65+35+6−29≤ \le \frac{-26}{5} + 2+ \frac{6}{5}≤5−26+2+56
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\(a,\left(\frac{31}{20}-\frac{26}{45}\right)\cdot\left(\frac{-36}{35}\right)< x< \left(\frac{51}{56}+\frac{8}{21}+\frac{1}{3}\right)\cdot\frac{8}{13}\)
\(taco:\left(\frac{31}{20}-\frac{26}{45}\right)\cdot\left(\frac{-36}{35}\right)=\frac{35}{36}\cdot\frac{-36}{35}=-1\)
\(\left(\frac{51}{56}+\frac{8}{21}+\frac{1}{3}\right)\cdot\frac{8}{13}=\frac{13}{8}\cdot\frac{8}{13}=1\)
\(=>x=0\)
\(b,\frac{-5}{6}+\frac{8}{3}+\frac{29}{-3}< x< \frac{-1}{2}+2+\frac{5}{2}\)(dau <co dau gach ngang o duoi nha)
\(taco:\frac{-5}{6}+\frac{8}{3}+\frac{29}{-3}=\frac{-5}{6}+\frac{8}{3}+\frac{-29}{3}=\frac{-5}{6}+\frac{16}{6}+\frac{-58}{6}=\frac{-47}{6}=-7,8\)
\(\frac{-1}{2}+2+\frac{5}{2}=\frac{3}{2}+\frac{5}{2}=4\)
tu do \(=>x=-7,8;...;0;1;2;3;4\)
1) $(-35)\cdot18+28\cdot35$
$=35\cdot(-18)+28\cdot35$
$=35\cdot(-18+28)$
$=35\cdot10$
$=350$
2) $53\cdot29-47\cdot(-29)$
$=53\cdot29-47\cdot(-1)\cdot29$
$=53\cdot29+47\cdot29$
$=29\cdot(53+47)$
$=29\cdot100$
$=2900$
3) $125\cdot(-24)+24\cdot225$
$=(-125)\cdot24+24\cdot225$
$=24\cdot(-125+225)$
$=24\cdot100$
$=2400$
4) $26\cdot(-121)-121\cdot(-36)$
$=-121\cdot[26+(-36)]$
$=-121\cdot(26-36)$
$=-121\cdot(-10)$
$=121\cdot10$
$=1210$
5) $65\cdot(-25)+25\cdot(-23)$
$=-65\cdot25+25\cdot(-23)$
$=25\cdot(-65-23)$
$=25\cdot(-88)$
$=-2200$
6) $237\cdot(-26)+26\cdot137$
$=-237\cdot26+26\cdot137$
$=26\cdot(-237+137)$
$=26\cdot(-100)$
$=-2600$
7) $30\cdot(-125)+25\cdot30$
$=30\cdot(-125+25)$
$=30\cdot(-100)$
$=-3000$
8) $(-37)\cdot69+(-31)\cdot37$
$=37\cdot(-69)+37\cdot(-31)$
$=37\cdot(-69-31)$
$=37\cdot(-100)$
$=-3700$
9) $31\cdot72-31\cdot70-31\cdot2$
$=31\cdot(72-70-2)$
$=31\cdot(2-2)$
$=31\cdot0$
$=0$
10) $(-12)\cdot47+(-12)\cdot52+(-12)$
$=-12\cdot(47+52+1)$
$=-12\cdot100$
$=-1200$
$\text{#}Toru$
Đề bài thiếu : \(x\inℤ\)
Ta có :
\(\frac{-5}{6}+\frac{8}{3}+\frac{-29}{6}\le x\le\frac{-1}{2}+2+\frac{5}{2}\)
\(\Leftrightarrow\)\(\frac{-5+16-29}{6}\le x\le\frac{-1+4+5}{2}\)
\(\Leftrightarrow\)\(\frac{-18}{6}\le x\le\frac{8}{2}\)
\(\Leftrightarrow\)\(-3\le x\le4\)
\(\Rightarrow\)\(x\in\left\{-3;-2;-1;0;1;2;3;4\right\}\)
Vậy \(x\in\left\{-3;-2;-1;0;1;2;3;4\right\}\)
Chúc bạn học tốt ~
\(a)\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{3}{5}+\frac{5}{7}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{-2}{5}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{-1}{4}+\frac{2}{7}+\frac{5}{7}+\frac{3}{5}\)
\(\Rightarrow\frac{2}{6}+\frac{1}{6}+\frac{-3}{5}\le x< -1+1+\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}+\frac{-3}{5}\le x< \frac{3}{5}\)
\(\Rightarrow\frac{-1}{10}\le x< \frac{6}{10}\)
\(\Rightarrow-1\le x< 6\)
\(\Rightarrow x\in\left\{-1;0;1;2;3;4;5\right\}\)
Bài b tương tự
Ta có : \(-\frac{5}{6}+\frac{8}{3}+\frac{29}{-6}=-3\) và \(\frac{1}{2}+2+\frac{5}{2}=5\)
Vậy -3 < x < 5. Do x \(\in\) Z nên x \(\in\) {-2; -1; 0; 1; 2; 3; 4}
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