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a) \(ĐKXĐ:\hept{\begin{cases}x\ne\pm2\\x\ne-3\end{cases}}\)
b) \(P=1+\frac{x+3}{x^2+5x+6}\div\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)
\(\Leftrightarrow P=1+\frac{x+3}{\left(x+3\right)\left(x+2\right)}:\left(\frac{8x^2}{4x^2\left(x-2\right)}-\frac{3x}{3\left(x^2-4\right)}-\frac{1}{x+2}\right)\)
\(\Leftrightarrow P=1+\frac{1}{x+2}:\left(\frac{2}{x-2}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{1}{x+2}\right)\)
\(\Leftrightarrow P=1+\frac{1}{x+2}:\frac{2x+4-x-x+2}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow P=1+\frac{1}{x+2}:\frac{6}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow P=1+\frac{\left(x-2\right)\left(x+2\right)}{6\left(x+2\right)}\)
\(\Leftrightarrow P=1+\frac{x-2}{6}\)
\(\Leftrightarrow P=\frac{x+4}{6}\)
c) Để P = 0
\(\Leftrightarrow\frac{x+4}{6}=0\)
\(\Leftrightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
Để P = 1
\(\Leftrightarrow\frac{x+4}{6}=1\)
\(\Leftrightarrow x+4=6\)
\(\Leftrightarrow x=2\)
d) Để P > 0
\(\Leftrightarrow\frac{x+4}{6}>0\)
\(\Leftrightarrow x+4>0\)(Vì 6>0)
\(\Leftrightarrow x>-4\)
1: \(B=\left(\dfrac{4x}{x+2}-\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\cdot\dfrac{4\left(x^2-2x+4\right)}{\left(x-2\right)\left(x+2\right)}\right):\dfrac{16}{x+2}\cdot\dfrac{\left(x+2\right)\left(x+1\right)}{x^2+x+1}\)
\(=\left(\dfrac{4x}{x+2}-\dfrac{4\left(x^2+2x+4\right)}{\left(x+2\right)^2}\right)\cdot\dfrac{x+2}{16}\cdot\dfrac{\left(x+2\right)\left(x+1\right)}{x^2+x+1}\)
\(=\dfrac{4x^2+8x-4x^2-8x-16}{\left(x+2\right)^2}\cdot\dfrac{\left(x+2\right)^2\cdot\left(x+1\right)}{16\left(x^2+x+1\right)}\)
\(=\dfrac{-\left(x+1\right)}{x^2+x+1}\)
2: Để B=0 thì -x-1=0
hay x=-1(nhận)
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) \(\dfrac{3x+5}{4x^3y}-\dfrac{5-15x}{4x^3y}\)
\(=\dfrac{3x+5-5+15x}{4x^3y}\)
\(=\dfrac{18x}{4x^3y}\)
\(=\dfrac{9}{2x^2y}\)
b) \(\dfrac{2}{x-2}+\dfrac{4}{x+2}+\dfrac{-6+5x}{4-x^2}\)
\(=\dfrac{2}{x-2}+\dfrac{4}{x+2}+\dfrac{6-5x}{x^2-4}\)
\(=\dfrac{2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{4\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{6-5x}{\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{2\left(x+2\right)+4\left(x-2\right)+6-5x}{\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{2x+4+4x-8+6-5x}{\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{x+2}{\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{1}{x-2}\)
a, \(\dfrac{3\left|x-4\right|}{3x^2-3x-36}=\dfrac{3\left|x-4\right|}{3\left(x^2-x-12\right)}=\dfrac{\left|x-4\right|}{x^2-x-12}\)
b, \(\dfrac{9-\left(x+5\right)^2}{x^2+4x+4}=\dfrac{\left(3-x-5\right)\left(3+x+5\right)}{\left(x+2\right)^2}\)
