\(f\left(x\right)+g\left(x\right)=x^2h\left(x\right)\)

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

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

chọn C

`#3107.101107`

Ta có:

`f(x) + g(x) = x^2h(x)`

`\Rightarrow g(x) = x^2h(x) - f(x)`

`g(x) = x^2 * (2x + 1) - (2x^3 - 3x^2 + 3x + 1)`

`= 2x^3 + x^2 - 2x^3 + 3x^2 - 3x - 1`

`= 4x^2 - 3x - 1`

Chọn C.

1C

2B

3B

4A

5D

6D

7A

8B

9D

10D

11B

12C 

13B 

14B

15D

16A

17A

18D

19D

 Xét đường thẳng BC, có AH, AB lần lượt là đường vuông góc và đường xiên kẻ từ A đến BC. Do đó \(AH< AB\).

 Chứng minh tương tự, ta được \(BK< BC\) và \(CL< CA\)

 Cộng theo vế 3 BĐT vừa tìm được, ta có:

 \(AH+BK+CL< AB+BC+CA\) (đpcm)

\(A=\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\\ =\dfrac{\left(2^2\right)^5\cdot\left(3^2\right)^4-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot20}\\ =\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\\ =\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\\ =\dfrac{2^{10}\cdot3^8\cdot\left(1-3\right)}{2^{10}\cdot3^8\cdot\left(1+5\right)}\\ =\dfrac{-2}{6}\\ =-\dfrac{1}{3}\)

\(B=4\cdot2^7:\left(3^{11}\cdot\dfrac{1}{9}\right)\\ =4\cdot2^7:\left(3^{11}\cdot\dfrac{1}{3^2}\right)\\ =4\cdot2^7:\dfrac{3^{11}}{3^2}\\ =4\cdot2^7:3^9\\ =\dfrac{2^2\cdot2^7}{3^9}\\ =\dfrac{2^9}{3^9}\\ =\left(\dfrac{2}{3}\right)^9\)

\(\left(2x-1\right)^{50}=2x-1\\ =>\left(2x-1\right)^{50}-\left(2x-1\right)=0\\ =>\left(2x-1\right)\left[\left(2x-1\right)^{49}-1\right]=0\)

TH1: 

\(2x-1\\ =>2x=1\\ =>x=\dfrac{1}{2}\)

TH2:

\(\left(2x-1\right)^{49}-1=0\\=>\left(2x-1\right)^{49}=1\\ =>\left(2x-1\right)^{49}=1^{49}\\ =>2x-1=1\\ =>2x=1+1=2\\ =>x=\dfrac{2}{2}\\ =>x=1\)

\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{23}+\dfrac{1}{6}\\ =\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\dfrac{1}{23}+\dfrac{1}{6}\\ =\dfrac{1}{6}+\dfrac{1}{23}+\dfrac{1}{6}\\ =\dfrac{2}{6}+\dfrac{1}{23}\\ =\dfrac{1}{3}+\dfrac{1}{23}\\ =\dfrac{26}{69}\)

\(3^{x-1}+5\cdot3^{x-1}=162\\ =>3^{x-1}\cdot\left(1+5\right)=162\\ =>3^{x-1}\cdot6=162\\ =>3^{x-1}=162:6\\ =>3^{x-1}=27\\ =>3^{x-1}=3^3\\ =>x-1=3\\ =>x=1+3=4\)

\(a)A\left(x\right)=x^7-3x^2-x^5+x^4-x^2+2x-7\\ =x^7-x^5+x^4-\left(3x^2+x^2\right)+2x-7\\ =x^7-x^5+x^4-4x^2+2x-7\\ B\left(x\right)=x-2x^2+x^4-x^5-x^7-4x^2-1\\ =-x^7-x^5+x^4-\left(2x^2+4x^2\right)+x-1\\ =-x^7-x^5+x^4-6x^2+x-1\) 

\(b)A\left(x\right)+B\left(x\right)\\ =\left(x^7-x^5+x^4-4x^2+2x-7\right)+\left(-x^7-x^5+x^4-6x^2+x-1\right)\\ =\left(x^7-x^7\right)-\left(x^5+x^5\right)+\left(x^4+x^4\right)-\left(4x^2+6x^2\right)+\left(2x+x\right)-\left(7+1\right)\\ =-x^5+2x^4-10x^2+3x-8\) 

\(A\left(x\right)-B\left(x\right)=\left(x^7-x^5+x^4-4x^2+2x-7\right)-\left(-x^7-x^5+x^4-6x^2+x-1\right)\\ =\left(x^7+x^7\right)+\left(x^5-x^5\right)+\left(x^4-x^4\right)-\left(4x^2-6x^2\right)+\left(2x-x\right)-\left(7-1\right)\\ =2x^7+2x^2+x-6\)

c) Thay x = -1  vào C(x) = A(x) + B(x) ta có:

\(C\left(x\right)=-\left(-1\right)^5+2\cdot\left(-1\right)^4-10\cdot\left(-1\right)^2+3\cdot\left(-1\right)-8\\ =-\left(-1\right)+2\cdot1-10\cdot1-3\cdot1-8\\ =1+2-10-3-8\\ =-18\)