Bài 2:

a: ĐKXĐ: \(x\notin\left\{2;5\right\}\)

\(\dfrac{6x+1}{x^2-7x+10}+\dfrac{5}{x-2}=\dfrac{3}{x-5}\)

=>\(\dfrac{6x+1}{\left(x-2\right)\left(x-5\right)}+\dfrac{5\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}=\dfrac{3\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}\)

=>6x+1+5x-25=3x-6

=>11x-24=3x-6

=>8x=18

=>x=9/4(nhận)

b: ĐKXĐ: \(x\notin\left\{0;2;-2\right\}\)

\(\dfrac{2}{x^2-4}-\dfrac{x-1}{x\left(x-2\right)}+\dfrac{x-4}{x\left(x+2\right)}=0\)

=>\(\dfrac{2x}{x\left(x-2\right)\left(x+2\right)}-\dfrac{\left(x-1\right)\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-4\right)\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}=0\)

=>2x-(x-1)(x+2)+(x-4)(x-2)=0

=>\(2x-\left(x^2+x-2\right)+x^2-6x+8=0\)

=>\(x^2-4x+8-x^2-x+2=0\)

=>-5x+10=0

=>x=2(loại)

c: ĐKXĐ: \(x\notin\left\{3;-1\right\}\)

\(\dfrac{1}{3-x}-\dfrac{1}{x+1}=\dfrac{x}{x-3}-\dfrac{\left(x-1\right)^2}{x^2-2x-3}\)

=>\(\dfrac{-1}{x-3}-\dfrac{1}{x+1}-\dfrac{x}{x-3}+\dfrac{\left(x-1\right)^2}{\left(x-3\right)\left(x+1\right)}=0\)

=>\(\dfrac{\left(-1-x\right)\left(x+1\right)-x+3}{\left(x-3\right)\left(x+1\right)}+\dfrac{\left(x-1\right)^2}{\left(x-3\right)\left(x+1\right)}=0\)

=>-(x+1)^2-x+3+(x-1)²=0

=>\(-x^2-2x-1-x+3+x^2-2x+1=0\)

=>-5x+3=0

=>\(x=\dfrac{3}{5}\left(nhận\right)\)

d: ĐKXĐ: \(x\notin\left\{2;-3\right\}\)

\(\dfrac{1}{x-2}-\dfrac{6}{x+3}=\dfrac{5}{6-x^2-x}\)

=>\(\dfrac{x+3-6\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}=\dfrac{-5}{\left(x+3\right)\left(x-2\right)}\)

=>x+3-6(x-2)=-5

=>x+3-6x+12+5=0

=>-5x+20=0

=>x=4(nhận)

e: ĐKXĐ: x<>-2

\(\dfrac{2}{x+2}-\dfrac{2x^2+16}{x^3+8}=\dfrac{5}{x^2-2x+4}\)

=>\(\dfrac{2}{x+2}-\dfrac{2x^2+16}{\left(x+2\right)\left(x^2-2x+4\right)}-\dfrac{5}{x^2-2x+4}=0\)

=>\(\dfrac{2\left(x^2-2x+4\right)-2x^2-16-5x-10}{\left(x+2\right)\left(x^2-2x+4\right)}=0\)

=>\(2x^2-4x+8-2x^2-5x-26=0\)

=>-9x-18=0

=>x=-2(loại)

f: ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

\(\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{2\left(x+2\right)^2}{x^6-1}\)

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

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

=>2(x^2-1)=2(x+2)^2

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

=>\(x^2+4x+4-x^2+1=0\)

=>4x+5=0

=>\(x=-\dfrac{5}{4}\left(nhận\right)\)

Bài 3:

c:

=>\(\dfrac{x}{x-1}+\dfrac{x}{x-2}+\dfrac{x}{x-3}=\dfrac{3x-12}{x-6}\)

=>

ĐKXĐ: \(x\notin\left\{1;2;\dfrac{3\pm\sqrt{7}}{2}\right\}\)

 \(\dfrac{4}{x^2-3x+2}-\dfrac{3}{2x^2-6x+1}+1=0\)

