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Câu 4:
a: a=1; b=-5; c=-7
Vì ac<0 nên phương trình có hai nghiệm trái dấu
b: \(x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2\)
\(=5^2-2\cdot\left(-7\right)=25+14=39\)
\(\dfrac{1}{x_1^2}+\dfrac{1}{x_2^2}=\dfrac{x_1^2+x_2^2}{\left(x_1\cdot x_2\right)^2}=\dfrac{39}{7^2}=\dfrac{39}{49}\)
Bài 2:
1.Thay m=3, ta có:
\(\left\{{}\begin{matrix}3x+2y=5\\2x+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=1\end{matrix}\right.\)
Bài 1:
\(\left\{{}\begin{matrix}\left|x+1\right|+\left|y-1\right|=5\\\left|x+1\right|-4y=-4\end{matrix}\right.\)
\(\Rightarrow\left|y-1\right|-4y=9\)\(\Leftrightarrow\left[{}\begin{matrix}y=-3,\left(3\right)\left(KTM\right)\left(ĐK:y\ge1\right)\\y=-1,6\left(TM\right)\left(ĐK:y< 1\right)\end{matrix}\right.\)
Thay y=-1,6 vào hpt, ta được:
\(\left\{{}\begin{matrix}\left|x+1\right|=2,4\\\left|x+1\right|=-10,4\left(vl\right)\end{matrix}\right.\)
Vậy pt vô nghiệm.
1.
a, \(\left\{{}\begin{matrix}2x-3y=3\\-4x=3x-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-3y=3\\-4x-3x=13\end{matrix}\right.\)\(\left\{{}\begin{matrix}-4x+6y=-6\\-4x-3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9y=-19\\-4x+6y=-6\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{3}\\y=-\dfrac{19}{9}\end{matrix}\right.\)
b, \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=3\\\dfrac{3}{x}+\dfrac{2}{y}=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}+\dfrac{3}{y}=9\\\dfrac{3}{x}+\dfrac{2}{y}=7\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=2\\\dfrac{3}{x}+\dfrac{3}{y}=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\left(TM\right)\\y=\dfrac{1}{2}\left(TM\right)\end{matrix}\right.\)
c, \(\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{5}{y}=1\\\dfrac{2}{x}+\dfrac{1}{y}=3\end{matrix}\right.\left(x,y\ne0\right)\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{5}{y}=1\\\dfrac{10}{x}+\dfrac{5}{y}=15\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{13}{x}=16\\\dfrac{10}{x}+\dfrac{5}{y}=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{13}{16}\left(TM\right)\\y=\dfrac{13}{7}\left(TM\right)\end{matrix}\right.\)
d, \(\left\{{}\begin{matrix}\sqrt{x+1}-3\sqrt{y-1}=-4\\2\sqrt{x+1}-\sqrt{y-1}=2\end{matrix}\right.\left(x\ge-1,y\ge1\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x+1}-6\sqrt{y-1}=-8\\2\sqrt{x+1}-\sqrt{y-1}=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}-5\sqrt{y-1}=-10\\2\sqrt{x+1}-6\sqrt{y-1}=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y-1}=2\\2\sqrt{x+1}-6\sqrt{y-1}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\left(TM\right)\\y=5\left(TM\right)\end{matrix}\right.\)
Bài 2:
a: \(\Leftrightarrow\left\{{}\begin{matrix}2-x+y-3x-3y=5\\3x-3y+5x+5y=-2\end{matrix}\right.\)
=>-4x-2y=3 và 8x+2y=-2
=>x=1/4; y=-2
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{y-1}=1\\\dfrac{1}{x-2}+\dfrac{1}{y-1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-1=5\\\dfrac{1}{x-2}=1-\dfrac{1}{5}=\dfrac{4}{5}\end{matrix}\right.\)
=>y=6 và x-2=5/4
=>x=13/4; y=6
c: =>x+y=24 và 3x+y=78
