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

b) 

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

\(\left\{{}\begin{matrix}x^2+y^2=2\\3x+3y+x^2y+xy^2=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-2xy=2\\3\left(x+y\right)+xy\left(x+y\right)=8\end{matrix}\right.\)

Đặt: \(\left\{{}\begin{matrix}x+y=S\\xy=P\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}S^2-2P=2\\3S+SP=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left(\dfrac{8}{P+3}\right)^2-2P=2\\S=\dfrac{8}{P+3}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{64}{P^2+6P+9}-2P=2\\S=\dfrac{8}{P+3}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}64-2P\left(P^2+6P+9\right)=2\left(P^2+6P+9\right)\\S=\dfrac{8}{P+3}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2P^3+14P^2+30P-46=0\\S=\dfrac{8}{P+3}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left(P-1\right)\left(2P^2+16P+46\right)=0\\S=\dfrac{8}{P+3}\end{matrix}\right.\) 

Ta có: \(P^2+8P+46=\left(P^2+8P+16\right)+30=\left(P+4\right)^2+30>0\forall P\) 

=> \(P-1=0\Leftrightarrow P=1\) 

\(\Rightarrow S=\dfrac{8}{1+3}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x+y=2\\xy=1\end{matrix}\right.\)

Khi đó x,y là nghiệm của pt: \(X^2-2X+1=0\\ \Leftrightarrow\left(X-1\right)^2=0\\ \Leftrightarrow X=1\)

Vậy x =1 và y = 1

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

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

đk x khác 1/2 

\(\dfrac{4x+3}{2x-1}-\dfrac{1}{2}\le0\Leftrightarrow\dfrac{8x+6-2x+1}{2\left(2x-1\right)}\le0\Leftrightarrow\dfrac{6x+7}{2\left(2x-1\right)}\le0\)

TH1 \(\left\{{}\begin{matrix}6x+7\ge0\\2\left(2x-1\right)\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{7}{6}\\x\le\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow-\dfrac{7}{6}\le x\le\dfrac{1}{2}\)

TH2 : \(\left\{{}\begin{matrix}x\le-\dfrac{7}{6}\\x\ge\dfrac{1}{2}\end{matrix}\right.\)loại 

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

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

c) 

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

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

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

c) 

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

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:

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

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 

Kẻ đường cao BD của tam giác ABC \(\left(D\in AC\right)\)

Khi đó \(AD=AB.cosA=c.cosA\)

\(BD=AB.sinA=c\sqrt{1-cos^2A}\)

\(CD=AC-AD=b-c.cosA\)

Tam giác BCD vuông tại D

\(\Rightarrow BC^2=CD^2+BD^2\)

\(\Leftrightarrow a^2=\left(b-c.cosA\right)^2+\left(c\sqrt{1-cos^2A}\right)^2\)

\(\Leftrightarrow a^2=b^2-2bc.cosA+c^2.cos^2A+c^2\left(1-cos^2A\right)\)

\(\Leftrightarrow a^2=b^2+c^2-2bc.cosA\)

Ta có đpcm.