Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\Leftrightarrow\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1+\frac{x+324}{5}+1+\frac{25}{5}=5\)
\(\Leftrightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Leftrightarrow\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
Vì: \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\ne0\)\(\)
\(\Rightarrow x=-329\)
Đặt : x+3 = a
=> x+5 = a+2
pt <=> a^4+(a+2)^4 = 16
<=> a^4+a^4+8a^3+24a^2+32a+16 = 16
<=> 2a^4+8a^3+24a^2+32a = 0
<=> a^4+4a^3+12a^2+16a = 0
<=> a.(a^3+4a^2+12a+16) = 0
<=> a.[(a^3+2a^2)+(2a^4+4a)+(8a+16)] = 0
<=> a.(a+2).(a^2+2a+8) = 0
<=> a.(a+2) = 0 ( vì a^2+2a+8 > 0 )
<=> a=0 hoặc a+2=0
<=> a=0 hoặc a=-2
<=> x+3=0 hoặc x+3=-2
<=> x=-3 hoặc x=-5
Vậy ..............
Tk mk nha
Ta có: \(\left(x+3\right)^4+\left(x+5\right)^4=16\left(1\right)\)
Đặt x + 4 = y thì phương trình (1) trở thành:
\(\left(y-1\right)^4+\left(y+1\right)^4=16\)
\(\Leftrightarrow y^4-4y^3+6y^2-4y+1+y^4+4y^3+6y^2+4y+1=16\)
\(\Leftrightarrow2y^4+12y^2+2=16\)
\(\Leftrightarrow2\left(y^4+6y^2+1\right)=16\)
\(\Leftrightarrow y^4+6y^2+1=8\)
\(\Leftrightarrow y^4+6y^2+1-8=0\)
\(\Leftrightarrow y^4+7y^2-y^2-7=0\)
\(\Leftrightarrow y^2\left(y^2-1\right)-7\left(y^2-1\right)=0\)
\(\Leftrightarrow\left(y^2-7\right)\left(y^2-1\right)=0\)
Vì \(y^2-7\ne0\)
\(\Rightarrow y^2-1=0\Rightarrow y^2=1\Rightarrow y=\pm1\)
Với y = 1 => x + 4 = y => x + 4 = 1 => x = -3
Với y = -1 => x + 4 = y => x + 4 = -1 => x = -5
Vậy x = {-3;-5}
a,\(\left(x-4-5\right)\left(x-4+5\right)=0\Leftrightarrow\left(x-9\right)\left(x+1\right)=0\Leftrightarrow x=9;x=-1\)
b, \(\left(x-3-x-1\right)\left(x-3+x+1\right)=0\Leftrightarrow2x-2=0\Leftrightarrow x=1\)
c, \(\left(x^2-4\right)\left(2x-3\right)-\left(x^2-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(2x-3-x+1\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-2\right)=0\Leftrightarrow x=-2;x=2\)
d, \(\left(3x-7\right)^2-\left(2x+2\right)^2=0\Leftrightarrow\left(3x-7-2x-2\right)\left(3x-7+2x+2\right)=0\)
\(\Leftrightarrow\left(x-9\right)\left(5x-5\right)=0\Leftrightarrow x=1;x=9\)
Ta có: \(x^3-5x^2+6x=0\)
\(\Leftrightarrow x\left(x^2-5x+6\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=3\end{matrix}\right.\)
Vậy: S={0;2;3}
\(x^3-5x^2+6x=0\)
\(\Leftrightarrow x^3-2x^2-3x^2+6x=0\)
\(\Leftrightarrow x^2\left(x-2\right)-3x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^2-3x\right)\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=3\end{matrix}\right.\)
\(S=\left\{0,2,3\right\}\)
https://olm.vn/hoi-dap/question/129065.html
tham khảo ở đó bn nha
chúc hok tốt
cái này khác mà bạn?