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Bài 2:a)<=>5(x+4) =125-38
<=>5(x+4)=87
<=>x+4=17,4
<=>x=17,4-4=13.4
b)<=> ( 3x – 24) . 73 = 2.74
<=> (3x – 16) = 2.74 : 73
<=> (3x – 16) =2.7
<=>3x – 16 = 14
<=> 3x = 30
<=> x = 10
c)=> xϵ ƯC (70, 84) và x > 8
Ta có: 70 = 2. 5. 7 => ƯCLN( 70,84) = 2. 7 = 14
84 = 22. 3. 7
=> ƯC (70, 84) = Ư(14) = { 1 ;2; 7; 14 }
Vì x > 8 => x = 14
d)=>xϵ BCNN (12, 25, 30) và 0 < x < 500
=>BC(12,25,30)=B(300) = { 0; 300; 600; …}
Vì 0 < x < 500
=> x = 300
giup minh nha cac ban . lam duoc minh se tick cho va ket ban . cam on cac ban
a, \(11^{25}\div11^{23}-3^5\div\left(1^{10}+2^3\right)-60\)
\(=11^{25}\div11^{23}-3^5\div\left(1+8\right)-60\)
\(=11^{25}\div11^{23}-3^5\div3^2-60\)
\(=11^2-3^3-60\)
\(=121-27-60\)
\(=94-60=34\)
b, \(2345-1000\div\left[19-2\left(21-18\right)^2\right]\)
\(=2345-1000\div\left[19-2.3^2\right]\)
\(=2345-1000\div\left[19-2.9\right]\)
\(=2345-1000\div1=2345-1000=1345\)
c, \(128-\left[68+8\left(37-35\right)^2\right]\div4\)
\(=128-\left[68+8.2^2\right]\div4\)
\(=128-\left[68+8.4\right]\div4\)
\(=128-\left[68+32\right]\div4\)
\(=128-100\div4=128-25=103\)
d, \(107-\left\{38+\left[7.3^2-24\div6+\left(9-7\right)^3\right]\right\}\div15\)
\(=107-\left\{38+\left[7.9-4+2^3\right]\right\}\div15\)
\(=107-\left\{38+\left[63-4+8\right]\right\}\div15\)
\(=107-\left\{38+\left[59+8\right]\right\}\div15\)
\(=107-\left\{38+67\right\}\div15\)
\(=107-105\div15\)
\(=107-7=100\)
e, \(50-\left[50-\left(2^3.5\right)\div2+3\right]\)
\(=50-\left[50-8.5\div2+3\right]\)
\(=50-\left[50-40\div2+3\right]\)
\(=50-\left[50-20+3\right]\)
\(=50-\left[30+3\right]\)
\(=50-33=17\)
Bài 2 :
a, \(5\left(x-9\right)=350-5^2\)
\(5\left(x-9\right)=350-25\)
\(5\left(x-9\right)=325\)
\(x-9=325\div5\)
\(x-9=65\)
\(\Rightarrow x=65+9\)
\(\Rightarrow x=74\)
Vậy x = 74
b, \(2\left(x-51\right)=2.2^3+20\)
\(2\left(x-51\right)=16+20\)
\(2\left(x-51\right)=36\)
\(x-51=36\div2\)
\(x-51=18\)
\(\Rightarrow x=18+51\)
\(\Rightarrow x=69\)
Vậy x = 69
c, \(2.3^x=162\)
\(\Rightarrow2.3^x=2.3^4\)
\(\Rightarrow3^x=3^4\)
\(\Rightarrow x=4\)
Vậy x = 4
1, a,b ko chia hết cho 3 nhưng có cùng số dư khi chia cho 3
=> a,b cùng chia 3 dư 1 hoặc 2
sau đó xét 2 TH;
=> ab chia 3 dư 1 => ab-1 là bội của 3 (ĐPCM)
\(=10-\left[\left(91-48\right).5+\left(8.10+8\right)\right]:28\)
\(=10-\left[43.5+88\right]:28\)
\(=10-\left[215+88\right]:28\)
\(=10-303:28\)
\(=10-10,821\)
\(=0,179\)
Ko chắc
\(a.\left(5+2\right).x=36-1\)
\(7.x=35\)
\(x=5\)
\(b.\left(5+1\right).x=75+3\)
\(6.x=78\)
\(x=13\)
\(c.\left(6+1\right).x=5^2+3\)
\(7.x=25+3\)
\(7.x=28\)
\(x=4\)
\(d.\left(7-1\right).x=5^2+3.4-1\)
\(6.x=25+12-1\)
\(6.x=36\)
\(x=6\)
ai bít được mới có lớp 4 thôi
lay hon