\(\left(x+16\right)^2=10^2-86\)

=>\(\left(x+16\right)^2=100-86=14\)

=>\(x+16=\pm\sqrt{14}\)

=>\(x=-16\pm\sqrt{14}\)

google luôn giải đáp cho bn=))))

a, 5x=-36+16=20

=>x=4

b, 4x=10-(-2)=12

=>x=3

c, x-5=-10 : 2= -5

=> x=0

d, 40+x=-80

=> x=-80-40=-120

e, 15-x=-40

=> x=15-(-40)=55

f, x=6+15+4x=21+4x

=>x-4x=21

=>-3x=21

=> x = -7

làm câu nào cũng đc gấp gấp!!!

2⁴.5 - [113 - (11 - 7)²]

= 16.5 - 113 + 4²

= 80 - 113 + 16

= -33 + 16

= -17

5 + 3^x+1 = 86

=> 3^x+1 = 86 - 5

=> 3^x+1 = 81 = 3⁴

=> x + 1 = 4

=> x = 4 - 1

=> x = 3

Bài 2:

a) \(\left(x-3\right)^3+27=0\)

\(\Leftrightarrow\left(x-3\right)^3=0-27\)

\(\Leftrightarrow\left(x-3\right)^3=-27\)

\(\Leftrightarrow\left(x-3\right)^3=\left(-3\right)^3\)

\(\Leftrightarrow x-3=-3\)

\(\Leftrightarrow x=\left(-3\right)+3\)

\(\Leftrightarrow x=0\)

b) \(-125-\left(x+1\right)^3=0\)

\(\Leftrightarrow\left(x+1\right)^3=-125-0\)

\(\Leftrightarrow\left(x+1\right)^3=-125\)

\(\Leftrightarrow\left(x+1\right)^3=\left(-5\right)^3\)

\(\Leftrightarrow x+1=-5\)

\(\Leftrightarrow x=\left(-5\right)-1\)

\(\Leftrightarrow x=-6\)

c) \(\left(2x-\dfrac{1}{4}\right)^2-\dfrac{1}{16}=0\)

\(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=0+\dfrac{1}{16}\)

\(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=\dfrac{1}{16}\)

\(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=\left(\dfrac{1}{4}\right)^2\)

\(\Leftrightarrow2x-\dfrac{1}{4}=\dfrac{1}{4}\)

\(\Leftrightarrow2x=\dfrac{1}{4}+\dfrac{1}{4}\)

\(\Leftrightarrow2x=\dfrac{1}{2}\)

\(\Leftrightarrow x=\dfrac{1}{2}:2\)

\(\Leftrightarrow x=\dfrac{1}{4}\)

d) \(2^x+2^{x+1}=24\)

\(\Leftrightarrow2^x+2^x.2=24\)

\(\Leftrightarrow2^x\left(1+2\right)=24\)

\(\Leftrightarrow2^x.3=24\)

\(\Leftrightarrow2^x=24:3\)

\(\Leftrightarrow2^x=8\)

\(\Leftrightarrow2^x=2^3\)

\(\Rightarrow x=3\)

e) \(\left|x+\dfrac{1}{5}\right|-\dfrac{1}{2}=1\)

\(\Leftrightarrow\left|x+\dfrac{1}{5}\right|=1+\dfrac{1}{2}\)

\(\Leftrightarrow\left|x+\dfrac{1}{5}\right|=\dfrac{3}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=-\dfrac{3}{2}\\x+\dfrac{1}{5}=\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{17}{10}\\x=\dfrac{13}{10}\end{matrix}\right.\)

g) \(\left|x-3\right|+2x=10\)

\(\Leftrightarrow\left|x-3\right|=10-2x\)

\(\Leftrightarrow\left|x-3\right|=2.5-2x\)

\(\Leftrightarrow\left|x-3\right|=2\left(5-x\right)\)

(không chắc có nên làm tiếp câu g không, thấy đề cứ là lạ, có j sai sai...)