\(=\dfrac{\left(-x-2\right)\left(x+8\right)}{\left(x+2\right)^2}=\dfrac{-\left(x+2\right)\left(x+8\right)}{\left(x+2\right)^2}\)
\(=\dfrac{-\left(x+8\right)}{x+2}=\dfrac{-x-8}{x+2}\)
c, \(\dfrac{x^2+5x+6}{x^2+4x+4}=1+\dfrac{x+2}{x^2+4x+4}=1+\dfrac{x+2}{\left(x+2\right)^2}\)
\(=1+\dfrac{1}{x+2}=\dfrac{x+3}{x+2}\)
a, tiếp:
+) Xét \(x\ge4\) có:
\(\dfrac{x-4}{x^2-x-12}=\dfrac{x-4}{x^2-4x+3x-12}=\dfrac{x-4}{x\left(x-4\right)+3\left(x-4\right)}\)
\(=\dfrac{x-4}{\left(x+3\right)\left(x-4\right)}=\dfrac{1}{x+3}\)
+) Xét x < 4 có:
\(\dfrac{4-x}{x^2-x-12}=\dfrac{4-x}{x^2-4x+3x-12}=\dfrac{4-x}{x\left(x-4\right)+3\left(x-4\right)}\)
\(=\dfrac{4-x}{\left(x+3\right)\left(x-4\right)}=\dfrac{-\left(x-4\right)}{\left(x+3\right)\left(x-4\right)}=\dfrac{-1}{x+3}\)
a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne-5\end{cases}}\)
\(P=\frac{x^2}{5x+25}+\frac{2x-10}{x}+\frac{50+5x}{x^2+5x}\)\(=\frac{x^2}{5\left(x+5\right)}+\frac{2\left(x-5\right)}{x}+\frac{5\left(x+10\right)}{x\left(x+5\right)}\)
\(=\frac{x^3}{5x\left(x+5\right)}+\frac{10\left(x-5\right)\left(x+5\right)}{5x\left(x+5\right)}+\frac{25\left(x+10\right)}{5x\left(x+5\right)}\)
\(=\frac{x^3+10\left(x-5\right)\left(x+5\right)+25\left(x+10\right)}{5x\left(x+5\right)}=\frac{x^3+10\left(x^2-25\right)+25x+250}{5x\left(x+5\right)}\)
\(=\frac{x^3+10x^2-250+25x+250}{5x\left(x+5\right)}=\frac{x^3+10x^2+25x}{5x\left(x+5\right)}\)\(=\frac{x\left(x^2+10x+25\right)}{5x\left(x+5\right)}\)\(=\frac{\left(x+5\right)^2}{5\left(x+5\right)}=\frac{x+5}{5}\)
b) \(x^2-3x=0\)\(\Leftrightarrow x\left(x-3\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
So sánh với ĐKXĐ, ta thấy \(x=0\)không thoả mãn
Thay \(x=3\)vào biểu thức ta được: \(P=\frac{3+5}{5}=\frac{8}{5}\)
c) Để \(P=-4\)thì \(\frac{x+5}{5}=-4\)\(\Leftrightarrow x+5=-20\)\(\Leftrightarrow x=-25\)( thoả mãn ĐKXĐ )
Vậy \(P=-4\)\(\Leftrightarrow x=-25\)
d) Để \(P\ge0\)thì \(\frac{x+5}{5}\ge0\)\(\Leftrightarrow x+5\ge0\)( vì \(5>0\))\(\Leftrightarrow x\ge-5\)
So sánh với ĐKXĐ, ta thấy x phải thoả mãn \(x>-5\)và \(x\ne0\)
Vậy \(P\ge0\)\(\Leftrightarrow\)\(x>-5\)và \(x\ne0\)
b: =>(x+5)(x-3)=0
=>x=3 hoặc x=-5
c: \(\Leftrightarrow x\left(x^2-4x+5\right)=0\)
=>x=0
d: \(\Leftrightarrow2\cdot2^x-10\cdot2^x=-16\)
\(\Leftrightarrow-8\cdot2^x=-16\)
\(\Leftrightarrow2^x=2\)
hay x=1
\(S=\dfrac{3x+6}{x^2-4x+4}-\dfrac{5x-16}{x^2+4x+4}\)
\(S=\dfrac{3x+6}{\left(x-2\right)^2}-\dfrac{5x-16}{\left(x+2\right)^2}\)
\(S=\dfrac{3x+6}{\left(x-2\right)^2}-\dfrac{5x-16}{-\left(x-2\right)^2}\)
\(S=\dfrac{3x+6}{\left(x-2\right)^2}-\dfrac{-(5x-16)}{\left(x-2\right)^2}\)
\(S=\dfrac{3x+6}{\left(x-2\right)^2}-\dfrac{-5x+16}{\left(x-2\right)^2}\)
\(S=\dfrac{3x+6-\left(-5x\right)+16}{\left(x-2\right)^2}\)
\(S=\dfrac{3x-\left(-5x\right)+6+16}{\left(x-2\right)^2}\)
\(S=\dfrac{8x+22}{\left(x-2\right)^2}\)