![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a)P\left(x\right)=5x^5+3x-4x^4-2x^3+6+4x^2\)
\(P\left(x\right)=5x^5-4x^4-2x^3+4x^2+3x+6\)
\(Q\left(x\right)=2x^4-x+3x^2-2x^3+\dfrac{1}{4}-x^5\)
\(Q\left(x\right)=-x^5+2x^4-2x^3+3x^2-x+\dfrac{1}{4}\)
\(a)P\left(x\right)-Q\left(x\right)=\left(5x^5-4x^4-2x^3+4x^2+3x+6\right)+\left(-x^5+2x^4-2x^3+3x^2-x+\dfrac{1}{4}\right)\)
\(P\left(x\right)-Q\left(x\right)=5x^5-4x^4-2x^3+4x^2+3x+6+x^5-2x^4+2x^3-3x^2+x-\dfrac{1}{4}\)
\(P\left(x\right)-Q\left(x\right)=\left(5x^5+x^5\right)+\left(-4x^4-2x^4\right)+\left(-2x^3+2x^3\right)+\left(4x^2-3x^2\right)+\left(3x+x\right)+\left(6-\dfrac{1}{4}\right)\)
\(P\left(x\right)-Q\left(x\right)=6x^5-6x^4+x^2+4x+\dfrac{23}{4}\)
\(\text{c)Thay x=-1 vào biểu thức P(x),ta được:}\)
\(P\left(x\right)=5.\left(-1\right)^5-4.\left(-1\right)^4-2.\left(-1\right)^3+4.\left(-1\right)^2+3.\left(-1\right)+6\)
\(P\left(x\right)=\left(-5\right)-4-\left(-2\right)+4+\left(-3\right)+6\)
\(P\left(x\right)=\left(-9\right)-\left(-2\right)+4+\left(-3\right)+6\)
\(P\left(x\right)=\left(-7\right)+4+\left(-3\right)+6\)
\(P\left(x\right)=\left(-3\right)+\left(-3\right)+6\)
\(P\left(x\right)=\left(-6\right)+6=0\)
\(\text{Vậy giá trị của P(x) tại x=-1 là:0}\)
\(\text{Vậy =-1 là nghiệm của P(x)}\)
\(\text{Thay x=-1 vào biểu thức Q(x),ta được:}\)
\(Q\left(x\right)=\left(-1\right).5+2.\left(-1\right)^4-2.\left(-1\right)^3+3.\left(-1\right)^2-\left(-1\right)+\dfrac{1}{4}\)
\(Q\left(x\right)=\left(-5\right)+2-\left(-2\right)+3-\left(-1\right)+\dfrac{1}{4}\)
\(Q\left(x\right)=\left(-3\right)-\left(-2\right)+3-\left(-1\right)+\dfrac{1}{4}\)
\(Q\left(x\right)=\left(-5\right)+3-\left(-1\right)+\dfrac{1}{4}\)
\(Q\left(x\right)=\left(-2\right)-\left(-1\right)+\dfrac{1}{4}\)
\(Q\left(x\right)=\left(-3\right)+\dfrac{1}{4}=\dfrac{-13}{4}\)
\(\text{Vậy x=-1 không phải là nghiệm của Q(x)}\)
\(\text{d)Thay x=-1 vào biểu thức }P\left(x\right)-Q\left(x\right),\text{ta được:}\)
\(P\left(x\right)-Q\left(x\right)=6.\left(-1\right)^5-6.\left(-1\right)^4+\left(-1\right)^2+4.\left(-1\right)+\dfrac{23}{4}\)
\(P\left(x\right)-Q\left(x\right)=\left(-6\right)-6+1+\left(-4\right)+\dfrac{23}{4}\)
\(P\left(x\right)-Q\left(x\right)=\left(-12\right)+1+\left(-4\right)+\dfrac{23}{4}\)
\(P\left(x\right)-Q\left(x\right)=\left(-11\right)+\left(-4\right)+\dfrac{23}{4}\)
\(P\left(x\right)-Q\left(x\right)=\left(-15\right)+\dfrac{23}{4}=\dfrac{-37}{4}\)
\(\text{Vậy giá trị của P(x)-Q(x) tại x=-1 là:}\dfrac{-37}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(1-x\right)^2=2003.\left(x-1\right)\)
\(\left(1-x\right)^2-2003\left(x-1\right)=0\)
\(\left(1-x\right)^2+2003\left(1-x\right)=0\)
\(\left(1-x\right)\left(1-x+2003\right)=0\)
\(\left(1-x\right)\left(2004-x\right)=0\)
\(TH1:1-x=0\)
\(x=1\)
\(TH2:2004-x=0\)
\(x=2004\)
vậy........
