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NH
0
C
29 tháng 6 2017
1) \(4x^2+4x+1=\left(2x+1\right)^2\)
2)\(9x^2-24xy+16y^2=\left(3x-4y\right)^2\)
3)\(-x^2+10x-25=-\left(x-5\right)^2\)
4)\(1+12x+36x^2=\left(1+6x\right)^2\)
5) \(\dfrac{x^2}{4}+2xy+4y^2=\left(\dfrac{x}{2}+2y\right)^2\)
6) \(4x^2+4xy+y^2=\left(2x+y\right)^2\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 2 2020
ĐKXĐ: \(x\ne\pm\frac{3}{2}\)
\(\frac{1}{\left(2x-3\right)^2}+\frac{3}{\left(2x-3\right)\left(2x+3\right)}-\frac{4}{\left(2x+3\right)^2}=0\)
\(\Leftrightarrow\frac{1}{\left(2x-3\right)^2}-\frac{1}{\left(2x-3\right)\left(2x+3\right)}+\frac{4}{\left(2x-3\right)\left(2x+3\right)}-\frac{4}{\left(2x-3\right)^2}=0\)
\(\Leftrightarrow\frac{1}{2x-3}\left(\frac{1}{2x-3}-\frac{1}{2x+3}\right)-\frac{4}{2x-3}\left(\frac{1}{2x-3}-\frac{1}{2x+3}\right)=0\)
\(\Leftrightarrow\left(\frac{1}{2x-3}-\frac{4}{2x+3}\right)\left(\frac{1}{2x-3}-\frac{1}{2x+3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=2x-3\left(vn\right)\\2x+3=4\left(2x-3\right)\Rightarrow x=\frac{5}{2}\end{matrix}\right.\)
ND
10 tháng 2 2018
d. ĐKXĐ: x khác 1, x khác 3
\(\dfrac{x+5}{x-1}=\dfrac{x+1}{\left(x-3\right)}-\dfrac{8}{x^2-4x+3}\)
\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+5\right)}{\left(x-1\right)\left(x-3\right)}=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x-3\right)}-\dfrac{8}{\left(x-1\right)\left(x-3\right)}\) \(\Leftrightarrow x^2+2x-15=x^2-1-8\)
\(\Leftrightarrow2x-15+1+8=0\)
\(\Leftrightarrow2x-6=0\)
\(\Leftrightarrow x=3\) (loại)
Vậy pt vô nghiệm
DL
5
MD
9 tháng 6 2017
f) \(4x^2-12x+9=0\)
<=> \(\left(2x-3\right)^2\) = 0
<=> \(2x-3=0\)
<=> \(2x=3\) <=> \(x=\dfrac{3}{2}\)
Vậy ...............
g) \(3x^2+7x+2=0\)
<=> \(\left(3x^2+6x\right)+\left(x+2\right)=0\)
<=> \(3x\left(x+2\right)+\left(x+2\right)=0\)
<=> \(\left(x+2\right)\left(3x+1\right)=0\)
<=> \(\left[{}\begin{matrix}x=-2\\x=\dfrac{-1}{3}\end{matrix}\right.\)
Vậy ........................
h) \(x^2-4x+1=0\)
<=> \(\left(x^2-4x+4\right)-3=0\)
<=> \(\left(x-2\right)^2=3\)
<=> \(\left[{}\begin{matrix}x+2=\sqrt{3}\\x+2=-\sqrt{3}\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=\sqrt{3}-2\\x=-\sqrt{3}-2\end{matrix}\right.\)
Vậy .........................
i) \(2x^2-6x+1=0\)
<=> \(2\left(x^2-3x+2,25\right)-3,5=0\)
<=> \(\left(x-1,5\right)^2=1,75\)
<=> \(\left[{}\begin{matrix}x-1,5=\sqrt{1,75}\\x-1,5=-\sqrt{1,75}\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=\sqrt{1,75}+1,5\\x=-\sqrt{1,75}+1,5\end{matrix}\right.\)
Vậy ...................
j) \(3x^2+4x-4=0\)
<=> \(\left(3x^2+6x\right)-\left(2x+4\right)=0\)
<=> \(3x\left(x+2\right)-2\left(x+2\right)\) = 0
<=> \(\left(x+2\right)\left(3x-2\right)=0\)
<=> \(\left[{}\begin{matrix}x=-2\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy ....................................
9 tháng 6 2017
f) \(4x^2-12x+9=0\)
\(\Rightarrow\left(2x-3\right)^2=0\)
\(\Rightarrow2x-3=0\)
\(\Rightarrow x=\dfrac{3}{2}\)
Vậy..
g) \(3x^2+7x+2=0\)
\(\Rightarrow3x^2+6x+x+2=0\)
\(\Rightarrow3x\left(x+2\right)+\left(x+2\right)=0\)
\(\Rightarrow\left(x+2\right)\left(3x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+2=0\\3x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{-1}{3}\end{matrix}\right.\)
Vậy..
h) \(x^2-4x+1=0\)
\(\Rightarrow x^2-4x+4-3=0\)
\(\Rightarrow\left(x-2\right)^2-3=0\)
\(\Rightarrow\left(x-2-\sqrt{3}\right)\left(x-2+\sqrt{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2-\sqrt{3}=0\\x-2+\sqrt{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2+\sqrt{3}\\x=2-\sqrt{3}\end{matrix}\right.\)
Vậy..
j) \(3x^2+4x-4=0\)
\(\Rightarrow3x^2+6x-2x-4=0\)
\(\Rightarrow3x\left(x+2\right)-2\left(x+2\right)=0\)
\(\Rightarrow\left(x+2\right)\left(3x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+2=0\\3x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy..
QN
0
ban con choi bangbang ko cho minh muon nick