Tính (x-2)(x+1)
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\(a,A=\dfrac{x^2-x-2}{x^2-1}+\dfrac{1}{x-1}-\dfrac{1}{x+1}\)
\(\Rightarrow A=\dfrac{x^2-x-2}{\left(x-1\right)\left(x+1\right)}+\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{x-1}{\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow A=\dfrac{x^2-x-2x+x+1-x+1}{\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow A=\dfrac{x^2-3x+2}{\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow A=\dfrac{x^2-2x-x+2}{\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow A=\dfrac{x\left(x-2\right)-\left(x-2\right)}{\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow A=\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow A=\dfrac{x-2}{x+1}\)
\(b,A=\dfrac{3}{4}\\ \Rightarrow\dfrac{x-2}{x+1}=\dfrac{3}{4}\\ \Rightarrow4\left(x-2\right)=3\left(x+1\right)\\ \Rightarrow4x-8=3x+3\\ \Rightarrow4x-8-3x-3=0\\ \Rightarrow x-11=0\\ \Rightarrow x=11\)
\(c,\left|x-3\right|=2\Rightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
Thay x=5 vào A ta có:
\(A=\dfrac{x-2}{x+1}=\dfrac{5-2}{5+1}=\dfrac{3}{6}=\dfrac{1}{2}\)
Thay x=1 vào A ta có:
\(A=\dfrac{x-2}{x+1}=\dfrac{1-2}{1+1}=\dfrac{-1}{2}\)
a: \(=\dfrac{6}{x+1}+\dfrac{4}{x-1}-\dfrac{10}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{6x-6+4x+4-10}{\left(x-1\right)\left(x+1\right)}=\dfrac{10x-12}{\left(x-1\right)\left(x+1\right)}\)
b: \(=\dfrac{1}{x}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+6}+\dfrac{6}{x}=\dfrac{1}{x}+\dfrac{6}{x}=\dfrac{7}{x}\)
Ta có : 1/x - 1/(x+1) = 1/x(x+1)
<=> pcm \(\frac{x+1}{x\left(x+1\right)}-\frac{x}{x\left(x+1\right)}=\frac{1}{x\left(x+1\right)}\)
<=> pcm \(\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x\left(x+1\right)}\)
<=> pcm 1/x(x+1) = 1/x(x+1)
Đây là điều luôn đúng nên ta có điều phải chứng minh
Chú ý : Chữ pcm là phải chứng minh
Ta có : \(\frac{1}{x^2+x}+\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}+\frac{1}{x+5}\)
\(=\frac{1}{x\left(x+1\right)}+\frac{1}{x^2+x+2x+2}+\frac{1}{x^2+2x+3x+6}+\frac{1}{x^2+3x+4x+12}+\frac{1}{x^2+4x+5x+20}+\frac{1}{x+5}\)
\(=\frac{1}{x\left(x+1\right)}+\frac{1}{x\left(x+1\right)+2\left(x+1\right)}+\frac{1}{x\left(x+2\right)+3\left(x+2\right)}+\frac{1}{x\left(x+3\right)+4\left(x+3\right)}\)
\(+\frac{1}{x\left(x+4\right)+5\left(x+4\right)}+\frac{1}{x+5}\)
\(=\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{x+5}\)
Áp dụng chứng minh trên ta có :
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}\)
=1/x
a) Ta có: \(A=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)
\(=x^2+2x+y^2-2y-2xy+37\)
\(=\left(x^2-2xy+y^2\right)+\left(2x-2y\right)+37\)
\(=\left(x-y\right)^2+2\left(x-y\right)+37\)
\(=\left(x-y\right)\left(x-y+2\right)+37\)(1)
Thay x-y=7 vào biểu thức (1), ta được:
\(A=7\cdot\left(7+2\right)+37=7\cdot9+37=100\)
Vậy: Khi x-y=7 thì A=100
b) Ta có: \(x+y=2\)
\(\Leftrightarrow\left(x+y\right)^2=4\)
\(\Leftrightarrow x^2+y^2+2xy=4\)
