A B C M N I K

a, Xét tam giác ABC ta co : 

M là trung điểm AB 

N là trung điểm AC

=> MN là đường trung bình tam giác ABC 

=> MN // BC và MN = 1/2 BC 

=> BMNC là hình bình hành 

b, Vì  AK cắt BC tại K

Mà MN // BC => AK cắt MN tại I 

=> MI = NI ( I là trung điểm )

=> AMKN là hình bình hành 

=>  AI = IK 

\(-ab\left(-bc\right)\left(-ac\right)=-\left(abc\right)^2\)

Ta có:\(a+b=2\)\(\Rightarrow a=2-b\)

Có:\(a.b=-2\)

\(\Rightarrow\left(2-b\right).b=-2\)

\(\Rightarrow2b-b^2=-2\)

\(\Rightarrow2b-b^2+2=0\)

\(\Rightarrow b^2-2b-2=0\)

\(\Rightarrow b^2-2b+1-3=0\)

\(\Rightarrow b^2-2b+1=3\)

\(\Rightarrow\left(b-1\right)^2=3\)

\(\Rightarrow\orbr{\begin{cases}b-1=\sqrt{3}\\b-1=-\sqrt{3}\end{cases}\Rightarrow\orbr{\begin{cases}b=\sqrt{3}+1\\b=-\sqrt{3}+1\end{cases}}}\)

\(\Rightarrow\orbr{\begin{cases}a=2-b=2-\left(\sqrt{3}+1\right)=1-\sqrt{3}\\a=2-b=2-\left(-\sqrt{3}+1\right)=1+\sqrt{3}\end{cases}}\)

Vậy \(\left(a;b\right)=\orbr{\begin{cases}\left(1-\sqrt{3};1+\sqrt{3}\right)\\\left(1+\sqrt{3};1-\sqrt{3}\right)\end{cases}}\)

\(3x-\frac{3x^2}{x+3}=\frac{4x}{x+3}+\frac{2}{5}\)

⇔ \(3x-\frac{3x^2}{x+3}-\frac{4x}{x+3}-\frac{2}{5}=0\)

⇔ \(\frac{15x\left(x+3\right)}{5\left(x+3\right)}-\frac{15x^2}{5\left(x+3\right)}-\frac{20x}{5\left(x+3\right)}-\frac{2\left(x+3\right)}{5\left(x+3\right)}=0\)

⇔ \(\frac{15x^2+45x}{5\left(x+3\right)}-\frac{15x^2}{5\left(x+3\right)}-\frac{20x}{5\left(x+3\right)}-\frac{2x+6}{5\left(x+3\right)}=0\)

⇔ \(\frac{15x^2+45x-15x^2-20x-2x-6}{5\left(x+3\right)}=0\)

⇔ \(\frac{23x-6}{5\left(x+3\right)}=0\)

⇔ 23x - 6 = 0

⇔ x = 6/23

\(3x-\frac{3x^2}{x+3}=\frac{4x}{x+3}+\frac{2}{5}\)

\(\Leftrightarrow\frac{3x\left(x+3\right)}{x+3}-\frac{3x^2}{x+3}=\frac{20x}{5\left(x+3\right)}+\frac{2\left(x+3\right)}{5\left(x+3\right)}\)

\(\Leftrightarrow\frac{3x^2+9x}{x+3}-\frac{3x^2}{x+3}=\frac{20x}{5\left(x+3\right)}+\frac{2x+6}{5\left(x+3\right)}\)

\(\Leftrightarrow\frac{3x^2+9x-3x^2}{x+3}=\frac{20x+2x+6}{5\left(x+3\right)}\)

\(\Leftrightarrow\frac{9x}{x+3}=\frac{22x+6}{5\left(x+3\right)}\)

\(\Leftrightarrow\frac{9x}{x+3}-\frac{22x-6}{5\left(x+3\right)}=0\)

\(\Leftrightarrow\frac{45x}{5\left(x+3\right)}-\frac{22x+6}{5\left(x+3\right)}=0\)

\(\Leftrightarrow\frac{45x-22x-6}{5\left(x+3\right)}=0\)

