\(\left(5x-1\right)^6=729\)

\(\Rightarrow\left(5x-1\right)^6=3^6\)

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

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

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

=>(5x-1)^6=3^6

=>5x-1=3 hoặc 5x-1=-3

=>x=4/5 hoặc x=-2/5

a)  | x | − 3 4 = 5 3 ⇒ | x | = 29 12 ⇒ x = 29 12 x = − 29 12

b)  x − 3 2 = 1 ⇒ x − 3 2 = 1 x − 3 2 = − 1 ⇒ x = 5 2 x = 1 2

1)A>B

2)A=B

3)A=B

1)A>B

2)A=B

3)A=B

a ) | x | − 3 4 = 5 3 = > | x | = 29 12 = > x = 29 12 x = − 29 12

b)  x − 3 2 = 1 = > x − 3 2 = 1 x − 3 2 = − 1 = > x = 5 2 x = 1 2

c)  x − 3 2 + 11 4 = 15 4 = > x − 3 2 = 1 = > x = 5 2 x = 1 2

d)  17 4 − 5 6 − x = − 7 4 = > 5 6 − x = 6 = > x = − 31 6 x = 41 6

e)

  x + 1 5 2 − 9 25 = 0 = > x + 1 5 2 = 9 25 = > x + 1 5 = 3 5 x + 1 5 = − 3 5 = > x = 2 5 x = − 4 5

f) x + 1 5 2 + 17 25 = 26 25 = > x + 1 5 2 = 9 25 = > x = 2 5 x = − 4 5

đề sai thì phải

\(a,3^{x+1}=27\)

\(3^{x+1}=3^3\)

\(x+1=3\)

\(x=3-1=2\)

\(b,6^{x-1}=36\)

\(6^{x-1}=6^2\)

\(x-1=2\)

\(x=2+1=3\)

\(c,3^{2x+1}=243\)

\(3^{2x+1}=3^5\)

\(2x+1=5\)

\(2x=5+1=6\)

\(x=6:2=3\)

Mình hong hiểu câu d :<

\(x^{50}=x\\ \Rightarrow x^{50}-x=0\\ \Rightarrow x\left(x^{49}-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^{49}-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^{49}=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Bài 1 

\(B=35x53+18\)

\(=34x53+18+53\)

\(C=35+53x34\) 

=>B>C

Còn C thì chắc chắn > 1 rồi bn nhé

ĐKXĐ: \(x\ge0;x\ne4\)

\(A=\dfrac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

b. \(x=36\Rightarrow A=\dfrac{\sqrt{36}}{\sqrt{36}-2}=\dfrac{6}{6-2}=\dfrac{3}{2}\)

c. \(A=-\dfrac{1}{3}\Rightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}=-\dfrac{1}{3}\Rightarrow3\sqrt{x}=2-\sqrt{x}\)

\(\Rightarrow4\sqrt{x}=2\Rightarrow\sqrt{x}=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{4}\)

d. \(A>0\Rightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}>0\Rightarrow\sqrt{x}-2>0\Rightarrow x>4\)

e. \(A=\dfrac{\sqrt{x}-2+2}{\sqrt{x}-2}=1+\dfrac{2}{\sqrt{x}-2}\in Z\Rightarrow\sqrt{x}-2=Ư\left(2\right)\)

\(\Rightarrow\sqrt{x}-2=\left\{-2;-1;1;2\right\}\)

\(\Rightarrow\sqrt{x}=\left\{0;1;3;4\right\}\Rightarrow x=\left\{0;1;9;16\right\}\)

a: Ta có: \(A=\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)

\(=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

b: Thay x=36 vào A, ta được:

\(A=\dfrac{6}{6-2}=\dfrac{6}{4}=\dfrac{3}{2}\)

c: Để \(A=-\dfrac{1}{3}\) thì \(3\sqrt{x}=-\sqrt{x}+2\)

\(\Leftrightarrow4\sqrt{x}=2\)

hay \(x=\dfrac{1}{4}\)

d: Để A>0 thì \(\sqrt{x}-2>0\)

hay x>4

e: Để A nguyên thì \(\sqrt{x}⋮\sqrt{x}-2\)

\(\Leftrightarrow\sqrt{x}-2\in\left\{-1;1;2;-2\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{1;3;4;0\right\}\)

hay \(x\in\left\{1;9;16;0\right\}\)

A.35 x 34 + 35 x 38 + 65 x 75 + 65 x 45

Câu này không tính hợp lí được

B. 12x 53 + 53 x 172 - 53 x 84

= 53 x (12 + 172 - 84)

= 53 x 100

= 5300

C. 36 x 28 + 36 x 82 + 64 x 69 + 64 x 41

= 36 x (28+82) + 64 x (69+41)

= 36 x 110 + 64 x 110

= (36+64) x 110

= 100 x 110

= 11 000

D. A = 1+2+3+...+100

= (1+ 100) x 50

= 101 x 50

= 5050

E. B =17+19+...+101+103

= (103 +17) x [(103 - 17):2 + 1]

= 120 x 44

=5280

G. D= 3x (12+13+14+15) + 3x(8+7+6+5)

= 3 x (5+6+7+8+12+13+14+15)

= 3 x (20 x 4)

= 3 x 80

=240

a) 10320

b) 5300

c) 11000

d) 5050

e) 2460

g)  240

a: 58/63=3190/3465

36/55=2268/3465

=>58/63>36/55

b: 27/53=1998/3922

36/74=1908/3922

=>27/53>36/74