tìm x là nghiệm nguyên dương của phương trình \(\frac{x^2\left(4x^6-2x^3+1\right)}{12^{x^2-4x+3}}=\frac{8x^9+1}{6^{x^2-4x+3}+8^{x^2-4x+3}+9^{x^2+4x+3}}\)
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R
30 tháng 8 2017
Xét x=2 , loại . \(=>x\in Z^+,x\ne2.\\ \)
\(=>a=x^2-4x+3\ge0,x\ne2.\\
\)
\(pt=>\left(\frac{1}{2}\right)^a+\left(\frac{2}{3}\right)^a+\left(\frac{3}{4}\right)^a=2x+\frac{1}{x^2},x\ne0\\
\)
BĐT nhỉ haha:V
HP
2 tháng 2 2021
1.
\(x^4-6x^2-12x-8=0\)
\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)
\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)
\(\Leftrightarrow x=1\pm\sqrt{5}\)
HP
2 tháng 2 2021
3.
ĐK: \(x\ge-9\)
\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)
\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)
Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)
\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)
\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)
\(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 6 2019
b/ \(3-100x+8x^2=8x^2+x-300\)
\(\Leftrightarrow-101x=-303\)
\(\Rightarrow x=3\)
c/ \(5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-150\)
\(\Leftrightarrow25x+10-80x+10=24x+12-150\)
\(\Leftrightarrow-79x=-158\)
\(\Rightarrow x=2\)
d/ \(3\left(3x+2\right)-\left(3x+1\right)=12x+10\)
\(\Leftrightarrow9x+6-3x-1=12x+10\)
\(\Leftrightarrow-6x=5\)
\(\Rightarrow x=-\frac{5}{6}\)
e/ \(30x-6\left(2x-5\right)+5\left(x+8\right)=210+10\left(x-1\right)\)
\(\Leftrightarrow30x-12x+30+5x+40=210+10x-10\)
\(\Leftrightarrow13x=130\)
\(\Rightarrow x=10\)
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 6 2019
\(A=x^2-4x+1=\left(x-2\right)^2-3\ge-3\)
\(\Rightarrow A_{min}=-3\) khi \(x=2\)
\(B=4x^2+4x+11=\left(2x+1\right)^2+10\ge10\)
\(\Rightarrow B_{min}=10\) khi \(x=-\frac{1}{2}\)
\(C=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(=\left(x^2+5x\right)^2-36\ge-36\)
\(\Rightarrow C_{min}=-36\) khi \(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
\(D=-x^2-8x-16+21=21-\left(x+4\right)^2\le21\)
\(\Rightarrow C_{max}=21\) khi \(x=-4\)
\(E=-x^2+4x-4+5=5-\left(x-2\right)^2\le5\)
\(\Rightarrow E_{max}=5\) khi \(x=2\)
4 tháng 5 2019
b, \(\frac{1}{x-1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\left(ĐKXĐ:x\ne\pm1;x\ne2\right)\)
\(\Leftrightarrow\)\(\frac{1}{x-1}+\frac{5}{2-x}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
\(\Leftrightarrow\)\(\frac{\left(x+1\right)\left(2-x\right)+5\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(2-x\right)\left(x-1\right)}=\frac{15\left(x-1\right)}{\left(x-1\right)\left(x+1\right)\left(2-x\right)}\)
Suy ra:
\(\Leftrightarrow\)(x+1)(2-x)+5(x-1)(x+1) = 15(x-1)
\(\Leftrightarrow\)2x-x2-x+2+5x2-5 = 15x-15
\(\Leftrightarrow\)2x-x2-x+5x2-15x = -15+5-2
\(\Leftrightarrow\)4x2-14x = -12
\(\Leftrightarrow4x^2-14x+12=0\)
\(\Leftrightarrow4x^2-8x-6x+12=0\)
\(\Leftrightarrow\)4x(x-2) - 6(x-2) = 0
