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Để (1) có 2 nghiệm dương \(\Rightarrow\left\{{}\begin{matrix}\Delta'=\left(m+3\right)^2-m-1\ge0\\x_1+x_2=2\left(m+3\right)>0\\x_1x_2=m+1>0\end{matrix}\right.\) \(\Rightarrow m>-1\)
\(P=\left|\dfrac{\sqrt{x_1}-\sqrt{x_2}}{\sqrt{x_1x_2}}\right|>0\Rightarrow P^2=\dfrac{\left(\sqrt{x_1}-\sqrt{x_2}\right)^2}{x_1x_2}\)
\(P^2=\dfrac{x_1+x_2-2\sqrt{x_1x_2}}{x_1x_2}=\dfrac{2\left(m+3\right)-2\sqrt{m+1}}{m+1}=\dfrac{4}{m+1}-\dfrac{2}{\sqrt{m+1}}+2\)
\(P^2=\left(\dfrac{2}{\sqrt{m+1}}-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\Rightarrow P\ge\dfrac{\sqrt{7}}{2}\)
Dấu "=" xảy ra khi \(\sqrt{m+1}=4\Rightarrow m=15\)
\(a+b+c=1-2\left(m+3\right)+2m+5=0\)
\(\Rightarrow\) phương trình luôn có 2 nghiệm: \(\left\{{}\begin{matrix}x_1=1\\x_2=2m+5\end{matrix}\right.\)
Để 2 nghiệm của pt thỏa mãn yêu cầu của đề bài \(\Rightarrow x_2>0\Rightarrow2m+5>0\Rightarrow m>\dfrac{-5}{2}\)
\(\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}=\dfrac{4}{3}\Rightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2m+5}}=\dfrac{4}{3}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{2m+5}}=\dfrac{1}{3}\Rightarrow2m+5=9\Rightarrow m=2\)
Bài 1)
ĐK: \(x\geq 0; x\neq -4\)
Ta có:
\(A=\frac{1}{\sqrt{x}+2}+\frac{1}{2+\sqrt{x}}-\frac{2\sqrt{x}}{x+4}\)
\(=\frac{2}{\sqrt{x}+2}-\frac{2\sqrt{x}}{x+4}=2\left(\frac{1}{\sqrt{x}+2}-\frac{\sqrt{x}}{x+4}\right)\)
\(=2.\frac{x+4-x-2\sqrt{x}}{(\sqrt{x}+2)(x+4)}=2.\frac{4-2\sqrt{x}}{(\sqrt{x}+2)(x+4)}=\frac{4(2-\sqrt{x})}{(\sqrt{x}+2)(x+4)}\)
\(B=(\sqrt{2}+\sqrt{3}).\sqrt{2}-\sqrt{6}+\frac{\sqrt{333}}{\sqrt{111}}\)
\(=2+\sqrt{6}-\sqrt{6}+\frac{\sqrt{3}.\sqrt{111}}{\sqrt{111}}=2+\sqrt{3}\)
Để \(A=B\Leftrightarrow \frac{4(2-\sqrt{x})}{(\sqrt{x}+2)(x+4)}=2+\sqrt{3}\)
PT rất xấu. Mình nghĩ bạn đã chép sai biểu thức A.
Bài 2 : Tọa độ điểm B ?
Bài 3:
Để pt có hai nghiệm thì \(\Delta'=(m-3)^2-(m^2-1)>0\)
\(\Leftrightarrow 10-6m>0\Leftrightarrow m< \frac{5}{3}\)
Áp dụng định lý Viete: \(\left\{\begin{matrix} x_1+x_2=2(m-3)\\ x_1x_2=m^2-1\end{matrix}\right.\)
Khi đó:
\(4=2x_1+x_2=x_1+(x_1+x_2)=x_1+2(m-3)\)
\(\Rightarrow x_1=10-2m\)
\(\Rightarrow x_2=2(m-3)-(10-2m)=4m-16\)
Suy ra: \(\Rightarrow x_1x_2=(10-2m)(4m-16)\)
\(\Leftrightarrow m^2-1=8(5-m)(m-4)\)
\(\Leftrightarrow m^2-1=8(-m^2+9m-20)\)
\(\Leftrightarrow 9m^2-72m+159=0\)
\(\Leftrightarrow (3m-12)^2+15=0\) (vô lý)
Vậy không tồn tại $m$ thỏa mãn điều kiện trên.
