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DP
4 tháng 6 2019
a, \(2x-\frac{1}{2}=\frac{2x+1}{4}-\frac{1-2x}{8}\)
\(\Leftrightarrow\frac{1}{2}\left(4x-1\right)=\frac{1}{8}\left(6x+1\right)\)
\(\Leftrightarrow4\left(4x-1\right)=6x+1\)
\(\Leftrightarrow10x=5\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy x = \(\frac{1}{2}\)
b, \(\frac{x-3}{13}+\frac{x-3}{14}=\frac{x-3}{15}+\frac{x-3}{16}\)
\(\Leftrightarrow\frac{x-3}{13}+\frac{x-3}{14}-\frac{x-3}{15}-\frac{x-3}{16}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{13}+\frac{1}{14}-\frac{1}{15}-\frac{1}{16}\right)=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
Vậy x = 3
KN
4 tháng 6 2019
\(\frac{x-3}{13}+\frac{x-3}{14}=\frac{x-3}{15}+\frac{x-3}{16}\)
\(\Leftrightarrow\frac{x-3}{13}+\frac{x-3}{14}-\frac{x-3}{15}-\frac{x-3}{16}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{13}+\frac{1}{14}-\frac{1}{15}-\frac{1}{16}\right)=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=0+3\)
\(\Leftrightarrow x=3\)
N
2 tháng 12 2018
1) \(A=\frac{2018x^2-2.2018x+2018^2}{2018x^2}=\frac{\left(x-2018\right)^2+2017x^2}{2018x^2}=\frac{\left(x-2018\right)^2}{2018x^2}+\frac{2017}{2018}\)
vì \(\frac{\left(x-2018\right)^2}{2018x^2}\ge0\Rightarrow\frac{\left(x-2018\right)^2}{2018x^2}+\frac{2017}{2018}\ge\frac{2017}{2018}\)
dấu = xảy ra khi x-2018=0
=> x=2018
Vậy Min A=\(\frac{2017}{2017}\)khi x=2018
2) \(B=\frac{3x^2+9x+17}{3x^2+9x+7}=\frac{3x^2+9x+7+10}{3x^2+9x+7}=1+\frac{10}{3x^2+9x+7}=1+\frac{10}{3.x^2+9x+7}\)
\(=1+\frac{10}{3.\left(x^2+9x\right)+7}=1+\frac{10}{3.\left[x^2+\frac{2.x.3}{2}+\left(\frac{3}{2}\right)^2\right]-\frac{9}{4}+7}=1+\frac{10}{3.\left(x+\frac{9}{2}\right)^2+\frac{1}{4}}\)
để B lớn nhất => \(3.\left(x+\frac{3}{2}\right)^2+\frac{1}{4}\)nhỏ nhất
mà \(3.\left(x+\frac{3}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}\)vì \(3.\left(x+\frac{3}{2}\right)^2\ge0\)
dấu = xảy ra khi \(x+\frac{3}{2}=0\)
=> x=\(-\frac{3}{2}\)
Vậy maxB=\(41\)khi x=\(-\frac{3}{2}\)
3) \(M=\frac{3x^2+14}{x^2+4}=\frac{3.\left(x^2+4\right)+2}{x^2+4}=3+\frac{2}{x^2+4}\)
để M lớn nhất => x2+4 nhỏ nhất
mà \(x^2+4\ge4\)(vì x2 lớn hơn hoặc bằng 0)
dấu = xảy ra khi x2 =0
=> x=0
Vậy Max M\(=\frac{7}{2}\)khi x=0
ps: bài này khá dài, sai sót bỏ qua =))
NN
29 tháng 11 2017
1) \(\dfrac{x}{x+1}-\dfrac{2x}{x-1}+\dfrac{x+3}{x^2-1}\)
\(=\dfrac{x}{x+1}-\dfrac{2x}{x-1}+\dfrac{x+3}{\left(x-1\right)\left(x+1\right)}\) MTC: \(\left(x-1\right)\left(x+1\right)\)
\(=\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{2x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{x+3}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x\left(x-1\right)-2x\left(x+1\right)+\left(x+3\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2-x-2x^2-2x+x+3}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x^2-2x+3}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x^2+x-3x+3}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-\left(x^2-x\right)-\left(3x-3\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x\left(x-1\right)-3\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x-1\right)\left(-x-3\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x-3}{x+1}\)
2) \(\dfrac{5}{x+1}-\dfrac{10}{x-x^2-1}-\dfrac{15}{x^3+1}\)
\(=\dfrac{5}{x+1}-\dfrac{10}{-\left(x^2-x+1\right)}-\dfrac{15}{x^3+1}\)
\(=\dfrac{5}{x+1}+\dfrac{10}{\left(x^2-x+1\right)}-\dfrac{15}{\left(x+1\right)\left(x^2-x+1\right)}\) MTC: \(\left(x+1\right)\left(x^2-x+1\right)\)
\(=\dfrac{5\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{10\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{15}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{5\left(x^2-x+1\right)+10\left(x+1\right)-15}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{5x^2-5x+5+10x+10-15}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{5x^2+5x}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{5x\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{5x}{x^2-x+1}\)
