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b: =>3|x-5|=8+4=12
=>|x-5|=4
=>x-5=4 hoặc x-5=-4
=>x=9 hoặc x=1
d: =>2x+6=3-3x-2
=>2x+6=1-3x
=>5x=-5
hay x=-1
e: \(\Leftrightarrow x-3\inƯC\left(70;98\right)\)
\(\Leftrightarrow x-3\in\left\{1;2;7;14\right\}\)
mà x>8
nên \(x\in\left\{10;17\right\}\)
a) 23 . 17 - 23 . 14
= 23(17-14)
=23. 3= 24
b) 80 - ( 4 . 5 - 3 . 23 )
= 80- 22. (5-3.2)= 80- 22.(-1)
= 80- 4.(-1)= 80- (-4)= 84
c) 35 - { 12 - [ -14 + ( -2) }
= 35 - { 12 - (-16) }
=35- 28= 7
d) 49 - ( -45) - 23
= 94-23= 71
e) 13 - 18 - ( -42) - 15
= -5 -(-42) -15
= -20+42= 22
g) -452 - ( -67 + 75 - 452 )
= -452- (8-452)
= -452 -(- 444)= -8
h) |31−17|−|13−52|
= 14- 39= -25
80 -(4.52 -3.23)
= 80- (4.25- 3.8)
= 80- (100-24)
= 80-76= 4
a)
4 . 25 – 12 . 25 + 170 : 10
= (4 . 25) – (12 . 25) + (170 : 10)
= 100 - 300 + 17
= -183
b)
(7 + 33 + 32) . 4 – 3
= (7 + 27 + 9) .4 – 3
= 43 . 4 – 3
= (43 . 4) – 3
= 45
c)
12 : {400 : [500 – (125 + 25 . 7)}
= 12 : {400 : [500 – (125 + 175)}
= 12 : (400: 200)
= 12 : 2
= 6
d)
168 + {[2.(24 + 32) - 2560] : 72}.
= 168 + [2 . (16 + 9) – 1] : 49
= 168 + 49: 49
= 168 + 1
= 167
a)
4 . 25 – 12 . 25 + 170 : 10
= (4 . 25) – (12 . 25) + (170 : 10)
= 100 - 300 + 17
= -183
b)
(7 + 33 + 32) . 4 – 3
= (7 + 27 + 9) .4 – 3
= 43 . 4 – 3
= (43 . 4) – 3
= 45
c)
12 : {400 : [500 – (125 + 25 . 7)}
= 12 : {400 : [500 – (125 + 175)}
= 12 : (400: 200)
= 12 : 2
= 6
d)
168 + {[2.(24 + 32) - 2560] : 72}.
= 168 + [2 . (16 + 9) – 1] : 49
= 168 + 49: 49
= 168 + 1
= 167
Bài 1 :
a, Ta có : \(\left(-123\right)+\left|-13\right|+\left(-7\right)\)
= \(\left(-123\right)+13+\left(-7\right)=\left(-117\right)\)
b, Ta có : \(\left|-10\right|+\left|45\right|+\left(-\left|-455\right|\right)+\left|-750\right|\)
= \(10+45-455+750=350\)
c, Ta có : \(-\left|-33\right|+\left(-15\right)+20-\left|45-40\right|-57\)
= \(\left(-33\right)+\left(-15\right)+20-5-57=-90\)
\(2.THPT\)
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{98.99}+\frac{9}{99.100}\)
\(A=9\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(A=9\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=9\left(1-\frac{1}{100}\right)\)
\(A=9.\frac{99}{100}\)
\(A=\frac{891}{100}\)
\(B=\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{93.95}\)
\(B=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{93}-\frac{1}{95}\)
\(B=\frac{1}{5}-\frac{1}{95}\)
\(B=\frac{18}{95}\)
\(D=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(D=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)
\(D=\frac{1}{2}-\frac{1}{28}\)
\(D=\frac{13}{28}\)
\(\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{59.61}\)
= \(\dfrac{2}{2}.\left(\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{59.61}\right)\)
= \(\dfrac{3}{2}.\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\right)\)
= \(\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\)
= \(\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{61}\right)\)
