Áp dụng bđt cauchy dạng engel ta có:

\(\frac{1}{a^2+b^2+1}+\frac{1}{b^2+c^2+1}+\frac{1}{c^2+a^2+1}\ge\frac{\left(1+1+1\right)^2}{a^2+b^2+b^2+c^2+c^2+a^2+1+1+1}\)

\(=\frac{9}{2\left(a^2+b^2+c^2\right)+3}\le\frac{9}{2\left(ab+bc+ca\right)+3}=\frac{9}{2.3+3}=1\left(đpcm\right)\)

Dấu "=" xảy ra khi a=b=c

Hình như bạn sai thì phải nhưng mình lỡ k r

1 bên \(\ge\)

1 bên \(\le\)

Sao so sánh đc

\(3\left(x-1\right)^2-3x\left(x-5\right)-2=0\)

\(3\left(x^2-2x+1\right)-3x^2+15x-2=0\)

\(3x^2-6x+3-3x^2+15x-2=0\)

\(9x+1=0\)

\(x=\frac{-1}{9}\)

\(3\left(x-1\right)^2-3x\left(x-5\right)-2=0\)

\(3.\left(x^2-2x+1\right)-3x\left(x-5\right)-2=0\)

\(3x^2-6x+3-3x^2+15x-2=0\)

\(9x=-1\)

\(x=\frac{-1}{9}\)

Đề là gì vậy bạn

a)x^3 - 3x^2 + 4x - 12 
= x^2(x - 3) + 4(x - 3) 
= (x - 3)(x^2 + 4) 

\(A=4x-x^2=4-\left(x^2-4x+4\right)=4-\left(x-2\right)^2\)

Vì \(\left(x-2\right)^2\ge0\forall x\)

\(\Rightarrow-\left(x-2\right)^2\le0\forall x\)\(\Rightarrow4-\left(X-2\right)^2\le4\forall X\)\(\Rightarrow A\le4\)

\(A=4\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Vậy: maxA=4\(\Leftrightarrow x=2\)

a) \(A=x^2+x+1\)

\(A=x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)

\(A=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x+\frac{1}{2}=0\Leftrightarrow x=\frac{-1}{2}\)

c) \(C=x^2\left(2-x^2\right)\)

\(C=2x^2-x^4\)

\(C=-\left(x^4-2x^2\right)\)

\(C=-\left[\left(x^2\right)^2-2\cdot x^2\cdot1+1^2-1\right]\)

\(C=-\left[\left(x^2-1\right)^2-1\right]\)

\(C=1-\left(x^2-1\right)^2\le1\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x^2-1=0\Leftrightarrow x=\left\{\pm1\right\}\)

Fe(NO₃)₂₊2HNO_3->Fe(NO₃)₃+NO₂₊H₂O

giup minh di 1 cau cx dc ma mih ca lam do

(x^2-x+2)^2+(x-2)^2 
= [(x^2-x+2)+(x-2)]^2-2[(x^2-x+2)*(x-2)] (áp dụng (a^2+b^2)=(a+b)^2-2ab 
=(x^2)^2- 2((x^3-3x^2+4x-4) 
=x^4-2x^3+6x^2-8x+8 
 giờ phân tích đa thức 
x^4-2x^3+6x^2+8x-8 
=(x^4-2x^3+2x^2)+(4x^2-8x+8) (cái này làm bài tập nhiêu nhìn ra nhanh) 
=[x^2(x^2-2x+2)]+4(x^2-2x+2) dẹp luôn 
=(x^2-2x+2)(x^2+4) 

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

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

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

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

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

1)I ...doesn't meet.....Paul today, but I........met.... him last Sunday.

(phần 1 bạn không cho từ trong ngoặc nhé)

2) Ellen .........always writes......with her left hand.(always/ to write)

Học tốt nhé

Bài làm

1) Doesn't meet    -   met

2) Always writes

P/s tham khảo nha