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Một cách thôi nha 2 cách lòi ruột đấy :
\(A=\left(5x+3\right)^3\)
\(=\left(5x\right)^3+3.\left(5x\right)^2.3+3.5x.9+3^3\)
\(=125x^3+225x^2+135x+27\)
\(B=\left(8x-5\right)^3\)
\(=\left(8x\right)^3-3.\left(8x\right)^2.5+3.8x.5^2-5^3\)
\(=512x^3-960x^2+600x-125\)
\(C=\left(5x-1\right)\left(25x^2-5x+1\right)\)
Sai rồi nha bạn phải là : \(\left(5x-1\right)\left(25x^2+5x+1\right)\)
\(=\left(5x\right)^3-1^3\)
\(=125x^3-1\)
\(D=\left(x+3\right)\left(x^2-3x+1\right)\)
\(=x^3+3^3\)
\(=x^3+27\)
a: \(A=\dfrac{2\cdot8^4\cdot27^2+44\cdot6^9}{2^7\cdot6^7+2^7\cdot40\cdot9^4}\)
\(=\dfrac{2\cdot2^{12}\cdot3^6+2^2\cdot11\cdot2^9\cdot3^9}{2^7\cdot3^7\cdot2^7+2^7\cdot2^3\cdot5\cdot3^8}\)
\(=\dfrac{2^{13}\cdot3^6+2^{11}\cdot3^9\cdot11}{2^{14}\cdot3^7+2^{10}\cdot5\cdot3^8}\)
\(=\dfrac{2^{11}\cdot3^6\left(2^2+3^3\cdot11\right)}{2^{10}\cdot3^7\left(2^4+5\cdot3\right)}\)
\(=\dfrac{2\cdot301}{3\cdot31}=\dfrac{602}{93}\)
\(\left\{{}\begin{matrix}3\left(x-1\right)=2\left(y-2\right)\\5\left(y-2\right)=4\left(z-3\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3\left(x-1\right)}{6}=\dfrac{2\left(y-2\right)}{6}\\\dfrac{5\left(y-2\right)}{20}=\dfrac{4\left(z-3\right)}{20}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=\dfrac{y-2}{3}\\\dfrac{y-2}{4}=\dfrac{z-3}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x-1}{8}=\dfrac{y-2}{12}\\\dfrac{y-2}{12}=\dfrac{z-3}{15}\end{matrix}\right.\Leftrightarrow\dfrac{x-1}{8}=\dfrac{y-2}{12}=\dfrac{z-3}{15}\Leftrightarrow\dfrac{2x-2}{16}=\dfrac{3y-6}{36}=\dfrac{z-3}{15}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2x-2}{16}=\dfrac{3y-6}{36}=\dfrac{z-3}{15}=\dfrac{2x-2+3y-6-z+3}{16+36-15}=\dfrac{\left(2x+3y-z\right)+\left(3-2-6\right)}{37}=\dfrac{79-5}{37}=\dfrac{74}{37}=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.8+1=17\\y=2.12+2=26\\z=2.15+3=33\end{matrix}\right.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{3a+b+c}{a}=\dfrac{a+3b+c}{b}=\dfrac{a+b+3c}{c}=\dfrac{3a+b+c+a+3b+c+a+b+3c}{a+b+c}=\dfrac{5a+5b+5c}{a+b+c}=\dfrac{5\left(a+b+c\right)}{a+b+c}=5\)\(\Rightarrow\left\{{}\begin{matrix}3a+b+c=5a\\a+3b+c=5b\\a+b+3c=5c\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b+c=2a\\a+c=2b\\a+b=2c\end{matrix}\right.\)
\(M=\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}=\dfrac{2c}{c}+\dfrac{2a}{a}+\dfrac{2b}{b}=2+2+2=6\)
Sử dụng hằng đẳng thức: \(x^3-y^3=\left(x-y\right)\left(x^2+xy+y\right)\)và \(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\)
Lưu ý: \(\left(x\pm y\right)=x^2\pm2xy+y^2\)