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\(P=-\dfrac{2019}{x^2}+\dfrac{m}{x}=-2019\left(\dfrac{1}{x^2}-2.\dfrac{m}{2.2019}.\dfrac{1}{x}\right)\)
\(=-2019\left(\dfrac{1}{x^2}-2.\dfrac{m}{4038}.\dfrac{1}{x}+\dfrac{m^2}{4038^2}-\dfrac{m^2}{4038^2}\right)=-2019\left(\dfrac{1}{x}-\dfrac{m}{4038}\right)^2+\dfrac{2019m^2}{4038^2}\le\dfrac{2019m^2}{4038^2}\)
\(\Rightarrow\dfrac{2019m^2}{4038^2}=2019\Rightarrow m=\pm4038\)
\(P=\dfrac{mx-2019}{x^2}\Rightarrow px^2-mx+2019=0\)
\(\Delta=m^2-4.2019P\ge0\)
\(\Leftrightarrow P\le\dfrac{m^x}{8076}\)
để \(\max\limits_P=2019\) thì \(\dfrac{m^2}{8076}=2019\)
\(\Leftrightarrow m^2=8076.2019\)
\(=2.2.2019.2019\)
\(\Leftrightarrow m=4038\)(vì m>0)
vậy m=4038
ĐKXĐ : \(x\ne0;x\ne\pm1\)
a) Bạn ghi lại rõ đề.
b) \(B=\dfrac{x-1}{x+1}+\dfrac{3x-x^2}{x^2-1}=\dfrac{x-1}{x+1}+\dfrac{3x-x^2}{\left(x-1\right).\left(x+1\right)}\)
\(=\dfrac{\left(x-1\right)^2+3x-x^2}{\left(x-1\right).\left(x+1\right)}=\dfrac{x+1}{\left(x-1\right).\left(x+1\right)}=\dfrac{1}{x-1}\)
c) \(P=A.B=\dfrac{x^2+x-2}{x.\left(x-1\right)}=\dfrac{\left(x-1\right).\left(x+2\right)}{x\left(x-1\right)}=\dfrac{x+2}{x}=1+\dfrac{2}{x}\)
Không tồn tại Min P \(\forall x\inℝ\)
1) Cho biểu thức A =
Bài 5:
a: Thay \(x=4+2\sqrt{3}\) vào E, ta được:
\(E=\dfrac{\sqrt{3}+1-1}{\sqrt{3}+1-3}=\dfrac{\sqrt{3}}{\sqrt{3}-2}=-3-2\sqrt{3}\)
b: Để E<1 thì E-1<0
\(\Leftrightarrow\dfrac{\sqrt{x}-1-\sqrt{x}+3}{\sqrt{x}-3}< 0\)
\(\Leftrightarrow\sqrt{x}-3< 0\)
hay x<9
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne1\end{matrix}\right.\)
c: Để E nguyên thì \(4⋮\sqrt{x}-3\)
\(\Leftrightarrow\sqrt{x}-3\in\left\{-2;1;2;4\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{4;5;7\right\}\)
hay \(x\in\left\{16;25;49\right\}\)
Câu 2:
a) Ta có \(x=4-2\sqrt{3}\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{3}-2\right)^2}=\sqrt{3}-2\)
Thay \(x=\sqrt{3}-1\) vào \(B\), ta được
\(B=\dfrac{\sqrt{3}-1-2}{\sqrt{3}-1+1}=\dfrac{\sqrt{3}-3}{\sqrt{3}}=1-\sqrt{3}\)
b) Để \(B\) âm thì \(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\) mà \(\sqrt{x}+1\ge1>0\forall x\) \(\Rightarrow\sqrt{x}-2< 0\Rightarrow\sqrt{x}< 2\Rightarrow x< 4\)
c) Ta có \(B=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=1-\dfrac{3}{\sqrt{x}+1}\)
Với mọi \(x\ge0\) thì \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+1\ge1\Rightarrow\dfrac{3}{\sqrt{x}+1}\le3\Rightarrow B=1-\dfrac{3}{\sqrt{x}+1}\ge-2\)
Dấu "=" xảy ra khi \(\sqrt{x}+1=1\Leftrightarrow x=0\)
Vậy \(B_{min}=-2\) khi \(x=0\)
a: Ta có: \(x^2=3-2\sqrt{2}\)
nên \(x=\sqrt{2}-1\)
Thay \(x=\sqrt{2}-1\) vào A, ta được:
\(A=\dfrac{\left(\sqrt{2}+1\right)^2}{\sqrt{2}-1}=\dfrac{3+2\sqrt{2}}{\sqrt{2}-1}=7+5\sqrt{2}\)
1. \(x=\frac{1}{9}\) thỏa mãn đk: \(x\ge0;x\ne9\)
Thay \(x=\frac{1}{9}\) vào A ta có:
\(A=\frac{\sqrt{\frac{1}{9}}+1}{\sqrt{\frac{1}{9}}-3}=-\frac{1}{2}\)
2. \(B=...\)
\(B=\frac{3\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{4x+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{3x-9\sqrt{x}+x+3\sqrt{x}-4x-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{-6\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
3. \(P=A:B=\frac{\sqrt{x}+1}{\sqrt{x}-3}:\frac{-6\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{\sqrt{x}+3}{-6}\)
Vì \(\sqrt{x}+3\ge3\forall x\)\(\Rightarrow\frac{\sqrt{x}+3}{-6}\le\frac{3}{-6}=-\frac{1}{2}\)
hay \(P\le-\frac{1}{2}\)
Dấu "=" xảy ra <=> x=0
a: Ta có: \(N=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=x-\sqrt{x}+1\)
Lời giải:
$\frac{\sqrt{x}+1}{\sqrt{x}+4}=\frac{\sqrt{x}+4-3}{\sqrt{x}+4}=1-\frac{3}{\sqrt{x}+4}$
Vì $\sqrt{x}\geq 0$ nên $\sqrt{x}+4\geq 4$
$\Rightarrow \frac{3}{\sqrt{x}+4}\leq \frac{3}{4}$
$\Rightarrow \frac{\sqrt{x}+1}{\sqrt{x}+4}=1-\frac{3}{\sqrt{x}+4}\geq 1-\frac{3}{4}=\frac{1}{4}$
Vậy $M=\frac{1}{4}$
------------------
$N=\frac{\sqrt{x}+5}{\sqrt{x}+2}=1+\frac{3}{\sqrt{x}+2}$
Do $\sqrt{x}\geq 0$ nên $\sqrt{x}+2\geq 2$
$\Rightarrow \frac{3}{\sqrt{x}+2}\leq \frac{3}{2}$
$\Rightarrow \frac{\sqrt{x}+5}{\sqrt{x}+2}\leq 1+\frac{3}{2}=\frac{5}{2}$
Vậy $N=\frac{5}{2}$
$\Rightarrow 2M+N =2.\frac{1}{4}+\frac{5}{2}=3$
Đáp án C.
\(A=\dfrac{x^2+mx+n}{x^2+2x+4}\)
\(\Leftrightarrow Ax^2+2Ax+4A=x^2+mx+n\)
\(\Leftrightarrow\left(A-1\right)x^2+\left(2A-m\right)x+\left(4A-n\right)=0\left(1\right)\)
A có cực trị khi (1) có nghiệm
\(\Leftrightarrow\Delta=\left(4A^2-4Am+m^2\right)-4\left[4A^2-A\left(n+4\right)+n\right]\ge0\)
\(\Leftrightarrow-12A^2-4A\left(m-n-4\right)+m^2-4n\ge0\) (1)
Mặt khác, theo gt, ta có: \(\left\{{}\begin{matrix}A\ge\dfrac{1}{3}\\A\le3\end{matrix}\right.\)
\(\Rightarrow\left(3A-1\right)\left(3-A\right)\ge0\)
\(\Leftrightarrow-3A^2+10A-3\ge0\)
\(\Leftrightarrow-12A^2+40A-12\ge0\) (2)
Từ (1) và (2) suy ra \(\left\{{}\begin{matrix}m-n-4=-10\\m^2-4n=-12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m+6=n\\m^2-4\left(m+6\right)=-12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}n=m+6\\\left(m-6\right)\left(m+2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}n=12\\n=4\end{matrix}\right.\\\left[{}\begin{matrix}m=6\\m=-2\end{matrix}\right.\end{matrix}\right.\)
Vậy \(\left(m;n\right)=\left(6;12\right);\left(-2;4\right)\)