=>-2x=-54 và x+y=24
=>x=27; y=-3
d: \(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x-1}-6\sqrt{y+2}=4\\2\sqrt{x-1}+5\sqrt{y+2}=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-11\sqrt{y+2}=-11\\\sqrt{x-1}=2+3\cdot1=5\end{matrix}\right.\)
=>y+2=1 và x-1=25
=>x=26; y=-1
Ta co:\(\Sigma\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}=\Sigma\frac{\left(y+\frac{1}{z}\right)^2}{z+\frac{1}{x}}\ge\frac{\left(x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}{x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}=x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)Ta lai co:
\(\Sigma x+\Sigma\frac{1}{x}=\Sigma\left(x+\frac{1}{4x}\right)+\frac{3}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge3+\frac{3}{4}.\frac{9}{x+y+z}\ge3+\frac{3}{4}.\frac{9}{\frac{3}{2}}=\frac{15}{2}\)
Dau '=' xay ra khi \(x=y=z=\frac{1}{2}\)
Vay \(P_{min}=\frac{15}{2}\)khi \(x=y=z=\frac{1}{2}\)
Câu 1:
a) Ta có: \(P=\left(\dfrac{4}{\sqrt{x}}-\dfrac{3}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}-2}{x+\sqrt{x}}+\dfrac{2}{\sqrt{x}}\right)\)
\(=\left(\dfrac{4\left(\sqrt{x}+1\right)-3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}-2+2\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{4\sqrt{x}+4-3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}-2+2\sqrt{x}+2}\)
\(=\dfrac{\sqrt{x}+4}{3\sqrt{x}}\)
b) Ta có: \(P=2\sqrt{x}-3\)
\(\Leftrightarrow\dfrac{\sqrt{x}+4}{3\sqrt{x}}=2\sqrt{x}-3\)
\(\Leftrightarrow\sqrt{x}+4=6x-9\sqrt{x}\)
\(\Leftrightarrow6x-9\sqrt{x}-\sqrt{x}-4=0\)
\(\Leftrightarrow6x-10\sqrt{x}-4=0\)
\(\Leftrightarrow6x-12\sqrt{x}+2\sqrt{x}-4=0\)
\(\Leftrightarrow6\sqrt{x}\left(\sqrt{x}-2\right)+2\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(6\sqrt{x}+2\right)=0\)
mà \(6\sqrt{x}+2>0\forall x>0\)
nên \(\sqrt{x}-2=0\)
\(\Leftrightarrow\sqrt{x}=2\)
hay x=4(thỏa ĐK)
Vậy: Để \(P=2\sqrt{x}-3\) thì x=4
Câu 2 :
\(\left\{{}\begin{matrix}x+y=2020\\2021x-y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=2020-x\\2021x-y=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=2020-x\\2021x-\left(2020-x\right)=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=2020-x\\2022x-2020=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=2020-x\\x=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2019\end{matrix}\right.\)
\(b.\)
\(\text{Vì (P) đi qua A(1,2) nên : }\)
\(2=\left(m-2\right)\cdot1\)
\(\Leftrightarrow m=4\left(1\right)\)
\(\text{Vì (d) đi qua A(1,2) nên : }\)
\(2=-2\cdot1+m^2-12\)
\(\Leftrightarrow m^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}m=4\\m=-4\end{matrix}\right.\left(2\right)\)
\(\text{Từ (1) , (2) : }\)\(m=4\)
\(\text{Khi đó : }\)
\(\left(d\right):y=-2x+4^2-12\)
\(\Leftrightarrow y=-2x+4\)
\(\left(P\right):\) \(y=\left(4-2\right)\cdot x^2\Leftrightarrow y=2x^2\)
\(\text{Phương trình hoành độ giao điểm: }\)
\(-2x+4=2x^2\)
\(\Leftrightarrow2x^2+2x-4=0\)
\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
\(\text{Với : }\) \(x=1\Rightarrow y=2x^2=2\cdot1=2\)
\(\text{Với : }\) \(x=-2\Rightarrow y=2x^2=2\cdot\left(-2\right)^2=8\)
\(B\left(-2,8\right)\)