\(48\div4+3^9\div3^7+5^0\)

\(=12+9+1\)

\(=22\)

\(\left(5^{19}\div5^{17}+3\right)\div7\)

\(=\left(25+3\right)\div7\)

\(=28\div7\)

\(=4\)

\(295-\left(32-2^2\times5\right)^2\)

\(=295-144\)

\(=151\)

\(6^2\div9+50\times2-3^3\times3\)

\(=36\div9+100-27\)

\(=4+100-27\)

\(=100-27\)

\(=73\)

\(48:4+3^9:3^7+5^0\)

\(=12+3^{9-7}+1\)

\(=12+9+1\)

\(=22\)

\(\left(5^{19}:5^{17}+3\right):7\)

\(=\left(5^{19-17}+3\right):7\)

\(=\left(5^2+3\right):7\)

\(=\left(25+3\right):7\)

\(=28:7\)

\(=4\)

\(295-\left(31-2^2\cdot5\right)^2\)

\(=295-\left(31-4\cdot5\right)^2\)

\(=295-\left(31-20\right)^2\)

\(=295-11^2\)

\(=295-121\)

\(=174\)

\(6^2:9+50\cdot2-3^3\cdot3\)

\(=36:9+50\cdot2-3^4\)

\(=4+100-81\)

\(=23\)

\(48:4+3^9:3^7+5^0=12+3^2+1=12+9+1=22\)

\(\left(5^{19}:5^{17}+3\right):7=\left(5^2+3\right):7=\left(25+3\right):7=28:7=4\)

\(295-\left(31-2^2.5^2\right)=295-\left(31-10^2\right)=295-\left(31-100\right)=295-\left(-69\right)=364\)

\(6^2:9+50.2-3^3.3=36:9+100-3^2=4+100-9=95\)

a,  10 x + 2 2 . 5 = 10 2

10x = 100–20

10x = 80

x = 8

b,  5 2 + 15 - x = 30

15–x = 5

x = 10

c,  2 2 . 5 2 - 25 + x = 40

100–(25+x) = 40

25+x = 60

x = 35

d,  7 2 x - 6 2 x = 13 . 2 3 - 26

49x–36x = 104–26

13x = 78

x = 6

\(a)\) \(5^{13}:5^{10}-25.2^2=5^3-25.4=125-100=25\)

\(b)\) \(20:2^2+5^9:5^8=20:4+5^1=5+5=10\)

\(c)\) \(100:5^2+7.3^2=100:25+7.9=4+36=40\)

\(d)\) \(84:4+3^9:3^7+5^0=84:4+3^2+1=21+9+1=31\)

\(e)\)

\(29-\left[16+3.\left(51-49\right)\right]=29-\left[16+3.2\right]=29-\left[16+6\right]=29-22=7\)

\(f)\) \(5.2^2+98:7^2=5.4+98:49=20+2=22\)

\(g)\) \(3^{11}:3^9-147:7^2=3^2-147:49=9-3=6\)

\(295-\left(31-2^2.5\right)^2=295-\left(31-4.5\right)^2=295-\left(31-20\right)^2=295-11^2=295-121=174\)

\(7^{18}:7^{16}+2^2.3^2=7^2+4.9=49+36=85\)

a: \(5^{13}:5^{10}-25\cdot2^2\)

\(=5^3-25\cdot4\)

=125-100

=25

b: \(20:2^2+5^9:5^8\)

\(=20:4+5\)

=5+5

=10

Vo câu lần trước của cậu xem câu của tớ có đúng ko

720 : [ 41 - (2\(x\) - 5)] = 22.5

           41 - (2\(x\) - 5) = 720 :110

                  2\(x\) - 5 = 41 - \(\dfrac{72}{11}\)

                  2\(x\) - 5 = \(\dfrac{379}{11}\)

                  2\(x\)      = \(\dfrac{379}{11}\) + 5

                 2\(x\)     = \(\dfrac{434}{11}\)

                   \(x\)     = \(\dfrac{434}{11}\) : 2

                   \(x\)     = \(\dfrac{217}{11}\)

15 - \(x\) = 7  - (-2) 