=>\(\dfrac{4\left(2x^2-6x+1\right)-3\left(x^2-3x+2\right)}{\left(x^2-3x+2\right)\left(2x^2-6x+1\right)}=-1\)

=>\(8x^2-24x+4-3x^2+9x-6=-\left(x^2-3x+2\right)\left[2\cdot\left(x^2-3x\right)+1\right]\)

=>\(5x^2-15x-2=-\left[2\left(x^2-3x\right)^2+5\left(x^2-3x\right)+2\right]\)

=>\(5\left(x^2-3x\right)-2+2\left(x^2-3x\right)^2+5\left(x^2-3x\right)+2=0\)

=>\(2\left(x^2-3x\right)^2+10\left(x^2-3x\right)=0\)

=>\(\left(x^2-3x\right)^2+5\left(x^2-3x\right)=0\)

=>\(\left(x^2-3x\right)\left(x^2-3x+5\right)=0\)

mà \(x^2-3x+5=\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}>=\dfrac{11}{4}>0\forall x\)

nên x(x-3)=0

=>\(\left[{}\begin{matrix}x=0\left(nhận\right)\\x=3\left(nhận\right)\end{matrix}\right.\)

a:

ĐKXĐ: \(x\notin\left\{8;9;10;11\right\}\)

 \(\dfrac{8}{x-8}+\dfrac{11}{x-11}=\dfrac{9}{x-9}+\dfrac{10}{x-10}\)

=>\(\left(\dfrac{8}{x-8}+1\right)+\left(\dfrac{11}{x-11}+1\right)=\left(\dfrac{9}{x-9}+1\right)+\left(\dfrac{10}{x-10}+1\right)\)

=>\(\dfrac{x}{x-8}+\dfrac{x}{x-11}-\dfrac{x}{x-9}-\dfrac{x}{x-10}=0\)

=>\(x\left(\dfrac{1}{x-8}+\dfrac{1}{x-11}-\dfrac{1}{x-9}-\dfrac{1}{x-10}\right)=0\)

=>x=0(nhận)

b:

ĐKXĐ: \(x\notin\left\{3;4;5;6\right\}\)

 \(\dfrac{x}{x-3}-\dfrac{x}{x-5}=\dfrac{x}{x-4}-\dfrac{x}{x-6}\)

=>\(\dfrac{x\left(x-5\right)-x\left(x-3\right)}{\left(x-3\right)\left(x-5\right)}=\dfrac{x\left(x-6\right)-x\left(x-4\right)}{\left(x-4\right)\left(x-6\right)}\)

=>\(\dfrac{-2x}{\left(x-3\right)\left(x-5\right)}=\dfrac{-2x}{\left(x-4\right)\left(x-6\right)}\)

=>\(x\left(\dfrac{1}{\left(x-3\right)\left(x-5\right)}-\dfrac{1}{\left(x-4\right)\left(x-6\right)}\right)=0\)

=>\(x\cdot\dfrac{\left(x-4\right)\left(x-6\right)-\left(x-3\right)\left(x-5\right)}{\left(x-3\right)\left(x-5\right)\left(x-4\right)\left(x-6\right)}=0\)

=>\(x\left(x^2-10x+24-x^2+8x-15\right)=0\)

=>x(-2x+9)=0

=>\(\left[{}\begin{matrix}x=0\left(nhận\right)\\x=\dfrac{9}{2}\left(nhận\right)\end{matrix}\right.\)

\(A=\left(\dfrac{x+1}{x^3-1}-\dfrac{1}{x-1}\right)\left(\dfrac{x+2}{x-1}-\dfrac{1}{x}\right)\left(x\ne1;0\right)\\ =\left[\dfrac{x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right]\left[\dfrac{x\left(x+2\right)}{x\left(x-1\right)}-\dfrac{x-1}{x\left(x-1\right)}\right]\\ =\dfrac{x+1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+2x-x+1}{x\left(x-1\right)}\\ =\dfrac{-x^2}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+x+1}{x\left(x-1\right)}\\ =\dfrac{-x}{\left(x-1\right)^2}\\ =\dfrac{-x}{x^2-2x+1}\)