\(\Leftrightarrow\left(x-2\right)\left(4x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\4x-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(kotm\right)\\x=\frac{3}{2}\left(tm\right)\end{matrix}\right.\)
Vậy pt có nghiệm duy nhất x = \(\frac{3}{2}\)
23 tháng 3 2019
\(\frac{4x}{x^2+4x+3}-1=6\left(\frac{1}{x+3}-\frac{1}{2x+2}\right)\) \(ĐK:x\ne-1;x\ne-3\)
\(\Leftrightarrow\frac{4x}{x^2+4x+3}-\frac{x^2+4x+3}{x^2+4x+3}=6\left[\frac{2\left(x+1\right)}{2\left(x+3\right)\left(x+1\right)}-\frac{x+3}{2\left(x+1\right)\left(x+3\right)}\right]\)
\(\Leftrightarrow\frac{4x-x^2-4x-3}{x^2+4x+3}=6\left[\frac{2\left(x+1\right)-x-3}{2\left(x+3\right)\left(x+1\right)}\right]\)
\(\Leftrightarrow\frac{-x^2-3}{x^2+4x+3}=6\left[\frac{2x+2-x-3}{2\left(x^2+4x+3\right)}\right]\)
\(\Leftrightarrow\frac{-x^2-3}{x^2+4x+3}=\frac{6\left(x-1\right)}{2\left(x^2+4x+3\right)}\)
\(\Leftrightarrow\frac{-x^2-3}{x^2+4x+3}=\frac{3\left(x-1\right)}{x^2+4x+3}\)
\(\Leftrightarrow-x^2-3=3x-3\)
\(\Leftrightarrow-x^2-3x=0\)
\(\Leftrightarrow-x\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-3\left(loại\right)\end{cases}}\)
Vậy x = 0
23 tháng 3 2019
\(ĐK:x\ne\frac{-1}{2};x\ne\frac{-3}{2}\)
\(\frac{3}{2x+1}=\frac{6}{2x+3}+\frac{8}{4x^2+8x+3}\)
\(\Leftrightarrow\frac{3}{2x+1}-\frac{6}{2x+3}=\frac{8}{4x^2+8x+3}\)
\(\Leftrightarrow\frac{3\left(2x+3\right)-6\left(2x+1\right)}{\left(2x+1\right)\left(2x+3\right)}=\frac{8}{4x^2+8x+3}\)
\(\Leftrightarrow\frac{6x+9-12x-6}{4x^2+8x+3}=\frac{8}{4x^2+8x+3}\)
\(\Leftrightarrow-6x+3=8\)
\(\Leftrightarrow x=-\frac{5}{6}\)
Vậy ...
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 4 2019
\(\frac{6}{x^2+2}+\frac{7}{x^2+3}+\frac{12}{x^2+8}-\frac{3\left(x^2+10\right)-14}{x^2+10}-1=0\)
\(\Leftrightarrow\frac{6}{x^2+2}+\frac{7}{x^2+3}+\frac{12}{x^2+8}+\frac{14}{x^2+10}-4=0\)
\(\Leftrightarrow\Leftrightarrow\frac{6}{x^2+2}-1+\frac{7}{x^2+3}-1+\frac{12}{x^2+8}-1+\frac{14}{x^2+10}-1=0\)
\(\Leftrightarrow\frac{4-x^2}{x^2+2}+\frac{4-x^2}{x^2+3}+\frac{4-x^2}{x^2+8}+\frac{4-x^2}{x^2+10}=0\)
\(\Leftrightarrow\left(4-x^2\right)\left(\frac{1}{x^2+2}+\frac{1}{x^2+3}+\frac{1}{x^2+8}+\frac{1}{x^2+10}\right)=0\)
\(\Leftrightarrow4-x^2=0\) (do \(\frac{1}{x^2+2}+\frac{1}{x^2+3}+\frac{1}{x^2+8}+\frac{1}{x^2+10}>0\))
\(\Rightarrow x=\pm2\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 4 2019
b/
\(2x\left(4x-1\right)\left(8x-1\right)^2=9\)
\(\Leftrightarrow\left(8x^2-2x\right)\left(64x^2-16x+1\right)-9=0\)
Đặt \(8x^2-2x=a\Rightarrow64x^2-16x=8a\)
\(a\left(8a+1\right)-9=0\)
\(\Leftrightarrow8a^2+a-9=0\)
\(\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{9}{8}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}8x^2-2x-1=0\\8x^2-2x+\frac{9}{8}=0\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{4}\end{matrix}\right.\)
dài thế để tôi nghĩ đã
x=+-10;x=1+431/1000;x=-1893/2500;x=-7543/10000;x=1