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a, \(Đkxđ:\hept{\begin{cases}x\ne1\\x\ne\pm3\end{cases}}\)
\(P=\left(1+\frac{1}{x-1}\right):\left(\frac{x^2-7}{x^2-4x+3}+\frac{1}{x-1}+\frac{1}{3-x}\right)\)
\(=\left(\frac{x-1}{x-1}+\frac{1}{x-1}\right):\left(\frac{x^2-7}{\left(x-1\right)\left(x-3\right)}+\frac{1}{x-1}-\frac{1}{x-3}\right)\)
\(=\left(\frac{x-1+1}{x-1}\right):\left(\frac{x^2-7+x-3-\left(x-1\right)}{\left(x-1\right)\left(x-3\right)}\right)\)
\(=\frac{x}{x-1}:\frac{x^2-7+x-3-x+1}{\left(x-1\right)\left(x-3\right)}\)
\(=\frac{x}{x-1}.\frac{\left(x-1\right)\left(x-3\right)}{x^2-9}\)
\(=\frac{x}{x-1}.\frac{\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x}{x+3}\)
b, \(|x+2|=5\)
\(\Rightarrow x+2=\hept{\begin{cases}5\Leftrightarrow x+2\ge0\Rightarrow x\ge-2\\-5\Leftrightarrow x+2< 0\Rightarrow x< -2\end{cases}}\)
Nếu \(x\ge-2\Rightarrow x+2=5\)
\(\Rightarrow x=3\)\(\left(ktmđkxđ\right)\)
Nếu \(x< -2\Rightarrow x+2=-5\)
\(\Rightarrow x=-7\)\(\left(tm\right)\)
Vậy \(x=-7\)
2/ \(\frac{1}{2}x2y5z3=\left(\frac{1}{2}.2.5.3\right)xyz\)\(=15xyz\)
\(\Rightarrow\frac{1}{2}x2y5z3\)có bậc là 3
3/ \(\frac{x}{4}=\frac{9}{x}\Leftrightarrow x^2=9.4\Rightarrow x^2=36\) mà \(x>0\Rightarrow x=6\)
4/ \(\left|2x-\frac{1}{2}\right|+\frac{3}{7}=\frac{38}{7}\Rightarrow\left|2x+\frac{1}{2}\right|=\frac{35}{7}=5\Rightarrow\hept{\begin{cases}2x+\frac{1}{2}=5\Rightarrow2x=\frac{9}{2}\Rightarrow x=\frac{9}{4}\\2x+\frac{1}{2}=-5\Rightarrow2x=\frac{-11}{2}\Rightarrow x=\frac{-11}{4}\end{cases}}\)
a)\(A=\frac{x+1}{x^2+2x+1}:\left(\frac{1}{x^2-x}+\frac{1}{x-1}\right)\left(ĐK:x\ne0;x\ne1\right)\)
\(=\frac{x+1}{\left(x+1\right)^2}:\frac{1+x}{x\left(x-1\right)}\)
\(=\frac{1}{x+1}\cdot\frac{x\left(x+1\right)}{x+1}=\frac{x}{x+1}\)
b)Có: \(x^2+x-2=0\\ \Leftrightarrow x^2-x+2x-2=0\\ \Leftrightarrow x\left(x-1\right)+2\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+2\right)\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x+2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\left(loại\right)\\x=-2\end{array}\right.\)
Thay x=-2 vào A ta có
\(A=\frac{-2}{-2+1}=\frac{-2}{-1}=2\)
1, \(=\left[\frac{\left(1-x\right)\left(1+x+x^2\right)}{1-x}-x\right]:\frac{1-x^2}{\left(1-x\right)-x^2\left(1-x\right)}\)
\(=\left(1+x+x^2-x\right):\frac{1-x^2}{\left(1-x\right)\left(1-x^2\right)}\)\(=\left(x^2+1\right)\left(1-x\right)\)
2, để B<0 <=> (x2+1)(1-x)<0
vì x^2+1 > 0 với mọi x
=> \(\hept{\begin{cases}x^2+1>0\\1-x< 0\end{cases}\Leftrightarrow x>1}\)
3, \(\left|x-4\right|=5\Leftrightarrow\orbr{\begin{cases}x=9\\x=-1\left(loại\right)\end{cases}}\)
Thay x=9 vào B ta có: B=(92+1)(1-9)=82.(-8)=-656
\(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\frac{2}{xy}-\frac{1}{z^2}\)
Khai triển cả 2 vế ta được \(\left(\frac{1}{y}+\frac{1}{z}\right)^2+\left(\frac{1}{x}+\frac{1}{z}\right)^2=0\)
=>\(\hept{\begin{cases}\frac{1}{y}+\frac{1}{z}=0\\\frac{1}{x}+\frac{1}{z}=0\end{cases}}\)=>\(\frac{1}{x}=\frac{1}{y}\Rightarrow x=y\)
=>\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{2}{x}+\frac{1}{z}=2\Rightarrow\frac{4}{x^2}+\frac{4}{xz}+\frac{1}{z^2}=4\)(1)
\(\frac{2}{xy}-\frac{1}{z^2}=\frac{2}{x^2}-\frac{1}{z^2}=4\)(2)
Từ (1) và (2) suy ra
\(\frac{2}{x^2}+\frac{4}{xz}+\frac{2}{z^2}=0\Rightarrow\frac{1}{x^2}+\frac{2}{xz}+\frac{1}{z^2}=0\Rightarrow\left(\frac{1}{x}+\frac{1}{z}\right)^2=0\)\(\Rightarrow\frac{1}{x}+\frac{1}{z}=0\Rightarrow x=y=-z\)
=> \(P=\left(x+2y+z\right)^{2019}=\left(2y\right)^{2019}\)
à thêm cái này nữa. Sorry viết thiếu
Vì x=y=-z\(\Rightarrow\frac{2}{x}-\frac{1}{x}=2\Rightarrow\frac{1}{x}=2\Rightarrow x=\frac{1}{2}.\)
lúc đó \(P=\left(2.\frac{1}{2}\right)^{2019}=1\)