ĐKXĐ: \(x\notin\left\{1;0\right\}\)

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

\(=\left(\dfrac{x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{1}{x-1}\right)\cdot\left(\dfrac{x\left(x+2\right)-x+1}{x\left(x-1\right)}\right)\)

\(=\dfrac{x+1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+x+1}{x\left(x-1\right)}\)

\(=\dfrac{-x^2}{\left(x-1\right)\cdot x\left(x-1\right)}=\dfrac{-x}{\left(x-1\right)^2}\)

a: ΔBDH vuông tại D

=>\(DH^2+DB^2=BH^2\)

=>\(DH^2=6^2-3,6^2=23,04=4,8^2\)

=>DH=4,8(cm)

Xét ΔAHB vuông tại H có HD là đường cao

nên \(HD^2=DA\cdot DB\)

=>\(DA=\frac{HD^2}{DB}=\frac{4.8^2}{3,6}=6,4\left(\operatorname{cm}\right)\)

AB=AD+DB=6,4+3,6=10(cm)

Xét ΔAHB vuông tại H có \(AH^2+HB^2=AB^2\)

=>\(AH^2=10^2-6^2=64=8^2\)

=>AH=8(cm)

Xét ΔAHB vuông tại H có \(\sin B=\frac{AH}{AB}=\frac{8}{10}=\frac45\)

nên \(\hat{B}\) ≃53 độ

b: Xét ΔAHB vuông tại H có HD là đường cao

nên \(AD\cdot AB=AH^2\left(1\right)\)

Xét ΔAHC vuông tại H có HE là đường cao

nên \(AE\cdot AC=AH^2\left(2\right)\)

Từ (1),(2) suy ra \(AD\cdot AB=AE\cdot AC\)

=>\(\frac{AD}{AC}=\frac{AE}{AB}\)

XétΔADE và ΔACB có

\(\frac{AD}{AC}=\frac{AE}{AB}\)

góc DAE chung

Do đó: ΔADE~ΔACB

c: \(\cot HAC+\cot ACB=\tan ACB+\cot ACB\)

\(=\frac{AH}{HC}+\frac{HC}{AH}=\frac{AH^2+HC^2}{HC\cdot AH}=\frac{AC^2}{AH\cdot HC}\)

\(=\frac{AC^2}{HE\cdot AC}=\frac{AC}{HE}\)

d: ΔADE~ΔACB

=>\(\frac{S_{ADE}}{S_{ACB}}=\left(\frac{AD}{AC}\right)^2=\frac{AD}{AC}\cdot\frac{AE}{AB}=\frac{AD}{AB}\cdot\frac{AE}{AC}\)

\(=\frac{AH^2}{AB\cdot AB}\cdot\frac{AH^2}{AC\cdot AC}=\left(\frac{AH}{AB}\right)^2\cdot\left(\frac{AH}{AC}\right)^2=\sin^2B\cdot\sin^2C\)

=>\(S_{ADE}=S_{ABC}\cdot\sin^2B\cdot\sin^2C\)

a: Xét (O) có

CM,CA là các tiếp tuyến

Do đó: CM=CA và OC là phân giác của góc MOA

Xét (O) có

DM,DB là các tiếp tuyến

Do đó: DM=DB và OD là phân giác của góc MOB

AC+BD

=CM+MD

=CD
b: \(\widehat{COD}=\widehat{COM}+\widehat{DOM}=\dfrac{1}{2}\cdot\widehat{MOA}+\dfrac{1}{2}\cdot\widehat{MOB}\)

\(=\dfrac{1}{2}\left(\widehat{MOA}+\widehat{MOB}\right)=\dfrac{1}{2}\cdot\widehat{AOB}=90^0\)

=>ΔCOD vuông tại O

c: Xét ΔCOD vuông tại O có OM là đường cao

nên \(OM^2=MC\cdot MD\)

giúp tôi ý d với bạn ơi

Bài 4:

d: 

ĐKXĐ: \(x\notin\left\{1;-1;2;-2\right\}\)

