Tìm x biết: \(\frac{9}{11}< \frac{x}{15}< \frac{10}{11}\)
cho mình biết cách giải nữa.
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Bài làm
\(\frac{11}{15}-\frac{9}{10}< x< \frac{11}{15}:\frac{9}{10}\)
\(\Rightarrow\frac{22}{30}-\frac{27}{30}< x< \frac{11}{15}.\frac{10}{9}\)
\(\Rightarrow-\frac{5}{30}< x< \frac{11}{3}.\frac{2}{9}\)
\(\Rightarrow-\frac{5}{30}< x< \frac{22}{27}\)
\(\Rightarrow x\in\left\{-4;-3;-2;-1;0;1;2;3;4;5;6;7;8;9;10;11;12;13;14;15;16;17;18;19;20;21\right\}\)
Vậy \(x\in\left\{-4;-3;-2;-1;0;1;2;3;4;5;6;7;8;9;10;11;12;13;14;15;16;17;18;19;20;21\right\}\)
~ Chắc z ~
# Chúc bạn học tốt #
Ta có:\(\frac{11}{15}-\frac{9}{10}< x< \frac{11}{15}:\frac{9}{10}\)
\(\Leftrightarrow\frac{110-135}{30}< x< \frac{11.10}{15.9}\)
\(\Leftrightarrow\frac{-15}{30}< x< \frac{22}{27}\)
(Vì x c Z)\(\Leftrightarrow-1< x< 1\Rightarrow x\in\left\{0\right\}\)
a)\(\frac{7}{11}< x-\frac{1}{7}< \frac{10}{3}\)
\(=>\frac{7}{11}+\frac{1}{7}< x< \frac{10}{3}+\frac{1}{7}\)
\(=>\frac{60}{77}< x< \frac{73}{21}\)
VẬY \(\frac{60}{77}< x< \frac{73}{21}\)
b) \(\frac{40}{99}>x>\frac{43}{176}\)
1) \(\left|x-\frac{3}{5}\right|< \frac{1}{3}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{3}{5}< \frac{1}{3}\\x-\frac{3}{5}< -\frac{1}{3}\end{cases}}\Rightarrow\orbr{\begin{cases}x< \frac{1}{3}+\frac{3}{5}\\x< \frac{-1}{3}+\frac{3}{5}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x< \frac{5}{15}+\frac{9}{15}\\x< \frac{-5}{15}+\frac{9}{15}\end{cases}}\Rightarrow\orbr{\begin{cases}x< \frac{14}{15}\\x< \frac{4}{15}\end{cases}}\)
vay \(\orbr{\begin{cases}x< \frac{14}{15}\\x< \frac{4}{15}\end{cases}}\)
2) \(\left|x+\frac{11}{2}\right|>\left|-5,5\right|\)
\(\left|x+\frac{11}{2}\right|>5,5\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{11}{2}>\frac{11}{2}\\x+\frac{11}{2}>-\frac{11}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x>\frac{11}{2}-\frac{11}{2}\\x>\frac{-11}{2}-\frac{11}{2}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x>0\\x>-11\end{cases}}\)
vay \(\orbr{\begin{cases}x>0\\x>-11\end{cases}}\)
3) \(\frac{2}{5}< \left|x-\frac{7}{5}\right|< \frac{3}{5}\)
\(\Rightarrow\left|x-\frac{7}{5}\right|>\frac{2}{5}\) va \(\left|x-\frac{7}{5}\right|< \frac{3}{5}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{7}{5}>\frac{2}{5}\\x-\frac{7}{5}>\frac{-2}{5}\end{cases}}\Rightarrow\orbr{\begin{cases}x>\frac{2}{5}+\frac{7}{5}\\x>\frac{-2}{5}+\frac{7}{5}\end{cases}}\)va \(\orbr{\begin{cases}x-\frac{7}{5}< \frac{3}{5}\\x-\frac{7}{5}< \frac{-3}{5}\end{cases}}\Rightarrow\orbr{\begin{cases}x< \frac{3}{5}+\frac{7}{5}\\x< \frac{-3}{5}+\frac{7}{5}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x>\frac{9}{5}\\x>1\end{cases}}\)va \(\orbr{\begin{cases}x< 2\\x< \frac{4}{5}\end{cases}}\)
vay ....