=\(\dfrac{3}{2}.\dfrac{56}{305}\)
= \(\dfrac{78}{305}\)
\(\left(x^2-4\right)\left(6-2x\right)=0\) ⇔ \(x^2-4=0\) hoặc \(6-2x=0\)
*Nếu \(x^2-4=0\)
⇒ x2 = 4
⇒ x ∈ {2 ; -2}
*Nếu \(6-2x=0\)
⇒2x = 6
⇒ x = 6 : 2 = 3
Vậy x ∈ { -2 ; 2 ; 3 }
2. a) \(3^{200}=\left(3^2\right)^{100}=9^{100}\)
\(2^{300}=\left(2^3\right)^{100}=8^{100}\)
Vì \(9^{100}>8^{100}\Rightarrow3^{200}>2^{300}\)
b) \(71^{50}=\left(71^2\right)^{25}=5041^{25}\)
\(37^{75}=\left(3^3\right)^{25}=27^{25}\)
Vì \(5041^{25}>27^{25}\Rightarrow71^{50}>37^{75}\)
c) \(\frac{201201}{202202}=\frac{201201:1001}{202202:1001}=\frac{201}{202}\)
\(\frac{201201201}{202202202}=\frac{201201201:1001001}{202202202:1001001}=\frac{201}{202}\)
Vì \(\frac{201}{202}=\frac{201}{202}\Rightarrow\frac{201201}{202202}=\frac{201201201}{202202202}\)
Câu 2:
a: \(\Leftrightarrow12x-60=7x-5\)
=>5x=55
=>x=11
b: \(\Leftrightarrow\left(2x-3\right)^{2010}\left[\left(2x-3\right)^2-1\right]=0\)
=>(2x-3)(2x-2)(2x-4)=0
hay \(x\in\left\{\dfrac{3}{2};1;2\right\}\)
Giải:
\(A=\left[\left(-8\right)+\left(-7\right)\right]+\left(-10\right)=\left(-15\right)+\left(-10\right)=\left(-25\right)\)
\(B=-\left(-299\right)+\left(-219\right)+\left(-401\right)+12=299-219-401+12=-309\)
\(C=555+\left(-100\right)+\left(-80\right)+\left|-333\right|=555-100-80+333=708\)
\(D=\left|-347\right|+\left(-40\right)+3150+\left(-307\right)=347-40+3150-307=2456\)
\(E=98.42-\left\{50.\left[\left(18-2^3\right):2+3^2\right]\right\}\)
\(E=4116-\left\{50.\left[\left(18-8\right):2+9\right]\right\}\)
\(E=4116-\left\{50.\left[10:2+9\right]\right\}\)
\(E=4116-\left\{50.\left[5+9\right]\right\}\)
\(E=4116-\left\{50.14\right\}\)
\(E=4116-700\)
\(E=3416\)
\(F=-80-\left[-130-\left(12-4^2\right)\right]+2008^0\)
\(F=-80-\left[-130-\left(12-16\right)\right]+1\)
\(F=-80-\left[-130-\left(-4\right)\right]+1\)
\(F=-80-\left(-126\right)+1\)
\(F=47\)
\(G=1000+\left(-670\right)+297+\left(-330\right)=1000-670+297-330=297\)
\(H=1024:2^4+140:\left(38+2^5\right)-7^{23}:7^{21}\)
\(H=2^{10}:2^4+140:\left(38+32\right)-7^2\)
\(H=2^6+140:70-49\)
\(H=64+2-49\)
\(H=17\)
\(I=\left|-129\right|-119+\left|2-31\right|=129-119+\left|-29\right|=10+29=39\)
\(K=219+573+381-173\)
\(K=\left(219+381\right)+\left(573-173\right)\)
\(K=600+400\)
\(K=1000\)
\(L=36.33-105.11+22.15\)
\(L=36.33-35.3.11+11.2.3.5\)
\(L=36.33-35.33+33.10\)
\(L=33.\left(36-35+10\right)\)
\(L=33.11\)
\(L=363\)
\(N=160-\left(2^3.5^2-6.25\right)\)
\(N=160-\left[25.\left(8-6\right)\right]\)
\(N=160-\left[25.2\right]\)
\(N=160-50\)
\(N=110\)
\(O=\left(44.52.60\right):\left(11.13.15\right)=137280:2145=64\)
\(P=\left(2^{17}+15^4\right).\left(3^9-2^{17}\right).\left(2^4-4^2\right)=\left(2^{17}+15^4\right).\left(3^9-2^{17}\right).0=0\)
\(Q=100+98+96+...+4+2-97-95-...-3-1\)
\(Q=100+\left(98-97\right)+\left(96-95\right)+...+\left(4-3\right)+\left(2-1\right)\)
\(Q=100+1+1+...+1+1\)
Vì có 98:2=49 số 1
\(\Rightarrow Q=100+49\)
\(Q=149\)