\(\dfrac{x+4}{x-1}+\dfrac{x-4}{x+1}=\dfrac{x+8}{x-2}+\dfrac{x-8}{x+2}+6\)

=>\(\dfrac{\left(x+4\right)\left(x+1\right)+\left(x-4\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x+8\right)\left(x+2\right)+\left(x-8\right)\left(x-2\right)+6\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

=>\(\dfrac{2x^2+8}{\left(x-1\right)\left(x+1\right)}=\dfrac{2x^2+32+6x^2-24}{\left(x-2\right)\left(x+2\right)}\)

=>\(\dfrac{2x^2+8}{x^2-1}=\dfrac{8x^2+8}{x^2-4}\)

=>\(\left(2x^2+8\right)\left(x^2-4\right)=\left(8x^2+8\right)\left(x^2-1\right)\)

=>\(2x^4-32=8x^4-8\)

=>\(-6x^4=24\)

=>\(x^4=-4\left(loại\right)\)

Vậy: Phương trình vô nghiệm

c:

ĐKXĐ: \(x\notin\left\{-1;-3;-8;-10\right\}\)

 \(\dfrac{2}{x^2+4x+3}+\dfrac{5}{x^2+11x+24}+\dfrac{2}{x^2+18x+80}=\dfrac{9}{52}\)

=>\(\dfrac{2}{\left(x+1\right)\left(x+3\right)}+\dfrac{5}{\left(x+3\right)\left(x+8\right)}+\dfrac{2}{\left(x+8\right)\left(x+10\right)}=\dfrac{9}{52}\)

=>\(\dfrac{1}{x+1}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+8}+\dfrac{1}{x+8}-\dfrac{1}{x+10}=\dfrac{9}{52}\)

=>\(\dfrac{1}{x+1}-\dfrac{1}{x+10}=\dfrac{9}{52}\)

=>\(\dfrac{9}{\left(x+1\right)\left(x+10\right)}=\dfrac{9}{52}\)

=>(x+1)(x+10)=52

=>\(x^2+11x-42=0\)

=>(x+14)(x-3)=0

=>\(\left[{}\begin{matrix}x=-14\left(nhận\right)\\x=3\left(nhận\right)\end{matrix}\right.\)

b: 

ĐXKĐ: \(x\notin\left\{-2;-3;-4;-5;-6\right\}\)\(\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}=\dfrac{1}{8}\)

=>\(\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)

=>\(\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{8}\)

=>\(\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)

=>\(\dfrac{x+6-x-2}{\left(x+2\right)\left(x+6\right)}=\dfrac{1}{8}\)

=>(x+2)(x+6)=32

=>\(x^2+8x-20=0\)

=>(x+10)(x-2)=0

=>\(\left[{}\begin{matrix}x=-10\left(nhận\right)\\x=2\left(nhận\right)\end{matrix}\right.\)

a: \(\dfrac{x^2}{x^2+2x+2}+\dfrac{x^2}{x^2-2x+2}-\dfrac{4x^2-20}{x^4+4}=\dfrac{322}{65}\)

=>\(\dfrac{x^2\left(x^2-2x+2\right)+x^2\left(x^2+2x+2\right)-4x^2+20}{\left(x^2+2x+2\right)\left(x^2-2x+2\right)}=\dfrac{322}{65}\)

=>\(\dfrac{x^4-2x^3+2x^2+x^4+2x^3+2x^2-4x^2+20}{x^4+4}=\dfrac{322}{65}\)

=>\(\dfrac{2x^4+20}{x^4+4}=\dfrac{322}{65}\)

=>\(322\left(x^4+4\right)=65\left(2x^4+20\right)\)

=>\(322x^4+1288-130x^4-1300=0\)

=>\(192x^4=12\)

=>\(x^4=\dfrac{1}{16}\)

=>\(\left[{}\begin{matrix}x=\dfrac{1}{2}\left(nhận\right)\\x=-\dfrac{1}{2}\left(nhận\right)\end{matrix}\right.\)

a: \(\sqrt{14-6\sqrt{5}}=\sqrt{9-2\cdot3\cdot\sqrt{5}+5}\)

