tìm phép tính đúng
a) A. 29.28=122 B.29.28=199 C.1999-(-99)=200 D. 200.(-199)=29800
b) A. 29.10=199 B.29.29=500 C.1999.(-987)=-1973013 D. 200.(-199)=199900
C) A. 29.10=199 B.399.499=199101 C.1999.(-988)=199999 D. 200.(-219)=199000
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\(a,\dfrac{199}{200}=1-\dfrac{1}{200};\dfrac{200}{201}=1-\dfrac{1}{201}\\ Vì:\dfrac{1}{200}>\dfrac{1}{201}\\ \Rightarrow1-\dfrac{1}{200}< 1-\dfrac{1}{201}\\ Vậy:\dfrac{199}{200}< \dfrac{200}{201}\\ b,\dfrac{2001}{2002}=1-\dfrac{1}{2002};\dfrac{2002}{2003}=1-\dfrac{1}{2003}\\ Vì:\dfrac{1}{2002}>\dfrac{1}{2003}\Rightarrow1-\dfrac{1}{2002}< 1-\dfrac{1}{2003}\\ Vậy:\dfrac{2001}{2002}< \dfrac{2002}{2003}\)
\(c,\dfrac{2021}{2020}=1+\dfrac{1}{2020};\dfrac{2020}{2019}=1+\dfrac{1}{2019}\\ Vì:\dfrac{1}{2020}< \dfrac{1}{2019}\\ Nên:1+\dfrac{1}{2020}< 1+\dfrac{1}{2019}\\ Vậy:\dfrac{2021}{2020}< \dfrac{2020}{2019}\\ d,\dfrac{199}{198}=1+\dfrac{1}{198};\dfrac{200}{199}=1+\dfrac{1}{199}\\ Vì:\dfrac{1}{198}>\dfrac{1}{199}\\ Nên:1+\dfrac{1}{198}>1+\dfrac{1}{199}\\ Vậy:\dfrac{199}{198}>\dfrac{200}{199}\)
a) Ta có \(\sqrt{17}>\sqrt{16}=4\)
\(\sqrt{26}>\sqrt{25}=5\)
Khi đó \(\sqrt{17}+\sqrt{26}+1>4+5+1=10\) (1)
Mà \(\sqrt{99}< \sqrt{100}=10\) (2)
Từ (1) và (2) suy ra \(\sqrt{17}+\sqrt{26}+1>\sqrt{99}\)
Vậy....
a: \(\left(2345+45\right)-2345\)
\(=2345+45-2345\)
\(=45+\left(2345-2345\right)\)
=45+0
=45
b: \(\left(-2010\right)-\left(119-2010\right)\)
\(=-2010-119+2010\)
\(=\left(-2010+2010\right)-119\)
=0-119
=-119
c: \(\left(18+29\right)+\left(158-18-29\right)\)
\(=18+29+158-18-29\)
\(=158+\left(18-18\right)+\left(29-29\right)\)
=158+0+0
=158
d: \(126+\left(-20\right)+2004+\left(-106\right)\)
\(=126-20+2004-106\)
\(=\left(126-20-106\right)+2004\)
=0+2004
=2004
e: \(\left(-199\right)+\left(-200\right)+\left(201\right)\)
\(=-199-200+201\)
\(=-199+\left(201-200\right)\)
=-199+1
=-198
f: \(99+\left(-100\right)+101\)
\(=\left(99+101\right)-100\)
=200-100
=100
g: \(217+\left[43+\left(-217\right)+\left(-23\right)\right]\)
\(=217+43-217-23\)
\(=\left(217-217\right)+\left(43-23\right)\)
=0+20
=20