a)

\(M=\sqrt{3\sqrt{2}+2\sqrt{3}}\times\sqrt{3\sqrt{2}-2\sqrt{3}}\)

\(M^2=\sqrt{2\times3}\left(\sqrt{3}+\sqrt{2}\right)\times\sqrt{2\times3}\left(\sqrt{3}-\sqrt{2}\right)\)

\(=6\)

\(\Rightarrow M=\sqrt{6}\)

b)

\(\left(1-\sqrt{2}+\sqrt{3}\right)\left(1+\sqrt{2}-\sqrt{3}\right)\)

\(=1-\left(\sqrt{3}-\sqrt{2}\right)^2\)

\(=1-3+2\sqrt{6}-2=-4+2\sqrt{6}\)

c)

\(\left(5+4\sqrt{2}\right)\left(3+2\sqrt{1+\sqrt{2}}\right)\left(3-2\sqrt{1+\sqrt{2}}\right)\)

\(=\left(5+4\sqrt{2}\right)\left(9-4-4\sqrt{2}\right)\)

\(=25-32=-7\)

M = 2x² + y² - 2xy + 10x - 6y

= (x² + 4x + 4) + (x² + 3² + y² + 6x - 2xy - 6y) - 13

= (x + 2)² + (x + 3 - y)² - 13 \(\ge\) - 13

Dấu "=" xảy ra khi x = - 2 và y = 1

\(M=9x^2+y^2-6x+3y+5\)

\(=\left(9x^2+6x+1\right)+\left(y^2+3x+\dfrac{9}{4}\right)+\dfrac{7}{4}\)

\(=\left(3x+1\right)^2+\left(y+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)

Dấu "=" xảy ra khi \(x=-\dfrac{1}{3}\) và \(y=-\dfrac{3}{2}\)

\(M=4\left(x-1\right)\left(x+1\right)-5x\left(x-2\right)+x^2\)

\(=4x^2-4-5x^2+10x+x^2\)

\(=10x-4\)

\(M=\left(y^2+2\right)\left(y-4\right)-\left(2y^2+1\right)\left(\dfrac{1}{2}y-2\right)\)

\(=\left(y^2+2\right)\left(y-4\right)-\dfrac{1}{2}\left(2y^2+1\right)\left(y-4\right)\)

\(=\left(y-4\right)\left(y^2+2-y^2-\dfrac{1}{2}\right)\)

\(=\dfrac{3}{2}y-6\)

c)

\(C=\left(3-2x\right)\left(x-2\right)-4\left(x-1\right)\left(x-3\right)-\left(x-2\right)\left(x+2\right)\)

= 3x - 6 - 2x² + 4x - 4x² + 12x + 4x - 12 - x² + 4

= - 7x² + 23x - 14

a)

5x² - 4x + 1 = 0

<=> \(5\left(x^2-\dfrac{4}{5}x+\dfrac{4}{25}\right)+\dfrac{1}{5}=0\)

\(\Leftrightarrow5\left(x-\dfrac{2}{5}\right)^2+\dfrac{1}{5}=0\)

mà \(5\left(x-\dfrac{2}{5}\right)^2+\dfrac{1}{5}\ge\dfrac{1}{5}>0\)

Vậy pt vô nghiệm.

b)

x² - x - 6 = 0

<=> (x - 3)(x + 2) = 0

<=> \(\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Vậy S = {3 ; - 2}

c)

2x² + x - 1 = 0

<=> (2x - 1)(x + 1) = 0

<=> \(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-1\end{matrix}\right.\)

Vậy S = {- 1 ; 0,5}

Kẻ đường cao DM và đường cao DN của \(\Delta ABD\) và \(\Delta ACD\).

Tam giác ABC vuông tại A

\(\Rightarrow BC^2=AB^2+AC^2\) (pytago)

=> AB = 9 (cm)

AMDN là hình chữ nhật \(\left(\widehat{DMA}=\widehat{MAN}=\widehat{AND}=90^0\right)\) có AD là tia phân giác

=> AMDN là hình vuông

\(\Rightarrow AD=\sqrt{2}DM\) và \(DM=DN\)

\(S_{ABC}=S_{ABD}+S_{ACD}\)

\(\dfrac{1}{2}AB.AC=\dfrac{1}{2}DM.AB+\dfrac{1}{2}DN.AC\)

=> DM = \(\dfrac{36}{7}\) (cm)

=> AD \(\approx7,27\) (cm)

b)

Áp dụng BĐT Cauchy Shwarz, ta có:

\(\left(1+1+1\right)\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)

\(\Leftrightarrow\dfrac{\left(a+b+c\right)^2}{3}\le a^2+b^2+c^2\)

Áp dụng BĐT Cauchy Shwarz dạng Engel, ta có:

\(\dfrac{a^3}{bc}+\dfrac{b^3}{ac}+\dfrac{c^3}{bc}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{3abc}\)

\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{\dfrac{\left(a+b+c\right)^3}{9}}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{\dfrac{\left(a+b+c\right)}{3}\times\left(a^2+b^2+c^2\right)}\)

\(=\dfrac{3\left(a^2+b^2+c^2\right)}{a+b+c}\) (đpcm)

Dấu "=" xảy ra khi a = b = c.

c)

Áp dụng BĐT Cauchy Shwarz, ta có:

\(\left(1+1+1\right)\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)

\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2\ge\dfrac{\left(a+b+c\right)^4}{9}\)

Áp dụng BĐT AM - GM, ta có:

\(a+b+c\ge3\sqrt[3]{abc}\Leftrightarrow3abc\le\dfrac{\left(a+b+c\right)^3}{9}\)

Áp dụng BĐT Cauchy Shwarz dạng Engel, ta có:

\(M=\dfrac{a^3}{bc}+\dfrac{b^3}{ac}+\dfrac{c^3}{ab}=\dfrac{a^4}{abc}+\dfrac{b^4}{abc}+\dfrac{c^4}{abc}\)

\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{3abc}\ge\dfrac{\dfrac{\left(a+b+c\right)^4}{9}}{\dfrac{\left(a+b+c\right)^3}{9}}=a+b+c\left(\text{đ}pcm\right)\)

Dấu "=" xảy ra khi a = b = c

ông cũng chơi bang bang ak tích tui nha

\(M=\left(\dfrac{3\sqrt{x}-1}{x-1}-\dfrac{1}{x+\sqrt{x}}\right)\div\dfrac{1}{x+\sqrt{x}}\)

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

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

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

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

\(2P-x=3\Leftrightarrow2\left(\dfrac{3x-2\sqrt{x}+1}{\sqrt{x}-1}\right)-x=3\)

\(\Leftrightarrow\dfrac{6x-4\sqrt{x}+2-x\sqrt{x}+x-3\sqrt{x}+3}{\sqrt{x}-1}=0\)

\(\Leftrightarrow7x-7\sqrt{x}-x\sqrt{x}+5=0\)

Đến đây sao số xấu quá (T^T)