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![](https://rs.olm.vn/images/avt/0.png?1311)
1 \(=\)\(\frac{46656}{216}\)\(=\)216
2\(=\)\(\frac{64}{1024}\)\(=\)\(\frac{1}{16}\)
3 \(=\)\(\frac{900}{-27000}\)\(=\)\(\frac{-1}{30}\)
4 \(=\)\(\frac{225}{-3375}\)\(=\)\(\frac{-1}{15}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài làm
a) Ta có:
\(P\left(x\right)=x^5-3x^2+7x^4-9x^3+x^2-\frac{1}{4}x\)
\(P\left(x\right)=x^5-2x^2+7x^4-9x^3-\frac{1}{4}x\)
\(P\left(x\right)=x^5+7x^4-9x^3-2x^2-\frac{1}{4}x\)
\(Q\left(x\right)=5x^4-x^5+x^2-2x^3+3x^2-\frac{1}{4}\)
\(Q\left(x\right)=5x^4-x^5-2x^3+4x^2-\frac{1}{4}\)
\(Q\left(x\right)=-x^5+5x^4-2x^3+4x^2-\frac{1}{4}\)
b) \(P\left(x\right)+Q\left(x\right)=x^5+7x^4-9x^3-2x^2-\frac{1}{4}x-x^5+5x^4-2x^3+4x^2-\frac{1}{4}\)
\(P\left(x\right)+Q\left(x\right)=12x^4-11x^3+2x^2-\frac{1}{4}x-\frac{1}{4}\)
Vậy \(P\left(x\right)+Q\left(x\right)=12x^4-11x^3+2x^2-\frac{1}{4}x-\frac{1}{4}\)
\(P\left(x\right)-Q\left(x\right)=x^5+7x^4-9x^3-2x^2-\frac{1}{4}x+x^5-5x^4+2x^3-4x^2+\frac{1}{4}\)
\(P\left(x\right)-Q\left(x\right)=2x^5-2x^4-7x^3-6x^2-\frac{1}{4}x-\frac{1}{4}\)
Vậy \(P\left(x\right)-Q\left(x\right)=2x^5-2x^4-7x^3-6x^2-\frac{1}{4}x-\frac{1}{4}\)
c) Ta có:
\(P\left(1\right)=1^5+7.1^4-9.1^3-2.1^2-\frac{1}{4}.1\)
\(P\left(1\right)=-\frac{13}{4}\)
Vậy giá trị của biểu thức P = -13/4 khi x = 1
\(Q\left(0\right)=-0^5+5.0^4-2.0^3+4.0^2-\frac{1}{4}\)
\(Q\left(0\right)=-\frac{1}{4}\)
Vậy \(Q\left(0\right)=-\frac{1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b, Ta có : \(\frac{2^7x9^2}{3^3x2^5}=\frac{2^52^2x3^23.3}{3.3^2x2^5}=\frac{2^2x3}{x}=12\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 273 : 32 = (33)3 : 32
= 39 : 32
= 37
b) (3/5)15 : (9/25)5 = (3/5)15 : [(3/5)2]5
= (3/5)15 : (3/5)10
= (3/5)2
\(9\). \(\left(\frac{-1}{3}\right)^3+\frac{1}{3}\)
= \(9\). \(\left(\frac{-1}{27}\right)+\frac{1}{3}\)
= \(\frac{-1}{3}+\frac{1}{3}\)
= \(0\)
\(\text{Giải :}\)
\(9.\left(\frac{-1}{3}\right)^3+\frac{1}{3}=9.\left(\frac{-1}{27}\right)+\frac{1}{3}\)
\(\frac{-1}{3}+\frac{1}{3}=0\)
\(\text{#Hok tốt!}\)