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a,9.3.4.25
= 27.100
= 2700
b, 12.125.54
= 3.4.125.54
= 125.4.3.54
= 500.162
= 81000
c: \(=72\cdot14+72\cdot7+72\cdot19=72\cdot40=2880\)
d: \(=1700\cdot7-1700\cdot6=1700\)
e: Số số hạng là (65-1)/4+1=17(số)
Tổng là: \(\dfrac{65+1}{2}\cdot17=33\cdot17=561\)
a = 5400
b = 81000
c = 7 000 000
d = 108800
e = 3600 nhé bạn
k cho mình nha , cảm ơn nhiều
a) 9 x 24 x 25
=216x25
=5400
b) 12 x 125 x 54
=1500x54
=81000
c) 64 x 125 x 875
=8000x875
=7000000
d) 425 x 7
a) 9 x 24 x 25
b) 12 x 125 x 54
c) 64 x 125 x 875
d) 425 x 7 x4 - 170 x 60
=2975x4 - 170 x 60
=11900- 170 x 60
=11900-10200
=1700
e) 8 x 9 x 14 + 6 x 7 x 12 + 19 x 4 x 18
=1008+504+1368
=2880
a,\(\frac{37.13-13}{24+37.12}=\frac{36.13}{39.12}=1\)
b,125.7.16.25.4=(125.16).(25.4).7=2000.100.7=1400000
c,8.9.14+6.17.12+19.4.18=2^3.3^2.2.7+2.3.17.2^2.3+19.2^2.3^2.2=2^3.3^2.(2.7+17+19)=72.50=3600
15*6*4*125*8=360000
14*25*6*7=14700
24*3*5*10=3600
18*26*25*9=105300
12*5*15*7=6300
15 x 6 x 4 x 125 x 8
= (15 x 4) x (125 x 8) x 6
= 60 x 1000 x 6
= 60000 x 6
= 360000
14 x 25 x 6 x 7
= (14 x 25) x 6 x 7
= 350 x 6 x 7
= 2100 x 7
= 14700
24 x 3 x 5 x 10
= (24 x 5) x 10 x 3
= 120 x 10 x 3
= 1200 x 3
= 3600
18 x 26 x 25 x 9
= (26 x 25) x 18 x 9
= 650 x 18 x 9
= 11700 x 9
= 105300
12 x 5 x 15 x 7
= (12 x 15) x 5 x 7
= 180 x 5 x 7
= 900 x 7
= 6300
Bài 46:
11: Ta có: \(-4\left|x-2\right|=-8\)
\(\Leftrightarrow\left|x-2\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=2\\x-2=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)
Vậy: x∈{0;4}
12: Ta có: \(5\left|x+2\right|=-10\cdot\left(-2\right)\)
\(\Leftrightarrow5\left|x+2\right|=20\)
\(\Leftrightarrow\left|x+2\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)
Vậy: x∈{-6;2}
13: Ta có: \(6\left|x-2\right|=18:\left(-3\right)\)
\(\Leftrightarrow6\left|x-2\right|=-6\)(1)
Ta có: \(\left|x-2\right|\ge0\forall x\)
\(\Rightarrow6\left|x-2\right|\ge0\forall x\)(2)
Ta có: -6<0(3)
Từ (1), (2) và (3) suy ra x∈∅
Vậy: x∈∅
14: Ta có:\(-7\left|x+4\right|=21:\left(-3\right)\)
\(\Leftrightarrow-7\left|x+4\right|=-7\)
\(\Leftrightarrow\left|x+4\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=1\\x+4=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)
Vậy: x∈{-5;-3}
15: Ta có: \(4\left|x+1\right|=8\left(-2\right)-8\left(-5\right)\)
\(\Leftrightarrow4\left|x+1\right|=-16-\left(-40\right)\)
\(\Leftrightarrow4\left|x+1\right|=24\)
\(\Leftrightarrow\left|x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=6\\x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-7\end{matrix}\right.\)
Vậy: x∈{-7;5}
16: Ta có: \(3\left|x+5\right|=-9\)(4)
Ta có: |x+5|≥0∀x
⇒3|x+5|≥0∀x(5)
Ta có: -9<0(6)
Từ (4), (5) và (6) suy ra x∈∅
Vậy: x∈∅
17: Ta có: \(-8\left|x-3\right|=24-16:2\)
\(\Leftrightarrow-8\left|x-3\right|=16\)
\(\Leftrightarrow\left|x-3\right|=-2\)
mà |x-3|≥0>-2∀x
nên x∈∅
Vậy: x∈∅
18: Ta có: \(-3\left|x+6\right|=6\cdot2-9\)
\(\Leftrightarrow-3\left|x+6\right|=3\)
\(\Leftrightarrow\left|x+6\right|=-1\)
mà |x+6|≥0>-1∀x
nên x∈∅
Vậy: x∈∅
19: Ta có: \(5-\left|x+7\right|=4\)
\(\Leftrightarrow\left|x+7\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x+7=-1\\x+7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=-6\end{matrix}\right.\)
Vậy: x∈{-8;-6}
20: Ta có: \(12-\left|x+8\right|=10\)
\(\Leftrightarrow\left|x+8\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=2\\x+8=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=-10\end{matrix}\right.\)
Vậy: x∈{-10;-6}
8x9x4+6x7x12+19+4x18
=(8x9)x4+(6x12)x7+19x(18x4)
=72x4+72x7+19x72
=72x(19+7+4)
=72x30
=2160
8x9x4+6x7x12+19x4x18=2096