Đại số lớp 6

TL

Bai 1:Thuc hien phep tinh sau:

a)1125:1123-35:(110+23)-60

b)2345-1000;[19-2(21-18)2]

c)128-[68+8(37-35)2]:4

d)107-{38+[7.32-24:6+(9-7)3]};15

e)50-[50-23.5):2+3]

Bai 2:Tim x \(\in\)N biet:

a)5.(x-9)=350-52                      b)2.(x-51)=2.23+20              c)2.3x=162

Help me ! Thank you !!!

TM
10 tháng 10 2016 lúc 20:13

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, \(50-\left[50-\left(2^3.5\right)\div2+3\right]\)

\(=50-\left[50-8.5\div2+3\right]\)

\(=50-\left[50-40\div2+3\right]\)

\(=50-\left[50-20+3\right]\)

\(=50-\left[30+3\right]\)

\(=50-33=17\)

Bình luận (0)
TM
10 tháng 10 2016 lúc 20:18

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

Bình luận (2)

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