\(\left(x-3\right)^3+\left(x+1\right)^3+8\left(1-x\right)^3=0\)

\(\Leftrightarrow x^3-9x^2+27x-27+x^3+3x^2+3x+1+8\left(1-3x+3x^2-x^3\right)=0\)

\(\Leftrightarrow2x^3-6x^2+30x-26+8-24x+24x^2-8x^3=0\)

\(\Leftrightarrow-6x^3+18x^2+6x-18=0\)

\(\Leftrightarrow-6\left(x^3-3x^2-x+3\right)=0\)

\(\Leftrightarrow x^3-3x^2-x+3=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\sqrt{1}\end{cases}}}\)

#H

(4x-3)(2x-5) +(3-4x)(x-1)=0

(4x-3)(2x-5)-(4x-3)(x-1)=0

(4x-3)(2x-5-x+1)=0

(4x-3)(x-4)=0

4x-3=0 hoặc x-4=0

x=\(\frac{3}{4}\)hoặc x=4

Ta có \(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}\)

> \(\frac{a}{a+b+c+d}+\frac{b}{a+b+c+d}+\frac{c}{a+b+c+d}+\frac{d}{a+b+c+d}=1\)(1)

Tương tự ta chứng minh được \(\frac{b}{a+b}+\frac{c}{b+c}+\frac{d}{c+d}+\frac{a}{a+d}>1\)(2)

mà \(\frac{a}{a+b}+\frac{b}{a+b}+\frac{c}{b+c}+\frac{b}{b+c}+\frac{d}{c+d}+\frac{c}{c+d}+\frac{a}{a+d}+\frac{d}{a+d}=4\)(3)

Từ (1) (2) (3) => \(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}=2\left(a;b;c;d\inℕ\right)\)

\(x^4-25=0\)

\(\Leftrightarrow x^4=25\)

\(\Leftrightarrow x^2=5\)

\(\Leftrightarrow x=\sqrt{5}\)

#H

x=\(\sqrt{5}\)

ai k cho mình

nhắn mình mình k lại!

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

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

Biểu đồ cho (x^​4+​x^​(2*​y^​2)+​y^​4)*​(x^​2-​y^​2)*​(x^​6+​y^​6)

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