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2)
\(A=\dfrac{5\sqrt{a}-3}{\sqrt{a}-2}+\dfrac{3\sqrt{a}+1}{\sqrt{a}+2}-\dfrac{a^2+2\sqrt{a}+8}{a-4}\)
\(=\dfrac{\left(5\sqrt{a}-3\right)\left(\sqrt{a}+2\right)+\left(3\sqrt{a}+1\right)\left(\sqrt{a}-2\right)-a^2-2\sqrt{a}-8}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=\dfrac{5a+10\sqrt{a}-3\sqrt{a}-6+3a-6\sqrt{a}+\sqrt{a}-2-a^2-2\sqrt{a}-8}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=\dfrac{-a^2+8a-16}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}=\dfrac{-\left(a-4\right)^2}{a-4}=4-a\)
1: Ta có: \(\left\{{}\begin{matrix}3x-y=2m-1\\x+y=3m+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x=5m+1\\x+y=3m+2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5m+1}{4}\\y=3m+2-x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5m+1}{4}\\y=\dfrac{12m+8-5m-1}{4}=\dfrac{7m+7}{4}\end{matrix}\right.\)
Ta có: \(x^2+2y^2=9\)
\(\Leftrightarrow\left(\dfrac{5m+1}{4}\right)^2+2\cdot\left(\dfrac{7m+7}{4}\right)^2=9\)
\(\Leftrightarrow\dfrac{25m^2+10m+1}{16}+\dfrac{2\cdot\left(49m^2+98m+49\right)}{16}=9\)
\(\Leftrightarrow25m^2+10m+1+98m^2+196m+98-144=0\)
\(\Leftrightarrow123m^2+206m-45=0\)
Đến đây bạn tự làm nhé, chỉ cần giải phương trình bậc hai bằng delta thôi
Hàm là bậc nhất khi:
a. \(3m-2\ne0\Rightarrow m\ne\dfrac{2}{3}\)
b. \(3-m>0\Rightarrow m< 3\)
c. \(\left\{{}\begin{matrix}2m-1\ne0\\m+2\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m\ne\dfrac{1}{2}\\m\ne-2\end{matrix}\right.\)
d. \(\left\{{}\begin{matrix}m^2-4=0\\m+2\ne0\end{matrix}\right.\) \(\Rightarrow m=2\)
a: ĐKXĐ: \(m\ne\dfrac{2}{3}\)
b: ĐKXĐ: \(m< 3\)
c: ĐKXĐ: \(\left[{}\begin{matrix}m\ge\dfrac{1}{2}\\m< -2\end{matrix}\right.\)
d: ĐKXĐ: \(m=2\)
Thay \(x=-1\) vào ta được:
\(\left(-1\right)^2-\left(3m+1\right)\left(-1\right)+m-5=0\)
\(\Leftrightarrow4m-3=0\Rightarrow m=\dfrac{3}{4}\)
Để đây là hàm số bậc nhất thì \(\dfrac{m^2}{3-4m}< >0\)
=>\(m\notin\left\{0;\dfrac{3}{4}\right\}\)
Để hàm số \(y=\dfrac{m^2}{3-4m}x+3m-2\) nghịch biến trên R thì
\(\dfrac{m^2}{3-4m}< 0\)
=>3-4m<0
=>-4m<-3
=>\(m>\dfrac{3}{4}\)
a) \(P=\dfrac{3m+\sqrt{9m}-3}{m+\sqrt{m}-2}-\dfrac{\sqrt{m}-2}{\sqrt{m}-1}+\dfrac{1}{\sqrt{m}+2}-1\)
\(=\dfrac{3m+3\sqrt{m}-3}{\left(\sqrt{m}-1\right)\left(\sqrt{m}+2\right)}-\dfrac{m-4}{\left(\sqrt{m}-1\right)\left(\sqrt{m}+2\right)}+\dfrac{\sqrt{m}-1}{\left(\sqrt{m}-1\right)\left(\sqrt{m}+2\right)}-\dfrac{m+\sqrt{m}-2}{\left(\sqrt{m}-1\right)\left(\sqrt{m}+2\right)}\)
