a) \(\left(2x-5\right)^2-\left(2x+3\right)\left(2x-3\right)=10\Leftrightarrow\left(4x^2-20x+25\right)-\left(4x^2-9\right)-10=0\)

\(\Leftrightarrow-20x+24=0\Leftrightarrow x=\frac{6}{5}\)

b) \(\left(4x-1\right)\left(x+2\right)-\left(2x+3\right)^2-5\left(x-1\right)=9\Leftrightarrow-10x-15=0\)

\(\Leftrightarrow x=\frac{-3}{2}\)

c) \(\left(x+1\right)^3-\left(x-1\right)^3-2=6\Leftrightarrow\left(x^3+3x^2+3x+1\right)-\left(x^3-3x^2+3x-1\right)-8=0\)

\(\Leftrightarrow6x^2-6=0\Leftrightarrow x=\pm1\)

d) \(\left(x+2\right)\left(x^2-2x+4\right)-\left(x+1\right)\left(x^2-x+1\right)-3\left(-x-2\right)=5\)

\(\Leftrightarrow\left(x^3+8\right)-\left(x^3+1\right)+3x+6=5\Leftrightarrow3x+8=0\Leftrightarrow x=\frac{-8}{3}\)

1. 2^x=16\(\Rightarrow\)X=4

2. 2^2x-1=2⁷

\(\Rightarrow\)2⁷=2^2.4-1

Vậy x =4

x=4 nha chị

a) \(4^n=4096\Rightarrow4^n=4^6\Rightarrow n=6\)

b) \(5^n=15625\Rightarrow5^n=5^6\Rightarrow n=6\)

c) \(6^{n+3}=216\Rightarrow6^{n+3}=6^3\Rightarrow n+3=3\Rightarrow n=0\)

d) \(x^2=x^3\Rightarrow x^3-x^2=0\Rightarrow x^2\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

e) \(3^{x-1}=27\Rightarrow3^{x-1}=3^3\Rightarrow x-1=3\Rightarrow x=4\)

f) \(3^{x+1}=9\Rightarrow3^{x+1}=3^2\Rightarrow x+1=2\Rightarrow x=1\)

g) \(6^{x+1}=36\Rightarrow6^{x+1}=6^2\Rightarrow x+1=2\Rightarrow x=1\)

h) \(3^{2x+1}=27\Rightarrow3^{2x+1}=3^3\Rightarrow2x+1=3\Rightarrow2x=2\Rightarrow x=1\)

i) \(x^{50}=x\Rightarrow x^{50}-x=0\Rightarrow x\left(x^{49}-1\right)=0\)

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

4ⁿ  =  4096 

4ⁿ = 2¹²

n = 12

5ⁿ = 15625 

5ⁿ = 5⁶

n   = 6

6ⁿ⁺³ = 216

6ⁿ⁺³ = 2³.3³

6ⁿ⁺³ = 6³

n + 3 = 3

4ⁿ = 4096

4ⁿ = 4⁶

n = 6 (nhận)

Vậy n = 6

--------------------

5ⁿ = 15625

5ⁿ = 5⁶

n = 6 (nhận)

Vậy n = 6

--------------------

4ⁿ⁻¹ = 1024

4ⁿ⁻¹ = 4⁵

n - 1 = 5

n = 6 (nhận)

Vậy n = 6

-------------------

6ⁿ⁺³ = 216

6ⁿ⁺³ = 6³

n + 3 = 3

n = 0  (nhận)

Vậy n = 0

--------------------

x² = x³

x³ - x² = 0

x(x² - 1) = 0

x = 0 (nhận) hoặc x² - 1 = 0

*) x² - 1 = 0

x² = 1

x = 1 (nhận) hoặc x = -1 (loại)

Vậy x = 0; x = 1 

--------------------

3ˣ⁻¹ = 27

3ˣ⁻¹ = 3³

x - 1 = 3

x = 3 + 1

x = 4 (nhận)

Vậy x = 4

---------------------

3ˣ⁺¹ = 9

3ˣ⁺¹ = 3²

x + 1 = 2

x = 2 - 1

x = 1 (nhận)

Vậy x = 1

--------------------

6ˣ⁺¹ = 36

6ˣ⁺¹ = 6²

x + 1 = 2

x = 2 - 1

x = 1 (nhận)

Vậy x = 1

--------------------

3²ˣ⁺¹ = 27

3²ˣ⁺¹ = 3³

2x + 1 = 3

2x = 3 - 1

2x = 2

x = 1 (nhận)

Vậy x = 1

--------------------

x⁵⁰ = x

x⁵⁰ - x = 0

x(x⁴⁹ - 1) = 0

x = 0 (nhận) hoặc x⁴⁹ - 1 = 0

*) x⁴⁹ - 1 = 0

x⁴⁹ = 1

x = 1 (nhận)

