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\(a,2.10^3+6.10^2+0.10+1=2.1000+6.100+0+1=2601\\ b,5.10^4+7.10^3+9.10^2+1.10+5\\ =5.10000+7.1000+9.100+10+5 =57915\)
a) \(...=2000+600+0+1=2601\)
b) \(...=50000+7000+900+10+5=57915\)
a) \(\left(\dfrac{3}{29}-\dfrac{1}{5}\right)\cdot\dfrac{29}{3}\)
\(=\dfrac{3}{29}\cdot\dfrac{29}{3}-\dfrac{1}{5}\cdot\dfrac{29}{3}\)
\(=1-\dfrac{29}{15}\)
\(=\dfrac{15-29}{15}\)
\(=-\dfrac{14}{15}\)
b) \(\dfrac{3}{4}\cdot\dfrac{7}{9}+\dfrac{1}{4}\cdot\dfrac{7}{9}\)
\(=\dfrac{7}{9}\cdot\left(\dfrac{1}{4}+\dfrac{3}{4}\right)\)
\(=\dfrac{7}{9}\cdot1\)
\(=\dfrac{7}{9}\)
c) \(\dfrac{1}{7}\cdot\dfrac{5}{9}+\dfrac{5}{9}\cdot\dfrac{1}{7}+\dfrac{5}{9}\cdot\dfrac{3}{7}\)
\(=\dfrac{5}{9}\cdot\left(\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)
\(=\dfrac{5}{9}\cdot\dfrac{5}{7}\)
\(=\dfrac{25}{63}\)
d) \(4\cdot11\cdot\dfrac{3}{4}\cdot\dfrac{9}{121}\)
\(=\left(4\cdot\dfrac{3}{4}\right)\cdot\left(11\cdot\dfrac{9}{121}\right)\)
\(=3\cdot\dfrac{9}{11}\)
\(=\dfrac{27}{11}\)
`@` `\text {Ans}`
`\downarrow`
`a)`
`(-134) + 51.134 + (-134).48`
`= 134*(-1 + 51 - 48)`
`= 134*2`
`= 268`
`b)`
`(-41) . ( 59+2 ) + 59.(41-2)`
`= (-41)*59 - 41*2 + 59*41 + 59*2`
`= [(-41)*59 + 41*59] + (-41*2 + 59*2)`
`= 0 + 2(-41+59)`
`= 18`
`c)`
`369.(-2) - 41.82`
`= 41*9*(-2) - 41*82`
`= -41*(9*2 +82)`
`= -41*(18+82)`
`= -41*100`
`= -4100`
`d)`
`( 135 - 35) . (-37)+ 37.(-42 - 58)`
`= 100*(-37) + 37*(-100)`
`= 37*(-100 - 100)`
`= 37*(-200)`
`= -7400`
`@` `\text {Kaizuu lv uuu}`
a: =-134(1-51+48)
=-134*(-2)=268
b: =-41*59-41*2+59*41-59*2
=-41*2-59*2=-200
d: =100*(-37)+37(-100)
=100(-37-37)
=-7400
Bài 3:
a: Ta có: \(23\left(42-x\right)=23\)
\(\Leftrightarrow42-x=1\)
hay x=41
b: Ta có: 15(x-3)=30
nên x-3=2
hay x=5
Bài 1:
a: 32+89+68=100+89=189
b: 64+112+236=300+112=412
c: \(1350+360+650+40=2000+400=2400\)
a) 2.3.12+4.6.42+8.27.3
=2.12.3+4.6.42+8.3.27
=24.3+24.42+24.27
=24.(3+42+27)
=24.72
=1728
=
a) 2 . 3 . 12 + 4 . 6 . 42 + 8 . 27 . 3
= 3 . 24 + 24 . 42 + 24 . 27
= 24 ( 3 + 42 + 27 )
= 24 . 72 = 3024
b) Số số hạng là :(100 – 2) : 2 + 1 = 50 (số hạng)
Tổng là: (100 + 2) x 50 : 2 = 2550
97; 95; ….; 1 có (97 – 1) : 2 + 1 = 49 (số hạng)
Tổng là: (97 + 1) x 49 : 2 = 2401
A=(100+98+96+...+2)-(97+95+...+1) = 2550 – 2401
A = 149
2A=1/2+1/6+1/12+1/20+1/30+1/42
=1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7
=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7
=1-1/7
=6/7
a. 96 - 3(x + 1) = 42
<=> 96 - 3x - 3 = 42
<=> 96 - 3 - 42 = 3x
<=> 3x = 51
<=> x = 17
b. (x - 37) : 12 = 9
<=> \(\dfrac{x}{12}-\dfrac{37}{12}=9\)
<=> \(\dfrac{x}{12}-\dfrac{37}{12}=\dfrac{108}{12}\)
<=> x - 37 = 108
<=> x = 108 + 37
<=> x = 145
c. 125 : (x - 3) = 5
<=> \(\dfrac{125}{x}-\dfrac{125}{3}=5\)
<=> \(\dfrac{375}{3x}-\dfrac{125x}{3x}=\dfrac{625x}{3x}\)
<=> 375 - 125x = 625x
<=> 375 = 625x + 125x
<=> 500x = 375
<=> x = \(\dfrac{375}{500}=\dfrac{3}{4}\)