a: \(\left[\left(10-x\right)\cdot2+51\right]:3-2=3\)

=>\(\left[2\left(10-x\right)+51\right]:3=5\)

=>\(\left[2\left(10-x\right)+51\right]=15\)

=>\(2\left(10-x\right)=15-51=-36\)

=>10-x=-36/2=-18

=>\(x=10-\left(-18\right)=10+18=28\)

b: \(\left(x-12\right)-15=20-\left(17+x\right)\)

=>\(x-12-15=20-17-x\)

=>\(x-27=3-x\)

=>\(2x=30\)

=>\(x=\dfrac{30}{2}=15\)

c: \(720-\left[41-\left(2x-5\right)\right]=2^3\cdot5\)

=>\(720-\left[41-2x+5\right]=8\cdot5=40\)

=>\(\left[41-2x+5\right]=720-40=680\)

=>-2x+46=680

=>-2x=680-46=634

=>\(x=\dfrac{634}{-2}=-317\)

a) 128 - 3(x + 4) = 23

             3(x + 4) = 128 -23

             3(x + 4) = 105

                x + 4  = 105 : 3

                x + 4  = 35

                x        = 35 - 4

                x        = 31

\(a.128-3\left(x+4\right)=23.\)

\(3\left(x+4\right)=128-23\)

\(3\left(x+4\right)=105\)

\(x+4=105:3=35\)

\(x=35-4=31\)

\(b.\left[\left(4x+28\right)\cdot3+55\right]:5=35\)

\(\left(4x+28\right)\cdot3+55=35\cdot5=175\)

\(\left(4x+28\right)\cdot3=175-55=120\)

\(4x+28=120:3=40\)

\(4x=40-28=12\)

\(x=12:4=3\)

a, \(128-3.\left(x+4\right)=23\)

\(\Rightarrow\) \(3.\left(x+4\right)=105\)

\(\Rightarrow\) \(x+4=35\)

\(\Rightarrow\) \(x=31\)

b, \(\left[\left(4x+28\right).3+55\right]:5=35\)

\(\Rightarrow\) \(\left(4x+28\right).3+55=175\)

\(\Rightarrow\) \(\left(4x+28\right).3=120\)

\(\Rightarrow\) \(4x+28=40\)

\(\Rightarrow\) \(4x=12\)

\(\Rightarrow\) \(x=3\)

\(-22x^3-\left(-21x^3+19x^2+23^0\right)-\left(-x^3-18x^2\right)+\left(x^2-23^1\right)\)

\(=-22x^3+21x^3-19x^2-1+x^3+18x^2+x^2-23\)

\(=\left(-22x^3+21x^3+x^3\right)+\left(-19x^2+18x^2+x^2\right)+\left(-1-23\right)\)

\(=0x^3+0x^2-24\)

\(=-24\)

Vậy biểu thức trên có giá trị không phụ thuộc vào biến.

a) \(2^{4x}-3^{2x}=144-164:41\)

\(\Leftrightarrow16^x-9^x=144-4\)

\(\Leftrightarrow\) ......

bjan tự tính tiếp đc òi                 

1: \(\dfrac{16^{11}\cdot5^{40}}{10^{41}}=\dfrac{2^{44}\cdot5^{40}}{2^{41}\cdot5^{41}}=\dfrac{2^3}{5^1}=\dfrac{8}{5}\)

2: \(\dfrac{3^7\cdot8^5}{6^6\cdot\left(-2\right)^{12}}=\dfrac{3^7\cdot2^{15}}{2^6\cdot3^6\cdot2^{12}}=\dfrac{3}{2^3}=\dfrac{3}{8}\)

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b: =>x-3=12

hay x=15

a. 4.(x+41) = 7

x + 41 = 7 : 4 = 1,75

x = 1,75 - 41 = -39,25

b. 4.(x-3) = 7² - 1¹⁰ = 49 - 1 = 48

x - 3 = 48 : 4 = 12

x = 12 + 3 = 15

a) \(4\left(x+41\right)=400\)
\(\Rightarrow x+41=400:4\)
\(\Rightarrow x+41=100\)
\(\Rightarrow x=100-41\)
\(\Rightarrow x=59\)

a/ \(4.\left(x+41\right)=400\)

\(x+41=\dfrac{400}{4}=100\)

\(\Rightarrow x=59\)

b/ \(4.\left(x-3\right)=7^2-1^{10}\)

\(4.\left(x-3\right)=49-1\)

\(4.\left(x-3\right)=48\)

\(x-3=\dfrac{48}{3}=16\)

\(\Rightarrow x=19\)

#AEZn8

a ) 4 . ( x + 41 ) = 400

             x + 41    = 400 : 4

             x + 41    = 100

             x             = 100 - 41

             x             = 59

b ) 4 . ( x - 3 ) = 7²- 1¹⁰

      4 . ( x - 3 ) = 49 - 1

      4 . ( x - 3 ) = 48

             x - 3    = 48 : 4

             x - 3     = 12

             x           = 12 + 3

             x           = 15