CÂU 6:

\(P=2x+3y+\frac{6}{x}+\frac{10}{y}\)

\(=\left(\frac{6}{x}+\frac{3}{2}x\right)+\left(\frac{10}{y}+\frac{5}{2}y\right)+\frac{1}{2}\left(x+y\right)\)

\(\ge2\sqrt{\frac{6}{x}.\frac{3}{2}x}+2\sqrt{\frac{10}{y}.\frac{5y}{2}}+\frac{1}{2}\left(x+y\right)\)( BĐT cô si )

\(\ge6+10+2=18\)( do \(x+y\ge4\)

Dấu "=" xảy ra <=>x=y=2

Vậy Min P=18 <=> x=y=2 

Kết quả câu a như nào nhỉ ?

a) đk: \(\hept{\begin{cases}a>b\\a< -b\end{cases}}\left(b>0\right)\) hoặc \(\hept{\begin{cases}a>-b\\a< b\end{cases}\left(b< 0\right)}\)

Ta có:
\(B=\frac{a}{\sqrt{a^2-b^2}}-\left(1+\frac{a}{\sqrt{a^2-b^2}}\right)\div\frac{b}{a-\sqrt{a^2-b^2}}\)

\(B=\frac{a}{\sqrt{a^2-b^2}}-\frac{a+\sqrt{a^2-b^2}}{\sqrt{a^2-b^2}}\cdot\frac{a-\sqrt{a^2-b^2}}{b}\)

\(B=\frac{a}{\sqrt{a^2-b^2}}-\frac{a^2-a^2+b^2}{b\sqrt{a^2-b^2}}\)

\(B=\frac{a}{\sqrt{a^2-b^2}}-\frac{b}{\sqrt{a^2-b^2}}=\frac{a-b}{\sqrt{a^2-b^2}}=\sqrt{\frac{a-b}{a+b}}\)

b) \(B< 1\Leftrightarrow\sqrt{\frac{a-b}{a+b}}< 1\Leftrightarrow\frac{a-b}{a+b}< 1\)

\(\Leftrightarrow\frac{-2b}{a+b}< 0\) ta xét 2TH:

Nếu \(b>0\Rightarrow a>-b\)

Nếu \(b< 0\Rightarrow a< -b\)

Vậy ...

Gọi vận tốc thực của cano là x( km/h) ĐK: \(x>4\)

Ta có: \(\hept{\begin{cases}V_{xuoidong}=x+4\left(km/h\right)\\V_{nguocdong}=x-4\left(km/h\right)\end{cases}}\)

Thời gian ca nô đi xuôi dòng là \(\frac{80}{x+4}\left(h\right)\)

Thời gian ca nô đi ngược dòng là\(\frac{80}{x-4}\left(h\right)\)

Theo bài ra ta có pt sau : \(\frac{80}{x+4}=\frac{80}{x-4}-\frac{1}{2}\)

\(\Leftrightarrow80\left(\frac{1}{x+4}-\frac{1}{x-4}\right)=\frac{-1}{2}\)

\(\Leftrightarrow\frac{-8}{x^2-16}=\frac{-1}{160}\)

\(\Rightarrow x^2-16=1280\)

\(\Leftrightarrow x^2=1296\)

\(\Leftrightarrow\orbr{\begin{cases}x=36\left(tm\right)\\x=-36\left(loai\right)\end{cases}}\)

Vậy vận tốc thực của ca nô là 36km/h 

"Ara-ara"

thank~you~

\(\Delta=\left(-2m\right)^2-4.\left(2m-3\right)\)

   \(=4m^2-8m+12\)

\(\Delta'=m^{^2}-2m+3\)

     \(=\left(m-1\right)^2+2\)

ĐKXĐ: x \(\ge\)1/2

Đặt: \(x+3=a\left(a>0\right)\)

  \(\sqrt{2x-1}=b\) (b \(\ge\)0)

=> 3a + b² = 3x + 9 + 2x - 1 = 5x + 8 => 5x - 1 = b² + 3a - 9 

Do đó, ta có: b² + 3a - ab  - 9 = 0

<=> (b - 3)(b + 3) - a(b - 3) = 0

<=> (b - 3)(b - a + 3) = 0

<=> \(\orbr{\begin{cases}b=3\\b-a+3=0\end{cases}}\)

Với b = 3=> \(\sqrt{2x-1}=3\)=> 2x - 1 = 9 => x = 5 (tm)

với b - a + 3 = 0 => \(\sqrt{2x-1}-x-3+3=0\)

<=> \(\sqrt{2x-1}=x\) (x \(\ge\)1/2)

