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2
25 tháng 10 2021
\(a,x^2-5x\)
\(=x\left(x-5\right)\)
\(b,5x\left(x+5\right)+4x+20\)
\(=5x\left(x+5\right)+4\left(x+5\right)\)
\(=\left(5x+4\right)\left(x+5\right)\)
\(c,7x\left(2x-1\right)-4x+2\)
\(=7x\left(2x-1\right)-2\left(2x-1\right)\)
\(=\left(7x-2\right)-\left(2x-1\right)\)
2
25 tháng 10 2021
\(d,x^2-16+2\left(x+4\right)\)
\(=x^2-16+2x+8\)
\(=x\left(x-2\right)-8\) ( Ý này thì k chắc lắm, sai thông cảm :)) )
\(e,x^2-10x+9\)
\(=x^2-x-9x+9\)
\(=x\left(x-1\right)-9\left(x-1\right)\)
\(=\left(x-9\right)\left(x-1\right)\)
\(f,\left(2x-1\right)^2-\left(x-3\right)^2=0\) ( mk đoán bài này là tìm x, sai thì bảo mk để mk sửa nhé )
\(\Rightarrow\left(2x-1\right)^2=\left(x-3\right)^2\)
\(\Leftrightarrow\pm\left(2x-1\right)=\pm\left(x-3\right)\)
\(\Rightarrow\hept{\begin{cases}2x-1=x-3\\-\left(2x-1\right)=-\left(x-3\right)\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}2x-1-x+3=0\\-2x+1-x+3=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x+2=0\\-3x+4=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\left(-2\right)\\x=\frac{4}{3}\end{cases}}\)
Vậy ...
HP
15 tháng 10 2020
1.
a, \(\left(x+3\right)\left(x-3\right)-\left(x-3\right)^2\)
\(=\left(x-3\right)\left(x+3-x+3\right)\)
\(=9\left(x-3\right)=9x-27\)
b, \(\left(2x+1\right)^2+2\left(2x+1\right)\left(x-1\right)+\left(x-1\right)^2\)
\(=\left(2x+1+x-1\right)^2=9x^2\)
c, \(x\left(x-3\right)\left(x+3\right)-\left(x^2+1\right)\left(x^2-1\right)\)
\(=x\left(x^2-9\right)-\left(x^4-1\right)\)
\(=x^3-9x-x^4+1=-x^4+x^3-9x+1\)
NV
8 tháng 1 2018
Bài 2: a) \(3x^3-3x=0\Leftrightarrow3x\left(x^2-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
b) \(x^2-x+\frac{1}{4}=0\Leftrightarrow x^2-2.\frac{1}{2}+\left(\frac{1}{2}\right)^2=0\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)
KT
26 tháng 4 2018
BÀI 1:
a) \(ĐKXĐ:\) \(\hept{\begin{cases}x-2\ne0\\x+2\ne0\end{cases}}\) \(\Leftrightarrow\)\(\hept{\begin{cases}x\ne2\\x\ne-2\end{cases}}\)
b) \(A=\left(\frac{2}{x-2}-\frac{2}{x+2}\right).\frac{x^2+4x+4}{8}\)
\(=\left(\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\right).\frac{\left(x+2\right)^2}{8}\)
\(=\frac{2x+4-2x+4}{\left(x-2\right)\left(x+2\right)}.\frac{\left(x+2\right)^2}{8}\)
\(=\frac{x+2}{x-2}\)
c) \(A=0\) \(\Rightarrow\)\(\frac{x+2}{x-2}=0\)
\(\Leftrightarrow\) \(x+2=0\)
\(\Leftrightarrow\)\(x=-2\) (loại vì ko thỏa mãn ĐKXĐ)
Vậy ko tìm đc x để A = 0
p/s: bn đăng từng bài ra đc ko, mk lm cho
15 tháng 7 2018
a) \(4.\left(x-1\right)^2-9=0\)
\(\Rightarrow4.\left(x-1\right)^2=9\)
\(\Rightarrow\left(x-1\right)^2=9:4=\dfrac{9}{4}=\left(\pm\dfrac{3}{2}\right)^2\)
\(\Rightarrow x-1=\pm\dfrac{3}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x-1=\dfrac{3}{2}\\x-1=\dfrac{-3}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
vậy\(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
b) \(\dfrac{1}{4}-9.\left(x-1\right)^2=0\)
\(\Rightarrow9.\left(x-1\right)^2=\dfrac{1}{4}\)
\(\Rightarrow\left(x-1^2\right)=\dfrac{1}{36}=(\pm\dfrac{1}{6})^2\)
\(\Rightarrow x-1=\pm\dfrac{1}{6}\)
\(\Rightarrow\left[{}\begin{matrix}x-1=\dfrac{1}{6}\\x-1=\dfrac{-1}{6}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=\dfrac{5}{6}\end{matrix}\right.\)
vậy \(\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=\dfrac{5}{6}\end{matrix}\right.\)
15 tháng 7 2018
e) \(\dfrac{1}{16}-\left(2x+\dfrac{3}{4}\right)^2=0\)
\(\Rightarrow\left(2x+\dfrac{3}{4}\right)^2=\dfrac{1}{16}=\left(\pm\dfrac{1}{4}\right)^2\)
\(\Rightarrow2x+\dfrac{3}{4}=\pm\dfrac{1}{4}\)
\(\Rightarrow\)\(\left[{}\begin{matrix}2x+\dfrac{3}{4}=\dfrac{1}{4}\\2x+\dfrac{3}{4}=\dfrac{-1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
vậy \(\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
a) x+5x2=0
<=> x(1+5x)=0
<=>\(\hept{\begin{cases}x=0\\1+5x=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=-\frac{1}{5}\end{cases}}}\)
b) x+1=(x+1)2
<=> x+1-(x+1)2=0
<=> (x+1)(1-x-1)=0
<=> -x(x+1)=0
<=> \(\hept{\begin{cases}-x=0\\x+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=-1\end{cases}}}\)
c) x3+x=0
<=> x(x2+1)=0
<=> x=0 (vì x2+1>0)