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dùng bddt bunhiacopsky là ra kq

ho ktoots

cố tử thần ♡๖ۣۜŦεαм♡❤Ɠ长♡ღ

Chị ơi dùng bđt BCS , dấu = xảy ra P =1 như thế có gọi là giá trị của P=1 không nhỉ ? 

x(x - 2) - 5(x - 2) = (x - 3)² 

<=> x^2 - 2x - 5x + 10 = x^2 - 6x + 9

<=> -7x + 10 = -6x + 9

<=> -6x + 7x = 10 - 9

<=> x = 1

<=> x² - 2x - 5x +10 = x² - 6x + 9

<=> x=1

Vậy phương trình có n^o duy nhất là 1

\(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)

\(=a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)\)

\(=a^2\left(b-c\right)-b^2\left[\left(b-c\right)+\left(a-b\right)\right]+c^2\left(a-b\right)\)

\(=a^2\left(b-c\right)-b^2\left(b-c\right)-b^2\left(a-b\right)+c^2\left(a-b\right)\)

\(=\left(b-c\right)\left(a^2-b^2\right)+\left(a-b\right)\left(c^2-b^2\right)\)

\(=\left(b-c\right)\left(a-b\right)\left(a+b\right)+\left(a-b\right)\left(c-b\right)\left(c+b\right)\)

\(=\left(b-c\right)\left(a-b\right)\left(a+b\right)-\left(a-b\right)\left(b-c\right)\left(b+c\right)\)

\(=\left(b-c\right)\left(a-b\right)\left(a+b-b-c\right)\)

\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(\frac{2x-1}{3}-\frac{5-3x}{2}=x-2\)

\(\Leftrightarrow\frac{2\left(2x-1\right)}{6}-\frac{3\left(5-3x\right)}{6}=\frac{6\left(x-2\right)}{6}\)

\(\Rightarrow2\left(2x-1\right)-3\left(5-3x\right)=6\left(x-2\right)\)

\(\Leftrightarrow4x-2-15+9x=6x-12\)

\(\Leftrightarrow4x-2-15+9x-6x+12=0\)

\(\Leftrightarrow7x-5=0\)

\(\Leftrightarrow7x=5\)

\(\Leftrightarrow x=\frac{5}{7}\)

CHÚC EM HỌC TỐT!!!