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16 tháng 11 2023

a)

\(\text{( 25 – 2x )³ : 5 – 3^2 = 4^2}\)

\(\text{( 25 – 2x )³ : 5 – 9 = 16}\)

\(\text{( 25 – 2x )³ : 5 = 16 + 9}\)

\(\text{( 25 – 2x )³ : 5 = 25}\)

\(\text{( 25 – 2x )³ = 25 . 5}\)

\(\text{( 25 – 2x )³ = 125}\)

\(\text{( 25 – 2x )³ = 5³}\)

\(\text{25 – 2x = 5}\)

\(\text{2x = 25 – 5}\)

\(\text{2x = 20}\)

\(\text{x = 10}\)

\(\text{________________________________________}\)

b)

\(\text{2.3^x = 10.3^12 + 8.27^4}\)

\(\text{2.3^x = 10.3^12 + 8.(3^3)^4}\)

\(\text{2.3^x = 3^12 . (10+8)}\)

\(\text{2.3^x = 3^12 . 18}\)

\(\text{3^x = 3^12 . 18:2}\)

\(\text{3^x = 3^12 . 9}\)

\(\text{3^x = 3^12 . 3^2}\)

\(\text{3^x = 3^14}\)

\(\text{=> x=14}\)

Bài 2: 

a: \(\Leftrightarrow\left(x-5\right)\left(x+5\right)-\left(x+5\right)=0\)

=>(x+5)(x-6)=0

=>x=-5 hoặc x=6

b: \(\Leftrightarrow4x^2-4x+1-4x^2+1=0\)

=>-4x+2=0

hay x=1/2

c: \(\Leftrightarrow\left(x^2+4\right)\left(x^2-1\right)=0\)

=>x=1 hoặc x=-1

10 tháng 8 2018

\(\left(9^{30}-27^{19}\right):3^{57}+\left(125^9-25^{12}\right):5^{24}\)

\(=\left(3^{60}-3^{57}\right):3^{57}+\left(5^{27}-5^{24}\right):5^{24}\)

\(=3^{57}\left(3^3-1\right):3^{57}+5^{24}\left(5^3-1\right):5^{24}\)

\(=3^3-1+5^3-1\)

\(=27-1+125-1\)

\(=150\)

2 )

\(x^2-25-\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-5\right)-\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-5-1\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=6\end{matrix}\right.\)

Vậy ...

b )

\(\left(2x-1\right)^2-\left(4x^2-1\right)=0\)

\(\Leftrightarrow4x^2-4x+1-4x^2+1=0\)

\(\Leftrightarrow2-4x=0\)

\(\Leftrightarrow4x=2\)

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

Vậy ...

c )

\(x^2\left(x^2+4\right)-x^2-4=0\)

\(\Leftrightarrow x^2\left(x^2+4\right)-\left(4+x^2\right)=0\)

\(\Leftrightarrow\left(x^2-1\right)\left(x^2+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=0\\x^2+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2=1\\x^2=-4\left(L\right)\end{matrix}\right.\)

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

Vậy ...

7 tháng 8 2018

a) \(\dfrac{10^{12}+5^{11}.2^9-5^{13}.2^8}{4.5^5.10^6}\)

\(=\dfrac{2^{12}.5^{12}+5^{11}.2^9-5^{13}.2^8}{2^2.5^5.2^6.5^6}\)

\(=\dfrac{2^{12}.5^{12}+5^{11}.2^9-5^{13}.2^8}{2^8.5^{11}}\)

\(=\dfrac{\left(2^8.5^{11}\right)\left(2^4.5+2-5^2\right)}{2^8.5^{11}}\)

\(=2^4.5+2-5^2\)

\(=57\)

b) \(\dfrac{\left[5\left(x-y\right)^4-3\left(x-y\right)^3+4\left(x-y\right)^2\right]}{\left(y-x\right)^2}\)

\(=\dfrac{\left(x-y\right)^2\left[5\left(x-y\right)^2-3\left(x-y\right)+4\right]}{\left(y-x\right)^2}\)

\(=\dfrac{\left(x^2+y^2-2xy\right)\left[5\left(x-y\right)^2-3\left(x-y\right)+4\right]}{\left(y^2+x^2-2xy\right)}\)

\(=5\left(x-y\right)^2-3\left(x-y\right)+4\)

c) \(\dfrac{\left(x+y\right)^5-2\left(x+y\right)^4+3\left(x+y\right)^3}{-5\left(x+y\right)^3}\)

