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a) \(B=3+3^2+3^3+...+3^{120}\)
\(B=3\cdot1+3\cdot3+3\cdot3^2+...+3\cdot3^{119}\)
\(B=3\cdot\left(1+3+3^2+...+3^{119}\right)\)
Suy ra B chia hết cho 3 (đpcm)
b) \(B=3+3^2+3^3+...+3^{120}\)
\(B=\left(3+3^2\right)+\left(3^3+3^4\right)+\left(3^5+3^6\right)+...+\left(3^{119}+3^{120}\right)\)
\(B=\left(1\cdot3+3\cdot3\right)+\left(1\cdot3^3+3\cdot3^3\right)+\left(1\cdot3^5+3\cdot3^5\right)+...+\left(1\cdot3^{119}+3\cdot3^{119}\right)\)
\(B=3\cdot\left(1+3\right)+3^3\cdot\left(1+3\right)+3^5\cdot\left(1+3\right)+...+3^{119}\cdot\left(1+3\right)\)
\(B=3\cdot4+3^3\cdot4+3^5\cdot4+...+3^{119}\cdot4\)
\(B=4\cdot\left(3+3^3+3^5+...+3^{119}\right)\)
Suy ra B chia hết cho 4 (đpcm)
c) \(B=3+3^2+3^3+...+3^{120}\)
\(B=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+\left(3^7+3^8+3^9\right)+...+\left(3^{118}+3^{119}+3^{120}\right)\)
\(B=\left(1\cdot3+3\cdot3+3^2\cdot3\right)+\left(1\cdot3^4+3\cdot3^4+3^2\cdot3^4\right)+...+\left(1\cdot3^{118}+3\cdot3^{118}+3^2\cdot3^{118}\right)\)
\(B=3\cdot\left(1+3+9\right)+3^4\cdot\left(1+3+9\right)+3^7\cdot\left(1+3+9\right)+...+3^{118}\cdot\left(1+3+9\right)\)
\(B=3\cdot13+3^4\cdot13+3^7\cdot13+...+3^{118}\cdot13\)
\(B=13\cdot\left(3+3^4+3^7+...+3^{118}\right)\)
Suy ra B chia hết cho 13 (đpcm)
(-4;-3;-2;-1;0;1;2;3;4)
Ko có dấu ngoặc nhọn nên mik xài ngoặc tròn nha
a
\(230+\left[16+\left(y-5\right)\right]=315\cdot1\)
\(230+\left[16+\left(y-5\right)\right]=315\)
\(16+\left(y-5\right)=315-230\)
\(16+\left(y-5\right)=85\)
\(y-5=85-16\)
\(y-5=69\)
\(y=69+5\)
\(y=74\)
b
\(707:\left[\left(2^y-5\right)+74\right]=16-9\)
\(707:\left[\left(2^y-5\right)+74\right]=7\)
\(\left(2^y-5\right)+74=707:7\)
\(\left(2^y-5\right)+74=101\)
\(2^y-5=101-74\)
\(2^y-5=27\)
\(2^y=27+5\)
\(2^y=32\)
\(2^y=2^5\)
\(\Rightarrow y=5\)
a) \(\frac{9}{20}\) c) \(\frac{-55}{4}\)
b) \(\frac{116}{75}\) d) \(\frac{-76}{45}\)
đúng hết đấy nhé mình tính kĩ lắm ko sai đâu
chúc may mắn
a) \(\left(-396\right)+2016+189+\left(-604\right)-109\)
\(=-396+2016+189-604-109\)
\(=-\left(396-2016\right)+189-604-109\)
\(=-\left(-1620\right)+189-604-109\)
\(=1620+189-604-109\)
\(=1809-604-109\)
\(=1205-109\)
\(=1096\)
b) \(10-\left|x+3\right|=-8-10\)
\(\Leftrightarrow10-\left|x+3\right|=-\left(8+10\right)\)
\(\Leftrightarrow10-\left|x+3\right|=-18\)
\(\Leftrightarrow\left|x+3\right|=10-\left(-18\right)\)
\(\Leftrightarrow\left|x+3\right|=10+18\)
\(\Leftrightarrow\left|x+3\right|=28\)
Xét trường hợp 1: \(x+3=28\)
\(\Rightarrow x=28-3\)
\(\Rightarrow x=25\)
Xét trường hợp 2: \(x+3=-28\)
\(\Rightarrow x=-28-3\)
\(\Rightarrow x=-\left(28+3\right)\)
\(\Rightarrow x=-31\)
Vậy \(x=25\) hoặc \(x=-31\)
a) 25.76+24.35 + \(5^3.2^3\)=x
25.76+24.35+ 125. 8=x
1900+ 840+1000=x
2740+840+1000=x
3580+1000=x
4580=x. Vậy x= 4580
b) 4x-166=\(3^3:3^2\)
4x-166=3
4x= 166+3
4x= 169
x= 169: 4
x= 42,25
c) x-4.\(5^2+3^2:2^4\)= 0
x- 4.25+ 9: 16=0
x- 100+0,5625=0
x - 100,5625=0
x=0+100,5625
x= 100,5625
d) 4x - 158 = \(2^3.3^2\)-50
4x-158= 8.9-50
4x-158= 72-50
4x-158= 22
4x= 158+22
4x=180
x= 180:4
x= 45
c: \(\left|\dfrac{7}{5}x+\dfrac{2}{3}\right|=\left|\dfrac{4}{3}x-\dfrac{1}{4}\right|\)
=>7/5x+2/3=4/3x-1/4 hoặc 7/5x+2/3=1/4-4/3x
=>1/15x=-11/12 hoặc 41/15x=-5/12
=>x=-55/4 hoặc x=-25/164
d: |7/8x+5/6|=|1/2x+5|
=>|42x+40|=|24x+240|
=>42x+40=24x+240 hoặc 42x+40=-24x-240
=>18x=200 hoặc 66x=-280
=>x=100/9 hoặc x=-140/33
a) x = 74
b) x = 5
c) x = 3
d) x = 14