mik nghi bang 6

nhg chắc ko phải

Bạn ơi ko có x nào thỏa mãn thì tính thế nào hả bn???

9x⁴-15x³-6x²+5

=3x²(3x²-5x-2)+5

=3x².0+5

=5

`15x^2-25x+18`

`=5.(3x^2-5x+6)+12`

`=5.2+12`

`=22`

\(9x^4-15x^3-6x^2+5\)

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

\(=3x^2.2-6x^2+5\)

\(=6x^2-6x^2+5\)

\(=5\)

\(9x^4-15x^3-6x^2+5\)

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

\(=3x^2.2-6x^2+5\)

\(=6x^2-6x^2+5\)

\(=5\)

1) \(4-25x^2=0\)

\(\Rightarrow\left(2-5x\right)\left(2+5x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}5x=2\\5x=-2\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\)

2) Tính thì phải cho giá trị của x.

\(A=x^3-3x^2+3x-1\)

\(=\left(x-1\right)^3\)

\(=\left[{}\begin{matrix}\left(\dfrac{5}{2}-1\right)^3=\dfrac{27}{8}\\\left(-\dfrac{5}{2}-1\right)^3=-\dfrac{343}{8}\end{matrix}\right.\)

\(3x^2-5x+3=2\)

\(5\left(3x^2-5x+3\right)=2.5\)

=>\(15x^2-25x+15=10\)

\(15x^2-25x+15+3=10+3\)

\(15x^2-25x+18=13\)

15x²-25x+18

=(3x².5-5.5x+15)+3

=5(3x²-5x+3)+3

=5.2+3

=10+3=13

\(E=33\left(\dfrac{2}{3}x-1\right)+\left(15x^2-10x\right):\left(-5x\right)-\left(3x-1\right)\)

\(=22x-33-3x+2-3x+1\)

\(=16x-30\)