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

\(=\left(1-x\right)-x^2\left(1-x\right)=\left(1-x^2\right)\left(1-x\right)\)

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

a, \(\left(2x-1\right)\left(x+3\right)-2x^2+5x=7\)

\(\Leftrightarrow2x^2+6x-x-3-2x^2+5x=7\)

\(\Leftrightarrow2x^2+5x-3-2x^2+5x=7\)

\(\Leftrightarrow10x-10=0\Leftrightarrow x=1\)

b, \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-4\right)\left(x+4\right)=54\)

\(\Leftrightarrow\left(x^3+27\right)-x\left(x^2-16\right)=54\)

\(\Leftrightarrow x^3+27-x^3+16x=54\)

\(\Leftrightarrow-27+16x=0\Leftrightarrow x=\frac{27}{16}\)

a, \(\left(x+1\right)^2-2\left(x+1\right)\left(y-3\right)+\left(y-3\right)^2=\left[\left(x+1\right)-\left(y-3\right)\right]^2\)

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

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

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

Ta có: P = (a^2+b^2+c^2-ab-bc-ca)/(a^2-c^2-2ab+2bc)

=1/2.(2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca)/(a^2 - 2ab + b^2 - b^2 +2bc  - c^2)

=1/2.[(a^2-2ab+b^2)+(b^2-2bc+c^2)+(a^2-2ac+c^2)]/[(a-b)^2-(b^2-2bc+c^2)]

=1/2.[(a-b)^2 + (b-c)^2 + (a-c)^2]/[(a-b)^2 - (b-c)^2

Lại có: a – b = 7; b – c = 3 ó a – b + b – c = 7 + 3 ó a – c = 10

Thay a - b = 7 ; b – c = 3; a - c  = 10 vào P, ta được:

P = 1/2 .(7^2 + 3^2 + 10^2)/(7^2 – 3^2)

= 1/2.(49 + 9 + 100)/(49 – 9)

= 1/2.158/40

= 158/80

= 79/40

# Chúc bạn học tốt!

\(a-b=7;b-c=3\text{ nên: }\left(a-b\right)+\left(b-c\right)=a-c=10\)

\(\text{tử P}=\frac{1}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\right]=\frac{1}{2}\left(3^2+7^2+10^2\right)=\frac{1}{2}.158=79\)

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

bạn ktra lại đề :)

             Bài làm :

Ta có phương trình hóa học :

hello bạn nha

Ta có x + y = 10

=> x = 10 - y

Khi đó P = xy + 1992

= (10 - y).y + 1992

= -y² + 10y + 1992

= -y² + 10y - 25 + 2017

= -(y² - 10y + 25) + 2017

= -(y - 5)² + 2017 \(\ge2017\)

Dấu "=" xảy ra <=> y - 5 = 0

=> y = 5

=> x = 5

Vậy Max P = 2017 <=> x = 5 ; y = 5

áp dụng định lý bơ du