12 x 11 + 123
a)TÍNH BẰNG CÁCH THUẬN TIỆN
123 x 54 + 27 x 123 - 12 x 123 + 123 x 31
b)TÌM n
121/27 x 54/11 < n < 100/21 : 25/126
\(a,123\times54+27\times123-12\times123+123\times31\)
\(=123\times\left(54+27-12+31\right)\)
\(=123\times100\)
\(=12300\)
\(b,\dfrac{121}{27}\times\dfrac{54}{11}< n< \dfrac{100}{21}:\dfrac{25}{126}\)
\(22< n< \dfrac{100}{21}\times\dfrac{126}{25}\)
\(22< n< 24\)
Vì \(23>22\) và \(23< 24\) nên \(n=23\)
a)TÍNH BẰNG CÁCH THUẬN TIỆN
123 x 54 + 27 x 123 - 12 x 123 + 123 x 31
b)TÌM n
121/27 x 54/11 < n < 100/21 : 25/126
a123x54+27x123-12x123+123x31
=123:54+27:123-12:123+123:31
=6642+27:123-12:123+123:31
=6642+3321-12:123+123:31
=6642+3321-1476+123:31
=6642+3321-1476+3813
=12300
b,121/27x54/11<n>100/21:25/126
=22<n>24
n=23
123 x 100
12 x 100
1100 + 11
123 x 100 = 12300
12 x 100 = 1200
1100 + 11 = 1111
hi
123 x 100
= 1230
12 x 100
= 1200
1100 + 11
= 1111
k cho mk nha pạn
tính bằng cách thuận tiện
2022 x 123 x 12 – 2022 x 11
giải giúp mik vs, mik đang cần gấp
`2022 \times 123 \times 12 - 2022 \times 11`
`= 2022 \times (123 \times 12 - 11)`
`= 2022 \times (1476 - 11)`
`= 2022 \times 1465`
`= 2962230`
123/56*3-12+65/12*22+11-22*22/33/5*123=? khó lắm
98 x 456 x 123 x 753 x 246 x 2222 x 11 x 11 x 11 x 25 x 36 x 25 x 12 x 13 x 159 x 487 x 258 x 12589 x 1524 x ( 36987 - 36987 ) = ?
98 x 456 x 123 x 753 x 246 x 2222 x 11 x 11 x 11 x 25 x 36 x 25 x 12 x 13 x 159 x 487 x 258 x 12589 x 1524 x ( 36987 - 36987 ) = 0
:)))
98 x 456 x 123 x 753 x 246 x 2222 x 11 x 11 x 11 x 25 x 36 x 25 x 12 x 13 x 159 x 487 x 258 x 12589 x 1524 x ( 36987 - 36987 ) =0
ht
k mk nha
98 x 456 x 123 x 753 x 246 x 2222 x 11 x 11 x 11 x 25 x 36 x 25 x 12 x 13 x 159 x 487 x 258 x 12589 x 1524 x ( 36987 - 36987 )
=>0
ht
Bài 1 Tìm x,biết
x.3/5=2/3
x.7/17=17/8
3/4 chia x=-7/12
3/8 - 1/6.x=1/4
1/3+1/2 chia x=-4
Bài 2 Tính
-6/11 chia [3/5 . 4/11]
7/12 + 5/12 chia 6 -11/36
[4/5 + 1/2] chia [3/13 - 8/13]
[2/3 - 1/4 + 5/11] chia [5/12 + 1 -7/11]
*Dấu . là dấu nhân nha mn*
Bài 1:
a: \(x=\dfrac{2}{3}:\dfrac{3}{5}=\dfrac{2}{3}\cdot\dfrac{5}{3}=\dfrac{10}{9}\)
b: \(x=\dfrac{17}{8}:\dfrac{7}{17}=\dfrac{17}{8}\cdot\dfrac{17}{7}=\dfrac{289}{56}\)
c: \(x=-\dfrac{3}{4}:\dfrac{7}{12}=\dfrac{-3}{4}\cdot\dfrac{12}{7}=\dfrac{-63}{28}=-\dfrac{9}{4}\)
d: \(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{1}{4}\)
hay \(x=\dfrac{1}{4}:\dfrac{1}{6}=\dfrac{3}{2}\)
e: \(\Leftrightarrow\dfrac{1}{2}:x=-4-\dfrac{1}{3}=-\dfrac{17}{3}\)
hay \(x=-\dfrac{1}{2}:\dfrac{17}{3}=\dfrac{-3}{34}\)
a)-6/35 :-54/39
b)-5/7x2/11+-5/7x9/11+12/7
c)(1/2+3/4-13/12) :(12/23+123/234+1234/2345)
a: =6/35x39/54=13/105
b: \(=\dfrac{-5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}=\dfrac{-5}{7}+\dfrac{12}{7}=1\)
Tim X
3) -12 + (2x – 9) + x= 0
4) 11 + (15 - x) = 1
5) 4 - (27 - 3) = x - (13 - 4)
6) 8 - (x - 10) = 23 - (- 4 +12)
7) 105 – 5(10 – 5x) = -20
8) (x -1)(8-2x)(3x+123) = 0
9) (x2 - 25)(x+ 10) = 0
10) x(x2+5) =
3) \(-12+2x-9+x=0\\ -21+3x=0\\ 3x=21\\ x=7\)
4)
\(11+\left(15-x\right)=1\)
\(15-x=1-11\)
\(15-x=-10\)
\(x=15-\left(-10\right)\)
\(x=25\)
5)
\(4-\left(27-3\right)=x-\left(13-4\right)\)
\(4-24=x-9\)
\(x-9=-20\)
\(x=-20+9\)
\(x=-11\)
\(3.-12+\left(2x-9\right)+x=0.\)
\(\Leftrightarrow-12+2x-9+x=0.\Leftrightarrow3x=21.\Leftrightarrow x=7.\)
Vậy \(x=7.\)
\(4.11+\left(15-x\right)=1.\Leftrightarrow11+15-x=1.\Leftrightarrow26-x=1.\Leftrightarrow x=25.\)
Vậy \(x=25.\)
\(5.4-\left(27-3\right)=x-\left(13-4\right).\Leftrightarrow4-24=x-9.\Leftrightarrow-20=x-9.\Leftrightarrow x=-11.\)
Vậy \(x=-11.\)
\(6.8-\left(x-10\right)=23-\left(-4+12\right).\Leftrightarrow8-x+10=23-8.\Leftrightarrow18-x=15.\Leftrightarrow x=3.\)
Vậy \(x=3.\)
\(7.105-5\left(10-5x\right)=-20.\Leftrightarrow105-50+25x=-20.\Leftrightarrow25x=-75.\Leftrightarrow x=-3.\)
Vậy \(x=-3.\)
\(8.\left(x-1\right)\left(8-2x\right)\left(3x+123\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0.\\8-2x=0.\\3x+123=0.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=4.\\x=-41.\end{matrix}\right.\)
Vậy \(x\in\left\{1;4;-41\right\}.\)
\(9.\left(x^2-25\right)\left(x+10\right)=0.\)
\(\Leftrightarrow\left(x-5\right)\left(x+5\right)\left(x+10\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0.\\x+5=0.\\x+10=0.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5.\\x=-5.\\x=-10.\end{matrix}\right.\)
Vậy \(x\in\left\{5;-5;-10\right\}.\)
\(10.x\left(x^2+5\right)=0.\Leftrightarrow x=0.\)