}x\right)=-\frac{7}{6}\)
\(\Leftrightarrow-\frac{2}{5}+\frac{5}{2}-\frac{4}{9}x=-\frac{7}{6}\)
\(\Leftrightarrow\frac{4}{9}x=-\frac{2}{5}+\frac{5}{2}+\frac{7}{6}\)
\(\Leftrightarrow\frac{4}{9}x=\frac{49}{15}\)
\(\Leftrightarrow x=\frac{49}{15}\div\frac{4}{9}\)
\(\Rightarrow x=\frac{147}{20}\)
Vậy \(x=\frac{147}{20}\)
Bài 2:
a) Ta có: \(F=\frac{3x-2}{x+3}=\frac{\left(3x+9\right)-11}{x+3}=3-\frac{11}{x+3}\)
Để F nguyên \(\Rightarrow\frac{11}{x+3}\inℤ\Leftrightarrow x+3\inƯ\left(11\right)=\left\{-11;-1;1;11\right\}\)
\(\Rightarrow x\in\left\{-14;-4;-2;8\right\}\)
Vậy \(x\in\left\{-14;-4;-2;8\right\}\)thì F nguyên
2b) Tách
\(G=\frac{x^2-2x+4}{x+1}=\frac{x^2+x-3x-3+7}{x+1}=\frac{x\left(x+1\right)-3\left(x+1\right)+7}{x+1}\)
\(=\frac{x\left(x+1\right)}{x+1}-\frac{3\left(x+1\right)}{x+1}+\frac{7}{x+1}=x-3+\frac{7}{x+1}\)
G là số nguyên <=> \(\frac{7}{x+1}\)là số nguyên <=> \(7⋮x+1\)<=> \(x+1\inƯ\left(7\right)=\left\{1;-1;7;-7\right\}\)
<=> \(x\in\left\{0;-2;6;-8\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1 :
\(a,2\left(\frac{3}{4}-5x\right)=\frac{4}{5}-3x\)
\(\Rightarrow\frac{3}{2}-10x=\frac{4}{5}-3x\)
\(\Rightarrow7x=\frac{3}{2}-\frac{4}{5}\)
\(\Rightarrow7x=\frac{7}{10}\)\(\Leftrightarrow x=0,1\)
\(b,\frac{3}{2}-4\left(\frac{1}{4}-x\right)=\frac{2}{3}-7x\)
\(\Rightarrow\frac{3}{2}-1+4x=\frac{2}{3}-7x\)
\(\Rightarrow11x=\frac{2}{3}+1-\frac{3}{2}\)
\(\Rightarrow11x=\frac{4+6-9}{6}-\frac{1}{6}\)
\(\Rightarrow x=\frac{1}{66}\)
Câu 2 :
\(a,\frac{2}{x-1}< 0\)
Vì \(2>0\Rightarrow\)để \(\frac{2}{x-1}< 0\)thì \(x-1< 0\Leftrightarrow x< 1\)
\(b,\frac{-5}{x-1}< 0\)
Vì \(-5< 0\)\(\Rightarrow\)để \(\frac{-5}{x-1}< 0\)thì \(x-1>0\Rightarrow x>1\)
\(c,\frac{7}{x-6}>0\)
Vì \(7>0\Rightarrow\)để \(\frac{7}{x-6}>0\)thì \(x-6>0\Rightarrow x>6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có:
\(\frac{4}{15}+\frac{1}{6}-\frac{4}{9}>\frac{2}{3}-x-\frac{1}{4}\\ \Rightarrow x+\frac{4}{15}+\frac{1}{6}-\frac{4}{9}>\frac{2}{3}-\frac{1}{4}\\ \Rightarrow x>\frac{2}{3}+\frac{4}{9}-\frac{1}{4}-\frac{1}{6}-\frac{4}{15}\\ \Rightarrow x>\left(\frac{6}{9}+\frac{4}{9}\right)-\left(\frac{15}{60}+\frac{10}{60}+\frac{16}{60}\right)\)
\(x>\frac{10}{9}-\frac{41}{60}\\ x>\frac{200-123}{180}\Rightarrow x>\frac{77}{180}\)
b) Bất đẳng thức kép
\(4-1\frac{1}{3}< x+\frac{1}{5}< 12\frac{2}{7}-3\frac{3}{8}\)
có nghĩa là ta phải có hai bất đẳng thức đồng thời:
\(x+\frac{1}{5}>4-1\frac{1}{3}\) và \(x+\frac{1}{5}< 12\frac{2}{7}-3\frac{3}{8}\)
Ta tìm các giá trị của x cần thỏa mãn bất đẳng thức thứ nhất:
\(x+\frac{1}{5}>4-1\frac{1}{3}\Rightarrow x>4-1\frac{1}{3}-\frac{1}{5}\\ \Rightarrow x>\frac{37}{15}\)
Từ bất đẳng thức thứ hai
\(x+\frac{1}{5}< 12\frac{2}{7}-3\frac{3}{8}\Rightarrow x< \frac{86}{7}-\frac{27}{8}-\frac{1}{5}\\ \Rightarrow x< \frac{2439}{280}.\)
