Ta có: Mx//Aa => \(\widehat{AMx}=\widehat{MAa}=40^0\)( 2 góc so le trong)(1)

Ta có Mx//Aa => \(\widehat{BMx}=\widehat{MBb}=50^0\)( 2 góc so le trong)(2)

Từ (1),(2) \(\Rightarrow\widehat{AMB}=\widehat{AMx}+\widehat{BMx}=40^0+50^0=90^0\Rightarrow MA\perp MB\)

\(x^4+4=5x^2\Rightarrow x^4-5x^2+4=0\Rightarrow\left(x-2\right)\left(x-1\right)\left(x+1\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=1\\x=-1\\x=-2\end{matrix}\right.\)

a) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\left(1\right)=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-15=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)

Đặt \(t=x^2+5x+4\)

(1) trở thành: \(t\left(t+2\right)-15=t^2+2t+1-16=\left(t+1\right)^2-4^2=\left(t-3\right)\left(t+5\right)\)

Thay t: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15=\left(x^2+5x+4-3\right)\left(x^2+5x+4+5\right)=\left(x^2+5x+1\right)\left(x^2+5x+9\right)\)

b) \(\left(2x+5\right)^2-\left(x-9\right)^2=\left(2x+5-x+9\right)\left(2x+5+x-9\right)=\left(x+14\right)\left(3x-4\right)\)

a)\(\dfrac{3\left(x-y\right)^4+2\left(x-y\right)^3-5\left(x-y\right)^2}{\left(y-x\right)^2}=\dfrac{\left(x-y\right)^2\left[3\left(x-y\right)^2+2\left(x-y\right)-5\right]}{\left(x-y\right)^2}=3x^2-6xy+3y^2+2x-2y-5\)

b) \(\dfrac{\left(x-2y\right)^3}{x^2-4xy+4y^2}=\dfrac{\left(x-2y\right)^3}{\left(x-2y\right)^2}=x-2y\)

c) \(\dfrac{x^3+y^3}{x+y}=\dfrac{\left(x+y\right)\left(x^2-xy+y^2\right)}{x+y}=x^2-xy+y^2\)

10) \(3H_2SO_4+2Al\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

11) \(2K+2H_2O\rightarrow2KOH+H_2\)

12) \(Ca+2H_2O\rightarrow Ca\left(OH\right)_2+H_2\)

13) \(Na_2SO_3+2HCl\rightarrow2NaCl+SO_2+H_2O\)

14) \(CH_4+2O_2\rightarrow CO_2+2H_2O\)

15) \(Na_2O+H_2O\rightarrow2NaOH\)

3) \(2Fe\left(OH\right)_3\rightarrow Fe_2O_3+3H_2O\)

4) \(Mg+CuSO_4\rightarrow MgSO_4+Cu\)

5) \(2SO_2+O_2\rightarrow2SO_3\)

6) \(CuO+2HCl\rightarrow CuCl_2+H_2O\)

7) \(BaCl_2+2AgNO_3\rightarrow Ba\left(NO_3\right)_2+2AgCl\)

8) \(2Al+3Cl_2\rightarrow2AlCl_3\)

9) \(2FeCl_2+Cl_2\rightarrow2FeCl_3\)

a) \(\dfrac{2x+1}{x-2}=3\Rightarrow2x+1=3x-6\Rightarrow x=7\)

b) \(\dfrac{2x-3}{x+1}=\dfrac{1}{2}\Rightarrow4x-6=x+1\Rightarrow3x=7\Rightarrow x=\dfrac{7}{3}\)

\(\left(x-2\right)^2-\left(x+3\right)^2-4\left(x+1\right)=5\Rightarrow x^2-4x+4-x^2-6x-9-4x-4=5\Rightarrow-14x=14\Rightarrow x=-1\)

Bài 1:

1c,2d,3b

a) \(\widehat{BAE}+\widehat{ABF}=96^0+84^0=180^0\)

Mà 2 góc là là 2 góc trong cùng phía

=> AE//BF

b)\(\widehat{EFB}=\widehat{xFt}=55^0\)( 2 góc đối đỉnh)

Ta có: AE//BF(cmt)

\(\Rightarrow\widehat{EFB}+\widehat{AEF}=180^0\Rightarrow\widehat{AEF}=180^0-\widehat{EFB}=180^0-55^0=125^0\)