Bài 2: Hệ hai phương trình bậc nhất hai ẩn. Luyện tập

\(\Leftrightarrow-2m+2+m^2-2-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}m=3\\m=-1\end{matrix}\right.\)

\(a,M=\dfrac{x+12+\sqrt{x}-2-4\sqrt{x}-8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\\ b,\dfrac{1}{M}=\dfrac{\sqrt{x}-1+3}{\sqrt{x}-1}=1+\dfrac{3}{\sqrt{x}-1}\in Z\\ \Leftrightarrow\sqrt{x}-1\inƯ\left(3\right)=\left\{-1;1;3\right\}\left(\sqrt{x}-1\ge-1\right)\\ \Leftrightarrow\sqrt{x}\in\left\{0;2;4\right\}\\ \Leftrightarrow x\in\left\{0;16\right\}\left(x\ne4\right)\\ c,M=\dfrac{\sqrt{x}+2-3}{\sqrt{x}+2}=1-\dfrac{3}{\sqrt{x}+2}< 1\left(\dfrac{3}{\sqrt{x}+2}>0\right)\\ d,M^2=-M\Leftrightarrow M^2+M=0\Leftrightarrow M\left(M+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}M=0\\M=-1\end{matrix}\right.\)

Với \(M=0\Leftrightarrow\sqrt{x}-1=0\Leftrightarrow x=1\left(tm\right)\)

Với \(M=-1\Leftrightarrow\sqrt{x}-1=-\sqrt{x}-2\Leftrightarrow2\sqrt{x}=-1\Leftrightarrow x\in\varnothing\)

a: \(P=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)

\(=\dfrac{-3\left(\sqrt{x}+1\right)}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\)

\(A=4\sqrt{7}-3-15=4\sqrt{7}-18\\ B=\left(2\sqrt{3}+3\sqrt{3}-12\sqrt{3}\right):\sqrt{3}=-7\sqrt{3}:\sqrt{3}=-7\\ C=\sqrt{5}-\sqrt{3}-\sqrt{5}+1+\sqrt{3}=1\\ D=\left[\dfrac{\sqrt{5}\left(\sqrt{3}-2\right)}{2-\sqrt{3}}+\dfrac{\sqrt{7}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\right]\left(\sqrt{7}-\sqrt{5}\right)=\left(-\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)=-\left(7-5\right)=-2\\ E=\dfrac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\dfrac{\sqrt{3}-1}{1-\sqrt{3}}=\sqrt{3}+\sqrt{2}+1\\ F=\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{\sqrt{5}}-\dfrac{2\left(\sqrt{5}+2\right)}{1}+\sqrt{\left(2-\sqrt{5}\right)^2}=\sqrt{5}-2-2\sqrt{5}-4+2-\sqrt{5}=-4-2\sqrt{5}\)

a) H/s là bậc nhất ⇔ m+5≠0 ⇔m ≠-5

b)  H/s đồng biến ⇔ m+5> 0 ⇔ m> -5

c)  H/s đi qua A( 2,3)    ⇔  2=(m+5).2 +2m -10    ⇔ 2m+ 2m +10 -10 =2

                                     ⇔ m= \(\dfrac{1}{2}\)

d) H/s cắt trục tung tại điểm có tung độ bằng 9

⇔ x=0 thì y=9       ⇔ (m+5).0 +2m -10 =9

                             ⇔m= \(\dfrac{19}{2}\)

e) H/s đi qua điểm 10 trên trục hoành ⇔ y=0, x=10 

⇔ 0= (m+5).10 +2m -10      ⇔m= \(\dfrac{-40}{12}\)

f) h/s song song với y=2x-1 

⇔ \(\left\{{}\begin{matrix}m+5=2\\2m-10\ne-1\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}m=-3\\m\ne\dfrac{9}{2}\end{matrix}\right.\)

⇔m=-3

giải pt bằng phương pháp thế 

\(a,\Leftrightarrow\left\{{}\begin{matrix}18-4y-3y=-3\\x=9-2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=9-2y\\7y-18=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=3\end{matrix}\right.\\ b,\Leftrightarrow\left\{{}\begin{matrix}x=y+1\\2y+2+3y=38\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y+1\\5y=36\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{41}{5}\\y=\dfrac{36}{5}\end{matrix}\right.\\ c,\Leftrightarrow\left\{{}\begin{matrix}x=-y-2\\-2y-4-y=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-y-2\\-3y=-9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=3\end{matrix}\right.\)

\(ĐK:x\ne-2;y\ne4\\ HPT\Leftrightarrow\left\{{}\begin{matrix}\dfrac{9}{y-1}-\dfrac{3}{x+2}=\dfrac{9}{4}\left(1\right)\\\dfrac{5}{y-1}+\dfrac{3}{x+2}=\dfrac{29}{12}\left(2\right)\end{matrix}\right.\\ \left(1\right)+\left(2\right)=\dfrac{14}{y-1}=\dfrac{14}{3}\\ \Leftrightarrow y-1=3\\ \Leftrightarrow y=4\left(tm\right)\\ \text{Thay vào }\left(2\right)\Leftrightarrow\dfrac{5}{3}+\dfrac{3}{x+2}=\dfrac{29}{12}\\ \Leftrightarrow\dfrac{3}{x+2}=\dfrac{3}{4}\Leftrightarrow x+2=4\Leftrightarrow x=2\left(tm\right)\)

Vậy \(\left(x;y\right)=\left(2;4\right)\)

\(HPT\Leftrightarrow\left\{{}\begin{matrix}\left(m+2\right)x+6x-9=4m-1\left(1\right)\\y=2x-3\end{matrix}\right.\\ \left(1\right)\Leftrightarrow x\left(m+8\right)=4m+8\\ \Leftrightarrow x=\dfrac{4m+8}{m+8}\Leftrightarrow A=\left(2x-3\right)^2+3x^2+8x=4x^2-12x+9+3x^2+8x\\ \Leftrightarrow A=7x^2-4x+9=7\left(x^2-2\cdot\dfrac{2}{7}x+\dfrac{9}{7}\right)\\ \Leftrightarrow A=7\left(x^2-\dfrac{4}{7}x+\dfrac{16}{49}+\dfrac{47}{49}\right)\\ \Leftrightarrow A=7\left(x-\dfrac{4}{7}\right)^2+\dfrac{47}{7}\ge\dfrac{47}{7}\\ A_{min}=\dfrac{47}{7}\Leftrightarrow x=\dfrac{4}{7}\Leftrightarrow\dfrac{4m+8}{m+8}=\dfrac{4}{7}\)

Từ đó bạn giải được \(m=-1\)

Câu 1:

a: \(\Leftrightarrow\left\{{}\begin{matrix}2x-4y=2\\2x+y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-5y=-5\\x-2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=1+2y=3\end{matrix}\right.\)