\(a,VT=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)

\(VT+3=\frac{a}{b+c}+1+\frac{b}{a+c}+1+\frac{c}{a+b}\)

\(VT+3=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}\)

\(VT+3=\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)

\(2\left(VT+3\right)=\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)

áp dụng bđt Cô si :

\(\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\ge3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\)

\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge3\sqrt[3]{\frac{1}{a+b}\cdot\frac{1}{b+c}\cdot\frac{1}{c+a}}\)

\(\Rightarrow2\left(VT+3\right)\ge3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\cdot3\sqrt[3]{\frac{1}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)

\(\Rightarrow2\left(VT+3\right)\ge9\) \(\Leftrightarrow VT\ge\frac{3}{2}\)

chưa có dấu = xảy ra kìa, khi a=b=c > 0

\(b,\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}=\frac{a^2}{a\left(b+2c\right)}+\frac{b^2}{b\left(c+2a\right)}+\frac{c^2}{c\left(a+2b\right)}\)

\(\Rightarrow vt\ge\frac{\left(a+b+c\right)^2}{3\left(ab+bc+ca\right)}\)

có : \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)

\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)

\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca\ge3\left(ab+bc+ca\right)\)

\(\Leftrightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

nên \(vt\ge\frac{3\left(ab+bc+ca\right)}{3\left(ab+cb+ca\right)}=1\)

dấu = xảy ra khi a=b=c

a.\(4xy-8x^3y=4xy\left(1-2x^2\right)\)

b.\(5\left(x^2-y^2\right)+16\left(x-y\right)=\left(x-y\right)\left[5\left(x+y\right)+16\right]\)

c.\(x^3-4x^2y+4xy^3=x\left(x-2y\right)^2\)

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

e.\(x^2y-2x^2+4y-8=\left(y-2\right)\left(x^2+4\right)\)

g. \(x^3+y^3+z^3-3xyz=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)

A = (2x –3)2+ (x –1)(x + 1)​

A = 5x^2-12x+8

rút gọn rồi vì ko biết đề bài 

nha bạn chúc bạn học tốt nha 

Ta có: 

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

\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)

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

Dấu \(=\)khi \(a=b=c=\frac{1}{\sqrt{3}}\). 

Trước tiên ta xét bđt phụ : a² + b² + c² ≥ ab + bc + ca

<=> 2a² + 2b² + 2c² - 2ab - 2bc - 2ca ≥ 0

<=> ( a - b )² + ( b - c )² + ( c - a )² ≥ 0 *đúng* . Dấu "=" <=> a=b=c

Áp dụng bđt Cauchy-Schwarz dạng Engel và bđt phụ trên : \(VT\ge\frac{\left(1+1+1\right)^2}{a^2+b^2+c^2+ab+bc+ca}\ge\frac{9}{2\left(a^2+b^2+c^2\right)}=\frac{9}{2\cdot1}=2\)

=> đpcm . Dấu "=" xảy ra <=> a=b=c=1/√3

ta có :

\(3^{75}+2^{100}=27^{25}+16^{25}\)chia hết cho \(27+16=43\)

vậy ta có đpcm

\(E=4x^4+52x^3+171x^2+13x-30\)

\(E=4x^4+24x^3+28x^3+168x^2+3x^2+18x-5x-30\)

\(E=4x^3\left(x+6\right)+28x^2\left(x+6\right)+3x\left(x+6\right)-5\left(x+6\right)\)

\(E=\left(4x^3+28x^2+3x-5\right)\left(x+6\right)\)

\(E=\left(4x^3+2x^2+26x^2+13x-10x-5\right)\left(x+6\right)\)

\(E=\left[2x^2\left(2x+1\right)+13x\left(2x+1\right)-5\left(2x+1\right)\right]\left(x+6\right)\)

\(E=\left(2x^2+13x-5\right)\left(2x+1\right)\left(x+6\right)\)

\(D=4\left(x^2+11x+30\right)\left(x^2+22x+120\right)-3x^2\)

\(D=4\left(x+5\right)\left(x+6\right)\left(x+10\right)\left(x+12\right)-3x^2\)

\(D=4\left(x^2+17x+60\right)\left(x^2+16x+60\right)-3x^2\)

đặt \(x^2+17x+60=a\)

\(D=4\left(a-x\right)a-3x^2\)

\(D=4a^2-4ax-3x^2\)

đến đây bí

ủa gòi câu d là câu nào 

bài 5 có câu d đâu bạn ơi

\(D=3\left(x^2-2\times\frac{2}{3}x+\frac{4}{9}\right)+\frac{11}{3}=3\left(x-\frac{2}{3}\right)^2+\frac{11}{3}\ge\frac{11}{3}\)

\(E=2\left(x^2+xy+\frac{y^2}{4}\right)+\frac{1}{2}\left(y^2-8y+16\right)+9=2\left(x+\frac{y}{2}\right)^2+\frac{1}{2}\left(y-4\right)^2+9\ge9\)

phân tích thành nhân tử: x^3/27+x^6/729+10x^3/9

= x^3/27+x^6/729+10x^3/9

= 10x^12/177147

nha bạn