25x 5+ 25x 6+ 25x 9

=25*(5+6+9)

=25*20

=500

Dấu * là nhân nhé bạn.

25x(5+6+9)= 25x20= 500

x^4 -25x^2 + 60x - 36 = 0

tách ra nhé.. Tớ có nghiệm này.. x=1,2,3,6 

\(x^4-10x^3+25x^2=36\)

➜\(x^4-10x^3=25x^2-36=0\)

➜\(x^3\left(x-3\right)-7x^2\left(x-3\right)+4x\left(x-3\right)+12\left(x-3\right)=0\)

➜\(\left(x-3\right)\left(x^3-7x^2+x+12\right)=0\)

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

➜\(\left[{}\begin{matrix}x-3=0\\x-2=0\\x+1=0\\x-6=0\end{matrix}\right.\text{➜}\left[{}\begin{matrix}x=3\\x=2\\x=-1\\x=6\end{matrix}\right.\)

Vậy..................................................

Ta có: \(x^4-10x^3+25x^2=36\Leftrightarrow x^4-10x^3+25x^2-36=0\Leftrightarrow x^4+x^3-11x^3-11x^2+36x^2-36=0\)

\(\Leftrightarrow x^3\left(x+1\right)-11x^2\left(x+1\right)+36\left(x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^3-11x^2+36x-36\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\\x=3\\x=6\end{matrix}\right.\)

đề bài quá ẩu

cho VT=0 thì nghiệm khá đẹp chắc là tìm nghiệm đa thức

https://h.vn/hoi-dap/question/238231.html?pos=815256

\(25x\left(35-47\right)-35x\left(25-47\right)\)

\(=5x5x\left(35-47\right)-5x7x\left(25-47\right)\)

\(=5x\left(175-235\right)-5x\left(175-329\right)\)

\(=5x\left(-60\right)-5x\left(-154\right)\)

\(=5x\left(-60+154\right)\)

\(=5x94=470\)

x=24 nên x+1=25

C=x^4-x^3(x+1)+x^2(x+1)-x(x+1)+30

=x^4-x^4-x^3+x^3+x^2-x^2-x+30

=-x+30

=30-24

=6

\(\Leftrightarrow6x^4+12x^3+13x^3+26x^2-14x-28x+3x+6=0\)

\(\Leftrightarrow\left(x+2\right)\left(6x^3+13x^2-14x+3\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(6x^3+18x^2-5x^2-13x+x+3\right)=0\)

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

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

hay \(x\in\left\{-2;-3;\dfrac{1}{2};\dfrac{1}{3}\right\}\)