Bài 1:

2^\(x\) = 4

2\(^x\) = 2²

 \(x=2\)

Vậy \(x=2\)

Bài 2:

2\(^x\) = 8

2\(^x\) = 2³

\(x=3\)

Vậy \(x=3\)

Bài 3

2\(^x\) = 16

2\(^x\) = 2⁴

  \(x=4\)

Vậy \(x=4\)

a)<=>

A,=(x+y)(x-y)=x^2-y^2

x=(-1/2)^5:(1/2)^4=-1/2

x^2=1/4

y=8^2/(-2)^5=-2

y^2=4

A=1/4-4=-15/4

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 \(x\in\left\{4;5;6\right\}\)

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

\(\Leftrightarrow\left(x^2+6x+9\right)-\left(x^2+4x-32\right)-1=0\)

\(\Leftrightarrow2x=-40\)

\(\Rightarrow x=-20\)

b) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=15\)

\(\Leftrightarrow x^3+27-x^3+4x=15\)

\(\Leftrightarrow4x=-12\)

\(\Rightarrow x=-3\)

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

\(\Leftrightarrow\left(x^2-4x+4\right)-\left(x^2+6x+9\right)-\left(4x+4\right)=5\)

\(\Leftrightarrow-14x=14\)

\(\Rightarrow x=-1\)

d) \(\left(2x-3\right)\left(2x+3\right)-\left(x-1\right)^2-3x\left(x-5\right)=-44\)

\(\Leftrightarrow4x^2-9-\left(x^2-2x+1\right)-\left(3x^2-15x\right)=-44\)

\(\Leftrightarrow17x=-34\)

\(\Rightarrow x=-2\)

e) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)

\(\Leftrightarrow24x=24\)

\(\Rightarrow x=1\)

f) \(5x\left(x-3\right)^2-5\left(x-1\right)^3+15\left(x+2\right)\left(x-2\right)=5\)

\(\Leftrightarrow5x^3-30x^2+45x-5x^3+15x^2-15x+5+15x^2-60=5\)

\(\Leftrightarrow30x=60\)

\(\Rightarrow x=2\)

g) \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)-3x^2=42\)

\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1-3x^2=42\)

\(\Leftrightarrow26x=14\)

\(\Rightarrow x=\frac{7}{13}\)

Câu a :

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

\(=4x^2+4x+1-4x^2+20\)

\(=4x+19\)

Câu b :

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

\(=x^2+6x+9-x^2-1\)

\(=6x-8\)

Câu c :

\(\left(x-5\right)^2-\left(x+2\right)^2\)

\(=\left(x-5-x-2\right)\left(x-5+x+2\right)\)

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

Bạn trình bày cho rõ ràng xem nào.

\(\text{b) }\left(x+3\right)^2-\left(x+1\right)\left(x-1\right)\\ =\left(x+3\right)^2-\left(x^2-1^2\right)\\ =x^2+2\cdot x\cdot3+3^2-x^2+1\\ =\left(x^2-x^2\right)+6x+\left(9+1\right)\\ =6x+10\\ \)

\(\text{c) }\left(x-5\right)^2-\left(x+2\right)^2\\ =\left(x^2-2\cdot x\cdot5+5^2\right)-\left(x^2+2\cdot x\cdot2+2^2\right)\\ =x^2-10x+25-x^2-4x-4\\ =\left(x^2-x^2\right)-\left(10x+4x\right)+\left(25-4\right)\\ =-14x+21\\ \)

\(\text{d) }\left(x+3\right)^2-\left(x-3\right)^2\\ =\left(x^2+2\cdot x\cdot3+3^2\right)-\left(x^2-2\cdot x\cdot3+3^2\right)\\ =x^2+6x+9-x^2+6x-9\\ =\left(x^2-x^2\right)+\left(6x+6x\right)+\left(9-9\right)\\ =12x\\ \)

\(\text{e) }2x\left(x+1\right)-\left(x+3\right)^2-x^2\\ =2x^2+2x-\left(x^2+2\cdot x\cdot3+3^2\right)-x^2\\ =2x^2+2x-x^2-6x-9-x^2\\ =\left(2x^2-x^2-x^2\right)+\left(2x-6x\right)-9\\ =-4x-9\\ \)

\(\text{g) }\left(x+3\right)^2+\left(x+2\right)^2-2\left(x+3\right)\left(x+2\right)\\ =\left[\left(x+3\right)-\left(x+2\right)\right]^2\\ =\left(x+3-x-2\right)^2\\ =1^2\\ =1\\ \)

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

may cai bai day ma cung khong biet oc cho

a) 89-(73-x)=20

=>73-x=89-20

=>73-x=69

=>x=73-69

=>x=4

Dài quá bn ơi, vs lại mấy cái này cx ko khó lắm đâu (nếu ko lm đc đăng mỗi lần một ít thui chứ dài quá mk đọc cn lười chứ đừng ns là làm)

Vũ Hồng Linh bạn check lại bài đầu dùm =_=" 

\(\left[-\frac{1}{3}\right]^3\cdot x=\frac{1}{81}\)

\(\Leftrightarrow x=\frac{1}{81}:\left[-\frac{1}{3}\right]^3\)

\(\Leftrightarrow x=\frac{1}{81}:\left[-\frac{1}{27}\right]\)

\(\Leftrightarrow x=\frac{1}{81}\cdot(-27)=-\frac{1}{3}\)

\(\left[x-\frac{1}{2}\right]^3=\frac{1}{27}\)

\(\Leftrightarrow\left[x-\frac{1}{2}\right]^3=\left[\frac{1}{3}\right]^3\)

=> Làm nốt 

Mấy bài kia cũng làm tương tự

(- \(\dfrac{1}{3}\))³.\(x\) = \(\dfrac{1}{81}\)

          \(x=\dfrac{1}{81}\) : (- \(\dfrac{1}{3}\))³

          \(x\) =  - (\(\dfrac{1}{3}\))⁴ :(\(\dfrac{1}{3}\))³

           \(x=-\dfrac{1}{3}\)

Vậy \(x=-\dfrac{1}{3}\)

(3\(x\) + 1)³ =  - 27

(3\(x\) + 1)³ = (-3)³

3\(x\) + 1 = -3

 3\(x\)       = - 3 - 1

 3\(x\) = -4

    \(x=-\dfrac{4}{3}\)

Vậy \(x=-\dfrac{4}{3}\)

\(\left(x+1\right)^3=27\)

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

\(\Rightarrow x+1=3\)

\(x=2\)

\(\left(x+1\right)^3=27\)

\(< =>\left(x+1\right)^3=3.3.3=3^3\)

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

\(\left(2x+3\right)^3=9.81\)

\(< =>\left(2x+3\right)^3=9.9.9\)

\(< =>\left(2x+3\right)^3=9^3\)

\(< =>2x+3=9< =>2x=6\)

\(< =>x=\frac{6}{2}=3\)

giúp mình bài này đi

21.73 + 27 . 21