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892+112+89.22
=892+112+89.11+11.89.
=89.(11+89)+11.(11+89)
=89.100+11.100
=8900+1100
=10000
Chúc em học tốt^^
892 + 112 + 89.22
= (892 + 89.11) + (112 + 89.11)
= 89.(89 + 11) + 11.(11 + 89)
= 89.100 + 11.100
= 100.(89 + 11)
= 100.100
= 10000
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
D = x\(^2\) + 2xy + y\(^2\) - z\(^2\) - 2zt - t\(^2\)
D = (x + y)\(^2\) - z\(^2\) + z\(^2\) - 2zt + t\(^2\) - t\(^2\)
D = (89 + 11)\(^2\) +(z - t)\(^2\) - z\(^2\) - t\(^2\)
D = 100\(^2\) + (60 - 30)\(^2\) - 60\(^2\) - 30\(^2\)
D = 10 000 + 900 - 3600 - 900
D = 6400
Học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{x+2}{89}+\dfrac{x+5}{86}>\dfrac{x+8}{83}+\dfrac{x+11}{80}\)
\(\Leftrightarrow\dfrac{x+2}{89}+1+\dfrac{x+5}{86}+1>\dfrac{x+8}{83}+1+\dfrac{x+11}{80}+1\)
\(\Leftrightarrow\dfrac{x+91}{89}+\dfrac{x+91}{86}>\dfrac{x+91}{83}+\dfrac{x+91}{80}\)
\(\Leftrightarrow\dfrac{x+91}{89}+\dfrac{x+91}{86}-\dfrac{x+91}{83}-\dfrac{x+91}{80}>0\)
\(\Leftrightarrow\left(x+91\right)\left(\dfrac{1}{89}+\dfrac{1}{86}-\dfrac{1}{83}-\dfrac{1}{80}\right)>0\)
Ta có: \(\dfrac{1}{89}+\dfrac{1}{86}+\dfrac{1}{83}+\dfrac{1}{80}< 0\)
\(\Leftrightarrow x+91< 0\)
\(\Leftrightarrow x< -91\)
Vậy...........
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892+112+22.89
=892+2.89..11+112
=(89+11)2
=1002
=10000
Tuy mình chưa học lớp 8 nhưng mình cũng có thể làm được bài này.
= 892 + 2.89.11 + 112 = (89 +11)2 = 1002 = 10000