Cho \(x^2-x-1=0\).Tính \(Q=\frac{x^6-3x^5+3x^4-x^3+2017}{x^6-x^3-3x^2-3x+2017}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
bạn phân tích đa thức thành nhân tử ở tử thức và mẫu thức sao cho chứa nhân tử chung là x2 - x - 1 . Còn lại 2013/2012
a/ ĐKXĐ: \(x\ge2\)
\(\Leftrightarrow2\sqrt{\left(x-2\right)\left(x+2\right)}-6\sqrt{x-2}+\sqrt{x+2}-3=0\)
\(\Leftrightarrow2\sqrt{x-2}\left(\sqrt{x+2}-3\right)+\sqrt{x+2}-3=0\)
\(\Leftrightarrow\left(2\sqrt{x-2}+1\right)\left(\sqrt{x+2}-3\right)=0\)
\(\Leftrightarrow\sqrt{x+2}-3=0\Rightarrow x=11\)
b/ ĐKXĐ: ....
Đặt \(\left\{{}\begin{matrix}\sqrt{x-2016}=a>0\\\sqrt{y-2017}=b>0\\\sqrt{z-2018}=a>0\end{matrix}\right.\)
\(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{4}-\frac{a-1}{a^2}+\frac{1}{4}-\frac{b-1}{b^2}+\frac{1}{4}-\frac{c-1}{c^2}=0\)
\(\Leftrightarrow\frac{\left(a-2\right)^2}{a^2}+\frac{\left(b-2\right)^2}{b^2}+\frac{\left(c-2\right)^2}{c^2}=0\)
\(\Leftrightarrow a=b=c=2\Rightarrow\left\{{}\begin{matrix}x=2020\\y=2021\\z=2022\end{matrix}\right.\)
a/ ĐK: \(x\ge0\)
\(\Leftrightarrow\sqrt{3+x}=x^2-3\)
Đặt \(\sqrt{3+x}=a>0\Rightarrow3=a^2-x\) pt trở thành:
\(a=x^2-\left(a^2-x\right)\)
\(\Leftrightarrow x^2-a^2+x-a=0\)
\(\Leftrightarrow\left(x-a\right)\left(x+a+1\right)=0\)
\(\Leftrightarrow x=a\) (do \(x\ge0;a>0\))
\(\Leftrightarrow\sqrt{3+x}=x\Leftrightarrow x^2-x-3=0\)
d/ ĐKXĐ: ...
\(\sqrt{6x^2+1}=\sqrt{2x-3}+x^2\)
\(\Leftrightarrow\sqrt{2x-3}-1+x^2+1-\sqrt{6x^2+1}\)
\(\Leftrightarrow\frac{2\left(x-2\right)}{\sqrt{2x-3}+1}+\frac{x^4+2x^2+1-6x^2-1}{\left(x^2+1\right)^2+\sqrt{6x^2+1}}=0\)
\(\Leftrightarrow\frac{2\left(x-2\right)}{\sqrt{2x-3}+1}+\frac{x^2\left(x+2\right)\left(x-2\right)}{\left(x^2+1\right)^2+\sqrt{6x^2+1}}=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{2}{\sqrt{2x-3}+1}+\frac{x^2\left(x+2\right)}{\left(x^2+1\right)^2+\sqrt{6x^2+1}}\right)=0\)
\(\Leftrightarrow x=2\) (phần trong ngoặc luôn dương với mọi \(x\ge\frac{3}{2}\))
\(\frac{x^6-3x^5+3x^4-x^3+2015}{x^6-x^3-3x^2-3x+2015}=\frac{x^6-3x^5+3x^4+3x^3+2015-4x^3}{x^6+3x^3-3x^2-3x+2015-4x^3}=\frac{x^6-3x^3\left(x^2-x-1\right)+2015-4x^3}{6+3x\left(x^2-x-1\right)+2015-4x^3}\)
Theo bài ra: \(x^2-x-1=0\)
\(\frac{x^6-3x^5+3x^4-x^3+2015}{x^6-x^3-3x^2-3x+2015}=\frac{x^6-3x^3\left(x^2-x-1\right)+2015-4x^3}{x^6+3x\left(x^2-x-1\right)+2015-4x^3}=\frac{x^6+2015-4x^3}{x^6+2015-4x^3}=1\)
Vậy:...
