a)Đặt \(T=\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ac+c+1}=1\) (*)

Từ \(abc=1\Rightarrow c=\frac{1}{ab}\).Thay vào (*) ta có:

\(T=\frac{1}{1+a+ab}+\frac{1}{1+b+\frac{1}{a}}+\frac{1}{1+\frac{1}{ab}+\frac{1}{b}}\)

\(=\frac{1}{1+a+ab}+\frac{1}{\frac{a+ab+1}{a}}+\frac{1}{\frac{ab+1+a}{ab}}\)

\(=\frac{1}{a+ab+1}+\frac{a}{a+ab+1}+\frac{ab}{a+ab+1}\)

\(=\frac{a+ab+1}{a+ab+1}=1=VP\) (Đpcm)

b)Áp dụng Bđt Cô-si ta có:

\(\frac{a^2}{b^2}+\frac{b^2}{c^2}\ge2\sqrt{\frac{a^2}{b^2}\cdot\frac{b^2}{c^2}}=\frac{2a}{c}\)

\(\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge2\sqrt{\frac{b^2}{c^2}\cdot\frac{c^2}{a^2}}=\frac{2b}{a}\)

\(\frac{a^2}{b^2}+\frac{c^2}{a^2}\ge2\sqrt{\frac{a^2}{b^2}\cdot\frac{c^2}{a^2}}=\frac{2c}{b}\)

Cộng theo vế ta có:

\(\frac{2a^2}{b^2}+\frac{2b^2}{c^2}+\frac{2c^2}{a^2}\ge\frac{2a}{c}+\frac{2b}{a}+\frac{2c}{b}\)

\(\Leftrightarrow2\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)\ge2\left(\frac{c}{b}+\frac{b}{a}+\frac{a}{c}\right)\)

\(\Leftrightarrow\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{c}{b}+\frac{b}{a}+\frac{a}{c}\) (Đpcm)

Dấu = khi a=b=c

Bài 1:ta có a+b+c=0

=> a+b=-c      ;     a+c=-b           ;           b+c=-a

M= a(a+b)(a+c)= a(-c)(-b)=abc

N = b(b+c)(b+a)=b(-a)(-c)=abc

P=c(c+a)(c+b)= c(-b)(-a)=abc

=> M=N=P

vế trái= \(\left(b+c\right)^2\)-a²=(a+b+c)(b+c-a) = 2p(2p-a-a)=4p(p-a)= VP

=> đpcm

\(a\left(a-1\right)\le0\Rightarrow a^2\le a\)

Tương tự với b,c ta có đpcm.

Bài 1:

a. A = x^2 - 5x - 1

\(=x^2-5x+\frac{25}{4}-\frac{29}{4}\)

\(=x^2-5x+\left(\frac{5}{2}\right)^2-\frac{29}{4}\)

\(=\left(x-\frac{5}{2}\right)^2-\frac{29}{4}\ge0-\frac{29}{4}=-\frac{29}{4}\)

Dấu = khi x=5/2

Vậy MinC=-29/4 khi x=5/2

2. Tìm x:
a. ( 2x - 3 )^2 - ( 4x + 1 )( 4x - 1 ) = ( 2x - 1 ).( 3 - 7x )

=>4x²-12x+9+1-16x²=-14x²+13x-3

=>-12x²-12x+10=-14x²+13x-3

=>2x²-25x+13=0

\(\Rightarrow2\left(x-\frac{25}{4}\right)^2-\frac{521}{8}=0\)

\(\Rightarrow\left(x-\frac{25}{4}\right)^2=\frac{521}{16}\)

\(\Rightarrow x-\frac{25}{4}=\pm\sqrt{\frac{521}{16}}\)

\(\Rightarrow x=\frac{25}{4}\pm\frac{\sqrt{521}}{4}\)

c. 4.( x - 3 ) - ( x + 2 ) = 0

=>4x-12-x-2=0

=>3x-14=0

=>3x=14

=>x=14/3

a, \(\left(a+b+c\right)^2=4\left(a^2+b^2+c^2\right)\)

\(\Rightarrow a^2+b^2+c^2+2\left(ab+ac+bc\right)=3a^2+3b^2+3c^2\)

\(\Rightarrow2a^2+2b^2+2c^2=2ab+2ac+2bc\)

\(\Rightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)

\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2c+c^2\right)+\left(b^2-2bc+c^2\right)=0\)

\(\Rightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\)

Vì \(\left(a-b\right)^2\ge0;\left(a-c\right)^2\ge0;\left(b-c\right)^2\ge0\Rightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(a-c\right)^2=0\\\left(b-c\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a-b=0\\a-c=0\\b-c=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=b\\a=c\\b=c\end{matrix}\right.\Rightarrow a=b=c\)

=> đpcm

b, \(\left(a+b+c\right)^2=3\left(ab+bc+ca\right)\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=3\left(ab+bc+ca\right)\)

\(\Rightarrow a^2+b^2+c^2=ab+bc+ca\Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)

Đếm đây bạn chuyển vế rồi làm như phần a

a) Ta có: \(a+b+c=0\)

\(\Rightarrow2abc\left(a+b+c\right)=0\)

\(\Rightarrow2a^2bc+2ab^2c+2abc^2=0\)

Ta lại có:

\(a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)^2\)  (cái này bạn tự chứng minh nha)

\(\Rightarrow a^4+b^4+c^4=2a^2b^2+2b^2c^2+2c^2a^2+4a^2bc+4ab^2c+4abc^2\)

\(\Rightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2\right)\)

\(\Rightarrow a^4+b^4+c^4=2\left(ab+bc+ca\right)^2\left(đpcm\right)\)

b) Ta có: \(a+b+c=0\)

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

\(\Rightarrow a^2=b^2+c^2+2bc\)

\(\Rightarrow a^2-b^2-c^2=2bc\)

\(\Rightarrow a^4+b^4+c^4-2a^2b^2-2a^2c^2+2b^2c^2=4b^2c^2\)

\(\Rightarrow a^4+b^4+c^4=4b^2c^2+2a^2b^2+2a^2c^2-2b^2c^2\)

\(\Rightarrow a^4+b^4+c^4=2a^2b^2+2a^2c^2+2b^2c^2\)

\(\Rightarrow a^4+b^4+c^4+a^4+b^4+c^4=a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2\)

\(\Rightarrow2\left(a^4+b^4+c^4\right)=\left(a^2+b^2+c^2\right)^2\)

\(\Rightarrow a^4+b^4+c^4=\frac{\left(a^2+b^2+c^2\right)^2}{2}\left(đpcm\right)\)

Chúc bạn học tốt và tíck cho mìk vs nhé!

Cảm ơn bạn 

a) 

a)   n²−3n+5 :  n−2 = n - 1 (R=3) . Để phép chia hết nên suy ra:  n-1 thuộc Ư(3) . Suy ra : n = { 4 ; -2 ; 0 ; 2 }

Bạn làm bài kiểm tra hả sao nhiều bài tek. Mk làm mất khá nhiều tg luôn đó

Có một số câu thì mình không làm được. Mong bạn thông cảm!!!