Ta có: \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)

\(\frac{\Rightarrow a}{x}=\frac{b}{y}=\frac{c}{z};a+b+c=1\)

ADTC dãy tỉ số bằng nhau ta có:

\(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}=\frac{a+b+c}{x+y+z}=\frac{1}{x+y+z}\)

\(\frac{\Leftrightarrow a^2}{x^2}=\frac{b^2}{y^2}=\frac{c^2}{z^2}=\frac{1}{\left(x+y+z\right)^2}\left(1\right)\)

Áp dụng tiếp tính chất dãy tỉ số bằng nhau ta đc:

\(\frac{a^2}{x^2}=\frac{b^2}{y^2}=\frac{c^2}{z^2}=\frac{a^2+b^2+c^2}{x^2+y^2+z^2}=\frac{1}{x^2+y^2+z^2}\left(2\right)\)

Từ (1) và (2) => (x+y+z)² = x² + y² + z²

=> (x+y+z)² - x² - y² - z² =0

=> 2.(xy+yz+xz) = 0

=> xy + yz + xz =0

Vậy.......

đpcm.

Ta có

\(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\Rightarrow\dfrac{x^2}{a^2}=\dfrac{y^2}{b^2}=\dfrac{z^2}{c^2}=\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}=x^2+y^2+z^2\) (1)

Ta có

\(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=\dfrac{x+y+z}{a+b+c}=x+y+z\)

\(\Rightarrow\dfrac{x^2}{a^2}=\dfrac{y^2}{b^2}=\dfrac{z^2}{c^2}=\left(x+y+z\right)^2=x^2+y^2+z^2+2\left(xy+yz+zx\right)\) (2)

Từ (1) và (2)

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

\(\Rightarrow xy+yz+zx=0\)

tôi ms lớp 7 

tick nhé mọi người

em mới lớp 5 nên khong bít

ai đó trả lời hộ tớ với

ta có: xy+x+y = 3

=> xy +x +y +1 =4

=> (x+1).(y+1) = 4 (1)

tương tự, ta có: (y+1).(z+1)= 9 (2)

(x+1).(z+1) = 16 (3)

Nhân (1);(2);(3) lại vs nhau

được: \([\left(x+1\right).\left(y+1\right).\left(z+1\right)]^2=576=24^2=\left(-24\right)^2.\)

TH1: (x+1).(y+1).(z+1) = 24

=> 4.(z+1)=24

=> z+1 = 6 => z = 5

mà yz +y +z = 8

=> 6y + 5 = 8 => y = 1/2

mà xz+z+x = 15

=> 6x + 5 = 15 => x = 5/3

=> P =  5/3 +1/2 + 5 = 43/6

TH2: (x+1).(y+1).(z+1) = -24

...

bn cũng lm tương tự như TH1 nha!

Áp dụng tính chất các dãy tỉ số bằng nhau ta có:

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

\(x=a\left(x+y+z\right)=x^2=a^2.\left(x+y+z\right)^2\)

\(y=b\left(x+y+z\right)=y^2=b^2\left(x+y+z\right)^2\)

\(z=c\left(x+y+z\right)=z^2=c^2.\left(x+y+z\right)^2\)

\(\Rightarrow x^2+y^2+z^2=a^2\left(x+y+z\right)^2+b^2\left(x+y+z\right)^2+c^2\left(x+y+z\right)^2\)

                         \(=\left(x+y+z\right)^2\left(a^2+b^2+c^2\right)=\left(x+y+z\right)^2\) (do \(a^2+b^2+c^2=1\))

lol

\(a,=\left(xy-1-x-y\right)\left(xy-1+x+y\right)\\ b,Sửa:a^3+2a^2+2a+1\\ =a^3+a^2+a^2+a+a+1=\left(a+1\right)\left(a^2+a+1\right)\\ c,=1-4a^2-a\left(a^2-4\right)=1-4a^2-a^3+4a\\ =\left(1-a\right)\left(1+a+a^2\right)+4a\left(1-a\right)\\ =\left(1-a\right)\left(1+5a+a^2\right)\\ d,=\left(a^2-a^2b^2\right)+\left(b^2-b\right)+\left(ab-a\right)\\ =a^2\left(1-b\right)\left(1+b\right)+b\left(b-1\right)+a\left(b-1\right)\\ =\left(b-1\right)\left(-a^2-ab+b+a\right)\\ =\left(b-1\right)\left(b-1\right)\left(a+b\right)\left(1-a\right)\)

\(e,=x^2y+xy^2-yz\left(y+z\right)+x^2z-xz^2\\ =\left(x^2y+x^2z\right)+\left(xy^2-xz^2\right)-yz\left(y+z\right)\\ =x^2\left(y+z\right)+x\left(y-z\right)\left(y+z\right)-yz\left(y+z\right)\\ =\left(y+z\right)\left(x^2+xy-xz-yz\right)\\ =\left(y+z\right)\left(x+y\right)\left(x-z\right)\)

\(f,=xyz-xy-yz-xz+x+y+z-1\\ =xy\left(z-1\right)-y\left(z-1\right)-x\left(z-1\right)+\left(x-1\right)\\ =\left(z-1\right)\left(xy-y-x+1\right)=\left(z-1\right)\left(x-1\right)\left(y-1\right)\)

Câu 1: 

\(a^2+b^2-a^2b^2+ab-a-b\)

\(=a^2\left(1-b^2\right)+b\left(b-1\right)+a\left(b-1\right)\)

\(=-a^2\left(b-1\right)\left(b+1\right)+\left(b-1\right)\left(a+b\right)\)

\(=\left(b-1\right)\left(-a^2b-a^2+a+b\right)\)

\(=\left(b-1\right)\cdot\left[-b\left(a^2-1\right)-a\left(a-1\right)\right]\)

\(=\left(b-1\right)\left(a-1\right)\left[-b\left(a+1\right)-a\right]\)