\(\frac{2a}{2x-3a}-1=0\)

\(\frac{2a-2x+3a}{2x-3a}=0\)

\(\frac{5a-2x}{2x-3a}=0\)

\(5a-2x=0\)

\(5a=2x\)

\(x=\frac{5a}{2}\)

Kết quả \(x=\frac{5a}{2}\left(đúng\right)\)

THANK YOU

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ăn lol nhé .com

a) (x - 2)(2x + 3) = 5(x - 2)

<=> 2x^2 + 3x - 4x - 6 = 5x - 10

<=> 2x^2 - x - 6 = 5x - 10

<=> 2x^2 - x - 5x = -10 + 6

<=> 2x^2 - 6x = -4

<=> 2x^2 - 6x + 4 = 0

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

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

<=> x - 2 = 0 hoặc x - 1 = 0

<=> x = 2 hoặc x = 1

b) 3x/x + 2 - 2 = x/x - 2

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

<=> x^2 + 6x + 8 = x^2 - 2x

<=> x^2 + 6x + 8 - x^2 + 2x = 0

<=> 8x + 8 = 0

<=> 8x = 0 - 8

<=> 8x = -8

<=> x = -1

áp dụng hằng đẳng thức (a+b)²=a²+2ab+b² vào (x+y)² ta được:

 \(\left(x+y\right)^2=x^2+2xy+y^2=x^2+y^2+2xy\)

ta có : xy=27

=>2xy=54 

thay 2xy=54 và x²+y²=29 vào bt x²+y²+2xy ta được

\(29+54=83\)

vậy giá trị của biểu thức (x+y)² tại  x²+y²=29 và xy=27 là 83

\(\frac{\sqrt{b^2+2a^2}}{ab}+\frac{\sqrt{c^2+2b^2}}{bc}+\frac{\sqrt{a^2+2c^2}}{ca}\ge\sqrt{3}\left(1\right)\)

Ta có ab+bc+ca=abc nên \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\)

\(\left(1\right)\Leftrightarrow\sqrt{\frac{1}{a^2}+\frac{2}{b^2}}+\sqrt{\frac{1}{b^2}+\frac{2}{c^2}}+\sqrt{\frac{1}{c^2}+\frac{2}{a^2}}\ge\sqrt{3}\)

Trong mặt phẳng với hệ tọa độ Oxy, với các Vecto

\(\overrightarrow{u}=\left(\frac{1}{a};\frac{\sqrt{2}}{b}\right);\left|\overrightarrow{u}\right|=\sqrt{\frac{1}{a^2}+\frac{2}{b^2}}\)

\(\overrightarrow{v}=\left(\frac{1}{b};\frac{\sqrt{2}}{c}\right)\Rightarrow\left|\overrightarrow{v}\right|=\sqrt{\frac{1}{b^2}+\frac{2}{c^2}}\)

\(\overrightarrow{w}=\left(\frac{1}{c};\frac{\sqrt{2}}{a}\right)\Rightarrow\left|\overrightarrow{w}\right|=\sqrt{\frac{1}{c^2}+\frac{2}{a^2}}\)

Ta có \(\overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c};2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\right)=\left(1;\sqrt{2}\right)\)

=> \(\left|\overrightarrow{u}\right|+\left|\overrightarrow{v}\right|+\left|\overrightarrow{w}\right|=\sqrt{1+2}=\sqrt{3}\)

Mặt khác \(\left|\overrightarrow{u}\right|+\left|\overrightarrow{v}\right|+\left|\overrightarrow{w}\right|\ge\left|\overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}\right|\)

\(\Rightarrow\frac{\sqrt{b^2+2a^2}}{ab}+\frac{\sqrt{c^2+2b^2}}{bc}+\frac{\sqrt{a^2+2c^2}}{ac}\ge\sqrt{3}\)

Dấu "=" xảy ra <=> a=b=c

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

x+2=0 hoặc 9-3x=0

x=-2 hoặc x=3

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

\(9x+18-3x^2-6x=0\)

\(3x+18-3x^2=0\)

\(3x-x^2-2x^2+18=0\)

\(x\left(3-x\right)-2\left(x^2-9\right)=0\)

\(-x\left(x-3\right)-2\left(x-3\right)\left(x+3\right)=0\)

\(\left(x-3\right)\left(-x-2x-6\right)=0\)

\(x\in\left\{3;-2\right\}\)