\(x+y+12=4\sqrt{x}-6\sqrt{y-1}\) (ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\y\ge1\end{matrix}\right.\))

\(\Leftrightarrow\left(x-4\sqrt{x}+4\right)+\left[\left(y-1\right)+6\sqrt{y-1}+9\right]=0\)

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

Vì \(\left\{{}\begin{matrix}\left(\sqrt{x}-2\right)^2\ge0\\\left(\sqrt{y-1}+3\right)^2\ge9\end{matrix}\right.\Rightarrow VT\ge9\)

Vậy pt vô nghiệm.

a)

\(\sqrt{4-2\sqrt{3}}-\sqrt{3}\)

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

\(=-1\)

b)

\(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}\)

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

\(=2\sqrt{2}\)

c)

\(\sqrt{10+2\sqrt{9}}-\sqrt{9}\)

\(=\sqrt{\left(\sqrt{9}+1\right)^2}-\sqrt{9}\)

\(=1\)

Áp dụng BĐT AM - GM, ta có:

\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

\(=1+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{a}+1+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{c}{b}+1\)

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

\(\ge3+2+2+2=9\)

Dấu "=" xảy ra khi a = b = c

Biến đổi tương đương:

\(\sqrt{\dfrac{a+b}{2}}\ge\dfrac{\sqrt{a}+\sqrt{b}}{2}\) (1)

\(\Leftrightarrow\dfrac{a+b}{2}\ge\dfrac{a+2\sqrt{ab}+b}{4}\)

\(\Leftrightarrow2a+2b-a-2\sqrt{ab}-b\ge0\)

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\) luôn đúng

=> (1) đúng

Dấu "=" xảy ra khi a = b

Muốn cách trình bày thì li-ke cho mk

16 : 15 = 4  

90 :15 =6

135 :15=9

vay so do la 15

Kẻ DM _I_ AC (M thuộc AC)

\(\sin\alpha=\dfrac{DK}{DO}=\dfrac{DK}{\dfrac{BD}{2}}=\dfrac{2DK}{BD}\)

\(\dfrac{1}{2}\times AC\times BD\times\sin\alpha\)

\(=\dfrac{1}{2}\times AC\times BD\times\dfrac{2DK}{BD}\)

\(=AC\times DK\)

\(=S_{ABCD}\)

\(\left(AC\times DK=2\times\dfrac{1}{2}AC\times DK=2S_{ACD}=S_{ABCD}\right)\)

a) x^2 + 2x - 35 = 0

<=> (x - 5)(x + 7) = 0

<=> x = 5 hoặc x = - 7

b) 4x^2 - 12x - 27 = 0

<=> (2x - 9)(2x + 3) = 0

<=> x = 4,5 hoặc x = - 1,5

c) 9x^2 + 24x + 7 = 0

<=> (3x + 1)(3x + 7) = 0

<=> x = - 1/3 hoặc x = - 7/3

d) x^2 + y^2 - 4x + 6y + 13 = 0

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

<=> x = 2 và y = - 3

e) 25x^2 - 10x - 24 = 0

<=> (5x - 6)(5x + 4) = 0

<=> x = 1,2 hoặc x = - 0,8

3x² + x - 10 = 0

<=> (x + 2)(3x - 5) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{5}{3}\end{matrix}\right.\)

2x² - x - 15 = 0

<=> (x - 3)(2x + 5) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)

7x² + 23x - 20 = 0

<=> (7x - 5)(x + 4) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{7}\\x=-4\end{matrix}\right.\)

6x² - 35x + 25 = 0

<=> (x - 5)(6x - 5) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{5}{6}\end{matrix}\right.\)

6x² - x - 2 = 0

<=> (3x - 2)(2x + 1) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

8x² + 6x - 35 = 0

<=> (4x - 7)(2x + 5) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{4}\\x=-\dfrac{5}{2}\end{matrix}\right.\)