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Lời giải:
a.
Nếu $m=3$ thì pt trở thành:
$x^2+4x-5=0$
$\Leftrightarrow (x-1)(x+5)=0$
$\Leftrightarrow x=1$ hoặc $x=-5$
b.
Để pt có 2 nghiệm pb $x_1,x_2$ thì:
$\Delta'=4+m^2-4>0\Leftrightarrow m^2>0\Leftrightarrow m\neq 0$
PT có 2 nghiệm $(-2+m, -2-m)$
Khi đó:
\(x_2=x_1^3+4x_2^2\Leftrightarrow \left[\begin{matrix} -2+m=(-2-m)^3+4(-2+m)^2\\ -2-m=(-2+m)^3+4(-2-m)^2\end{matrix}\right.\)
\(\Leftrightarrow \left[\begin{matrix} -m^3+2m^2-29m+10=0\\ m^3-2m^2+29m+10=0\end{matrix}\right.\)
Nghiệm khá xấu, cảm giác đề cứ sai sai bạn ạ.
a: Thay a=-2 vào pt, ta được:
\(-2x^2-2\cdot\left(-2-1\right)x-2+1=0\)
\(\Leftrightarrow-2x^2+6x-1=0\)
\(\Leftrightarrow2x^2-6x+1=0\)
\(\text{Δ}=\left(-6\right)^2-4\cdot2\cdot1=36-8=28>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{6-2\sqrt{7}}{2}=3-\sqrt{7}\\x_2=3+\sqrt{7}\end{matrix}\right.\)
b: Để phương trình có hai nghiệm phân biệt thì
\(\left\{{}\begin{matrix}\left(-2a+2\right)^2-4a\left(a+1\right)>0\\a< >0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4a^2-8a+4-4a^2-4a>0\\a< >0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-12a>-4\\a< >0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a< >0\\a< \dfrac{1}{3}\end{matrix}\right.\)
\(A=\dfrac{2\sqrt{x}+17}{\sqrt{x+5}}=\dfrac{2\sqrt{x}+10}{\sqrt{x}+5}+\dfrac{7}{\sqrt{x}+5}=2+\dfrac{7}{\sqrt{x}+5}\)
Để \(A\) ∈ \(Z\) thì \(\dfrac{7}{\sqrt{x}+5}\) phải ∈ \(Z\)
=> \(\sqrt{x}+5\) ∈ \(Ư\left(7\right)=\left\{-7;-1;1;7\right\}\)
# Với \(\sqrt{x}+5=-7=>\sqrt{x}=-12\)(Loại)
#Với \(\sqrt{x}+5=-1=>\sqrt{x}=-6\)(Loại)
#Với \(\sqrt{x}+5=1=>\sqrt{x}=-4\left(Loại\right)\)
#Với \(\sqrt{x}+5=7=>\sqrt{x}=2< =>x=4\left(Nhận\right)\)
Vậy \(x=4\) thì \(A\)∈\(Z\)
\(\sqrt[3]{\dfrac{a^4}{b^2\left(a^2-ab+b^2\right)}}+\sqrt[3]{\dfrac{b^4}{c^2\left(b^2-bc+c^2\right)}}\sqrt[3]{\dfrac{c^4}{a^2\left(c^2-ac+b^2\right)}}\) \(\text{≥}3\)
\(Ta\) \(Có\) : \(\sqrt[3]{\dfrac{a^4}{b^2\left(a^2-ab+b^2\right)}}=\sqrt[3]{\dfrac{a^6}{ab.ab\left(a^2-ab+b^2\right)}}=\dfrac{a^2}{\sqrt[3]{ab.ab.\left(a^2-ab+b^2\right)}}\)
\(Áp\) \(dụng\) \(bđt\) \(AM-GM\)
\(\sqrt[3]{ab.ab\left(a^2-ab+b^2\right)}\text{≤}\) \(\dfrac{ab+ab+a^2-ab+b^2}{3}\)
\(=>\dfrac{a^2}{\sqrt[3]{ab.ab\left(a^2-ab+b^2\right)}}\) \(\text{≥}\) \(\dfrac{3a^2}{a^2+ab+b^2}\) \(Hay\) \(\sqrt[3]{\dfrac{a^4}{b^2\left(a^2-ab+b^2\right)}}\text{≥}\dfrac{3a^2}{a^2+ab+b^2}\)
Tương tự ta cũng có :
\(\sqrt[3]{\dfrac{b^4}{c^2\left(b^2-bc+c^2\right)}}\text{≥}\dfrac{3b^2}{b^2+bc+c^2}\)
\(\sqrt[3]{\dfrac{c^4}{a^2\left(c^2-ac+a^2\right)}}\text{≥}\dfrac{3c^2}{a^2+ac+c^2}\)
\(=>\text{}\text{}\)\(\sqrt[3]{\dfrac{a^4}{b^2\left(a^2-ab+b^2\right)}}+\sqrt[3]{\dfrac{b^4}{c^2\left(b^2-bc+c^2\right)}}\sqrt[3]{\dfrac{c^4}{a^2\left(c^2-ac+b^2\right)}}\) \(\text{≥}\) \(3\left(\dfrac{a^2}{a^2+ab+b^2}+\dfrac{b^2}{b^2+bc+c^2}+\dfrac{c^2}{a^2+ac+c^2}\right)\)
