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: \(\dfrac{a}{2}=b=c=d=e\)

\(P=\dfrac{2a+4}{a\sqrt{a}-1}+\dfrac{\sqrt{a}+2}{a+\sqrt{a}+1}-\dfrac{2}{\sqrt{a}-1}\left(a\ne1;a\ge0\right)\\ =\dfrac{2a+4}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}+\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}-\dfrac{2\left(a+\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\\ =\dfrac{2a+4+\left(a+2\sqrt{a}-\sqrt{a}-2\right)-2\left(a+\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\\ =\dfrac{2a+4+a+\sqrt{a}-2-2a-2\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\\ =\dfrac{a-\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\\ =\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\\ =\dfrac{\sqrt{a}}{a+\sqrt{a}+1}\) 

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

\(\Rightarrow P=\dfrac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\left(3-2\sqrt{2}\right)+\sqrt{\left(\sqrt{2}-1\right)^2}+1}=\dfrac{\sqrt{2}-1}{3-2\sqrt{2}+\sqrt{2}-1+1}=\dfrac{\sqrt{2}-1}{3-\sqrt{2}}=\dfrac{2\sqrt{2}-1}{7}\)

Tính giá trị của P khi A=3-2 căn2

\(a)\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\\ =\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{2}}-\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\\ =\dfrac{\sqrt{1^2+2\cdot1\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}\right)^2-2\cdot\sqrt{3}\cdot1+1^2}}{\sqrt{2}}\\ =\dfrac{\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}}\\ =\dfrac{\sqrt{3}+1-\sqrt{3}+1}{\sqrt{2}}\\ =\dfrac{2}{\sqrt{2}}\\ =\sqrt{2}\) 

b) 

\(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}\\ =\dfrac{\sqrt{6+2\sqrt{5}}}{\sqrt{2}}-\dfrac{\sqrt{6-2\sqrt{5}}}{\sqrt{2}}-\sqrt{2}\\ =\dfrac{\sqrt{\left(\sqrt{5}\right)^2+2\cdot\sqrt{5}\cdot1+1^2}}{\sqrt{2}}-\dfrac{\sqrt{\left(\sqrt{5}\right)^2-2\cdot\sqrt{5}\cdot1+1^2}}{\sqrt{2}}-\sqrt{2}\\ =\dfrac{\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{2}}-\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{2}}-\sqrt{2}\\ =\dfrac{\sqrt{5}+1-\sqrt{5}+1-2}{\sqrt{2}}\\ =\dfrac{0}{\sqrt{2}}\\ =0\)

\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{45}\\\dfrac{y}{2}-\dfrac{x}{2}=28\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{45}\\\dfrac{y}{2}=\dfrac{x}{2}+28\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{x+56}=\dfrac{1}{45}\\y=x+56\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}45\left(x+56\right)+45x=x\left(x+56\right)\\y=x+56\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}90x+2520=x^2+56x\\y=x+56\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x^2-34x-2520=0\\y=x+56\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=70\\x=-36\end{matrix}\right.\\y=x+56\end{matrix}\right.\)

Khi x = 70 => y = 70 + 56 = 126

Khi x = -36 => y = (-36) + 56 = 20

Sửa đề: B là giao điểm có hoành độ dương của (P) và (d)

Phương trình hoành độ giao điểm của (P) và (d):

−x² = x − 2

x² + x − 2 = 0

x² − x + 2x − 2 = 0
(x² − x) + (2x − 2) = 0

x(x − 1) + 2(x− 1) = 0

(x − 1)(x + 2) = 0

x − 1 = 0 hoặc x + 2 = 0

*) x − 1 = 0

x = 1

y = −1² = −1

B(1; −1)

*) x + 2 = 0

x = −2

y = −(−2)² = −4

A(−2; −4)

* Phương trình đường thẳng OB:

Gọi (d'): y = ax + b là phương trình đường thẳng OB

Do (d') đi qua O nên b = 0

=> (d'): y = ax

Do (d') đi qua B(1; −1) nên:

a = −1

=> (d'): y = −x

Gọi (d''): y = a'x + b' là đường thẳng đi qua A(−2; −4)

Do (d'') // (d') nên a' = −1

=> (d''): y = −x + b

Do (d'') đi qua A(−2; −4) nên:

−(−2) + b = −4

b = −4 − 2

b = −6

=> (d''): y = −x − 6