a: Khi m = -4 thì:
\(x^2-5x+\left(-4\right)-2=0\)
\(\Leftrightarrow x^2-5x-6=0\)
\(\Delta=\left(-5\right)^2-5\cdot1\cdot\left(-6\right)=49\Rightarrow\sqrt{\Delta}=\sqrt{49}=7>0\)
Pt có 2 nghiệm phân biệt:
\(x_1=\dfrac{5+7}{2}=6;x_2=\dfrac{5-7}{2}=-1\)
a:
Sửa đề: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{3x+3}{9-x}\right)\cdot\left(\dfrac{\sqrt{x}-7}{\sqrt{x}+1}+1\right)\)
\(P=\left(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\right)\cdot\dfrac{\sqrt{x}-7+\sqrt{x}+1}{\sqrt{x}+1}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\cdot\dfrac{2\sqrt{x}-6}{\sqrt{x}+1}\)
\(=\dfrac{-3\sqrt{x}-3}{\sqrt{x}+3}\cdot\dfrac{2}{\sqrt{x}+1}=\dfrac{-6}{\sqrt{x}+3}\)
b: P>=1/2
=>P-1/2>=0
=>\(\dfrac{-6}{\sqrt{x}+3}-\dfrac{1}{2}>=0\)
=>\(\dfrac{-12-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>=0\)
=>\(-\sqrt{x}-15>=0\)
=>\(-\sqrt{x}>=15\)
=>căn x<=-15
=>\(x\in\varnothing\)
c: căn x+3>=3
=>6/căn x+3<=6/3=2
=>P>=-2
Dấu = xảy ra khi x=0
1: \(B=\dfrac{2\sqrt{x}-x-2\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\)
\(=\dfrac{-x}{\left(\sqrt{x}-2\right)\cdot\sqrt{x}}\)
\(=\dfrac{-\sqrt{x}}{\sqrt{x}-2}\)
a: \(P=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\cdot\dfrac{\sqrt{x}-7+\sqrt{x}+1}{\sqrt{x}+1}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\cdot\dfrac{2\sqrt{x}-6}{\sqrt{x}+1}\)
\(=\dfrac{-3\sqrt{x}-3}{\sqrt{x}+1}\cdot\dfrac{2}{\sqrt{x}+3}=-\dfrac{6}{\sqrt{x}+3}\)
b: P>=-1/2
=>P+1/2>=0
=>\(\dfrac{-6}{\sqrt{x}+3}+\dfrac{1}{2}>=0\)
=>\(\dfrac{-12+\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}>=0\)
=>căn x-9>=0
=>x>=81
c: căn x+3>=3
=>6/căn x+3<=6/3=2
=>-6/căn x+3>=-2
Dấu = xảy ra khi x=0
\(a,P=\dfrac{\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{2-\sqrt{x}}{\sqrt{x}}=\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{-2}{\sqrt{x}+2}\\ P=-\dfrac{3}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\\ \Leftrightarrow3\sqrt{x}+6=10\Leftrightarrow\sqrt{x}=\dfrac{4}{3}\Leftrightarrow x=\dfrac{16}{9}\left(tm\right)\)
\(P=-\dfrac{3}{5}\) sao suy ra đc \(\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\) thế
a) Ta có: \(M=\left(1-\dfrac{x-3\sqrt{x}}{x-9}\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}-\dfrac{\sqrt{x}-3}{2-\sqrt{x}}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\left(1-\dfrac{x-3\sqrt{x}}{x-9}\right):\left(\dfrac{9-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\left(1-\dfrac{x-3\sqrt{x}}{x-9}\right):\left(\dfrac{9-x+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\dfrac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\left(1-\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\dfrac{-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}+3-\sqrt{x}}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}+3}{-\left(\sqrt{x}-2\right)}\)
\(=\dfrac{-3}{\sqrt{x}-2}\)
Câu 1 :
\(P=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
Câu 2 :
Ta có :
\(\Delta=m^2+16>0\)
\(=>\) phương trình có 2 nghiệm phân biệt .
Theo định lý vi-ét ta có :
\(\left\{{}\begin{matrix}x_1+x_2=m\\x_1.x_2=-4\end{matrix}\right.\)
Thay vào ta được :
\(\dfrac{2m+7}{m^2+8}\ge-\dfrac{1}{8}\)
\(\Leftrightarrow16m+56\ge-m^2-8\)
\(\Leftrightarrow m^2+16m+64\ge0\)
\(\Leftrightarrow\left(m+8\right)^2\ge0\) ( đúng )