\(=\dfrac{3m+3\sqrt{m}-3-m+4+\sqrt{m}-1-m-\sqrt{m}+2}{\left(\sqrt{m}-1\right)\left(\sqrt{m}+2\right)}\)
\(=\dfrac{m+3\sqrt{m}+2}{\left(\sqrt{m}-1\right)\left(\sqrt{m}+2\right)}=\dfrac{\left(\sqrt{m}+1\right)\left(\sqrt{m}+2\right)}{\left(\sqrt{m}-1\right)\left(\sqrt{m}+2\right)}=\dfrac{\sqrt{m}+1}{\sqrt{m}-1}\)
b) Đk: \(\left\{{}\begin{matrix}m\ge0\\m\ne1\end{matrix}\right.\)
\(\left|P\right|=2\Leftrightarrow\left|\dfrac{\sqrt{m}+1}{\sqrt{m}-1}\right|=2\Leftrightarrow\left[{}\begin{matrix}\dfrac{\sqrt{m}+1}{\sqrt{m}-1}=-2\\\dfrac{\sqrt{m}+1}{\sqrt{m}-1}=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{m}+1=-2\sqrt{m}+2\\\sqrt{m}+1=2\sqrt{m}-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}3\sqrt{m}=1\\\sqrt{m}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=\dfrac{1}{9}\left(N\right)\\m=9\left(N\right)\end{matrix}\right.\)
c) \(P=\dfrac{\sqrt{m}+1}{\sqrt{m}-1}=\dfrac{\sqrt{m}-1+2}{\sqrt{m}-1}=1+\dfrac{2}{\sqrt{m}-1}\)
\(P\in N\Rightarrow\dfrac{2}{\sqrt{m}-1}\in Z\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{m}-1=-2\\\sqrt{m}-1=-1\\\sqrt{m}-1=1\\\sqrt{m}-1=2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{m}=-1\left(VN\right)\\\sqrt{m}=0\left(1\right)\\\sqrt{m}=2\left(VN,m\ne N\right)\\\sqrt{m}=3\left(2\right)\end{matrix}\right.\)
(1) \(\Leftrightarrow m=0\left(loại,P\notin N\right)\)
(2) \(\Leftrightarrow m=9\left(N\right)\)
Kl: a) \(P=\dfrac{\sqrt{m}+1}{\sqrt{m}-1}\)
b) m=1/9 , m = 9
c) m =9
Để hàm số đã cho là bậc nhất thì:
a/ \(m-\sqrt{3}\ne0\Rightarrow m\ne\sqrt{3}\)
b/ \(m-5>0\Rightarrow m>5\)
c/ \(m^2+m\ne0\Rightarrow\left\{{}\begin{matrix}m\ne0\\m\ne-1\end{matrix}\right.\)
d/ \(\left\{{}\begin{matrix}m^2-4m+3=0\\m^2-6m+5\ne0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m=1\\m=3\end{matrix}\right.\\\left[{}\begin{matrix}m\ne1\\m\ne5\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow m=3\)
a: Để C là số nguyên thì \(m^2-2m-m+2-5⋮m-2\)
\(\Leftrightarrow m-2\in\left\{1;-1;5;-5\right\}\)
hay \(m\in\left\{3;1;7;-3\right\}\)
c: Để E là số nguyên thì \(m+2⋮m^2-1\)
\(\Leftrightarrow m^2-1-3⋮m^2-1\)
\(\Leftrightarrow m^2-1\in\left\{1;-1;3;-3\right\}\)
hay \(m\in\left\{\sqrt{2};-\sqrt{2};0;2;-2\right\}\)
d: Để G là số nguyên thì \(3m+2⋮m^2-1\)
\(\Leftrightarrow9m^2-4⋮m^2-1\)
\(\Leftrightarrow m^2-1\in\left\{1;-1;5;-5\right\}\)
hay \(m\in\left\{\sqrt{2};-\sqrt{2};0;\sqrt{6};-\sqrt{6}\right\}\)