\(a)\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\left(x\ne\pm\dfrac{1}{3}\right)\\ \Leftrightarrow\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}=\dfrac{\left(1-3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}-\dfrac{\left(1+3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}\\ \Leftrightarrow12=\left(1-3x\right)^2-\left(1+3x\right)^2\\ \Leftrightarrow1-6x+9x^2-\left(1+6x+9x^2\right)=12\\ \Leftrightarrow1-6x+9x^2-1-6x-9x^2=12\\ \Leftrightarrow-12x=12\\ \Leftrightarrow x=-1\left(tm\right)\)

\(b)\dfrac{2}{x^2-4}-\dfrac{x-1}{x\left(x-2\right)}+\dfrac{x-4}{x\left(x+2\right)}=0\left(x\ne0;\pm2\right)\\ \Leftrightarrow\dfrac{2x}{x\left(x-2\right)\left(x+2\right)}-\dfrac{\left(x-1\right)\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-4\right)\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}=0\\ \Leftrightarrow2x-\left(x-1\right)\left(x+2\right)+\left(x-4\right)\left(x-2\right)=0\\ \Leftrightarrow2x-\left(x^2+2x-x-2\right)+\left(x^2-2x-4x+8\right)=0\\ \Leftrightarrow2x-x^2-x+2+x^2-6x+8=0\\ \Leftrightarrow-5x+10=0\\ \Leftrightarrow-5x=-10\\ \Leftrightarrow x=2\left(ktm\right)\)

c) 

\(\dfrac{16}{x^2-16}+\dfrac{2}{x+4}-\dfrac{1}{x-4}=0\left(x\ne\pm4\right)\\ \Leftrightarrow\dfrac{16}{\left(x+4\right)\left(x-4\right)}+\dfrac{2\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}-\dfrac{x+4}{\left(x+4\right)\left(x-4\right)}=0\\ \Leftrightarrow16+2\left(x-4\right)-\left(x-4\right)=0\\ \Leftrightarrow16+2x-8-x+4=0\\ \Leftrightarrow x+12=0\\ \Leftrightarrow x=-12\left(tm\right)\)

9: \(\int\dfrac{x^2-x+4}{x^3-3x^2+2x}dx\)

\(=\int\dfrac{\left(x-1\right)\cdot x+4}{x\left(x-1\right)\left(x-2\right)}dx=\int\left(\dfrac{1}{x-2}+\dfrac{4}{x\left(x-1\right)\left(x-2\right)}\right)dx\)

\(=\dfrac{1}{\left|x-2\right|}+\int\dfrac{4}{x\left(x-1\right)\left(x-2\right)}dx\)

Đặt \(\dfrac{4}{x\left(x-1\right)\left(x-2\right)}=\dfrac{A}{x}+\dfrac{B}{x-1}+\dfrac{C}{x-2}\)

=>\(\dfrac{4}{x\left(x-1\right)\left(x-2\right)}=\dfrac{A\left(x^2-3x+2\right)+B\left(x^2-2x\right)+C\left(x^2-x\right)}{x\left(x-1\right)\left(x-2\right)}\)

=>\(x^2\left(A+B+C\right)+x\left(-3A-2B-C\right)+2A=4\)

=>\(\left\{{}\begin{matrix}A+B+C=0\\-3A-2B-C=0\\2A=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}A=2\\B+C=-A=-2\\3A+2B+C=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}A=2\\B+C=-2\\2B+C=-3A=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}A=2\\B=-4\\C=2\end{matrix}\right.\)

=>\(\dfrac{4}{x\left(x-1\right)\left(x-2\right)}=\dfrac{2}{x}+\dfrac{-4}{x-1}+\dfrac{2}{x-2}\)

=>\(\int\dfrac{4}{x\left(x-1\right)\left(x-2\right)}dx=\dfrac{2}{\left|x\right|}+\dfrac{-4}{\left|x-1\right|}+\dfrac{2}{\left|x-2\right|}\)

=>\(\int\dfrac{x^2-x+4}{x^3-3x^2+2x}dx=\dfrac{1}{\left|x-2\right|}+\dfrac{2}{\left|x\right|}+\dfrac{-4}{\left|x-1\right|}+\dfrac{2}{\left|x-2\right|}=\dfrac{3}{\left|x-2\right|}+\dfrac{2}{\left|x\right|}-\dfrac{4}{\left|x-1\right|}\)

