\(f\left(x\right)=x^{312}-x^{313}+x^{314}-x^{315}+x^{446}\)

\(=\left(x^{312}-1\right)-\left(x^{313}-x\right)+\left(x^{314}-1\right)-\left(x^{315}-x\right)+\left(x^{446}-1\right)-2x+3\)

\(=\left[\left(x^2\right)^{156}-1\right]-x\left[\left(x^2\right)^{156}-1\right]+\left[\left(x^2\right)^{157}-1\right]-x\left[\left(x^2\right)^{157}-1\right]+\left[\left(x^2\right)^{223}-1\right]-2x+3\)

\(=\left(x^2-1\right)A_{\left(x\right)}-x\left(x^2-1\right)B_{\left(x\right)}+\left(x^2-1\right)C_{\left(x\right)}-x\left(x^2-1\right)D_{\left(x\right)}+\left(x^2-1\right)E_{\left(x\right)}-2x+3\)

Vậy số dư là -2x+3

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

để A = 0 => \(x^2+4x+4=0\)

\(x^2+4x+4=0\Rightarrow x^2+2x+2x+4=0\)

\(\Rightarrow x.\left(x+2\right)+2.\left(x+2\right)=0\)

\(\left(x+2\right)^2=0\Rightarrow x+2=0\Rightarrow x=-2\)

vậy để A=0 => x=-2

sorry nha

vì 

x+2 khác 0

=> x khác -2 

=> ko có giá trị x thỏa mãn A=0

1 cơ

đúng 1 thích

sai ko thích (ko t sai )

có 3 loại cơ

Ta có : 2x - 2 - 3x²

= -3x² + 2x - 2

= -3(x² + \(\frac{2}{3}\)x + \(\frac{2}{3}\))

= -3(x² + \(\frac{2}{3}\)x + \(\frac{1}{9}\) + \(\frac{5}{9}\))

= -3(x + \(\frac{1}{3}\) )² - \(\frac{15}{9}\)

Vì : -3(x + \(\frac{1}{3}\) )² \(\le0\)

=> -3(x + \(\frac{1}{3}\) )² - \(\frac{15}{9}\)\(\le-\frac{15}{9}\)

Vậy GTLN là : \(-\frac{15}{9}\)

Đặt \(A=2x-2-3x^2\)

\(A=-3\left(x^2-\frac{2}{3}x+\frac{2}{3}\right)\)

\(A=-3\left[x^2-2\cdot x\cdot\frac{1}{3}+\left(\frac{1}{3}\right)^2+\frac{5}{9}\right]\)

\(A=-3\left[\left(x-\frac{1}{3}\right)^2+\frac{5}{9}\right]\)

\(A=-3\left(x-\frac{1}{3}\right)^2-\frac{5}{3}\le\frac{-5}{3}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x-\frac{1}{3}=0\Leftrightarrow x=\frac{1}{3}\)

✓ ℍɠŞ ✓ sai dấu dòng thứ 2 nhé

TH1: a+b+c khác 0(a,b,c khác 0)

ta có:

\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\)

\(2+\frac{a+b-c}{c}=2+\frac{b+c-a}{a}=2+\frac{c+a-b}{b}\)

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

\(\Rightarrow a=b=c\)

thay a=b=c vào M ta có:

\(M=\left(1+\frac{a}{a}\right).\left(1+\frac{c}{c}\right).\left(1+\frac{b}{b}\right)\)

\(\Rightarrow M=\left(1+1\right).\left(1+1\right).\left(1+1\right)=2^3=8\)

TH2: a+b+c=0(a,b,c khác 0)

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

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

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

thay các giá trị của a,b,c vào bt M ta có:

\(M=\left(1+\frac{-c-a}{a}\right).\left(1+\frac{-b-a}{b}\right).\left(1+\frac{-c-b}{c}\right)\)

\(M=\frac{-c}{a}.\left(-\frac{a}{b}\right).\left(-\frac{b}{c}\right)\)

\(M=-1\)

vậy ...

p/s: bài này cô linh giải đúng rồi, nhưng ngắn quá => hơi khó hiểu

còn bn Darwin Watterson thiếu 1 trường hợp 

\(^{\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}=\frac{a+b-c+b+c-a+c+a-b}{a+b+c}=\frac{a+b+c}{a+b+c}=1}\)

Ta có: \(\frac{a+b-c}{c}=1\Rightarrow a+b-c=c\)

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

\(\frac{b+c-a}{a}=1\Rightarrow b+c-a=a\)

\(b+c=2a\)

\(\frac{c+a-b}{b}=1\Rightarrow c+a-b=b\)

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

\(M=\left(1+\frac{a}{b}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=\frac{a+b}{a}.\frac{c+a}{c}.\frac{b+c}{b}\)( Quy đồng lên như bình thường )

Thay từ trên vào biểu thức, ta có: \(=\frac{2a.2b.2c}{abc}=\frac{8\left(abc\right)}{abc}=8\)

Từ \(x^2+y=y^2+x\Leftrightarrow x^2-y^2-x+y=0\)

<=>(x-y)(x+y)-(x-y)=0

<=>(x-y)(x+y-1)=0

Vì x khác y => x+y-1=0

<=>x+y=1

<=> (x+y)²=1

<=> x²+y²=1-2xy

Thay vào A ta được: \(A=\frac{1-2xy+xy}{xy-1}=\frac{1-xy}{xy-1}=\frac{-\left(xy-1\right)}{xy-1}=-1\)

x#y sao lại suy ra x+y-1=0

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Leftrightarrow\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\Leftrightarrow\left(\frac{1}{x}+\frac{1}{y}\right)^3=\left(-\frac{1}{z}\right)^3\)

\(\Leftrightarrow\frac{1}{x^3}+\frac{3}{x^2y}+\frac{3}{xy^2}+\frac{1}{y^3}=\frac{-1}{z^3}\Leftrightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{3}{xy}\left(\frac{1}{x}+\frac{1}{y}\right)=\frac{-1}{z^3}\)

\(\Leftrightarrow\frac{1}{x^3}+\frac{1}{y^3}-\frac{3}{xyz}=-\frac{1}{z^3}\Leftrightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\)

Thay vào A ta đc: \(A=xyz\cdot\frac{3}{xyz}=3\)