Pt \(\Leftrightarrow\left(x^2+3x+2x+6\right)\left(x+4\right)-x^3+5x^2-6=0\)

\(\Leftrightarrow\left(x^2+5x+6\right)\left(x+4\right)-x^3+5x^2-6=0\)

\(\Leftrightarrow x^3+4x^2+5x^2+20x+6x+24-x^3+5x^2-6=0\)

\(\Leftrightarrow14x^2+26x+18=0\)

\(\Leftrightarrow14x^2+13x+13x+\dfrac{169}{14}+\dfrac{83}{14}=0\)

\(\Leftrightarrow x\left(14x+13\right)+\dfrac{13}{14}\left(14x+13\right)+\dfrac{83}{14}=0\)

\(\Leftrightarrow\left(x+\dfrac{13}{14}\right)\left(14x+13\right)+\dfrac{83}{14}=0\)

\(\Leftrightarrow\dfrac{1}{14}\left(14x+13\right)^2+\dfrac{81}{14}=0\)

Có \(\dfrac{1}{14}\left(14x+13\right)^2+\dfrac{81}{14}\ge0+\dfrac{81}{14}=\dfrac{81}{14}>0\)

\(\Rightarrow\dfrac{1}{14}\left(14x+13\right)^2+\dfrac{81}{14}=0\)  vô nghiệm

Vậy pt vô nghiệm

Đỗ Thanh Hải :>

b)\(\left\{{}\begin{matrix}x+y=-1+m\left(1\right)\\2x-y=2m\end{matrix}\right.\)

\(\Rightarrow3x=-1+3m\)

\(\Leftrightarrow x=\dfrac{-1+3m}{3}\) 

Thay \(x=\dfrac{-1+3m}{3}\) vào (1) có:

\(\dfrac{-1+3m}{3}+y=-1+m\)\(\Leftrightarrow y=-1+m-\dfrac{-1+3m}{3}=-\dfrac{2}{3}\)

Suy ra với mọi m hệ luôn có nghiệm duy nhất \(\left(x;y\right)=\left(\dfrac{-1+3m}{3};-\dfrac{2}{3}\right)\)

\(xy=\left(\dfrac{-1+3m}{3}\right).\left(-\dfrac{2}{3}\right)=10\)

\(\Leftrightarrow m=-\dfrac{44}{3}\)

Vậy...

I miss you Được em, hoặc trực tiếp nhóm thành HĐT, một vế là tổng các bình phương, vế còn lại bằng 0

\(A=\dfrac{1-\sqrt{x}}{\sqrt{x}+2}=\dfrac{3-\left(\sqrt{x}+2\right)}{\sqrt{x}+2}=\dfrac{3}{\sqrt{x}+2}-1\)

Có \(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+2\ge2\Leftrightarrow\dfrac{3}{\sqrt{x}+2}\le\dfrac{3}{2}\)\(\Leftrightarrow\dfrac{3}{\sqrt{x}+2}-1\le\dfrac{1}{2}\)\(\Leftrightarrow A\le\dfrac{1}{2}\)

Dấu "=" xảy ra khi x=0 (tm)

Vậy \(A_{max}=\dfrac{1}{2}\)

Bài 2:

Đk: \(x\ge3;y\ge5;z\ge4\)

Pt\(\Leftrightarrow\sqrt{x-3}+\dfrac{4}{\sqrt{x-3}}+\sqrt{y-5}+\dfrac{9}{\sqrt{y-5}}+\sqrt{z-4}+\dfrac{25}{\sqrt{z-4}}=20\)

Áp dụng AM-GM có:

\(\sqrt{x-3}+\dfrac{4}{\sqrt{x-3}}\ge2\sqrt{\sqrt{x-3}.\dfrac{4}{\sqrt{x-3}}}=4\)

\(\sqrt{y-5}+\dfrac{9}{\sqrt{y-5}}\ge6\)

\(\sqrt{z-4}+\dfrac{25}{\sqrt{z-4}}\ge10\)

Cộng vế với vế \(\Rightarrow VT\ge20\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\sqrt{x-3}=\dfrac{4}{\sqrt{x-3}}\\\sqrt{y-5}=\dfrac{9}{\sqrt{y-5}}\\\sqrt{z-4}=\dfrac{25}{\sqrt{z-4}}\end{matrix}\right.\)\(\Leftrightarrow x=7;y=14;z=29\) (tm)

Vậy...