\(a)\dfrac{4x-5}{x-1}=2+\dfrac{x}{x-1}\left(x\ne1\right)\\ \Leftrightarrow\dfrac{4x-5}{x-1}-\dfrac{x}{x-1}=2\\ \Leftrightarrow\dfrac{3x-5}{x-1}=2\\ \Leftrightarrow3x-5=2\left(x-1\right)\\ \Leftrightarrow3x-5=2x-2\\ \Leftrightarrow x=-2+5\\ \Leftrightarrow x=3\left(tm\right)\)

\(b)\dfrac{7}{x+2}=\dfrac{3}{x-5}\left(x\ne-2;x\ne5\right)\\ \Leftrightarrow7\left(x-5\right)=3\left(x+2\right)\\\Leftrightarrow7x-35=3x+6\\ \Leftrightarrow 7x-3x=6+35\\ \Leftrightarrow4x=41\\ \Leftrightarrow x=\dfrac{41}{4}\left(tm\right)\)

c) 

\(\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\left(x\ne-5;x\ne0\right)\\ \Leftrightarrow\dfrac{2x+5}{2x}=\dfrac{x}{x+5}\\ \Leftrightarrow\left(2x+5\right)\left(x+5\right)=2x\cdot x\\ \Leftrightarrow2x^2+10x+5x+25=2x^2\\ \Leftrightarrow15x+25=0\\ \Leftrightarrow x=\dfrac{-25}{15}\\ \Leftrightarrow x=\dfrac{-5}{3}\left(tm\right)\)

đề bài đâu?

\(-5x^3+xy^2z^3\)có bậc 6 vì  6 > 3 

Giải:

+ Xét hạng  tử thứ nhất là: 5\(x^3\) vậy hạng tử này có bậc là 3

+ Xét hạng tử thứ hai là: \(xy^2z^3\)

     \(x\)  có bậc là 1

     y² có bậc là 2

     z³ có bậc là 3 

Vậy hạng tử \(xy^2z^3\) có bậc là: 1 + 2 + 3 = 6

+ Bậc của hạng tử \(xy^2z^3\) lớn hơn bậc của hạng tử - 5\(x^3\) nên đó là bậc của đa thức vì vậy bậc của đa thức là 6

2:

a: Xét ΔBHA vuông tại H và ΔBKC vuông tại K có

\(\widehat{HBA}\) chung

Do đó: ΔBHA~ΔBKC

=>\(\dfrac{BH}{BK}=\dfrac{BA}{BC}\)(2)

=>\(\dfrac{BH}{BA}=\dfrac{BK}{BC}\)

=>\(BH\cdot BC=BK\cdot BA\)

b: Xét ΔBHK và ΔBAC có

\(\dfrac{BH}{BA}=\dfrac{BK}{BC}\)

\(\widehat{HBK}\) chung

Do đó: ΔBHK~ΔBAC
=>\(\widehat{BHK}=\widehat{BAC}=70^0\)

c: Xét ΔBKH có BI là phân giác

nên \(\dfrac{IH}{IK}=\dfrac{BH}{BK}\left(1\right)\)

Xét ΔBAC có BD là phân giác

nên \(\dfrac{DA}{DC}=\dfrac{BA}{BC}\left(3\right)\)

Từ (1),(2),(3) suy ra \(\dfrac{IH}{IK}=\dfrac{DA}{DC}\)

=>\(IH\cdot DC=DA\cdot IK\)

Bài 2:

a) 

\(4x-2=m\left(mx-1\right)\\ \Leftrightarrow m^2x-m-4x+2=0\\ \Leftrightarrow x\left(m^2-4\right)-\left(m-2\right)=0\\ \Leftrightarrow x\left(m^2-4\right)=m-2\)

Nếu: \(m^2-4=0\Leftrightarrow m=\pm2\) 

+) m = 2 => pt trở thành 0 = 0 => pt vô số nghiệm 

+) m = -2 => pt trở thành 0 = -4 => pt vô nghiệm 

Nếu \(m^2-4\ne0\Leftrightarrow m\ne\pm2\) thì pt có nghiệm là:

\(x=\dfrac{m-2}{m^2-4}=\dfrac{1}{m+2}\) 

b) 

\(m^2x-3=4x-\left(m-1\right)\\ \Leftrightarrow m^2x-3-4x+\left(m-1\right)=0\\ \Leftrightarrow x\left(m^2-4\right)+m-4=0\\ \Leftrightarrow x\left(m^2-4\right)=4-m\)