Như vậy các số hữu tỉ x cần thỏa mãn:
\(\frac{37}{15}< x< \frac{2439}{280}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/ a/\(-\frac{7}{18}=\left(-\frac{7}{2}\right)\left(\frac{1}{9}\right)\)
b/\(-\frac{7}{18}=\left(-\frac{7}{9}\right):2\)
2/
a/\(\frac{7}{15}\cdot\left(-\frac{3}{8}-\frac{3}{7}\right)=\frac{7}{15}\cdot\left(-\frac{45}{56}\right)=-\frac{3}{8}\)
b/\(\left(-\frac{3}{4}+\frac{2}{5}\right):\frac{3}{7}+\left(\frac{3}{5}+-\frac{4}{4}\right):\frac{3}{7}\)
\(=\left(-\frac{7}{20}\right):\frac{3}{7}+\left(-\frac{2}{5}\right):\frac{3}{7}\)
\(=\left(-\frac{49}{60}\right)+\left(-\frac{14}{15}\right)=-\frac{7}{4}\)
c/\(\frac{2}{3}\cdot\left(-\frac{5}{2}\right)+\frac{10}{15}\cdot\left(-\frac{3}{7}\right)-\frac{2}{3}\cdot\left(-\frac{5}{3}\right)\)
\(=\frac{2}{3}\cdot\left(-\frac{5}{2}-\frac{3}{7}+\frac{5}{3}\right)=-\frac{53}{63}\)
3/
\(2-\left(3-x\right)=-\frac{3}{2}\)
\(2-3+x=-\frac{3}{2}\)
\(x=-\frac{3}{2}+3-2=-\frac{1}{2}\)
4/
a/ Ta có 2 trường hợp:
TH1: \(x-3,5=7,5\)
\(x=7,5+3,5=11\)
TH2: \(x-3,5=-7,5\)
\(x=-7,5+3,5=-4\)
b/ Ta có 2 trường hợp:
TH1:\(x-0,4=3,6\)
\(x=4\)
TH2: \(x-0,4=-3,6\)
\(x=-3.2\)
c/ Ta có 2 trường hợp:
TH1:\(x+\frac{4}{5}=\frac{3}{2}\)
\(x=\frac{7}{10}\)
TH2:\(x+\frac{4}{5}=-\frac{3}{2}\)
\(x=-\frac{32}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 11:
Ta có: \(x=\dfrac{-101}{a+7}\) nguyên khi \(-101⋮a+7\)
Vậy: \(a+7\inƯ\left(101\right)\)
\(Ư\left(101\right)=\left\{101;1;-101;-1\right\}\)
\(a+7\in\left\{101;1;-101;-1\right\}\)
\(\Rightarrow a\in\left\{94;-108;-6;-8\right\}\)
Vậy x sẽ nguyên khi \(a\in\left\{94;-108l-6;-8\right\}\)
Bài 12:
Ta có: \(t=\dfrac{3x+8}{x-5}=\dfrac{3x+15-7}{x-5}=\dfrac{3\left(x+5\right)-7}{x-5}=3+\dfrac{7}{x-5}\)
t nguyên khi \(\dfrac{7}{x+5}\) nguyên tức là \(x-5\inƯ\left(7\right)\)
\(Ư\left(7\right)=\left\{-7;7;-1;1\right\}\)
\(\Rightarrow x-5\in\left\{-7;7;-1;1\right\}\)
\(\Rightarrow x\in\left\{12;-2;4;6\right\}\)
Vậy t sẽ nguyên khi \(x\in\left\{12;-2;4;6\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{3}-\left(\frac{2}{3}-x+\frac{5}{4}\right)=\frac{7}{12}-\left(\frac{5}{2}-\frac{13}{6}\right)\)
\(\frac{1}{3}-\left(\frac{2}{3}-x+\frac{5}{4}\right)=\frac{7}{12}-\frac{1}{3}\)
\(\frac{1}{3}-\left(\frac{2}{3}-x+\frac{5}{4}\right)=\frac{1}{4}\)
\(\frac{2}{3}-x+\frac{5}{4}=\frac{1}{3}-\frac{1}{4}\)
\(\frac{2}{3}-x+\frac{5}{4}=\frac{1}{12}\)
\(\frac{2}{3}-x=\frac{1}{12}-\frac{5}{4}\)
\(\frac{2}{3}-x=-\frac{7}{6}\)
\(x=\frac{2}{3}-\left(-\frac{7}{6}\right)\)
\(x=\frac{2}{3}+\frac{7}{6}\)
\(x=\frac{11}{6}\)
\(\Leftrightarrow\dfrac{5}{6}x=-\dfrac{4}{15}+\dfrac{2}{5}=\dfrac{2}{15}\\ \Leftrightarrow x=\dfrac{2}{15}\cdot\dfrac{6}{5}=\dfrac{4}{25}\)
Giúp vs ạ đúng mình tích cho ạ