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\(P=\frac{x\sqrt{x}-8}{x+2\sqrt{x}+4}+3\left(1-\sqrt{x}\right).\)
\(=\frac{\sqrt{x^3}-2^3}{x+2\sqrt{x}+4}+3-3\sqrt{x}\)
\(=\frac{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}{x+2\sqrt{x}+4}+3-3\sqrt{x}\)
\(=\sqrt{x}-2+3-3\sqrt{x}=-2\sqrt{x}+1\)
\(Q=\frac{2P}{1-P}=\frac{2\left(-2\sqrt{x}+1\right)}{1-\left(-2\sqrt{x}+1\right)}\)
\(=\frac{-4\sqrt{x}+2}{1+2\sqrt{x}-1}=\frac{-2\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{-2\sqrt{x}}{\sqrt{x}}+\frac{1}{\sqrt{x}}=-2+\frac{1}{\sqrt{x}}\)
\(Q\in Z\Leftrightarrow-2+\frac{1}{\sqrt{x}}\in Z\Rightarrow\frac{1}{\sqrt{x}}\in Z\)
\(\Rightarrow1\)\(⋮\)\(\sqrt{x}\)\(\Rightarrow\sqrt{x}\inƯ_1\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=1\\\sqrt{x}=-1\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x\in\varnothing\end{cases}}}\)
Vậy \(Q\in Z\Leftrightarrow x=1\)
đầu tiên ta chứng minh với x,y,z,t bất kì thì:
\(\sqrt{x^2+y^2}+\sqrt{z^2+t^2}\ge\sqrt{\left(x+z\right)^2+\left(y+t\right)^2}\) (*)
thật vậy bđt (*) tương đương với:
\(x^2+y^2+z^2+t^2+2\sqrt{\left(x^2+y^2\right)\left(z^2+t^2\right)}\ge x^2+2xz+z^2+y^2+2yt+t^2\)
\(\Leftrightarrow\sqrt{\left(x^2+y^2\right)\left(z^2+t^2\right)}\ge xz+yt\)
bđt trên đúng vì theo bđt bunhia cốp xki
\(\sqrt{\left(x^2+y^2\right)\left(z^2+t^2\right)}\ge\sqrt{\left(xz+yt\right)^2}=|xz+yt|\ge xz+yt\)
Áp dụng (*) ta có:
\(P=\sqrt{4+x^4}+\sqrt{4+y^4}+\sqrt{4+z^4}\ge\sqrt{\left(2+2\right)^2+\left(x^2+y^2\right)^2}+\sqrt{4+z^2}\)
\(\ge\sqrt{\left(2+2+2\right)^2+\left(x^2+y^2+z^2\right)^2}=\sqrt{36+\left(x^2+y^2+z^2\right)^2}\)
Ta có:
\(\left(x-1\right)^2+\left(y-1\right)^2+\left(z-1\right)^2+\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)
\(\Rightarrow3x^2+3y^2+3z^2+3\ge2x+2y+2z+2xy+2yz+2zx=2.6=12\)
\(\Rightarrow x^2+y^2+z^2\ge3\Rightarrow P\ge\sqrt{36+3}=3\sqrt{5}\)
Dấu bằng xảy ra khi x=y=z=1
a) DK : x > 0; x khác 1
\(P=\sqrt{x}\left(\sqrt{x}-1\right)-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\)
\(=x-\sqrt{x}+1\)
c ) \(Q=\frac{2\sqrt{x}}{P}=\frac{2\sqrt{x}}{x-\sqrt{x}+1}\)
<=> \(xQ-\left(Q+2\right)\sqrt{x}+Q=0\)(1)
TH1: Q = 0 => x = 0 loại
TH2: Q khác 0
(1) là phương trình bậc 2 với tham số Q ẩn x.
(1) có nghiệm <=> \(\left(Q+2\right)^2-4Q^2\ge0\)
<=> \(-3Q^2+4Q+4\ge0\)
<=> \(-\frac{2}{3}\le Q\le2\)
Vì Q nguyên và khác 0 nên Q = 1 hoặc Q = 2
Với Q = 1 => \(x-3\sqrt{x}+1=0\)
<=> \(\sqrt{x}=\frac{3}{2}\pm\frac{\sqrt{5}}{2}\)----> Tìm được x
Với Q = 2 => \(2x-4\sqrt{x}+1=0\Leftrightarrow\sqrt{x}=1\pm\frac{1}{\sqrt{2}}\)-----> tìm đc x.
Tự làm tiếp nhé! Kiểm tra lại đề bài câu b.
Áp dụng bđt AM-GM ta có
\(\sqrt{3x\left(2x+y\right)}+\sqrt{3y\left(2y+x\right)}\le\frac{3x+2x+y}{2}+\frac{3y+2y+x}{2}=\frac{6\left(x+y\right)}{2}=3\left(x+y\right)\)
\(\Rightarrow P\ge\frac{x+y}{3\left(x+y\right)}=\frac{1}{3}\)
Dấu "=" xảy ra khi x=y
http://olm.vn/hoi-dap/question/104313.html
coi hỉu j ko tui đang mò
a, \(A=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\left[\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{-3}{\sqrt{x}+3}\)
b, \(A\in Z\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}\in Z\)
\(\Leftrightarrow\sqrt{x}+3\inƯ_3=\left\{\pm1;\pm3\right\}\)
\(\Leftrightarrow\sqrt{x}=0\)
\(\Leftrightarrow x=0\)
\(a,A=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\left(x\ge0;x\ne9\right)\\ A=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\\ A=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\\ A=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\)
\(b,A\in Z\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}\in Z\Leftrightarrow-3⋮\sqrt{x}+3\\ \Leftrightarrow\sqrt{x}+3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{-6;-4;-2;0\right\}\)
Mà \(\sqrt{x}\ge0\)
\(\Leftrightarrow x\in\left\{0\right\}\)
Vậy \(x=0\) thì A nguyên