\(\Leftrightarrow\frac{23x-6}{5\left(x+3\right)}=0\)

\(\Leftrightarrow23x-6=0\)

\(\Leftrightarrow23x=6\)

\(\Leftrightarrow x=\frac{6}{23}\)

A B C D M N O

Xét tứ giác AMND có góc \(A=D=M=90^0\), do đó AMND là hình chữ nhật.

do AMND là hình chữ nhật nên \(AM=ND=NC\) mà AM//NC

do đó AMCN là hình bình hành

do đó AC cắt MN tại trung điểm của mỗi đường, do đó ta có đpcm

rồi tại sao am song song với nc

cái này nó hơi khó 1 tí nên chú ý chút khác lên lever :>

a, \(A=\left(\frac{4x}{x^2+2x}+\frac{2}{x-2}-\frac{6-5x}{4-x^2}\right):\frac{x+1}{x-2}\)ĐK : x khác 0 ; 2 ; -2

\(=\left(\frac{4x}{x\left(x+2\right)}+\frac{2}{x-2}-\frac{6-5x}{\left(2-x\right)\left(x+2\right)}\right):\frac{x+1}{x-2}\)

\(=\left(\frac{4x\left(x-2\right)}{MTC}+\frac{2x\left(x+2\right)}{MTC}+\frac{\left(6-5x\right)x}{MTC}\right):\frac{x+1}{x-2}\)

\(=\left(\frac{4x^2-8x+2x^2+4x+6x-5x^2}{MTC}\right):\frac{x+1}{x-2}\)

\(=\frac{x^2+2x}{x\left(x+2\right)\left(x-2\right)}.\frac{x-2}{x+1}=\frac{1}{x+1}\)

b, Ta có : \(x^2-2x=8\Leftrightarrow x^2-2x-8=0\)

\(\left(x-4\right)\left(x+2\right)=0\)<=> \(x=4;-2\)

TH1 : Thay x = 4 ta được : \(\frac{1}{4+1}=\frac{1}{5}\)

TH2 : Thay x = -2 ta được : ( ktmđkxđ ) 

\(A=\left(\frac{4x}{x^2+2x}+\frac{2}{x-2}-\frac{6-5x}{4-x^2}\right)\div\frac{x+1}{x-2}\)

a)\(=\left(\frac{4x}{x\left(x+2\right)}+\frac{2}{x-2}+\frac{6-5x}{x^2-4}\right)\times\frac{x-2}{x+1}\)

\(=\left(\frac{4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{6-5x}{\left(x-2\right)\left(x+2\right)}\right)\times\frac{x-2}{x+1}\)

\(=\left(\frac{4x-8+2x+4+6-5x}{\left(x-2\right)\left(x+2\right)}\right)\times\frac{x-2}{x+1}\)

\(=\frac{x+2}{\left(x-2\right)\left(x+2\right)}\times\frac{x-2}{x+1}\)

\(=\frac{1}{x+1}\)

b) x² - 2x = 8

<=> x² - 2x - 8 = 0

<=> x² - 4x + 2x - 8 = 0

<=> x( x - 4 ) + 2( x - 4 ) = 0

<=> ( x - 4 )( x + 2 ) = 0

<=> x = 4 ( tm ) hoặc x = -2 ( ktm )

Với x = 4 ( tm ) => A = 1/5

Với x = -2 ( ktm ) => A không xác định

\(\frac{x^2+6x}{x-7}\div\frac{x^2-36}{x^2-14x+49}\)

\(=\frac{x\left(x+6\right)}{x-7}\times\frac{\left(x-7\right)^2}{\left(x-6\right)\left(x+6\right)}\)

\(=\frac{x\left(x-7\right)}{x-6}\)

\(\frac{x^2+6x}{x-7}\div\frac{x^2-36}{x^2-14x+49}\)

\(=\frac{x^2+6x}{x-7}.\frac{x^2-14x+49}{x^2-36}\)

\(=\frac{x\left(x+6\right)\left(x-7\right)^2}{\left(x-7\right)\left(x-6\right)\left(x+6\right)}\)

\(=\frac{x\left(x-7\right)}{x-6}\)