\(=\sqrt{\left(3-\sqrt{5}\right)^2}=\left|3-\sqrt{5}\right|=3-\sqrt{5}\)

b: \(\sqrt{7-4\sqrt{7}+4}\)

\(=\sqrt{\left(\sqrt{7}\right)^2-2\cdot\sqrt{7}\cdot2+2^2}\)

\(=\sqrt{\left(\sqrt{7}-2\right)^2}=\left|\sqrt{7}-2\right|=\sqrt{7}-2\)

ĐKXĐ: \(x\notin\left\{1;7\right\}\)

\(\dfrac{x-8}{x-7}=8+\dfrac{1}{1-x}\)

=>\(\dfrac{x-8}{x-7}=\dfrac{8-8x+1}{1-x}\)

=>\(\dfrac{x-8}{x-7}=\dfrac{-8x+9}{1-x}\)

=>\(\dfrac{x-8}{x-7}=\dfrac{8x-9}{x-1}\)

=>\(\left(8x-9\right)\left(x-7\right)=\left(x-8\right)\left(x-1\right)\)

=>\(8x^2-65x+63-x^2+9x-8=0\)

=>\(7x^2-56x+55=0\)

\(\text{Δ}=\left(-56\right)^2-4\cdot7\cdot55=1596>0\)

=>Phương trình có hai nghiệm phân biệt là:

\(\left[{}\begin{matrix}x=\dfrac{56-2\sqrt{399}}{2\cdot7}=\dfrac{28-\sqrt{399}}{7}\left(nhận\right)\\x=\dfrac{28+\sqrt{399}}{7}\left(nhận\right)\end{matrix}\right.\)

a: \(\sqrt{3-2\sqrt{3}+1}\)

\(=\sqrt{\left(\sqrt{3}\right)^2-2\cdot\sqrt{3}\cdot1+1^2}\)

\(=\sqrt{\left(\sqrt{3}-1\right)^2}=\left|\sqrt{3}-1\right|=\sqrt{3}-1\)

b: \(\sqrt{5-2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}\right)^2-2\cdot\sqrt{5}\cdot1+1^2}\)

\(=\sqrt{\left(\sqrt{5}-1\right)^2}=\left|\sqrt{5}-1\right|=\sqrt{5}-1\)

c: \(\sqrt{1-2\sqrt{2}+2}=\sqrt{1^2-2\cdot1\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}\)

\(=\sqrt{\left(1-\sqrt{2}\right)^2}=\left|1-\sqrt{2}\right|=\sqrt{2}-1\)

a: Xét (O) có

ΔMNQ nội tiếp

MQ là đường kính

Do đó: ΔMNQ vuông tại N

b: Xét (O) có

ΔMPQ nội tiếp

MQ là đường kính

Do đó ΔMPQ vuông tại P

=>MP\(\perp\)AQ tại P

Ta có: ΔMNQ vuông tại N

=>QN\(\perp\)AM

Xét ΔAMQ có

QN,MP là các đường cao

QN cắt MP tại H

Do đó: H là trực tâm của ΔAMQ

=>AH\(\perp\)MQ

Sửa: 

\(a^2+b^2+c^2+d^2+e^2\ge a\left(b+c+d+e\right)\\ \Leftrightarrow a^2+b^2+c^2+d^2+e^2\ge ab+ac+ad+ae\\ \Leftrightarrow\left(\dfrac{a^2}{4}+ab+b^2\right)+\left(\dfrac{a^2}{4}-ac+c^2\right)+\left(\dfrac{a^2}{4}-ad+d^2\right)+\left(\dfrac{a^2}{4}-ae+e^2\right)\ge0\\ \Leftrightarrow\left(\dfrac{a}{2}-b\right)^2+\left(\dfrac{a}{2}-c\right)^2+\left(\dfrac{a}{2}-d\right)^2+\left(\dfrac{a}{2}-e\right)^2\ge0\)  

Dấu: "=" xảy ra: \(\dfrac{a}{2}=b=c=d=e\)