Bài 1

a) \(x=x^5\)

\(x^5-x=0\)

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

\(x=0\) hoặc \(x^4-1=0\)

* \(x^4-1=0\)

\(x^4=1\)

\(x=1\)

Vậy x = 0; x = 1

b) \(x^4=x^2\)

\(x^4-x^2=0\)

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

\(x^2=0\) hoặc \(x^2-1=0\)

*) \(x^2=0\)

\(x=0\)

*) \(x^2-1=0\)

\(x^2=1\)

\(x=1\)

Vậy \(x=0\); \(x=1\)

c) \(\left(x-1\right)^3=x-1\)

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

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

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

*) \(x-1=0\)

\(x=1\)

*) \(\left(x-1\right)^2-1=0\)

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

\(x-1=1\) hoặc \(x-1=-1\)

**) \(x-1=1\)

\(x=2\)

**) \(x-1=-1\)

\(x=0\)

Vậy \(x=0\); \(x=1\); \(x=2\)

  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 \(x=4\)

Lời giải:
a. $x^3=4^3\Rightarrow x=4$

b. $x^2=49=7^2=(-7)^2$

$\Rightarrow x=7$ hoặc $x=-7$

c. $x^3+1=28$

$x^3=28-1=27=3^3$

$\Rightarrow x=3$

d. $2^x=16=2^4$

$\Rightarrow x=4$

e. $2^4.2^x=2^6$

$\Rightarrow 2^{4+x}=2^6$

$\Rightarrow 4+x=6$

$\Rightarrow x=2$

g.

$5^x=25.5^3=5^2.5^3=5^5$

$\Rightarrow x=5$

Lần sau bạn lưu ý viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để đề được rõ ràng hơn nhé.

a.  64.4^x=4⁵

4³.4^x=4⁵

4^x=4⁵:4³

4^x=4²

x=2

có ai biết hoặc là fan gacha life không làm ơn kết bạn với mình đi team gacha life của mình đang thiếu người

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

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

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

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

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

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

`@` `\text {Ans}`

`\downarrow`

`a,`

\(f(x) -3x^2 + x - 1 + x^4 - x^3 - x^2 + 3x^4 + 2x^3\)

`= (x^4 +3x^4) + (-x^3 +2x^3) + (-3x^2 - x^2) + x - 1`

`= 4x^4 + x^3 -4x^2 + x -1`

\(g(x) = x^4 + x^2 - x^3 + x - 5 + 5x^3 - x^2 - 3x^4\)

`= (x^4-3x^4) + (-x^3+5x^3) + (x^2 - x^2) + x -5`

`= -2x^4 + 4x^3 +x - 5`

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.

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

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

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

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

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

Thêm nữa câu a) Tính: M(x) + N(x)+ P(x)

B) Tính M(x) - N (x) - P(x)

ok rồi giúp mình với nha