Vậy x = 0; x = 1

1. \(6x^3-8=40\\ 6x^3=48\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2

2. \(4x^5+15=47\\ 4x^5=32\\ x^5=8\\ \Rightarrow x\in\varnothing\left(\text{vì }x\in N\right)\)Vậy x ∈ ∅

3. \(2x^3-4=12\\ 2x^3=16\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2

4. \(5x^3-5=0\\ 5x^3=5\\ x^3=1\\ \Rightarrow x=1\)Vậy x = 1

5. \(\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)Vậy \(x\in\left\{5;6\right\}\)

6. \(\left(3x-2\right)^{20}=\left(3x-1\right)^{20}\\ \Rightarrow3x-2=3x-1\\ 3x-3x=2-1\\ 0=1\left(\text{vô lí}\right)\)Vậy x ∈ ∅

7. \(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\\ \left(3x-1\right)^{10}=\left[\left(3x-1\right)^2\right]^{10}\\ \Rightarrow\left(3x-1\right)^2=3x-1\\ \left(3x-1\right)^2-\left(3x-1\right)=0\\ \left(3x-1\right)\left[\left(3x-1\right)-1\right]=0\\ \left(3x-1\right)\left(3x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x-1=0\\3x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=1\\3x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\left(\text{loại vì }x\in N\right)\\x=\frac{2}{3}\left(\text{loại vì }x\in N\right)\end{matrix}\right.\)Vậy x ∈ ∅

8. \(\left(2x-1\right)^{50}=2x-1\\ \left(2x-1\right)^{50}-\left(2x-1\right)=0\\ \left(2x-1\right)\left[\left(2x-1\right)^{49}-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\\left(2x-1\right)^{49}=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=1\\2x-1=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\left(\text{loại vì }x\in N\right)\\x=1\left(t/m\right)\end{matrix}\right.\)Vậy x = 1

9. \(\left(\frac{x}{3}-5\right)^{2000}=\left(\frac{x}{3}-5\right)^{2008}\\ \left(\frac{x}{3}-5\right)^{2008}-\left(\frac{x}{3}-5\right)^{2000}=0\\ \left(\frac{x}{3}-5\right)^{2000}\left[\left(\frac{x}{3}-5\right)^8-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(\frac{x}{3}-5\right)^{2000}=0\\\left(\frac{x}{3}-5\right)^8=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}-5=0\\\frac{x}{3}-5=1\\\frac{x}{3}-5=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}=5\\\frac{x}{3}=6\\\frac{x}{3}=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\cdot3=15\\x=6\cdot3=18\\x=4\cdot3=12\end{matrix}\right.\)Vậy \(x\in\left\{15;18;12\right\}\)

\(1.6x^3-8=40\\ \Leftrightarrow6x^3=48\\ \Leftrightarrow x^3=8\Leftrightarrow x^3=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{2;-2\right\}\)

\(2.4x^3+15=47\) (T nghĩ đề là mũ 3)

\(\Leftrightarrow4x^3=32\Leftrightarrow x^3=8=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{2;-2\right\}\)

Câu 3, 4 tương tự nhé.

\(5.\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\\ \Leftrightarrow\left(x-5\right)^{2018}-\left(x-5\right)^{2016}=0\\ \Leftrightarrow\left(x-5\right)^{2016}\left[\left(x-5\right)^2-1\right]=0\\ \Leftrightarrow\left(x-5\right)^{2016}\left(x-5-1\right)\left(x-5+1\right)=0\\ \Leftrightarrow\left(x-5\right)^{2016}\left(x-6\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-5\right)^{2016}=0\\x-6=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\x=6\\x=4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\\x=4\end{matrix}\right.\)

Vậy \(x\in\left\{4;5;6\right\}\)

(2x - 1)⁶ = (2x - 1)⁴

=> (2x - 1)⁶ - (2x - 1)⁴ = 0

=> (2x - 1)⁴.[(2x - 1)² - 1] = 0

=> \(\orbr{\begin{cases}\left(2x-1\right)^4=0\\\left(2x-1\right)^2-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}2x-1=0\\\left(2x-1\right)^2=1\end{cases}}\Rightarrow\orbr{\begin{cases}2x-1=0\\2x-1=\pm1\end{cases}}\)

Khi 2x - 1 = 0 => x = 1/2

Khi 2x - 1 = -1 => x = 0

Khi 2x - 1 = 1 => x = 1

Vậy \(x\in\left\{\frac{1}{2};0;1\right\}\)là giá trị cần tìm

( 2x - 1 )⁶ = ( 2x - 1 )⁴

<=> ( 2x - 1 )⁶ - ( 2x - 1 )⁴ = 0

<=> ( 2x - 1 )⁴[ ( 2x - 1 )² - 1 ] = 0

<=> \(\orbr{\begin{cases}\left(2x-1\right)^4=0\\\left(2x-1\right)^2-1=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=\frac{1}{2}\\x=1\\x=0\end{cases}}\)( thay = dấu hoặc hộ nhé )

