\(\left(x+2014\right)^2=64\left(X+2007\right)^3\)

Đặt x + 2007 = a ta được

\(\Leftrightarrow\left(a+7\right)^2=64a^3\)

\(\Leftrightarrow64a^3-a^2-14a-49=0\)

\(\Leftrightarrow\left(64a^3-64a^2\right)+\left(63a^2-63a\right)+\left(49a-49\right)=0\)

\(\Leftrightarrow\left(a-1\right)\left(64a^2+63a+49\right)=0\)

\(\Leftrightarrow a=1\)

\(\Leftrightarrow x+2007=1\)

\(\Leftrightarrow\)x = - 2006

sai de roi 2007 chu ko phai la 2017 sao ma dua nao cung viet sai de giong minh

Thế vào ta được

\(M=\frac{3.\frac{7^2}{3^2}b^2+5b^2+\frac{7}{3}b^2}{2.\frac{7^2}{3^2}b^2+4b^2-3.\frac{7}{3}b^2}\)

\(=\frac{\frac{49+15+7}{3}}{\frac{98+36-63}{9}}=\frac{\frac{71}{3}}{\frac{71}{9}}=3\)

Ta có: \(6a^2+ab=35b^2\)

\(\Leftrightarrow\left(6a^2-14ab\right)+\left(15ab-35b^2\right)=0\)

\(\Leftrightarrow\left(3a-7b\right)\left(2a+5b\right)=0\)

\(\Rightarrow3a=7b\Rightarrow a=\frac{7b}{3}\)

\(\Rightarrow M=3\)

Từ \(6a^2+ab=35b^2\)\(\Rightarrow6a^2+ab-35b^2=0\)

\(\Rightarrow6a^2+15ab-14ab-35b^2=0\)

\(\Rightarrow3a\left(2a+5b\right)-7b\left(2a+5b\right)=0\)

\(\Rightarrow\left(3a-7b\right)\left(2a+5b\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3a=7b\\2a=-5b\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}a=\frac{7b}{3}\\a=-\frac{5b}{2}\end{cases}}\)

Thay vao tinh....

\(\Leftrightarrow\) \(6x^5-12x^4-17x^4+34x^3-7x^3+14x^2+13x^2-26x-3x+\)6 =0

\(6x^5-29x^4+27x^3+27x^2-29x+6=0\)

\(\Leftrightarrow\left(6x^5-18x^4\right)+\left(-11x^4+33x^3\right)+\left(-6x^3+18x^2\right)+\left(9x^2-27x\right)+\left(-2x+6\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(6x^4-11x^3-6x^2+9x-2\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(\left(6x^4-12x^3\right)+\left(x^3-2x^2\right)+\left(-4x^2+8x\right)+\left(x-2\right)\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(6x^3+x^2-4x+1\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(\left(6x^3+6x^2\right)+\left(-5x^2-5x\right)+\left(x+1\right)\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(x+1\right)\left(6x^2-5x+1\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(x+1\right)\left(\left(6x^2-3x\right)+\left(-2x+1\right)\right)=0\)

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

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

Kiểm tra giùm nha Nguyễn Ngọc Đạt , hihi

A = x mũ 3 - y mũ 2 = xy

   =  x mũ 2 . ( x - 1 ) + x mũ 2 - y mũ 2 = xy

  =  ( x mũ 2 . ( x -1) + xy )  +  ( x mũ 2 + y mũ 2 - 2xy )  - 2y mũ 2  = 0

  =  ( x mũ 2 . ( x -1) + xy ) + ( x - y ) mũ 2 - 2y mũ 2 = 0

Vậy chỉ có nghiệm là 0

bạn biết rùi còn gì nữa

tk nha@@@@@@@@@@@@@@@@@@@2\

LOL

hihi