Bài 1:

a) \(2^7+2^9⋮10\)

Ta có: \(2^7+2^9=2^{4.1}.2^3+2^{4.2}.2\)

\(\Leftrightarrow\overline{A6}.2^3+\overline{B6}.2\)

\(\Leftrightarrow\overline{A6}.8+\overline{B6}.2\)

\(\Leftrightarrow\overline{C8}+\overline{D2}\)

\(\Leftrightarrow\overline{E0}\)

Mà \(\overline{E0}⋮10\) \(\Rightarrow2^7+2^9⋮10\)

b) \(8^{24}.25^{10}⋮2^{36}.5^{20}\)

Ta có: \(8^{24}.25^{10}=\left(2^3\right)^{24}.\left(5^2\right)^{10}\)

\(\Leftrightarrow2^{72}.5^{20}\)

Do \(2^{72}⋮2^{36}\) và \(5^{20}⋮5^{20}\) \(\Rightarrow8^{24}.25^{10}⋮2^{36}.5^{20}\)

c) \(3^{10}+3^{12}⋮30\)

Ta có: \(3^{10}+3^{12}=3^{4.2}.3^2+3^{4.3}\)

\(\Leftrightarrow\overline{A1}.3^2+\overline{B1}\)

\(\Leftrightarrow\overline{A1}.9+\overline{B1}\)

\(\Leftrightarrow\overline{C9}+\overline{B1}\)

\(\Leftrightarrow\overline{D0}⋮10\)

(Chứng minh chia hết cho 10 rồi chứng minh chia hết cho 3, mình chưa tìm được cách làm, chờ chút)

1, x^2 . x^3 = 3^7 : 3^2

       x^5     = 3^5

=> x = 3

2, x - 18 : 3 = 12

    x - 6       = 12

    x            = 12 + 6

    x            = 18

3, (x - 18) : 2 = 12

     x - 18       = 12 . 2

     x - 18       = 24

     x             = 24 + 18

     x             = 42

4, x^2 = 16

    x^2 = 4^2

=> x = 4

x=4 nha bn

k k nha

mk k lại cho!

Lời giải:

a) \((5x-1)^6=729=3^6=(-3)^6\)

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

b)

\(\frac{8}{25}=\frac{2^x}{5^{x-1}}=\frac{2^x}{5^x:5}=5.(\frac{2}{5})^x\)

\(\Rightarrow \frac{8}{125}=(\frac{2}{5})^x\)

\(\Rightarrow (\frac{2}{5})^3=(\frac{2}{5})^x\Rightarrow x=3\)

c)

\((\frac{1}{16})^x=(\frac{1}{2})^{10}\)

\(\Rightarrow (\frac{1}{2^4})^x=(\frac{1}{2})^{10}\)

\(\Rightarrow (\frac{1}{2})^{4x}=(\frac{1}{2})^{10}\Rightarrow 4x=10\Rightarrow x=\frac{5}{2}\)

d)

\(9^{x}:3^x=3\Rightarrow (\frac{9}{3})^x=3\)

\(\Rightarrow 3^x=3^1\Rightarrow x=1\)

10 - { [ ( x : 3 + 17 ) : 10 + 3 : 2⁴ ] : 10 } = 5

[ ( x : 3 + 17 ) : 10 + 3 : 2⁴ ] : 10 = 10 - 5 = 5

( x : 3 + 17 ) : 10 + 3 : 2⁴ = 5 x 10

( x : 3 + 17 ) : 10 + 48 = 50

( x : 3 + 17 ) : 10 = 50 - 48

( x : 3 + 17 ) : 10 = 2

x : 3 + 17 = 2 x 10

x : 3 + 17 = 20

x : 3 = 20 - 17 = 3

x = 3 x 3 = 9

a) [(2x+14) : 4 - 3] : 2 = 1

(2x+14) : 4 - 3 = 1/2

(2x+14) : 4  = 1/2 + 3

(2x+14) : 4  = 7/2

2x+14 = 7/2 . 1/4

2x = 7/8 - 1/4

2x = 5/8

x= 5/8.1/2

x= 5/16

1/4 : x=(-2)-3/4

1/4 : x=-11/4

x=1/4 :( -11/4)

x=-1/11