a/ \(\frac{15}{x}-\frac{1}{3}=\frac{28}{51}\)
\(\frac{15}{x}=\frac{28}{51}+\frac{1}{3}\)
\(\frac{15}{x}=\frac{15}{17}\)
\(x=15:\frac{15}{17}\)
\(x=17\)
b) \(\frac{x}{20}-\frac{2}{5}=10\)
\(\frac{x}{20}=10+\frac{2}{5}\)
\(\frac{x}{20}=\frac{52}{5}\)
\(x=\frac{52}{5}\cdot20\)
\(x=208\)
c) \(x+\frac{18}{23}=2\frac{1}{3}\)
\(x+\frac{18}{23}=\frac{7}{3}\)
\(x=\frac{7}{3}-\frac{18}{23}\)
\(x=\frac{107}{69}\)
d) \(\frac{7}{11}< x-\frac{1}{7}< \frac{10}{13}\)
\(\Rightarrow\frac{7}{11}+\frac{1}{7}< x< \frac{10}{13}\)
\(\frac{60}{77}< x< \frac{60}{78}\)
Đến đây .....bí!
e) Tớ bỏ luôn đc ko.
D) 7/11<X-1/7<10/13
<=> 7/11+1/7<x< 10/13+1/7
<=> 60/77< x< 83/91
<=> 5460/1001 <x< 6391/1001
vậy X thuộc tập hợp các phÂN số lớn hơn 5460/1001 và bé hơn 913/1001
vd : Y/1001 trong đó y là 5461;5462;5463...6389;6390
1)
a)
\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)
\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)
\(\frac{-20}{5}< x< \frac{-3}{10}\)
\(\frac{-40}{10}< x< \frac{-3}{10}\)
\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)
a) (x-3)+(x-2)+(x-1)+....+10+11=11
(x-3)+(x-2)+(x-1)+....+10 =0
gọi số hạng của tổng vế trái là n
(x-3+10).\(\frac{n}{2}\)=0
(x+7).n:2=0
(x+7) =0
\(\Rightarrow\)x+7=0 (do n\(\ne\)0)
x=0-7
x=-7
b) \(\frac{2}{3}\left[\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right]<=x<=4\frac{1}{3}.\left[\frac{1}{2}-\frac{1}{6}\right]\)
\(\frac{2}{3}.\frac{11}{12}<=x<=\frac{13}{3}.\frac{1}{3}\)
\(\frac{11}{18}<=x<=\frac{13}{9}\)
do x\(\in\)z nên x=1
vậy x=1
\(\frac{3}{5}.x-x=\frac{-4}{15}\)
<=>\(x.\left(\frac{3}{5}-1\right)=\frac{-4}{15}\)
<=>\(\frac{-2}{5}.x=\frac{-4}{15}\)
<=>\(x=\frac{-4}{15}:\frac{-2}{5}\)
<=>\(x=\frac{2}{3}\)
vậy \(x=\frac{2}{3}\)
Ta có:\(\frac{3}{5}.x-x=-\frac{4}{15}\)
\(\left(\frac{3}{5}-1\right).x=-\frac{4}{15}\)
\(\left(\frac{3}{5}-\frac{5}{5}\right).x=-\frac{4}{15}\)
\(\left(-\frac{2}{5}\right).x=-\frac{4}{15}\)
\(x=-\frac{4}{15}:\left(-\frac{2}{5}\right)\)
\(x=-\frac{4}{15}.\left(-\frac{5}{2}\right)\)
\(x=\frac{\left(-4\right).\left(-5\right)}{15.2}\)
\(x=\frac{20}{30}=\frac{2}{3}\)
\(\frac{9}{11}< \frac{x}{15}< \frac{10}{11}\)
\(\frac{135}{165}< \frac{11x}{165}< \frac{150}{165}\)
135 < 11x < 150
Chỉ có 11x = 143 là thích hợp.
Vậy x = 143 : 11 = 13.
Ta có: \(\frac{9}{11}< \frac{x}{15}< \frac{10}{11}\)
=> \(\frac{135}{165}< \frac{x.11}{165}< \frac{150}{165}\)
=> 135 < x.11 < 150
=> x.11 E (thuộc) {136; 137;...;149}
Và x.11 phải chia hết cho 11 nên x = 143
Vậy: x = 143