(chứng minh rằng\) x y 3 −1 - Online Math

Ta có \(y^3-1=\left(y-1\right)\left(y^2+y+1\right)=-x\left(y^2+y+1\right)\)

(vì \(xy\ne0\Rightarrow x,y\ne0\))

\(\Rightarrow x-1\ne0;y-1\ne0\)

\(\Rightarrow\frac{x}{y^3-1}=\frac{-1}{y^2+y+1}\)

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

\(\Rightarrow\frac{x}{y^3-1}+\frac{y}{x^3-1}=\frac{-1}{y^2+y+1}+\frac{-1}{x^2+x+1}\)

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

\(=-\frac{4-2xy}{x^2y^2+3}\Rightarrow\frac{x}{y^3-1}+\frac{y}{x^3-1}-\frac{2\left(xy-2\right)}{x^2y^2+3}=0\)

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

\(=\frac{\left(x-1\right)\left(y+2\right)}{\left(x-1\right)\left(y-1\right)}=\frac{y+2}{y-1}\)

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

Phải đề thế này không

\(6^{5x+2}=36^{3x-4}\Leftrightarrow6^{5x+2}=6^{6x-8}\)

\(\Leftrightarrow5x+2=6x-8\)

\(\Leftrightarrow x=10\)

Ta có

\(\frac{a}{x}+\frac{a+1}{x-1}+\frac{a+2}{x-2}=\frac{\left(3a+3\right)x^2-\left(6a+4\right)x+2a}{x^3-3x^2+2x}=\frac{9x^2-16x+4}{x^3-3x^2+2x}\)

Dấu = xảy ra khi

\(\hept{\begin{cases}3a+3=9\\-6a-4=-16\\2a=4\end{cases}\Leftrightarrow a=2}\)

Ta có

\(4a^2+b^2=5ab\Leftrightarrow\left(4a^2-4ab\right)+\left(b^2-ab\right)=0\)

\(\Leftrightarrow\left(a-b\right)\left(4a-b\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}b=a\\b=4a\end{cases}}\)

Thế b = a vào M ta được

\(M=\frac{a.a}{4a^2-a^2}=\frac{1}{3}\)

Thế b = 4a vào M được

\(M=\frac{a.4a}{4a^2-16a^2}=-\frac{1}{3}\)

Ta có

\(M=\frac{16a^2-40ab}{8a^2-24ab}=\frac{16.\frac{a^2}{b^2}-40.\frac{a}{b}}{8.\frac{a^2}{b^2}-24.\frac{a}{b}}\)

\(=\frac{16.\left(\frac{10}{3}\right)^2-40.\frac{10}{3}}{8.\left(\frac{10}{3}\right)^2-24.\frac{10}{3}}=5\)

?????

2a²b²+ 2b²c²+ 2c²a²- a⁴- b⁴- c⁴

=4a²b²-(a⁴+2a²b²+b⁴)+(2b²c²+2a²c²)-c⁴

=2(ab)²-(a+b)²+2c²(a²+b²)-c⁴

=2(ab)²-[(a+b)²-2c²(a²+b²)+c⁴]

=2(ab)²-(b²+a²-c²)²

=(2ab+b²+a²-c²)(2ab-b²-a²+c²)

=[(a+b)²-c²][-(a-b)²+c²]

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

Vì a,b,c là 3 cạnh 1 tam giác nên:

a+b>c suy ra b+a-c>0

a+c>b suy ra a-b+c>0

a,b,c>0 suy ra a+b+c>0

b+c>a suy ra b+c-a>0

Vậy ta có điều phải chứng minh

dấu = thứ hai là (2ab)²- (a²+b²)²+2c²(a²+b²)-c⁴