\(=\dfrac{\left(x+y\right)^3\left[5\left(x+y\right)^2-2\left(x+y\right)+3\right]}{-5\left(x+y\right)^3}\)

\(=\dfrac{5\left(x+y\right)^2-2\left(x+y\right)+3}{-5}\)

9 tháng 7 2023

a) 25x² - 16

= (5x)² - 4²

= (5x - 4)(5x + 4)

b) 16a² - 9b²

= (4a)² - (3b)²

= (4a - 3b)(4a + 3b)

c) 8x³ + 1

= (2x)³ + 1³

= (2x + 1)(4x² - 2x + 1)

d) 125x³ + 27y³

= (5x)³ + (3y)³

= (5x + 3y)(25x² - 15xy + 9y²)

e) 8x³ - 125

= (2x)³ - 5³

= (2x - 5)(4x² + 10x + 25)

g) 27x³ - y³

= (3x)³ - y³

= (3x - y)(9x² + 3xy + y²)

9 tháng 7 2023

a) \(25x^2-16=\left(5x-4\right)\left(5x+4\right)\)

b) \(16a^2-9b^2=\left(4a-3b\right)\left(4a+3b\right)\)

c) \(8x^3+1=\left(2x+1\right)\left(4x^2-2x+1\right)\)

d) \(125x^3+27y^3=\left(5x+3y\right)\left(25x^2-15xy+9y^2\right)\)

e) \(8x^3-125=\left(2x-5\right)\left(4x^2-10x+25\right)\)

g) \(27x^3-y^3=\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)

21 tháng 9 2020

a) ( x - 3 )2 - 4 = 0

<=> ( x - 3 )2 - 22 = 0

<=> ( x - 3 - 2 )( x - 3 + 2 ) = 0

<=> ( x - 5 )( x - 1 ) = 0

<=> x = 5 hoặc x = 1

b( 2x + 3 )2 - ( 2x + 1 )( 2x - 1 ) = 22

<=> 4x2 + 12x + 9 - ( 4x2 - 1 ) = 22

<=> 4x2 + 12x + 9 - 4x2 + 1 = 22

<=> 12x + 10 = 22

<=> 12x = 12

<=> x = 1

c) ( 4x + 3 )( 4x - 3 ) - ( 4x - 5 )2 = 16

<=> 16x2 - 9 - ( 16x2 - 40x + 25 ) = 16

<=> 16x2 - 9 - 16x2 + 40x - 25 = 16

<=> 40x - 34 = 16

<=> 40x = 50

<=> x = 50/40 = 5/4

d) x3 - 9x2 + 27x - 27 = -8

<=> ( x - 3 )3 = -8

<=> ( x - 3 )3 = (-2)3

<=> x - 3 = -2

<=> x = 1 

e) ( x + 1 )3 - x2( x + 3 ) = 2

<=> x3 + 3x2 + 3x + 1 - x3 - 3x2 = 2

<=> 3x + 1 = 2

<=> 3x = 1

<=> x = 1/3

f) ( x - 2 )3 - x( x - 1 )( x + 1 ) + 6x2 = 5

<=> x3 - 6x2 + 12x - 8 - x( x2 - 1 ) + 6x2 = 5

<=> x3 + 12x - 8 - x3 + x = 5

<=> 13x - 8 = 5

<=> 13x = 13

<=> x = 1

21 tháng 9 2020

a) \(\left(x-3\right)^2-4=0\)

=> \(\left(x-3\right)^2-2^2=0\)

=> \(\left(x-3-2\right)\left(x-3+2\right)=0\)

=> \(\left(x-5\right)\left(x-1\right)=0\)

=> \(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)

b) \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)

=> \(\left(2x+3\right)^2-\left[\left(2x\right)^2-1^2\right]=22\)

=> \(\left(2x+3\right)^2-\left(4x^2-1\right)=22\)

=> \(\left(2x\right)^2+2\cdot2x\cdot3+3^2-4x^2+1=22\)

=> \(4x^2+12x+9-4x^2+1=22\)

=> \(12x+9+1=22\)

=> \(12x+10=22\)

=> 12x = 12

=> x = 1

c) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=16\)

=> \(\left(4x\right)^2-3^2-\left[\left(4x\right)^2-2\cdot4x\cdot5+5^2\right]=16\)

=> \(16x^2-9-\left(16x^2-40x+25\right)=16\)

=> \(16x^2-9-16x^2+40x-25=16\)

=> \(-9+40x-25=16\)

=> \(40x=16+25-\left(-9\right)=16+25+9=50\)

