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2:
a: \(A=\dfrac{x_1+x_2}{x_1x_2}=\dfrac{-6}{3}=-2\)
b: \(B=\dfrac{\left(x_1+x_2\right)^2-3x_1x_2}{1-x_1x_2}=\dfrac{36-3\cdot3}{1-3}=\dfrac{36-9}{-2}=-\dfrac{27}{2}\)
c: \(C=\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}\)
\(=\sqrt{\left(-6\right)^2-4\cdot3}=2\sqrt{6}\)
d: \(D=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)-3x_1x_2\)
\(=\left(-6\right)^3-3\cdot3\cdot\left(-6\right)-3\cdot3\)
=261
a) \(\sqrt{4x}+\sqrt{\dfrac{x}{4}}+\dfrac{1}{2}\sqrt{49x}=6\left(x\ge0\right)\)
\(\Rightarrow2\sqrt{x}+\dfrac{1}{2}\sqrt{x}+\dfrac{7}{2}\sqrt{x}=6\Rightarrow6\sqrt{x}=6\Rightarrow\sqrt{x}=1\Rightarrow x=1\)
b) ĐKXĐ: \(x\ge\dfrac{1}{2}\)
\(\sqrt{18x-9}-0,5\sqrt{2x-1}+\dfrac{1}{2}\sqrt{25\left(2x-1\right)}+\sqrt{49\left(2x-1\right)}=24\)
\(\Rightarrow\sqrt{9\left(2x-1\right)}-0,5\sqrt{2x-1}+\dfrac{5}{2}\sqrt{2x-1}+7\sqrt{2x-1}=24\)
\(\Rightarrow3\sqrt{2x-1}-0,5\sqrt{2x-1}+\dfrac{5}{2}\sqrt{2x-1}+7\sqrt{2x-1}=24\)
\(\Rightarrow12\sqrt{2x-1}=24\Rightarrow\sqrt{2x-1}=2\Rightarrow2x-1=4\Rightarrow x=\dfrac{5}{2}\)
c) \(\sqrt{x^2-2x+1}-7=0\Rightarrow\sqrt{\left(x-1\right)^2}=7\Rightarrow\left|x-1\right|=7\)
\(\Rightarrow\left[{}\begin{matrix}x-1=7\\x-1=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\\x=-6\end{matrix}\right.\)
d) \(\dfrac{1}{2}\sqrt{\dfrac{49x}{x+2}}-3\sqrt{\dfrac{x}{4x+8}}-2\sqrt{\dfrac{x}{x+2}}-\sqrt{5}=0\left(\dfrac{x}{x+2}\ge0,x\ne-2\right)\)
\(\Rightarrow\dfrac{7}{2}\sqrt{\dfrac{x}{x+2}}-3\sqrt{\dfrac{x}{4\left(x+2\right)}}-2\sqrt{\dfrac{x}{x+2}}=\sqrt{5}\)
\(\Rightarrow\dfrac{7}{2}\sqrt{\dfrac{x}{x+2}}-\dfrac{3}{2}\sqrt{\dfrac{x}{x+2}}-2\sqrt{\dfrac{x}{x+2}}=\sqrt{5}\)
\(\Rightarrow0=\sqrt{5}\) (vô lý) \(\Rightarrow\) pt vô nghiệm
a) \(\sqrt{4x}+\sqrt{\dfrac{x}{4}}+\dfrac{1}{2}\sqrt{49x}=6\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\2\sqrt{x}+\dfrac{\sqrt{x}}{2}+\dfrac{7}{2}\sqrt{x}=6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\left(2+\dfrac{1}{2}+\dfrac{7}{2}\right)=6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\6\sqrt{x}=6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt{x}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x=1\end{matrix}\right.\) \(\Leftrightarrow x=1\)
Vậy \(S=\left\{1\right\}\)
b) \(\sqrt{18x-9}-0.5\sqrt{2x-1}+\dfrac{1}{2}\sqrt{25\left(2x-1\right)}+\sqrt{49\left(2x-1\right)}=24\)
\(\Leftrightarrow3\sqrt{2x-1}-0,5\sqrt{2x-1}+\dfrac{5}{2}\sqrt{2x-1}+7\sqrt{2x-1}=24\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1\ge0\\\sqrt{2x-1}\left(3-0.5+\dfrac{5}{2}+7\right)=49\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\12\sqrt{2x-1}=24\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\\sqrt{2x-1}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\2x-1=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\) \(\Leftrightarrow x=\dfrac{5}{2}\)
Vậy \(S=\left\{\dfrac{5}{2}\right\}\)
c) \(\sqrt{x^2-2x+1}-7=0\) (*)
Ta có \(x^2-2x+1=\left(x-1\right)^2\ge0\forall x\) \(\Rightarrow\sqrt{x^2-2x+1}\ge0\forall x\)