Đk: \(x^2-3x+7\ge0\)

\(\Leftrightarrow x^2-2.\dfrac{3}{2}x+\dfrac{9}{4}+\dfrac{19}{4}\ge0\)

\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2+\dfrac{19}{4}\ge0\) (lđ với mọi x)

Vậy biểu thức luôn xác định với mọi x

Gọi hai phân số cần tìm là \(\dfrac{a}{8};\dfrac{a+1}{8};a\in N\)

Có \(\dfrac{a}{8}< \dfrac{5}{7}< \dfrac{a+1}{8}\)

\(\Leftrightarrow\dfrac{7a}{56}< \dfrac{40}{56}< \dfrac{7a+7}{56}\)

\(\Leftrightarrow7a< 40< 7a+7\)

\(\Leftrightarrow\left\{{}\begin{matrix}a< \dfrac{40}{7}\\a>\dfrac{33}{7}\end{matrix}\right.\) mà a nguyên \(\Rightarrow a=5\)

Vậy hai pso cần tìm là \(\dfrac{5}{8};\dfrac{6}{8}\) 

Phân số có dạng \(\dfrac{7}{a};a\ne0\)

Có \(\dfrac{10}{13}< \dfrac{7}{a}< \dfrac{10}{11}\)

\(\Leftrightarrow\dfrac{70}{91}< \dfrac{70}{10a}< \dfrac{70}{77}\)\(\Leftrightarrow91>10a>77\)\(\Leftrightarrow9,1>a>7,7\)

\(\Rightarrow a\in\left\{9;8\right\}\)

Vậy phân số cần tìm là \(\dfrac{7}{9};\dfrac{7}{8}\)

a) 

\(P=\left(\dfrac{b-a}{\sqrt{b}-\sqrt{a}}-\dfrac{a\sqrt{a}-b\sqrt{b}}{a-b}\right):\dfrac{\left(\sqrt{b}-\sqrt{a}\right)^2+\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\)

\(=\left[\sqrt{b}+\sqrt{a}-\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\right]:\dfrac{b-\sqrt{ab}+a}{\sqrt{a}+\sqrt{b}}\)

\(=\left(\sqrt{b}+\sqrt{a}-\dfrac{a+\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}\right).\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}\)

\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2-a-\sqrt{ab}-b}{\sqrt{a}+\sqrt{b}}.\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}\)

\(=\dfrac{\sqrt{ab}}{\sqrt{a}+\sqrt{b}}.\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}\)\(=\dfrac{\sqrt{ab}}{a-\sqrt{ab}+b}\)

b) \(P=\dfrac{\sqrt{ab}}{a-\sqrt{ab}+b}=\dfrac{\sqrt{ab}}{\left(\sqrt{a}-\dfrac{1}{2}\sqrt{b}\right)^2+\dfrac{3}{4}b}\)

Vì \(\left(\sqrt{a}-\dfrac{1}{2}\sqrt{b}\right)^2+\dfrac{3}{4}b>0;\forall a\ge0;b\ge0;a\ne b\)

\(\sqrt{ab}\ge0\)\(\forall a\ge0;b\ge0\)

\(\Rightarrow P=\dfrac{\sqrt{ab}}{\left(\sqrt{a}-\dfrac{1}{2}\sqrt{b}\right)^2+\dfrac{3}{4}b}\ge0\)

Vậy...