=> x = 50/40 = 5/4

d) \(x^3-9x^2+27x-27=-8\)

=> \(x^3-3\cdot x^2\cdot3+3\cdot x\cdot3^2-3^3=8\)

=> \(\left(x-3\right)^3=-8\)

=> \(\left(x-3\right)^3=\left(-2\right)^3\)

=> x - 3  = -2 => x = 1

e) \(\left(x+1\right)^3-x^2\left(x+3\right)=2\)

=> \(x^3+3x^2+3x+1-x^3-3x^2=2\)

=> \(3x+1=2\)

=> \(3x=1\)=> x = 1/3

f) \(\left(x-2\right)^3-x\left(x-1\right)\left(x+1\right)+6x^2=5\)

=> \(x^3-3\cdot x^2\cdot2+3\cdot x\cdot2^2-2^3-x\left(x^2-1\right)+6x^2=5\)

=> \(x^3-6x^2+12x-8-x^3+x+6x^2=5\)

=> \(\left(12x+x\right)-8=5\)

=> 13x  = 13

=> x = 1

25 tháng 4 2020

1.(x -5)^2 - 25 =0

=> (x - 5)^2 = 25

=> x - 5 = 5 hoặc x - 5 = -5

=> x = 10 hoặc x = 0

vậy_

2. (x -2)^3 =27

=> x - 2 = 3

=> x = 5

vậy_

3. 3(x -7) + 2x(x+2) = 2x^2

=> 3x - 21 + 2x^2 + 4x = 2x^2

=> 7x - 21 = 0

=> 7x = 21

=> x = 3

vậy_

4. (x^2 - 4) (x +8) =0

=> x^2 - 4 = 0 hoặc x + 8 = 0

=> x^2 = 4 hoặc x = -8

=> x = 2 hoặc x = -2 hoặc x = -8

vậy_

5. x^ 2 + 3x = 0

=> x(x + 3) = 0 

=> x = 0 hoặc x + 3 = 0

=> x = 0 hoặc x = -3

vậy_

6. 3x^3 - 3x = 0

=> 3x(x^2 - 1) = 0

=> 3x(x - 1)(x + 1) = 0

=> x = 0 hoặc x = 1 hoặc x = -1

vậy_

7. (x +1)^2 = ( 2x +3)^2

=> (x + 1 + 2x + 3)(x + 1 - 2x - 3) = 0

=> (3x + 3)(-x - 2) = 0

=> x = -1 hoặc x = -2

vậy_

Bài làm

1) ( x - 5 )2 - 25 = 0

<=> ( x - 5 - 5 )( x - 5 + 5 ) = 0

<=> x( x - 10 ) = 

<=> \(\orbr{\begin{cases}x=0\\x-10=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=10\end{cases}}}\)

Vậy S = { 0; 10 }

2) \(\left(x-2\right)^3=27\)

\(\Leftrightarrow\left(x-2\right)^3=3^3\)

\(\Leftrightarrow x-2=3\)

\(\Leftrightarrow x=5\)

Vậy x = 5 là nghiệm phương trình.

3) \(3\left(x-7\right)+2x\left(x+2\right)=2x^2\)

\(\Leftrightarrow3x+2x^2+4x-2x^2=21\)

\(\Leftrightarrow7x=21\)

\(\Leftrightarrow x=\frac{21}{7}=3\)

Vậy x = 3 là nghiệm phương trình

4) \(\left(x^2-4\right)\left(x+8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-4=0\\x+8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=\pm2\\x=-8\end{cases}}}\)

Vậy S = { 2; -2; -8 }

5) \(x^2+3x=0\)

\(\Leftrightarrow x\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-3\end{cases}}}\)

Vậy S = { 0; -3 } 

6) \(3x^3-3x=0\)

\(\Leftrightarrow3x\left(x^2-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x=0\\x^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}}\)

Vậy S = { +1; 0 }

7) \(\left(x+1\right)^2=\left(2x+3\right)^2\)

\(\Leftrightarrow\left(x+1\right)^2-\left(2x+3\right)^2=0\)

\(\Leftrightarrow\left(x+1-2x-3\right)\left(x+1+2x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}-x-2=0\\3x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{4}{3}\end{cases}}}\)

Vậy S = { -2; -4/3 }

# Học tốt #

8 tháng 8 2023

a. x mũ 2 - 2x + 1 = 25 

= x^2 + 2.x.1 + 1^2

= ( x + 1 ) ^2

ko bt có đúng ko nữa, mấy câu kia tui ko bt lm

8 tháng 8 2023

Sos