Cần c/m \(\left(\dfrac{a^2}{a^2+ab+b^2}+\dfrac{b^2}{b^2+bc+c^2}+\dfrac{c^2}{a^2+ac+c^2}\right)\) ≥ \(1\)
Ta có : \(\dfrac{a^2}{a^2+ab+b^2}\text{≥}\dfrac{1}{3}\)
\(< =>3a^2\text{≥}a^2+ab+b^2\) \(< =>2a^2-b\left(a+b\right)\text{≥}0\) (1)
Lại có : \(a^2\text{≥}-b\left(a+b\right)\) (2)
Từ (1) và (2) => \(\dfrac{a^2}{a^2+ab+b^2}\text{≥}\dfrac{1}{3}\)
Tương tự ta cũng có :
\(\dfrac{b^2}{b^2+bc+c^2}\text{≥}\dfrac{1}{3}\)
\(\dfrac{c^2}{a^2+ac+c^2}\text{≥}\dfrac{1}{3}\)
Do đó \(\dfrac{a^2}{a^2+ab+b^2}+\dfrac{b^2}{b^2+bc+c^2}+\dfrac{c^2}{a^2+ac+c^2}\text{≥}1\)
Suy ra : \(\sqrt[3]{\dfrac{a^4}{b^2\left(a^2-ab+b^2\right)}}+\sqrt[3]{\dfrac{b^4}{c^2\left(b^2-bc+c^2\right)}}\sqrt[3]{\dfrac{c^4}{a^2\left(c^2-ac+b^2\right)}}\) \(\text{≥}\) \(3\)
Đẳng thức xảy ra <=> \(a=b=c=1\)
1.2 với \(x\ge0,x\in Z\)
A=\(\dfrac{2\sqrt{x}+7}{\sqrt{x}+2}=2+\dfrac{3}{\sqrt{x}+2}\in Z< =>\sqrt{x}+2\inƯ\left(3\right)=\left(\pm1;\pm3\right)\)
*\(\sqrt{x}+2=1=>\sqrt{x}=-1\)(vô lí)
*\(\sqrt{x}+2=-1=>\sqrt{x}=-3\)(vô lí
*\(\sqrt{x}+2=3=>x=1\)(TM)
*\(\sqrt{x}+2=-3=\sqrt{x}=-5\)(vô lí)
vậy x=1 thì A\(\in Z\)
Bài 1.2
\(A=\dfrac{2\sqrt{x}+7}{\sqrt{x}+2}=2+\dfrac{3}{\sqrt{x}+2}\)
C1:Bạn dùng pp chặn như bài 2.2
C2: (Gợi ý)\(\sqrt{x}+2\ge2\) và \(\sqrt{x}+2\inƯ\left(3\right)\)\(\Rightarrow\sqrt{x}+2=3\Leftrightarrow x=1\)
Vậy x=1 thì A nguyên
Bài 2.2
\(A=\dfrac{\sqrt{x}+7}{\sqrt{x}+2}=1+\dfrac{5}{\sqrt{x}+2}\)
Do \(\sqrt{x}\ge0;\forall x\)\(\Rightarrow\sqrt{x}+2\ge2\) \(\Rightarrow\dfrac{5}{\sqrt{x}+2}\le\dfrac{5}{2}\)\(\Rightarrow A\le\dfrac{7}{2}\) (1)
mà \(\dfrac{5}{\sqrt{x}+2}>0;\forall x\Rightarrow A>1\) (2)
Từ (1) (2) \(\Rightarrow1< A\le\dfrac{7}{2}\) mà A nguyên
\(\Rightarrow\left[{}\begin{matrix}A=2\\A=3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}1+\dfrac{5}{\sqrt{x}+2}=2\\1+\dfrac{5}{\sqrt{x}+2}=3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+2=5\\\sqrt{x}+2=\dfrac{5}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy...
Bài 3.2
\(A=\dfrac{-x-2\sqrt{x}-5}{\sqrt{x}+2}\)\(=\dfrac{-\sqrt{x}\left(\sqrt{x}+2\right)-5}{\sqrt{x}+2}=-\sqrt{x}-\dfrac{5}{\sqrt{x}+2}\)
\(=2-\left(\sqrt{x}+2+\dfrac{5}{\sqrt{x}+2}\right)\)
Áp dụng bđt cosi: \(\sqrt{x}+2+\dfrac{5}{\sqrt{x}+2}\ge2\sqrt{\left(\sqrt{x}+2\right).\dfrac{5}{\sqrt{x}+2}}=2\sqrt{5}\)
\(\Rightarrow A\le2-2\sqrt{5}\)
Dấu = xảy ra \(\Leftrightarrow\sqrt{x}+2=\dfrac{5}{\sqrt{x}+2}\Leftrightarrow x=9-4\sqrt{5}\)
b: |x1|+|x2|=5
x1+x2=3 và x1*x2=-m^2-1
|x1|+|x2|=5
=>(x1+x2)^2-2*x1x2+2|x1x2|=25
=>3^2+2(m^2+1)+2(m^2+1)=25
=>4m^2=12
=>m^2=3
=>\(m=\pm\sqrt{3}\)
\(B=\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}+2+\sqrt{x}-2-2\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{1}{x-4}\)
\(B=-1\Leftrightarrow\dfrac{1}{x-4}=-1\Rightarrow4-x=1\Rightarrow x=3\)