![](https://rs.olm.vn/images/avt/0.png?1311)
\(C=\dfrac{-1}{5}+\left(\dfrac{1}{-5}\right)^2+\left(-\dfrac{1}{5}\right)^3+...+\left(-\dfrac{1}{5}\right)^{99}\)
=>\(5\cdot C=-1+\left(-\dfrac{1}{5}\right)+\left(-\dfrac{1}{5}\right)^2+...+\left(-\dfrac{1}{5}\right)^{98}\)
=>\(5\cdot C-C=\left(-1\right)-\left(-\dfrac{1}{5}\right)^{99}\)
=>\(4C=-1+\dfrac{1}{5^{99}}=\dfrac{-5^{99}+1}{5^{99}}\)
=>\(C=\dfrac{-5^{99}+1}{4\cdot5^{99}}\)
(x-3y)^2006+(y+4)^2008=0
=>x-3y=0 và y+4=0
=>x=3y và y=-4
=>x=3*(-4)=-12 và y=-4
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(\left(-3\right)^{x+3}=-\frac{1}{27}\)
\(\left(-3\right)^{x+3}=\left(-\frac{1}{3}\right)^3\)
\(\left(-3\right)^{x+3}=\left(-\frac{3^0}{3^1}\right)^3\)
\(\left(-3\right)^{x+3}=\left(-3^{-1}\right)^3\)
\(\left(-3\right)^{x+3}=\left(-3\right)^{-3}\)
\(\Rightarrow x+3=-3\)
\(\Rightarrow x=-6\)
b)\(\left(-6\right)^{2x+2}=\frac{1}{36}\)
\(\left(-6\right)^{2x+2}=\left(-\frac{1}{6}\right)^2\)
\(\left(-6\right)^{2x+2}=\left(-\frac{6^0}{6^1}\right)^2\)
\(\left(-6\right)^{2x+2}=\left(-6^{-1}\right)^2\)
\(\left(-6\right)^{2x+2}=\left(-6\right)^{-2}\)
\(\Rightarrow2x+2=-2\)
\(\Rightarrow2x=-4\)
\(\Rightarrow x=-2\)
c)\(\left(-3\right)^{x+5}=\frac{1}{81}\)
\(\left(-3\right)^{x+5}=\left(-\frac{1}{3}\right)^4\)
\(\left(-3\right)^{x+5}=\left(-\frac{3^0}{3^1}\right)^4\)
\(\left(-3\right)^{x+5}=\left(-3^{-1}\right)^4\)
\(\left(-3\right)^{x+5}=\left(-3\right)^{-4}\)
\(\Rightarrow x+5=-4\)
\(\Rightarrow x=-9\)
d)\(\left(\frac{1}{9}\right)^x=\left(\frac{1}{27}\right)^6\)
\(\left[\left(\frac{1}{3}\right)^2\right]^x=\left[\left(\frac{1}{3}\right)^3\right]^6\)
\(\left(\frac{1}{3}\right)^{2x}=\left(\frac{1}{3}\right)^{18}\)
\(\Rightarrow2x=18\)
\(\Rightarrow x=9\)
e)\(\left(\frac{4}{9}\right)^x=\left(\frac{8}{27}\right)^6\)
\(\left[\left(\frac{2}{3}\right)^2\right]^x=\left[\left(\frac{2}{3}\right)^3\right]^6\)
\(\left(\frac{2}{3}\right)^{2x}=\left(\frac{2}{3}\right)^{18}\)
\(\Rightarrow2x=18\)
\(\Rightarrow x=9\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(9^x:3^x=3^7\)
\(\Rightarrow9:3^x=3^7\)
\(\Rightarrow3^x=3^7\)
\(\Rightarrow x=7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Cau a la 1
Cau b la 1215
Cau c la 768
Cau d la \(\frac{4185}{13}\)
\(\frac{4^2.4^3}{2^{10}}=\frac{4^{2+3}}{\left(2^2\right)^5}=\frac{4^5}{4^5}=1\)
\(\frac{\left(0,6\right)^5}{\left(0,2\right)^6}=\frac{\left(0,2.3\right)^5}{\left(0,2\right)^6}=\frac{\left(0,2\right)^5.3^5}{\left(0,2\right)^6}=\frac{3^5}{0,2}=1215\)
\(\frac{2^7.9^3}{6^5.8^2}=\frac{2^2.2^5.\left(3^2\right)^3}{\left(2.3\right)^5.\left(2^3\right)^2}=\frac{2^2.2^5.3^6}{2^5.3^5.2^6}=\frac{3}{2^4}=\frac{3}{16}\)
\(\frac{6^3+3.6^2+3^3}{-13}=\frac{2^3.3^3+3.\left(2.3\right)^2+3^3}{-13}=\frac{2^3.3^3+3.2^2.3^2+3^3}{-13}=\frac{3^3.\left(2^3+2^2+1\right)}{-13}=\frac{3^3.13}{-13}=\left(-3\right)^3=-27\)
\(x-\dfrac{2}{9}=\dfrac{2}{36}\)
\(x=\dfrac{2}{36}+\dfrac{2}{9}\)
\(x=\dfrac{5}{18}\)