15 - \(x\) = 7 + 2

15 - \(x\) = 9

       \(x\) = 15 -  9

       \(x\) = 6

[230 - (15 - 5\(x\))]. 3 = 390

230 - (15 - 5\(x\)) = 390 : 3

230 - 15 + 5\(x\) = 130

215 + 5\(x\)       = 130

         5\(x\)        = 215 - 130

        5\(x\)        = -85

          \(x\)       = -85 : 5

          \(x\)      =  - 17

Đặt x/3 = y/4 = z/5=k

=>x=3k, y=14k,z=5k

Ta có x.y.z = 22,5

=>3k.14k.5k= 22,5

=>210.k^3=22,5

=>k^3=3/28

E tự tính nhé chắc đề sai r -.- chj tính đến đó thôi 

Bài 1:

a) \(24 - (-15) - 2\)

\(=39-2\)

\(=37\)

b) \((-85) + 10 - (-85) - 50\)

\(=[(-85)-(-85)]+10-50\)

\(=0+10-50\)

\(=10-50\)

\(=-40\)

c) \(71 - (-30) - (+18) + (-30)\)

\(=[(-30)-(-30)]+71-(+18)\)

\(=0+71-18\)

\(=71-18\)

\(=53\)

d) \(-(30) - (+37) + (+37) + (-85)\)

\(=[-(+37)+(+37)]-(30)+(-85)\)

\(=0-(30)+(-85)\)

\(=(-30)+(-85)\)

\(=-115\)

e) \((35-815) - (795-65)\)

\(=(-780)-730\)

\(=-1510\)

g) \((2002-79+15) + (-79+15)\)

\(=1938+(-64)\)

\(=1874\)

Bài 2:

a) \(25 - (30+x) = x - (27-8)\)

\(25-30-x=x-27+8\)

\(x+x=25-30+27-8\)

\(2x=14\)

\(x=14\div2\)

\(x=7\)

b) \((x-12) - 15 = (20-17) - (18+x) \)

\(x-12-15=13-18-x\)

\(x-27=-5-x\)

\(x+x=-5+27\)

\(2x=22\)

\(x=22\div2\)

\(x=11\)

c) \(15 - x = 7 - (-2)\)

\(15-x=9\)

\(x=15-9\)

\(x=6\)

d) \(x - 35 = (-12) - 3\)

\(x-35=-15\)

\(x=-15+35\)

\(x=20\)

e) \(\left|5-x\right|-26=-15\)

\(\left|5-x\right|=-15+26\)

\(\left|5-x\right|=11\)

Từ đây ta có:

*Nếu \(5-x=11\)

\(x=5-11\)

\(x=-6\)

*Nếu \(5-x=-11\)

\(x=5-(-11)\)

\(x=16\)

Vậy \(x=-6;x=16\)

d) Ta có: \(32\%-0.25:x=-\dfrac{17}{5}\)

\(\Leftrightarrow0.25:x=\dfrac{8}{25}+\dfrac{17}{5}=\dfrac{93}{25}\)

hay \(x=\dfrac{25}{372}\)

Vậy: \(x=\dfrac{25}{372}\)

e) Ta có: \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)

\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{2}{5};-\dfrac{4}{5}\right\}\)

f) Ta có: \(-\dfrac{32}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)

\(\Leftrightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-8}{27}\)

\(\Leftrightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)

\(\Leftrightarrow3x=\dfrac{1}{9}\)

hay \(x=\dfrac{1}{27}\)

g) Ta có: \(60\%\cdot x+0.4x+x:3=2\)

\(\Leftrightarrow\dfrac{4}{3}x=2\)

hay \(x=\dfrac{3}{2}\)

Vậy: \(x=\dfrac{3}{2}\)

h) PT \(\Leftrightarrow\left|\dfrac{20}{9}-x\right|=\dfrac{2}{9}\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{20}{9}-x=\dfrac{2}{9}\\x-\dfrac{20}{9}=\dfrac{2}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{22}{9}\end{matrix}\right.\)

  Vậy ...

i) PT \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}x=\dfrac{16}{5}\) \(\Leftrightarrow\dfrac{2}{5}x=\dfrac{8}{5}\) \(\Leftrightarrow x=4\)

  Vậy ...