Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(9x^4-15x^3-6x^2+5\)
\(=3x^2\left(3x^2-5x\right)-6x^2+5\)
\(=3x^2.2-6x^2+5\)
\(=6x^2-6x^2+5\)
\(=5\)
\(9x^4-15x^3-6x^2+5\)
\(=3x^2\left(3x^2-5x\right)-6x^2+5\)
\(=3x^2.2-6x^2+5\)
\(=6x^2-6x^2+5\)
\(=5\)
Bài 1:
Ta có:
\(A=9x^4-15x^3-6x^2+5=3x^2\left(3x^2-5x\right)-6x^2+5=3x^2.2-6x^2+5=6x^2-6x^2+5=5\)
Vậy, \(A=5\)
Bài 2: Ta có:
\(3^{15}+3^{16}+3^{17}=3^{15}+3^{15}.3+3^{15}.3^2=3^{15}.\left(1+3+3^2\right)=3^{15}.13\)
\(\Rightarrow3^{15}.13\) chia hết cho \(13\)
Do đó: \(3^{15}+3^{16}+3^{17}\) chia hết cho \(13\)
\(1,\left(x+2\right)\left(x^2-2x+4\right)+\left(x+2\right)^2=0\)
\(\Rightarrow\left(x+2\right)^3+\left(x+2\right)^2=0\)
\(\Rightarrow\left(x+2\right)^2.\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x+2\right)^2=0\\x+3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+2=0\\x+3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)
Vậy....
a/15.91,5+150.0.85
=15.91,5+15.3,5
=15(91,5+8,5)
=15.100=1500
b/52.143-52.39-8.26
=52.143-52.39-4.52
=52(143-39-4)
=52.100=5200
c/9x^4-15x^3-6x^2+5 tại 3x^2-5x=2
Ta có :9x4-15x3-6x2+5=3x2(3x2-5x-2)+5 (1)
3x2-5x=2=>3x2-5x-2=0 (2)
thay(2) vào (1), ta được : 3x2.0+5=5
Bài 1:
a: \(\left(\dfrac{1}{3}x+2\right)\left(3x-6\right)\)
\(=x^2-3x+6x-12\)
\(=x^2+3x-12\)
b: \(\left(x+3\right)\left(x^2-3x+9\right)=x^3+27\)
c: \(\left(-2xy+3\right)\left(xy+1\right)\)
\(=-2x^2y^2-2xy+3xy+3\)
\(=-2x^2y^2+xy+3\)
d: \(x\left(xy-1\right)\left(xy+1\right)\)
\(=x\left(x^2y^2-1\right)\)
\(=x^3y^2-x\)
Bài 2:
a: Ta có: \(M=\left(3x+2\right)\left(9x^2-6x+4\right)\)
\(=27x^3+8\)
\(=27\cdot\dfrac{1}{27}+8=9\)
b: Ta có: \(N=\left(5x-2y\right)\left(25x^2+10xy+4y^2\right)\)
\(=125x^3-8y^3\)
\(=125\cdot\dfrac{1}{125}-8\cdot\dfrac{1}{8}\)
=0
9x4-15x3-6x2+5
=3x2(3x2-5x-2)+5
=3x2.0+5
=5