(*) \(\Leftrightarrow\sqrt{\left(x-1\right)^2}-7=0\)
\(\Leftrightarrow\left|x-1\right|-7=0\)
\(\Leftrightarrow x-1-7=0\)
\(\Leftrightarrow x=8\)
Vậy \(S=\left\{8\right\}\)
\(\)d) \(\dfrac{1}{2}\sqrt{\dfrac{49x}{x+2}}-3\sqrt{\dfrac{x}{4x+8}}-2\sqrt{\dfrac{x}{x+2}}-\sqrt{5}=0\) (**)
\(\Leftrightarrow\dfrac{7}{2}\sqrt{\dfrac{x}{x+2}}-\dfrac{3}{2}\sqrt{\dfrac{x}{x+2}}-2\sqrt{\dfrac{x}{x+2}}=\sqrt{5}\)
ĐKXĐ: \(\dfrac{x}{x+2}\ge0\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x+2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x\le0\\x+2< 0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x>-2\end{matrix}\right.\\\left\{{}\begin{matrix}x\le0\\x< -2\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ge0\\x< -2\end{matrix}\right.\)
(**) \(\Leftrightarrow\sqrt{\dfrac{x}{x+2}}\left(\dfrac{7}{2}-\dfrac{3}{2}-2\right)=\sqrt{5}\)
\(\Leftrightarrow0\sqrt{\dfrac{x}{x+2}}=\sqrt{5}\)
\(\Leftrightarrow0=\sqrt{5}\) ( vô lý )
Vậy phương trình trên vô nghiệm
a) \(A=\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}-1\right)^2}=\left|\sqrt{5}-1\right|=\sqrt{5}-1\)
b) \(B=\sqrt{4-\sqrt{12}}=\sqrt{\left(\sqrt{3}-1\right)^2}=\left|\sqrt{3}-1\right|=\sqrt{3}-1\)
c) \(C=\sqrt{19-8\sqrt{3}}=\sqrt{\left(4-\sqrt{3}\right)^2}=\left|4-\sqrt{3}\right|=4-\sqrt{3}\)
d) \(D=\sqrt{5-2\sqrt{6}}=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\left|\sqrt{3}-\sqrt{2}\right|=\sqrt{3}-\sqrt{2}\)
a) \(A=\sqrt{6-2\sqrt{5}}=\sqrt{5}-1\)
b) \(B=\sqrt{4-\sqrt{12}}=\sqrt{3}-1\)
c) \(C=\sqrt{19-8\sqrt{3}}=4-\sqrt{3}\)
d) \(D=\sqrt{5-2\sqrt{6}}=\sqrt{3}-\sqrt{2}\)
2:
a: A=căn 3-1-2-căn 3=-3
b: =căn 3+căn 2-căn 3+căn 2=2*căn 2
d: =(căn 7/2+căn 5/2)*(căn 7-căn 5)=2/2=1
e: =3-căn 5+2căn 5+2-căn 5+2
=7
1:
a: =12/10-7/10=5/10=1/2
b: \(=\dfrac{4}{13}-\dfrac{4}{13}+\dfrac{-5}{11}-\dfrac{6}{11}=-\dfrac{11}{11}=-1\)
2:
a: x+2/7=-11/7
=>x=-11/7-2/7=-13/7
b: (x+3)/4=-7/2
=>x+3=-14
=>x=-17
\(\left(1\right)\Leftrightarrow\left(x^2-2y\right)\left(x^2+y^2+2\right)=0\)
\(\Leftrightarrow y=\frac{x^2}{2}\)
Thê vô (2) được
\(2x^2+\left(\frac{x^2}{2}\right)^2+x=14\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+2x^2+12x+28\right)=0\)
Bài 1:
Gọi vận tốc và thời gian dự định là a,b
Theo đề, ta có hệ phương trình:
\(\left\{{}\begin{matrix}\left(a+3\right)\left(b-2\right)=ab\\\left(a-3\right)\left(b+3\right)=ab\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2a+3b=6\\3a-3b=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=15\\b=6\end{matrix}\right.\)
Vậy: Chiều dài khúc sông là 90km
Bài 3 :
\(\Delta'=\left(m-1\right)^2-\left(m-2\right)\left(m+1\right)=m^2-2m+1-m^2+m+2=-m+3\)
Để pt có 2 nghiệm
\(\Delta'=3-m\ge0\Leftrightarrow m\le3\)
Theo Vi et : \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{2\left(m-1\right)}{m+1}\\x_1x_2=\dfrac{m-2}{m+1}\end{matrix}\right.\)
Thay vào ta được :
\(\dfrac{6\left(m-1\right)}{m+1}=\dfrac{5\left(m-2\right)}{m+1}\Leftrightarrow\dfrac{6m-6-5m+10}{m+1}=0\)
\(\Leftrightarrow\dfrac{m+4}{m+1}=0\Leftrightarrow m+4=0\Leftrightarrow m=-4\)(tmđk)
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