\(\Leftrightarrow2xy+10=4\)
\(\Leftrightarrow2xy=-6\)
\(\Leftrightarrow xy=-3\)
Ta có: \(A=x^3+y^3\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)\)(2)
Thay x+y=2; \(x^2+y^2=10\) và xy=-3 vào biểu thức (2), ta được:
\(A=2\cdot\left(10+3\right)=2\cdot13=26\)
Vậy: Khi x+y=2 và \(x^2+y^2=10\) thì A=26
\(\Rightarrow A=x^2+2x+y^2-2y-2xy+37=x^2-2xy+y^2+2\left(x-y\right)+37=\left(x-y\right)^2+2\left(x-y\right)+37=7^2+2\cdot7+37=100\)
\(\Rightarrow A=x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)=\left(x+y\right)\left[x^2+y^2-\dfrac{\left(x+y\right)^2-\left(x^2+y^2\right)}{2}\right]=2\cdot\left[10+3\right]=2\cdot13=26\) \(\Rightarrow\left\{{}\begin{matrix}x+y=-z\\x+z=-y\\y+z=-x\end{matrix}\right.\) \(\Rightarrow P=\left(\dfrac{x+y}{y}\right)\left(\dfrac{y+z}{z}\right)\left(\dfrac{x+z}{x}\right)=-\dfrac{z}{y}\cdot\dfrac{-x}{z}\cdot-\dfrac{y}{x}=-1\)
Bài 4:
1: \(\left(x-1\right)\left(x^2+x+1\right)-x^3-6x=11\)
=>\(x^3-1-x^3-6x=11\)
=>-6x-1=11
=>-6x=11+1=12
=>\(x=\dfrac{12}{-6}=-2\)
2: \(16x^2-\left(3x-4\right)^2=0\)
=>\(\left(4x\right)^2-\left(3x-4\right)^2=0\)
=>\(\left(4x-3x+4\right)\left(4x+3x-4\right)=0\)
=>(x+4)(7x-4)=0
=>\(\left[{}\begin{matrix}x+4=0\\7x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{4}{7}\end{matrix}\right.\)
3: \(x^3-x^2-3x+3=0\)
=>\(\left(x^3-x^2\right)-\left(3x-3\right)=0\)
=>\(x^2\left(x-1\right)-3\left(x-1\right)=0\)
=>\(\left(x-1\right)\left(x^2-3\right)=0\)
=>\(\left[{}\begin{matrix}x-1=0\\x^2-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x^2=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\)
4: \(\dfrac{x-1}{x+2}=\dfrac{x+2}{x+1}\)(ĐKXĐ: \(x\notin\left\{-2;-1\right\}\))
=>\(\left(x+2\right)^2=\left(x-1\right)\left(x+1\right)\)
=>\(x^2+4x+4=x^2-1\)
=>4x+4=-1
=>4x=-5
=>\(x=-\dfrac{5}{4}\left(nhận\right)\)
5: ĐKXĐ: \(x\notin\left\{0;-1\right\}\)
\(\dfrac{1}{x}+\dfrac{2}{x+1}=0\)
=>\(\dfrac{x+1+2x}{x\left(x+1\right)}=0\)
=>3x+1=0
=>3x=-1
=>\(x=-\dfrac{1}{3}\left(nhận\right)\)
6: ĐKXĐ: \(x\notin\left\{0;3\right\}\)
\(\dfrac{9-x^2}{x}:\left(x-3\right)=1\)
=>\(\dfrac{-\left(x^2-9\right)}{x\left(x-3\right)}=1\)
=>\(\dfrac{-\left(x-3\right)\left(x+3\right)}{x\left(x-3\right)}=1\)
=>\(\dfrac{-x-3}{x}=1\)
=>-x-3=x
=>-2x=3
=>\(x=-\dfrac{3}{2}\left(nhận\right)\)
Tính:
a, \(\dfrac{x-1}{x+2}+\dfrac{x+5}{x+2}=\dfrac{x-1+x+5}{x-2}=\dfrac{2x+4}{x-2}\) = \(\dfrac{2\left(x+2\right)}{x-2}\)
b, \(\dfrac{1}{x\left(x-1\right)}-\dfrac{2-x}{x-1}=\dfrac{1}{x\left(x-1\right)}-\dfrac{x\left(2-x\right)}{x\left(x-1\right)}=\dfrac{1-2x+x^2}{x\left(x-1\right)}=\dfrac{x^2-2x+1}{x\left(x-1\right)}=\dfrac{\left(x-1\right)^2}{x\left(x-1\right)}=\dfrac{x-1}{x}\)
\(x^2-2x+x-2=0\)
\(x^2-x-2=0\)
\(x^2+x-2x-2=0\)
\(x\left(x+1\right)-2\left(x+1\right)=0\)
\(\left(x+1\right)\left(x-2\right)=0\)
\(\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}< =>\orbr{\begin{cases}x=-1\\x=2\end{cases}}}\)
Dap an : x = - 1 , x = 2