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1/ Đặt \(\hept{\begin{cases}\sqrt{x-2013}=a\\\sqrt{x-2014}=b\end{cases}}\)
Thì ta có:
\(\frac{\sqrt{x-2013}}{x+2}+\frac{\sqrt{x-2014}}{x}=\frac{a}{a^2+2015}+\frac{b}{b^2+2014}\)
\(\le\frac{a}{2a\sqrt{2015}}+\frac{b}{2b\sqrt{2014}}=\frac{1}{2\sqrt{2015}}+\frac{1}{2\sqrt{2014}}\)
2/ \(\frac{x}{2x+y+z}+\frac{y}{x+2y+z}+\frac{z}{x+y+2z}\)
\(\le\frac{1}{4}\left(\frac{x}{x+y}+\frac{x}{x+z}+\frac{y}{y+x}+\frac{y}{y+z}+\frac{z}{z+x}+\frac{z}{z+y}\right)\)
\(=\frac{3}{4}\)
điều kiện: \(x\ge\frac{1}{2}\)
ta có \(x^2+8x-4-4x\sqrt{2x-1}=2x-1\)
\(\Leftrightarrow\left(x-2\sqrt{2x-1}\right)^2=2x-1\Leftrightarrow\orbr{\begin{cases}x-2\sqrt{2x-1}=\sqrt{2x-1}\\x-2\sqrt{2x-1}=-\sqrt{2x-1}\end{cases}}\)
\(\) hay \(\orbr{\begin{cases}x=3\sqrt{2x-1}\\x=\sqrt{2x-1}\end{cases}}\)
TH1: \(x=3\sqrt{2x-1}\Leftrightarrow x^2=18x-9\Leftrightarrow x=9\pm6\sqrt{2}\)
TH2: \(x=\sqrt{2x-1}\Leftrightarrow x^2=2x-1\Leftrightarrow x=1\)
( về cơ bản nó không khác cách e đặt ẩn phụ là mấy, chỉ có điều e liên hợp kiểu gì nhỉ)
ĐKXĐ: \(x\ge\dfrac{3}{4}\)
\(\Leftrightarrow\sqrt{5x^2+5x}=\sqrt{8x^2+10x-12}\) (1)
\(\Leftrightarrow\left(\sqrt{5x^2+5x}\right)^2=\left(\sqrt{8x^2+10x-12}\right)^2\)
\(\Leftrightarrow5x^2+5x=8x^2+10x-12\)
\(\Leftrightarrow5x^2+5x-\left(8x^2+10x-12\right)=8x^2+10x-12-\left(8x^2+10x-12\right)\)
\(\Leftrightarrow-3x^2-5x+12=0\)
\(\Leftrightarrow\left(-3x+4\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-3x+4=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-3x=-4\\x=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\left(OK\right)\\x=-3\left(loại\right)\end{matrix}\right.\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{4}{3}\right\}\)
\(\text{Ta có: }x=\sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}}}=\sqrt{\frac{\left(3-\sqrt{5}\right)^2}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}}=\frac{3-\sqrt{5}}{\sqrt{9-5}}=\frac{3-\sqrt{5}}{2}.\)
\(A=x^5-6x^4+12x^3-4x^2-13x+2020\)
\(=\left(x^5-3x^4+x^3\right)-\left(3x^4-9x^3+3x^2\right)+\left(2x^3-6x^2+2x\right)+\left(5x^2-15x+5\right)+2015\)
\(=x^3\left(x^2-3x+1\right)-3x^2\left(x^2-3x+1\right)+2x\left(x^2-3x+1\right)+5\left(x^2-3x+1\right)+2015\)
\(=\left(x^2-3x+1\right)\left(x^3-3x^2+2x+5\right)+2015\)
Thay x vào A ta có:
\(A=\left[\left(\frac{3-\sqrt{5}}{2}\right)^2-3.\frac{3-\sqrt{5}}{2}+1\right]\left(.....\right)+2015\)
\(=\left(\frac{14-6\sqrt{5}}{4}-\frac{9-3\sqrt{5}}{2}+1\right)\left(....\right)+2015\)
\(=0\cdot\left(......\right)+2015=2015\)
Vậy.....
\(x^2+6x-3=4x\sqrt{2x-1}\left(1\right)\) ĐK: \(x\ge\frac{1}{2}\)
Đặt \(\sqrt{2x-1}=a\ge0\)
\(\Rightarrow6x-3=3a^2\)
=> (1) <=> x^2 +3a^2 = 4ax
<=> x^2 -4ax +3a^2 =0
<=> x^2 -ax - 3ax + 3a^2 =0
<=> x(x-a) -3a(x-a) =0
<=> (x-a) ( x-3a ) =0
\(\Leftrightarrow\orbr{\begin{cases}x=a\\x=3a\end{cases}}\)
TH1: x=a
\(\Rightarrow x=\sqrt{2x-1}\)\(\left(x\ge0\right)\)
\(\Leftrightarrow x^2=2x-1\)
\(\Leftrightarrow\left(x-1\right)^2=0\)
<=> x=1 (tm)
TH2: x= 3a
\(\Rightarrow x=3\sqrt{2x-1}\left(x\ge0\right)\)
\(\Leftrightarrow x^2=18x-9\)
\(\Leftrightarrow x^2-18x+9=0\)
\(\Delta=288\)
=> pt có 2 nghiệm pb \(\orbr{\begin{cases}x=\frac{18+12\sqrt{2}}{2}=9+6\sqrt{2}\left(tm\right)\\x=\frac{18-12\sqrt{2}}{2}=9-6\sqrt{2}\left(tm\right)\end{cases}}\)
Vậy ...