NN
29 tháng 11 2017
3) \(\dfrac{2}{2x+1}-\dfrac{1}{2x-1}-\dfrac{2}{1-4x^2}\)
\(=\dfrac{2}{2x+1}-\dfrac{1}{2x-1}+\dfrac{2}{4x^2-1}\)
\(=\dfrac{2}{2x+1}-\dfrac{1}{2x-1}+\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}\) MTC: \(\left(2x-1\right)\left(2x+1\right)\)
\(=\dfrac{2\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}-\dfrac{2x+1}{\left(2x-1\right)\left(2x+1\right)}+\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}\)
\(=\dfrac{2\left(2x-1\right)-\left(2x+1\right)+2}{\left(2x-1\right)\left(2x+1\right)}\)
\(=\dfrac{4x-2-2x-1+2}{\left(2x-1\right)\left(2x+1\right)}\)
\(=\dfrac{2x-1}{\left(2x-1\right)\left(2x+1\right)}\)
\(=\dfrac{1}{2x+1}\)
4) \(\dfrac{3x^2+5x+14}{x^3+1}+\dfrac{x-1}{x^2-x+1}-\dfrac{4}{x+1}\)
\(=\dfrac{3x^2+5x+14}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{x-1}{x^2-x+1}-\dfrac{4}{x+1}\) MTC: \(\left(x+1\right)\left(x^2-x+1\right)\)
\(=\dfrac{3x^2+5x+14}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{4\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{\left(3x^2+5x+14\right)+\left(x-1\right)\left(x+1\right)-4\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{3x^2+5x+14+x^2-1-4x^2+4x-4}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{9x+9}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{9\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{9}{x^2-x+1}\)
17 tháng 6 2022
b: \(\Leftrightarrow4x^2-8x+4=x^2+2x+1+3\left(x^2+x-6\right)\)
\(\Leftrightarrow3x^2-10x+3=3x^2+3x-18\)
=>-13x=-21
hay x=21/13
c: \(\Leftrightarrow\left(\dfrac{x-90}{10}-1\right)+\left(\dfrac{x-76}{12}-2\right)+\left(\dfrac{x-58}{14}-3\right)+\left(\dfrac{x-36}{16}-4\right)+\left(\dfrac{x-15}{17}-5\right)=0\)
=>x-100=0
hay x=100
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 3 2020
a/ĐKXĐ: \(y\ne4\)
Đặt \(y-4=x\)
\(1+\frac{45}{x^2}=\frac{14}{x}\Leftrightarrow x^2-14x+45=0\Rightarrow\left[{}\begin{matrix}x=9\\x=5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}y-4=9\\y-4=5\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}y=13\\y=9\end{matrix}\right.\)
b/ ĐKXĐ: \(x\ne1\)
Đặt \(x-1=y\)
\(\frac{5}{y}-\frac{4}{3y^2}=3\Leftrightarrow9y^2=15y-4\)
\(\Leftrightarrow9y^2-15y+4=0\Rightarrow\left[{}\begin{matrix}y=\frac{4}{3}\\y=\frac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-1=\frac{4}{3}\\x-1=\frac{1}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{7}{3}\\x=\frac{4}{3}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 3 2020
c/ ĐKXĐ: \(x\ne5\)
\(\Leftrightarrow2x-5=3x-15\)
\(\Leftrightarrow x=10\)
d/ ĐKXĐ: \(x\ne0\)
\(\Leftrightarrow2\left(x^2-12\right)=2x^2+3x\)
\(\Leftrightarrow3x=-24\Rightarrow x=-8\)
e/ ĐKXĐ: \(x\ne2\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\left(l\right)\\x=1\end{matrix}\right.\)
f/ DKXĐ: \(x\ne-\frac{1}{2}\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=8\)
\(\Leftrightarrow4x^2-1=8\)
\(\Leftrightarrow x^2=\frac{9}{4}\Rightarrow x=\pm\frac{3}{2}\)
LB
28 tháng 4 2018
câu nào cũng ghi lại đề nha
a) \(x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
b)\(x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
c) \(\left(x+1\right)\left(x+2\right)+\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+1+x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{2}\end{matrix}\right.\)
LB
28 tháng 4 2018
d) \(\dfrac{1}{x-2}+3-\dfrac{3-x}{x-2}=0\)
\(\Leftrightarrow\dfrac{1+3\left(x-2\right)-\left(3-x\right)}{x-2}=0\)
\(\Leftrightarrow\dfrac{1+3x-6-3+x}{x-2}=0\) ( đk \(x\ne2\) )
\(\Leftrightarrow4x-8=0\Rightarrow x=2\)
đ) \(\dfrac{8-x}{x-7}-8-\dfrac{1}{x-7}=0\)
\(\Leftrightarrow\dfrac{8-x-8\left(x-7\right)-1}{x-7}=0\) (đk \(x\ne7\))
\(\Leftrightarrow8-x-8x+56-1=0\)
\(\Leftrightarrow-9x+63=0\)
\(\Leftrightarrow x=7\)
\( \left(x-1\right)\left(x+3\right)+\left(x-2\right)-2x^2=14\)
<=> \(x^2+2x-3+x-2-2x^2=14\)
<=> \(-x^2+3x-5=14\)
<=> \(-x^2+3x-19=0\)
<=> \(x^2-3x+19=0\) vô lý
Vậy pt vô nghiệm