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b)\(\left\{{}\begin{matrix}x+y=-1+m\left(1\right)\\2x-y=2m\end{matrix}\right.\)
\(\Rightarrow3x=-1+3m\)
\(\Leftrightarrow x=\dfrac{-1+3m}{3}\)
Thay \(x=\dfrac{-1+3m}{3}\) vào (1) có:
\(\dfrac{-1+3m}{3}+y=-1+m\)\(\Leftrightarrow y=-1+m-\dfrac{-1+3m}{3}=-\dfrac{2}{3}\)
Suy ra với mọi m hệ luôn có nghiệm duy nhất \(\left(x;y\right)=\left(\dfrac{-1+3m}{3};-\dfrac{2}{3}\right)\)
\(xy=\left(\dfrac{-1+3m}{3}\right).\left(-\dfrac{2}{3}\right)=10\)
\(\Leftrightarrow m=-\dfrac{44}{3}\)
Vậy...
\(\left\{{}\begin{matrix}x+y=m-1\\2x-y=2m\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}2x+2y=2m-2\\2x-y=2m\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}3y=-2\\x=m-1-y\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}y=\dfrac{-2}{3}\\x=m-\dfrac{1}{3}\end{matrix}\right.\)
Ta có :
\(x.y=10\text{⇔}\left(m-\dfrac{1}{3}\right).\dfrac{-2}{3}=10\)
\(\text{⇔}m=\dfrac{-44}{3}\)
Câu 2:
Ta có: \(x^2-2\left(m+1\right)x+m^2+4=0\)
\(\text{Δ}=\left(2m+2\right)^2-4\left(m^2+4\right)\)
\(=4m^2+8m+4-4m^2-16\)
\(=8m-12\)
Để phương trình có hai nghiệm phân biệt thì Δ>0
\(\Leftrightarrow8m>12\)
hay \(m>\dfrac{3}{2}\)
Áp dụng hệ thức Vi-et, ta được:
\(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)=2m+2\\x_1x_2=m^2+4\end{matrix}\right.\)
Vì x1 là nghiệm của phương trình nên ta có: \(x_1^2-2\left(m+1\right)x_1+m^2+4=0\)
\(\Leftrightarrow x_1^2=2\left(m+1\right)x_1-m^2-4\)
Ta có: \(x_1^2+2x_2\left(m+1\right)=2m^2+20\)
\(\Leftrightarrow2\left(m+1\right)\cdot x_1-m^2-4+2x_2\left(m+1\right)-2m^2-20=0\)
\(\Leftrightarrow2\left(m+1\right)\left(x_1+x_2\right)-3m^2-24=0\)
\(\Leftrightarrow4\left(m+1\right)\cdot\left(m+1\right)-3m^2-24=0\)
\(\Leftrightarrow4m^2+8m+4-3m^2-24=0\)
\(\Leftrightarrow m^2+8m-20=0\)
\(\Leftrightarrow\left(m+10\right)\left(m-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}m=-10\left(loại\right)\\m=2\left(nhận\right)\end{matrix}\right.\)
1: \(P=\left(\dfrac{\sqrt{x-2}\left(3-\sqrt{x-2}\right)}{9-x+2}+\dfrac{x+7}{11-x}\right):\left(\dfrac{3\sqrt{x-2}+1-\sqrt{x-2}+3}{\sqrt{x-2}\left(\sqrt{x-2}-3\right)}\right)\)
\(=\dfrac{3\sqrt{x-2}-x+2+x+7}{11-x}:\dfrac{2\sqrt{x-2}+4}{\sqrt{x-2}\left(\sqrt{x-2}-3\right)}\)
\(=\dfrac{3\sqrt{x-2}+9}{11-x}\cdot\dfrac{\sqrt{x-2}\left(\sqrt{x-2}-3\right)}{2\sqrt{x-2}+4}\)
\(=\dfrac{-3\left(\sqrt{x-2}+3\right)}{2\left(\sqrt{x-2}+2\right)}\cdot\dfrac{\sqrt{x-2}}{\sqrt{x-2}+3}\)
\(=\dfrac{-3\sqrt{x-2}}{2\sqrt{x-2}+4}\)
2: Đặt căn x-2=a(a>=0)
=>P=-3a/(2a+4)
P nguyên
=>-3a chia hết cho 2a+4
=>-6a chia hết cho 2a+4
=>-6a-12+12 chia hết cho 2a+4
=>2a+4 thuộc {1;-1;2;-2;3;-3;4;-4;6;-6;12;-12}
=>a thuộc {1;4}
=>x-2=1 hoặc x-2=16
=>x=3 hoặc x=18
\(x^4+3x^2=0\)
Có \(x^4\ge0;\forall x\); \(3x^2\ge0;\forall x\)
=> VT\(\ge0;\forall x\)
Dấu = xảy ra <=> x=0
Ý C
9: Ta có: \(\left(1+\sqrt{2}-\sqrt{3}\right)\left(1+\sqrt{2}+\sqrt{3}\right)\)
\(=\left(\sqrt{2}+1\right)^2-3\)
\(=3+2\sqrt{2}-3=2\sqrt{2}\)
10: Ta có: \(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{6}-\sqrt{2}}+2\sqrt{10}\)
\(=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{2}\left(\sqrt{3}-1\right)}+2\sqrt{10}\)
\(=\dfrac{\sqrt{10}}{2}+\dfrac{4\sqrt{10}}{2}=\dfrac{5\sqrt{10}}{2}\)
13)\(\dfrac{2\sqrt{10}+\sqrt{30}-2\sqrt{2}-\sqrt{6}}{2\sqrt{10}-2\sqrt{2}}=\dfrac{\sqrt{10}\left(2+\sqrt{3}\right)-\sqrt{2}\left(2+\sqrt{3}\right)}{2\left(\sqrt{10}-\sqrt{2}\right)}\)\(=\dfrac{\left(\sqrt{10}-\sqrt{2}\right)\left(2+\sqrt{3}\right)}{2\left(\sqrt{10}-\sqrt{2}\right)}=\dfrac{2+\sqrt{3}}{2}\)
14)sai đề? phải là \(\sqrt{3-\sqrt{5}}\)
\(=\dfrac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{2\sqrt{10}-2\sqrt{2}}=\dfrac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{2}\left(2\sqrt{10}-2\sqrt{2}\right)}\)
\(=\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}\left(3+\sqrt{5}\right)}{4\left(\sqrt{5}-1\right)}=\dfrac{\left|\sqrt{5}-1\right|\left(3+\sqrt{5}\right)}{4\left(\sqrt{5}-1\right)}\)
\(=\dfrac{3+\sqrt{5}}{4}\)
15)\(\sqrt{\left(1-\sqrt{2016}\right)^2}.\sqrt{2017+2\sqrt{2016}}=\left|1-\sqrt{2016}\right|\sqrt{1+2\sqrt{2016}+2016}\)
\(=\left(\sqrt{2016}-1\right)\sqrt{\left(1+\sqrt{2016}\right)^2}=\left(\sqrt{2016}-1\right)\left(1+\sqrt{2016}\right)\)
\(=2015\)