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1.
a) \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
b) \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
Bài 1:
a, \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow x\left(x+4\right)+\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
Vậy \(x=-4\) hoặc \(x=-1\)
b, \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x=3\) hoặc \(x=-2\)
a/ \(\dfrac{x^3}{x^2+1975}\cdot\dfrac{2x+1954}{x+1}+\dfrac{x^3}{x^2+1975}\cdot\dfrac{21-x}{x+1}=\dfrac{x^3\left(2x+1954\right)+x^3\left(21-x\right)}{\left(x^2+1975\right)\left(x+1\right)}=\dfrac{2x^4+1954x^3+21x^3-x^4}{\left(x^2+1975\right)\left(x+1\right)}=\dfrac{x^4+1975x^3}{\left(x^2+1975\right)\left(x+1\right)}\)
b/ \(\dfrac{19x+8}{x-7}\cdot\dfrac{5x-9}{x+1945}+\dfrac{19x+8}{x^2+1945}\cdot\dfrac{x-2}{x-7}=\dfrac{\left(19x+8\right)\left(5x-9\right)+\left(19x+8\right)\left(x-2\right)}{\left(x-7\right)\left(x+1945\right)}=\dfrac{\left(19x+8\right)\left(5x-9+x-2\right)}{\left(x-7\right)\left(x+1945\right)}=\dfrac{114x^2-209x+40x-88}{\left(x-7\right)\left(x+1945\right)}=\dfrac{114x^2-169x-88}{x^2+1938x-13615}\)
c/ \(\dfrac{x+1}{x^2-2x-8}\cdot\dfrac{4-x}{x^2+x}=\dfrac{\left(x+1\right)\left(4-x\right)}{x\left[x^2-4x+2x-8\right]\left(x+1\right)}=-\dfrac{x-4}{x\left(x-4\right)+2\left(x-4\right)}=-\dfrac{x-4}{\left(x-4\right)\left(x+2\right)}=-\dfrac{1}{x+2}\)
\(\frac{x^2+3x+9}{2x+10}.\frac{x+5}{x^3-27}\)
\(=\frac{x^2+3x+9}{2\left(x+5\right)}.\frac{x+5}{\left(x-3\right)\left(x^2+3x+9\right)}\)
\(=\frac{\left(x+5\right)\left(x^2+3x+9\right)}{2\left(x+5\right)\left(x-3\right)\left(x^2+3x+9\right)}\)
\(=\frac{1}{2\left(x-3\right)}\)
\(\left(\frac{6x+1}{x^2-6x}+\frac{6x-1}{x^2+6x}\right)\left(\frac{x^2-36}{x^2+1}\right)\)
\(=\left[\frac{6x+1}{x\left(x-6\right)}+\frac{6x-1}{x\left(x+6\right)}\right]\left[\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\right]\)
\(=\frac{\left(6x+1\right)\left(x+6\right)+\left(6x-1\right)\left(x-6\right)}{x\left(x-6\right)\left(x+6\right)}.\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)
\(=\frac{6x^2+36x+x+6+6x^2-36x-x+6}{x\left(x-6\right)\left(x+6\right)}.\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)
\(=\frac{12x^2+12}{x\left(x-6\right)\left(x+6\right)}.\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)
\(=\frac{12\left(x^2+1\right).\left(x-6\right)\left(x+6\right)}{x\left(x-6\right)\left(x+6\right)\left(x^2+1\right)}\)
\(=\frac{12}{x}\)
a) \(\dfrac{x}{x-3}+\dfrac{9-6x}{x^2-3x}=\dfrac{x^2}{x\left(x-3\right)}+\dfrac{9-6x}{x\left(x-3\right)}=\dfrac{x^2-6x+9}{x\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)
dễ thì giải cho người ta đi,bạn thông minh hơn thì thay vì ns người khác thì giúp người khác sẽ tốt hơn đó
a/ \(\dfrac{2x^2-20x+50}{3x+3}\cdot\dfrac{x^2-1}{4\left(x-5\right)^2}=\dfrac{2\left(x^2-10x+25\right)\cdot\left(x^2-1\right)}{3\left(x+1\right)\cdot4\left(x-5\right)^2}=\dfrac{2\left(x-5\right)^2\left(x-1\right)\left(x+1\right)}{12\left(x+1\right)\left(x-5\right)^2}=\dfrac{x+1}{6}\)
b/ \(\dfrac{6x-3}{5x^2+x}\cdot\dfrac{25x^2+10x+1}{1-8x^2}=-\dfrac{3\left(1-2x\right)\cdot\left(5x+1\right)^2}{x\left(5x+1\right)\left(1-2x\right)\left(1+2x+4x^2\right)}=\dfrac{3\left(5x+1\right)}{x\left(4x^2+2x+1\right)}\)
c/ \(\dfrac{3x^2-x}{x^2-1}\cdot\dfrac{1-x^4}{\left(1-3x\right)^3}=\dfrac{x-3x^2}{1-x^2}\cdot\dfrac{\left(1-x^2\right)\left(1+x^2\right)}{\left(1-3x\right)^3}=\dfrac{x\left(1-3x\right)\left(1-x^2\right)\left(1+x^2\right)}{\left(1-x^2\right)\left(1-3x\right)^3}=\dfrac{x\left(x^2+1\right)}{\left(1-3x\right)^3}\)
a: \(=\dfrac{x}{y\left(x-y\right)}+\dfrac{2x-y}{y\left(x-y\right)}=\dfrac{x+2x-y}{y\left(x-y\right)}=\dfrac{3x-y}{y\left(x-y\right)}\)
b: \(=\dfrac{x\left(x+3\right)}{\left(x+3\right)^2}+\dfrac{3}{x-3}-\dfrac{6x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x}{x+3}+\dfrac{3}{x-3}-\dfrac{6x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x^2-3x+3x+9-6x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{x-3}{x+3}\)
c: \(=\dfrac{x+9}{\left(x-3\right)\left(x+3\right)}-\dfrac{3}{x\left(x+3\right)}\)
\(=\dfrac{x^2+9x-3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x^2+9x-3x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{x+3}{x-3}\)
d: \(=\dfrac{x^2-1-x^2+4}{x+1}=\dfrac{3}{x+1}\)
\(a,\)
\(A=\dfrac{x^2-2x}{x^2+x-6}.\dfrac{x^2-9}{x^2-3x}\)
\(A=\dfrac{x\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}.\dfrac{\left(x-3\right)\left(x+3\right)}{x\left(x-3\right)}\)
\(A=\dfrac{x}{\left(x+3\right)}.\dfrac{\left(x+3\right)}{x}=1\)
\(b,\)
\(B=\dfrac{2x^2+5x-3}{x^2-9}.\dfrac{4x^2-1}{x^2-6x+9}\)
\(B=\dfrac{\left(x+3\right)\left(2x-1\right)}{\left(x-3\right)\left(x+3\right)}.\dfrac{\left(2x-1\right)\left(2x+1\right)}{\left(x-3\right)^2}\)
\(B=\dfrac{\left(2x-1\right)}{\left(x-3\right)}.\dfrac{\left(2x-1\right)\left(2x+1\right)}{\left(x-3\right)^2}\)
\(B=\dfrac{\left(2x-1\right)^2.\left(2x+1\right)}{\left(x-3\right)^3}\)