\(a^2+b^2+c^2+d^2+e^2\ge ab+ac+ad+ae\left(1\right)\)

\(\Leftrightarrow a^2+b^2+c^2+d^2+e^2-a\left(b-c-d-e\right)\ge0\)

\(\Leftrightarrow\left(b^2-ab+\frac{1}{4}a^2\right)+\left(c^2-ac+\frac{1}{4}a^2\right)+\left(d^2-ad+\frac{1}{4}a^2\right)+\left(e^2-ae+\frac{1}{4}a^2\right)\ge0\)

\(\Leftrightarrow\left(b+\frac{1}{2}a\right)^2+\left(c+\frac{1}{2}a\right)^2+\left(d+\frac{1}{2}a\right)^2+\left(e+\frac{1}{2}a\right)^2\ge0\left(2\right)\)

( 2 ) đúng => ( 1 ) đúng 

a.(b-c)-a.(b+d)=-a.(c+d)

a.b-a.c-a.b+a.d=-a.(c+d)

(a.b-a.b)-(a.c+a.d)=-a.(c+d)

0-a.(c+d)=-a.(c+d)

-a.(c+d)=-a.(c+d)

Vậy a.(b-c)-a.(b+d)=-a.(c+d).

https://olm.vn/hoi-dap/question/41860.html

bn vào đây tham khảo nha

Đặt a/b=c/d=k

=>a=bk; c=dk

a: \(\dfrac{3a+2b}{a}=\dfrac{3bk+2b}{bk}=\dfrac{3k+2}{k}\)

\(\dfrac{3c+2d}{c}=\dfrac{3dk+2d}{dk}=\dfrac{3k+2}{k}\)

Do đó: \(\dfrac{3a+2b}{a}=\dfrac{3c+2d}{c}\)

b: \(\dfrac{2a-3b}{b}=\dfrac{2bk-3b}{b}=2k-3\)

\(\dfrac{2c-3d}{d}=\dfrac{2dk-3d}{d}=2k-3\)

Do đó: \(\dfrac{2a-3b}{b}=\dfrac{2c-3d}{d}\)

c: \(\dfrac{a}{a-2b}=\dfrac{bk}{bk-2b}=\dfrac{k}{k-2}\)

\(\dfrac{c}{c-2d}=\dfrac{dk}{dk-2d}=\dfrac{k}{k-2}\)

Do đó: \(\dfrac{a}{a-2b}=\dfrac{c}{c-2d}\)

Có VT = a² + 2ab + b² + a² - 2ab + b² 

= 2a² + 2b² = 2(a² + b²) (= VP)

Vậy  (a + b)² + (a - b)² = 2(a² + b²)

\(-\left(-a+b+c\right)+\left(b-c-1\right)=\left(b-c+6\right)-\left(7-a+b\right)+c\)

\(a-b-c+b-c-1=b-c+6-7+a-b+c\)

\(a-2c-1=a-1\)

\(-2c\ne0\)hay đẳng thức ko xảy ra 

Trieu Trong Thai

 CM a3+b3+c2 >= ab+bc+ac (*) 
2a^2 +2b^2 +2c^2 - 2ab -2bc -2ac = (a-b)^2 + (b-c)^2 + (a-c)^2 >= 0 

từ * => a^2 +b^2+c^2 +2ab+2bc+2ac >= 3ab+3bc+3ac <=> (a+b+c)^2 >= 3ab +3ac+3bc 
từ * => 2ab +2ac+2bc+ a^2+b^2+c^2 =< 3a^2+3b^2+3c^2 <=> (a+b+c)^2 =< ... 

de bai sai sua lai la

\(a^3-b^3+ab\left(b-a\right)=\left(a-b\right)\left(a+b\right)^2\)

bien doi ve phai ta co:

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

\(=a^3+ab^2-a^2b-b^3\)

\(=a^3-b^3+ab\left(b-a\right)\)= ve trai

vay \(a^3-b^3+ab\left(b-a\right)=\left(a-b\right)\left(a+b\right)^2\)