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33000 = ( 33 )1000 = 271000
42000 =( 42 )1000 = 161000
Vì 271000 > 161000 nên 33000 > 42000
Ta co : 33000 = (3.3)1000 = 91000
42000 = (4.2)1000 = 81000
Vì : 9 > 8 nên 91000 > 81000.
Vay 33000 > 42000
k mk nha bn
a) Ta có: \(\widehat{xOy}+\widehat{yOz}=180^0\)(hai góc kề bù)
\(\Leftrightarrow\widehat{zOy}+140^0=180^0\)
hay \(\widehat{yOz}=40^0\)
Vậy: \(\widehat{yOz}=40^0\)
Câu 1:
a. $132-468-(-215)=132-468+215=(132+215)-468$
$=347-468=-121$
b.
$(144-97)-156=47-156=-109$
c.
$(-145)-(18-145)=-145-18+145=(-145+145)-18=0-18=-18$
d.
$115+(-15+27)=115-15+27=100+27=127$
e.
$(24+614)-(386-76)=638-310=328$
Câu 2:
a.
$(-105)+x=-217$
$x-105=-217$
$x=-217+105=-112$
b.
$x-67=-23$
$x=-23+67=44$
c.
$4(3-x)=28$
$3-x=28:4=7$
$x=3-7=-4$
d.
$5[x+(-81)]=400$
$x-81=400:5=80$
$x=80+81=161$
e.
$140:(x+108)=7$
$x+108=140:7=20$
$x=20-108=-88$
f.
$-27+(-x)=-104$
$-28-x=-104$
$x=-28-(-104)=-28+104=76$
c) \(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{12}\le x\le\dfrac{7}{10}+\dfrac{27}{6}\)
\(\Leftrightarrow\dfrac{2}{3}\le x\le\dfrac{26}{5}=5,2\), mà \(x\in Z\)
\(\Rightarrow x\in\left\{1;2;3;4;5\right\}\)
d) \(-\dfrac{31}{14}+\dfrac{115}{131}+\dfrac{111}{74}\le x\le\dfrac{6}{36}+\dfrac{9}{27}+\dfrac{48}{96}\)
\(\Leftrightarrow\dfrac{150}{917}\le x\le1\) , mà \(x\in Z\)
\(\Rightarrow x=1\)
a) \(\left(3x-2^4\right).7^3=2.7^4\)\(\Leftrightarrow3x-2^4=2.7^4:7^3\)
\(\Leftrightarrow3x-16=2.7\)\(\Leftrightarrow3x-16=14\)\(\Leftrightarrow3x=30\)
\(\Leftrightarrow x=10\)
Vậy \(x=10\)
b) \(3x+4x=\left|-75\right|+23\)\(\Leftrightarrow7x=75+23\)
\(\Leftrightarrow7x=98\)\(\Leftrightarrow x=14\)
Vậy \(x=14\)
a) \(\left(3x-2^4\right)\cdot7^3=2\cdot7^4\)
=> \(3x\cdot7^3-2^4\cdot7^3=2\cdot7\cdot7^3\)
=> \(3x\cdot7^3=14\cdot7^3+16\cdot7^3\)
=> \(3x\cdot7^3=\left(14+16\right)\cdot7^3\)
=> \(3x\cdot7^3=30\cdot7^3\)
=> \(3x=30\)(bỏ hai vế 73)
=> \(x=10\)
Vậy x = 10
b) \(3x+4x=\left|-75\right|+23\)
=> \(7x=75+23\)
=> \(7x=98\)
=> \(x=14\)
Vậy x = 14
a) 7/4 + 3/2 + ( - 9/16 ) = 13/4 + ( -9/16 ) = 41/112
b) - 2/7 + 3/5 + 9/7 + ( -18/5 )= ( - 2/7 + 9/7 ) + [( 3/5 + ( - 18/5 )] = 1 + ( - 3/1 ) = - 2/1
< tách ra ik >
a) 7/4 + 3/2 + ( - 9/16 ) = 13/4 + ( -9/16 ) = 41/112
b) - 2/7 + 3/5 + 9/7 + ( -18/5 )= ( - 2/7 + 9/7 ) + [( 3/5 + ( - 18/5 )] = 1 + ( - 3/1 ) = - 2/1
< tách ra ik >
Câu 4:
\(A=7^{2022}-7^{2021}+7^{2020}-7^{2019}+...+7^2-7\\ \Rightarrow7A=7^{2023}-7^{2022}+7^{2021}-7^{2020}+...+7^3-7^2\\ \Rightarrow7A+A=7^{2023}-7^{2022}+...+7^3-7^2+7^{2022}-7^{2021}+...+7^2-7\\ \Rightarrow8A=7^{2023}-7\\ \Rightarrow A=\dfrac{7^{2023}-7}{8}\)
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