a) |x-1|-2x=5

\(\Leftrightarrow\) x - 1 - 2x = 5 (đk: x\(\ge\)1 ) (1)

hoặc -x + 1 - 2x = 5 ( đk: x < 1) (2)

(1): \(\Leftrightarrow\) x - 2x = 5 + 1

\(\Leftrightarrow\) - x = 6

\(\Leftrightarrow\) x = -6 (ko thỏa mãn)

(2): \(\Leftrightarrow\) -x + 1 - 2x = 5

\(\Leftrightarrow\) -3x = 4

\(\Leftrightarrow\) x = \(\frac{-4}{3}\) (thỏa mãn)

Vậy: x = \(\frac{-4}{3}\)

b) |9-7x|=5x-3

\(\Leftrightarrow\) 9 - 7x = 5x - 3 (đk: \(\le\) \(\frac{9}{7}\) ) (1)

hoặc -9 + 7x = 5x - 3 (đk: x > \(\frac{9}{7}\) ) (2)

(1): \(\Leftrightarrow\) -7x-5x = -3-9 \(\Leftrightarrow\) -12x = -12 \(\Leftrightarrow\) x = 1 ( thỏa mãn)

(2): \(\Leftrightarrow\) 7x - 5x = -3 + 9 \(\Leftrightarrow\) 2x = 6 \(\Leftrightarrow\) x = 3 ( thỏa mãn)

Vậy: x= 1; 3

c) 5^x+2 = 625

\(\Leftrightarrow\) 5^x . 5² = 5⁴

\(\Leftrightarrow\) 5^x = \(\frac{5^4}{5^2}\)= 5²

^{\(\Leftrightarrow\)} x = 2

d) (2x - 3)² = 36

\(\Leftrightarrow\) \(\left[{}\begin{matrix}2x-3=6\left(đk:x\ge\frac{3}{2}\right)\text{​​}\Leftrightarrow x=4,5\left(TMĐK\right)\\2x-3=-6\left(đk:x< \frac{3}{2}\right)\Leftrightarrow x=-1,5\left(TMĐK\right)\end{matrix}\right.\)

Vậy: x = 4,5; -1,5

a) x = 4                

b) x = 7     

c) x = 2                

d) x = 5

e) x = 2                

f) x= 1. 

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a) x = 4

b) x = 7

c) x = 2

d) x = 5

e) x = 2

 f) x= 1

`@` `\text {Ans}`

`\downarrow`

`(5x+2)^3 = -125?`

`=> (5x+2)^3 = (-5)^3`

`=> 5x+2=-5`

`=> 5x = -5-2`

`=> 5x = -7`

`=> x = -7/5`

Vậy, `x = -7/5`

Hoặc nếu đề ntnay

`(5x+2)^4 = -625`

`=> (5x+2)^4 = -(5)^4`

`=> 5x+2= -5`

`=> 5x = -7`

`=> x = -7/5`

a: =>|7x-9|=5x-3

\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{3}{5}\\\left(7x-9-5x+3\right)\left(7x-9+5x-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{3}{5}\\\left(2x-6\right)\left(12x-12\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{3;1\right\}\)

b: =>|17x-5|=|17x+5|

=>17x-5=17x+5(vô lý) hoặc 17x-5=-17x-5

=>34x=0

hay x=0

c: =>|3x+4|=|4x-18|

=>4x-18=3x+4 hoặc 4x-18=-3x-4

=>x=22 hoặc 7x=14

=>x=22 hoặc x=2

phân tích đa thức thành nhân tử

a) 100a^2-(a^2+25)^2

= -(a-5)^2(a+5)^2

b) -5x^3y^3-5x^3y^3

= 

c) 16+2x^3y^3

= 2(xy+2)(x^2y^2-2xy+4)

Trả lời:

a, \(100a^2-\left(a^2+25\right)^2=\left(10a\right)^2-\left(a^2+25\right)^2=\left(10a-a^2-25\right)\left(10a+a^2+25\right)\)

b, \(-5x^3y^3-5x^3y^3=-10x^3y^3\)

c, \(16+2x^3y^3=2\left(8+x^3y^3\right)=2\left[2^3+\left(xy\right)^3\right]=2\left(2+xy\right)\left(4-2xy+x^2y^2\right)\)

a) 3^3x+1 . 5 = 10935

=> 3^3x . 3 . 5 = 10935

=> 27^x . 3 . 5 = 10935

=> 27^x = 729

=> 27^x = 27²

=> x = 2

b) 2^{5x + 1} . 3 = 6144

=> 2^5x . 2 . 3 = 6144

=> 32^x . 6 = 6144

=> 32^x = 1024

=> 32^x = 32²

=> x = 2

c) (2x + 3)⁴ = 625

=> (2x + 3)⁴ = (\(\pm\)5)⁴

=> \(\orbr{\begin{cases}2x+3=5\\2x+3=-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-4\end{cases}}\)

d) (3x + 1)³ = 343

=> (3x + 1)³ = 7³

=> 3x + 1 = 7 => 3x = 6 => x = 2

a) 3^{3x + 1}. 5 = 10935

3^3x+1 = 10935 : 5

3^3x+1 = 2187

3^3x+1 = 3⁷

=> 3x + 1 = 7

      3x = 7 - 1

      3x = 6

        x = 6:3

   => x = 2

a) \(3^{3x+1}\cdot5=10935\)

\(\Leftrightarrow3^{3x+1}=2187=3^7\)

\(\Rightarrow3x+1=7\)

\(\Leftrightarrow3x=6\)

\(\Rightarrow x=2\)

b) \(2^{5x+1}\cdot3=6144\)

\(\Leftrightarrow2^{5x+1}=2048=2^{11}\)

\(\Leftrightarrow5x+1=11\)

\(\Leftrightarrow5x=10\)

\(\Rightarrow x=2\)