ơ thiếu nghiệm :v

\(x=\left\{\frac{1}{2};1;0\right\}\)nhé ._.

a) 2( x - 1 )² - 4( 3 + x )² + 2x( x - 5 )

= 2( x² - 2x + 1 ) - 4( 9 + 6x + x² ) + 2x² - 10x

= 2x² - 4x + 2 - 36 - 24x - 4x² + 2x² - 10x

= ( 2x² - 4x² + 2x² ) + ( -4x - 24x - 10x ) + ( 2 - 36 )

= -38x - 34

b) 2( 2x + 5 )² - 3( 4x + 1 )( 1 - 4x )

= 2( 4x² + 20x + 25 ) + 3( 4x + 1 )( 4x - 1 )

= 8x² + 40x + 50 + 3( 16x² - 1 )

= 8x² + 40x + 50 + 48x² - 3

= 56x² + 40x + 47

c) ( x - 1 )³ - x( x - 3 )² + 1

= x³ - 3x² + 3x - 1 - x( x² - 6x + 9 ) + 1

= x³ - 3x² + 3x - x³ + 6x² - 9x

= 3x² - 6x

d) ( x + 2 )³ - x²( x + 6 ) 

= x³ + 6x² + 12x + 8 - x³ - 6x²

= 12x + 8

e) ( x - 2 )( x + 2 ) - ( x + 1 )³ - 2x( x - 1 )²

= x² - 4 - ( x³ + 3x² + 3x + 1 ) - 2x( x² - 2x + 1 )

= x² - 4 - x³ - 3x² - 3x - 1 - 2x³ + 4x² - 2x

= -3x³ + 2x² - 5x - 5 

f) ( a + b - c )² - ( b - c )² - 2a( b - c )

= [ ( a + b ) - c ]² - ( b² - 2bc + c² ) - 2ab + 2ac

= [ ( a + b )² - 2( a + b )c + c² ] - b² + 2bc - c² - 2ab + 2ac

= a² + 2ab + b² - 2ac - 2bc + c² - b² + 2bc - c² - 2ab + 2ac

= a²

a) \(2\left(x-1\right)^2-4\left(3+x\right)^2+2x\left(x-5\right)\)

Dùng hẳng đẳng thức thứ nhất + hai :

= \(2\left(x^2-2\cdot x\cdot1+1^2\right)-4\left(3^2+2\cdot3\cdot x+x^2\right)+2x^2-10x\)

= \(2\left(x^2-2x+1\right)-4\left(9+6x+x^2\right)+2x^2-10x\)

= \(2x^2-4x+2-36-24x-4x^2+2x^2-10x\)

= \(-38x-34\)

b) 2(2x + 5)² - 3(4x + 1)(1 - 4x)

Dùng đẳng thức thứ 1 + 3

= 2[(2x)² + 2.2x.5 + 5² ] - (-3)[(4x)² - 1² ]

= 2(4x² + 20x + 25) - (-3).(16x² - 1)

= 8x² + 40x + 50 - (3 - 48x²)

= 8x² + 40x + 50 - 3 + 48x²

= 56x² + 40x + 47

c) (x - 1)³ - x(x - 3)² + 1

Dùng đẳng thức 2 + 5:

= x³ - 3.x².1 + 3.x.1² - 1³ - x(x² - 2.x.3 + 3²) + 1

= x³ - 3x² + 3x - 1 - x³ + 6x² - 9x + 1

= (x³ - x³) + (-3x² + 6x²) + (3x - 9x) + (-1 + 1)

= 3x² - 6x

d) (x + 2)³ - x²(x + 6)

= x³ + 3.x².2 + 3.x.2² + 2³ - x³ - 6x²

= x³ + 6x² + 12x + 8 - x³ - 6x²

= (x³ - x³) + (6x² - 6x²) + 12x + 8 = 12x + 8

e) Dùng đẳng thức thứ 3,4 và 2

= x² - 4 - (x³ + 3.x².1 + 3.x.1² + 1³) - 2x(x² - 2.x.1 + 1²)

= x² - 4 - (x³ + 3x² + 3x + 1) - 2x³ + 4x² - 2x

= x² - 4 - x³ - 3x² - 3x - 1 - 2x³ + 4x² - 2x

= (x² - 3x² + 4x²) + (-4 - 1) + (-x³ - 2x³) + (-3x - 2x)

= 2x² - 5 - 3x³ - 5x

f) Đặt \(a+b-c=A\)

\(b-c=B\)

= \(A^2-B^2-2AB\)

= \(A^2-2AB+\left(-B\right)^2\)

\(=A^2-2AB+B^2\)

= (A - B)²

= (a + b - c - (b - c))²

= (a + b - c - b + c)²

= a²

a: =>9(2x+1)=6(3-x)

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

=>6x+3=6-2x

=>8x=3

=>x=3/8

b: =>-3x^2-2+3x^2-18x=-26

=>-18x=-24

=>x=4/3

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 #