Nếu: \(m^2-4=0\Leftrightarrow m=\pm2\) 

+) m = 2 => pt trở thành 0 = 2 => pt vô nghiệm 

+) m = -2 => pt trở thành 0 = 6 => pt vô nghiệm  

Nếu: \(m^2-4\ne0\Leftrightarrow m\ne\pm2\) thì pt có nghiệm là:

\(x=\dfrac{4-m}{m^2-4}\)

657 x 8 -1423

=5256 - 1423

=3833

\(657\times8-1423\)

\(=526-1423\)

\(=3833\)

=> E = {12; 14; 16; ... ; 100} 

657 x 8 -1423

=5256 - 1423

=3833

Bài 5:

a) Để hpt có nghiệm duy nhất thì \(\dfrac{m}{2}\ne\dfrac{2}{m}\Leftrightarrow m\ne\pm2\) 

\(\left\{{}\begin{matrix}mx+2y=m+1\\2x+my=2m-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{m-mx+1}{2}\\2x+m\cdot\dfrac{m-mx+1}{2}=2m-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{m-mx+1}{2}\\2x+\dfrac{m^2-m^2x+m}{2}=2m-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{m-mx+1}{2}\\4x+m^2-m^2x+m=4m-2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{m-mx+1}{2}\\\left(m^2-4\right)x=m^2-3m+2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{m-m\cdot\dfrac{m-1}{m+2}+1}{2}=\dfrac{\dfrac{m\left(m+2\right)-m\left(m-1\right)+m+2}{m+2}}{2}=\dfrac{2m+1}{m+2}\\x=\dfrac{m^2-3m+2}{m^2-4}=\dfrac{m-1}{m+2}\end{matrix}\right.\) 

Để x,y nguyên thì \(\dfrac{m-1}{m+2};\dfrac{2m+1}{m+2}\) phải nguyên 

+) Ta có: \(\dfrac{m-1}{m+2}=\dfrac{m+2-3}{m+2}=1-\dfrac{3}{m+2}\)

=> m + 2 ∈ Ư(3) = {1; -1; 3; -3}

=> m ∈ {-1; -3; 1; -5} (1)

+) Ta có: \(\dfrac{2m+1}{m+2}=\dfrac{2m+4-3}{m+2}=2-\dfrac{3}{m+2}\)

=> m + 2 ∈ Ư(3) = {1; -1; 3; -3}

=> m ∈ {-1; -3; 1; -5} (2) 

Từ (1) và (2) => m ∈ {1; -1; 3; -3} 

Bài 4 

a, \(\left\{{}\begin{matrix}-2\sqrt{3}x+3\sqrt{5}y=-21\\4x-2\sqrt{3}y=2\sqrt{3}\left(2+\sqrt{5}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{21-3\sqrt{5}y}{-2\sqrt{3}}\\\dfrac{4\left(21-3\sqrt{5}y\right)}{-2\sqrt{3}}-2\sqrt{3}y=2\sqrt{3}\left(2+\sqrt{5}\right)\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Rightarrow84-21\sqrt{5}y+12y=-12\left(2+\sqrt{5}\right)\)

\(\Leftrightarrow84+y\left(-21\sqrt{5}+12\right)=-24-12\sqrt{5}\Leftrightarrow y=\dfrac{-108-12\sqrt{5}}{-21\sqrt{5}+12}\)

\(\Rightarrow x=\dfrac{\dfrac{\left(21-3\sqrt{5}\right).\left(-108-12\sqrt{5}\right)}{-21\sqrt{5}+12}}{-2\sqrt{3}}\)

b, \(\left\{{}\begin{matrix}\left(x-1\right)^2+\left(y-2\right)^2=\left(x+1\right)^2+1+\left(y+1\right)^2\\\left(x-y-3\right)^2=\left(x-y-1\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2-\left(x+1\right)^2=1+\left(y+1\right)^2-\left(y-2\right)^2\\\left(x-y-3-x+y+1\right)\left(x-y-3+x-y-1\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-4x-1=-\left(2y-1\right)\\-2\left(2x-2y-4\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-4x+2y=2\\x-y-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-x+y=1\\x=y+2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-y-2+y=1\\x=y+2\end{matrix}\right.\)( vô lí ) 

Vậy hpt vô nghiệm