a) Ta có:

\(\widehat{ADE}=\widehat{ABC}\left(=45^o\right)\)

Mà hai góc này ở vị trí đồng vị

=> DE//BC

b) Ta có: 

\(\widehat{FEC}=\widehat{ECB}\left(gt\right)\)

Mà hai góc này ở vị trí so le trong 

=> EF//BC

c) Ta có: DE//BC

=> \(\widehat{DEC}+\widehat{ECB}=180^o\) (trong cùng phía) 

Mà: \(\widehat{FEC}=\widehat{ECB}\left(gt\right)\)

\(=>\widehat{FEC}+\widehat{ECB}=180^o\)

\(=>\widehat{DEF}\) là góc bẹt

=> D, E, F thẳng hàng

a) Ta có: 

\(\widehat{MAB}=\widehat{ABC}\left(=55^o\right)\)

Mà hai góc này ở vị trí so le trong 

=> AM//BC 

b) Ta có:
\(\widehat{NAC}=\widehat{ACB}\left(=40^o\right)\)

Mà hai góc này ở vị trí so le trong

=> AN//BC 

c) Xét tam giác ABC có:

\(\widehat{BAC}+\widehat{ABC}+\widehat{ACB}=180^o\\ =>\widehat{BAC}=180^o-\widehat{ABC}-\widehat{ACB}\\ =>\widehat{BAC}=180^o-55^o-40^o=85^o\) 

\(\widehat{MAB}+\widehat{BAC}+\widehat{NAC}=55^o+85^o+40^o=180^o\) 

=> \(\widehat{MAN}\) là góc bẹt => M, A, N thẳng hàng

\(\left(8+2\dfrac{1}{3}-\dfrac{3}{5}\right):\left(5-\dfrac{1}{4}-\dfrac{5}{8}\right)\\ =\left(8+2+\dfrac{1}{3}-\dfrac{3}{5}\right):\left(5-\dfrac{2}{8}-\dfrac{5}{8}\right)\\ =\left(10+\dfrac{1}{3}-\dfrac{3}{5}\right):\left(5-\dfrac{7}{8}\right)\\ =\left(\dfrac{150}{15}+\dfrac{5}{15}-\dfrac{9}{15}\right):\left(\dfrac{40}{8}-\dfrac{7}{8}\right)\\ =\dfrac{146}{15}:\dfrac{33}{8}\\ =\dfrac{146}{15}\cdot\dfrac{8}{33}\\ =\dfrac{1168}{495}\)

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

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

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

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:

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