<=> 2x - 1 = x²  <=> (x - 1)² = 0 <=> x = 1 (tm)

Vậy S = {1; 5}

hmmmmmmmmmmmmmmmmmmmmmmmm..................

a³ + b³ \(\ge\frac{1}{4}\)

<=> (a + b)(a² - ab + b²) \(\ge\frac{1}{4}\)

<=> a² - ab + b² \(\ge\frac{1}{4}\)

<=> 4a² - 4ab + 4b² \(\ge1\)

<=> 4a² - 4a(1 - a) + 4(1 - a)² \(\ge\)1

<=> 8a² - 4a + 4(a² - 2a + 1) \(\ge\)1

<=> 12a² - 12a + 3 \(\ge\)0

<=> 3(4a² - 4a + 1) \(\ge0\)

<=> (2a - 1)² \(\ge\)0 (đúng)

Dấu "=" xảy ra <=> \(a=b=\frac{1}{2}\)

b) Vì \(a^3+b^3\ge\frac{1}{4}\Rightarrow\frac{1}{a^3+b^3}\ge4\)

Khi đó \(\frac{1}{a^3+b^3}+\frac{3}{ab}\ge16\)

<=> \(\frac{3}{ab}\ge12\)

<=> ab \(\ge\frac{1}{4}\)

<=> 4ab \(\ge1\)

<=> 4a(1 - a) \(\ge1\)

<=> (2a - 1)² \(\ge0\)(đúng)

=> ĐPCM

Ta có: \(\frac{x^2}{\sqrt{1-x^2}}+\frac{y^2}{\sqrt{1-y^2}}+\frac{z^2}{\sqrt{1-z^2}}\)

\(=\frac{x^3}{x\sqrt{1-x^2}}+\frac{y^3}{y\sqrt{1-y^2}}+\frac{z^3}{z\sqrt{1-z^2}}\)

\(\ge\frac{x^3}{\frac{x^2+1-x^2}{2}}+\frac{y^3}{\frac{y^2+1-y^2}{2}}+\frac{z^3}{\frac{z^2+1-z^2}{2}}\)

\(=2x^3+2y^3+2z^3=2\left(x^3+y^3+z^3\right)=2\cdot\frac{3\sqrt{2}}{4}=\frac{3\sqrt{2}}{2}\)

Dấu "=" xảy ra khi: \(x=y=z=\frac{1}{\sqrt{2}}\)

đk: \(x\ge-1\)

Ta có: \(2x=\left[\left(x+3\right)-\left(x+1\right)\right]x=\left(\sqrt{x+3}-\sqrt{x+1}\right)\left(\sqrt{x+3}+\sqrt{x+1}\right)x\)

\(PT\Leftrightarrow\left(\sqrt{x+3}-\sqrt{x+1}\right)\left(x^2+\sqrt{x^2+4x+3}-x\sqrt{x+3}-x\sqrt{x+1}\right)=0\)

\(\Leftrightarrow\left(\sqrt{x+3}-\sqrt{x+1}\right)\left[x\left(x-\sqrt{x+3}\right)-\left(x-\sqrt{x+3}\right)\sqrt{x+1}\right]=0\)

\(\Leftrightarrow\left(\sqrt{x+3}-\sqrt{x+1}\right)\left(x-\sqrt{x+1}\right)\left(x-\sqrt{x+3}\right)=0\)

Nếu \(\sqrt{x+3}-\sqrt{x+1}=0\Leftrightarrow\sqrt{x+3}=\sqrt{x+1}\)

\(\Leftrightarrow x+3=x+1\Leftrightarrow2=0\left(ktm\right)\)

Nếu \(x-\sqrt{x+1}=0\Leftrightarrow x=\sqrt{x+1}\left(x\ge0\right)\)

\(\Leftrightarrow x^2=x+1\Leftrightarrow x^2-x-1=0\Leftrightarrow\orbr{\begin{cases}x=\frac{1+\sqrt{5}}{2}\left(tm\right)\\x=\frac{1-\sqrt{5}}{2}\left(ktm\right)\end{cases}}\)

Nếu \(x-\sqrt{x+3}=0\Leftrightarrow x=\sqrt{x+3}\left(x\ge0\right)\)

\(\Leftrightarrow x^2=x+3\Leftrightarrow x^2-x-3=0\Leftrightarrow\orbr{\begin{cases}x=\frac{1+\sqrt{13}}{2}\left(tm\right)\\x=\frac{1-\sqrt{13}}{2}\left(ktm\right)\end{cases}}\)

Vậy \(x\in\left\{\frac{1+\sqrt{5}}{2};\frac{1+\sqrt{13}}{2}\right\}\)