Mk nhầm đoạn số 6 bạn sửa lại thành x^6 nhé:
\(\frac{x^6-3x^5+3x^4-x^3+2015}{x^6-x^3-3x^2-3x+2015}=\frac{x^6-3x^5+3x^4+3x^3+2015-4x^3}{x^6+3x^3-3x^2-3x+2015-4x^3}=\frac{x^6-3x^3\left(x^2-x-1\right)+2015-4x^3}{x^6+3x\left(x^2-x-1\right)+2015-4x^3}\)
Theo bài ra: \(x^2-x-1=0\)
\(\Rightarrow\frac{x^6-3x^5+3x^4-x^3+2015}{x^6-x^3-3x^2-3x+2015}=\frac{x^6-3x^3\left(x^2-x-1\right)+2015-4x^3}{x^6+3x\left(x^2-x-1\right)+2015-4x^3}=\frac{x^6+2015-4x^3}{x^6+2015-4x^3}=1\)
Vậy:......
\(2x-2=8-3x\)
\(\Leftrightarrow\)\(2x+3x=8+2\)
\(\Leftrightarrow\)\(5x=10\)
\(\Leftrightarrow\)\(x=2\)
Vậy...
\(x^2-3x+1=x+x^2\)
\(\Leftrightarrow\)\(x^2-3x-x-x^2=-1\)
\(\Leftrightarrow\)\(-4x=-1\)
\(\Leftrightarrow\)\(x=\frac{1}{4}\)
Vậy...
mấy cái này bấm máy tính là đc òi. giải mất thời gian lắm :))
Ta có : \(Q=\frac{x^6-3x^5+3x^4-x^3+2020}{x^6-x^3-3x^2-3x+2020}\)
=> \(Q=\frac{\left(x^6-x^5-x^4\right)+\left(-2x^5+2x^4+2x^3\right)+\left(2x^4-2x^3-2x^2\right)+\left(-x^3+x^2+x\right)+\left(x^2-x-1\right)+2021}{\left(x^6-x^5-x^4\right)+\left(x^5-x^4-x^3\right)+\left(2x^4-2x^3-2x^2\right)+\left(2x^3-2x^2-2x\right)+\left(x^2-x-1\right)+2021}\)
=> \(Q=\frac{x^4\left(x^2-x-1\right)-2x^3\left(x^2-x-1\right)+2x^2\left(x^2-x-1\right)-x\left(x^2-x-1\right)+\left(x^2-x-1\right)+2021}{x^4\left(x^2-x-1\right)+x^3\left(x^2-x-1\right)+2x^2\left(x^2-x-1\right)+\left(x^2-x-1\right)+2021}\)
=> \(Q=\frac{x^4.0-2x^3.0+2x^2.0-x.0+0+2021}{x^4.0+x^3.0+2x^2.0+0+2021}\)
=> \(Q=\frac{2021}{2021}=1\)
( 3x - 24 ) . 75 = 2.76 .1/20170
( 3x - 24 ) . 75 =235298
( 3x - 24 ) = 235298 : 75
( 3x - 24 ) =14
3x = 14 + 24
3x = 30
x = 0
dung 100%
\(\left(3x-2^4\right).7^5=2.7^6.\frac{1}{2017^0}\)
\(\Leftrightarrow\left(3x-16\right).7^5=2.7^6.1\)
\(\Leftrightarrow3x-16=\frac{2.7^6}{7^5}\)
\(\Leftrightarrow3x-16=2.7\)
\(\Leftrightarrow3x-16=14\)
\(\Leftrightarrow3x=30\)
\(\Leftrightarrow x=10\)
Ta có \(x^2-x-1=0\Rightarrow x^2-x=1\Rightarrow\left(x^2-x\right)^3=1\)
\(\Rightarrow x^6-3x^5+3x^4-x^3=1\)
Mặt khác \(x^2-x-1-0\Rightarrow x^2=x+1\)
\(\Rightarrow x^6=\left(x+1\right)^3=x^3+2=3x^2+3x+1\)
\(\Rightarrow P=\frac{1+2017}{1+2017}=1\)