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a, ĐKXĐ : \(x-1\ne0\)
=> \(x\ne1\)
TH1 : \(x-2\ge0\left(x\ge2\right)\)
=> \(\left|x-2\right|=x-2=1\)
=> \(x=3\left(TM\right)\)
- Thay x = 3 vào biểu thức P ta được :
\(P=\frac{3+2}{3-1}=\frac{5}{2}\)
TH2 : \(x-2< 0\left(x< 2\right)\)
=> \(\left|x-2\right|=2-x=1\)
=> \(x=1\left(KTM\right)\)
Vậy giá trị của P là \(\frac{5}{2}\) .
a) \(P=\frac{x+2}{x-1}\) \(\left(ĐKXĐ:x\ne1\right)\)
Ta có: \(\left|x-2\right|=1\text{⇔}\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\text{⇔}\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\) (loại x = 1 vì x ≠ 1)
Thay \(x=3\) vào P, ta có:
\(P=\frac{3+2}{3-2}=\frac{5}{1}=5\)
Vậy P = 5 tại x = 3.
b) \(Q=\frac{x-1}{x}+\frac{2x+1}{x^2+x}=\frac{x-1}{x}+\frac{2x+1}{x\left(x+1\right)}=\frac{x^2-1}{x\left(x+1\right)}+\frac{2x+1}{x\left(x+1\right)}\) (ĐKXĐ: x ≠ 0, x ≠ -1)
\(=\frac{x^2+2x}{x\left(x+1\right)}=\frac{x\left(x+2\right)}{x\left(x+1\right)}=\frac{x+2}{x+1}\)
Trả lời:
a, \(-xy.\left(x^2+2xy-3\right)=-x^3y-2x^2y^2+3xy\)
b, \(\left(12x^6y^5-3x^3y^4+4x^2y\right):6x^2y\)
\(=12x^6y^5:6x^2y^2-3x^3y^4:6x^2y+4x^2y+6x^2y\)
\(=2x^4y^3-\frac{1}{2}xy^3+\frac{2}{3}\)
a.\(\left(-xy\right)\left(x^2+2xy-3\right)=-x^3y-2x^2y^2+6xy\)
b.\(\left(12x^6y^5-3x^3y^4+4x^2y\right):6x^2y=2x^4y^4-\frac{1}{2}xy^3+\frac{2}{3}\)
Câu 2:
a) \(ĐKXĐ:x\ne1\)
\(A=\left(\frac{1}{x-1}-\frac{2x}{x^3+x-x^2-1}\right)\div\left(1-\frac{2x}{x^2+1}\right)\)
\(\Leftrightarrow A=\left(\frac{1}{x-1}-\frac{2x}{\left(x-1\right)\left(x^2+1\right)}\right)\div\frac{x^2-2x+1}{x^2+1}\)
\(\Leftrightarrow A=\frac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}\div\frac{\left(x-1\right)^2}{x^2+1}\)
\(\Leftrightarrow A=\frac{\left(x-1\right)^2\left(x^2+1\right)}{\left(x-1\right)\left(x^2+1\right)\left(x-1\right)^2}\)
\(\Leftrightarrow A=\frac{1}{x-1}\)
b) Để A > 0
\(\Leftrightarrow x-1>0\)(Vì\(1>0\))
\(\Leftrightarrow x>1\)
Answer:
Câu 1:
\(\left(5x-x-\frac{1}{2}\right)2x\)
\(=\left(4x-\frac{1}{2}\right)2x\)
\(=4x.2x-\frac{1}{2}.2x\)
\(=8x^2-x\)
\(\left(x^3+4x^2+3x+12\right)\left(x+4\right)\)
\(=x\left(x^3+4x^2+3x+12\right)+4\left(x^3+4x^2+3x+12\right)\)
\(=x^4+4x^3+3x^2+12x+4x^3+16x^2+12x+48\)
\(=x^4+\left(4x^3+4x^3\right)+\left(3x^2+16x^2\right)+\left(12x+12x\right)+48\)
\(=x^4+8x^3+19x^2+24x+48\)
Ta thay \(x=99\) vào phân thức \(\frac{x^2+1}{x-1}\): \(\frac{\left(99\right)^2+1}{99-1}=\frac{9802}{98}=\frac{4901}{49}\)
Ta thay \(x=4\) vào phân thức \(\frac{x^2-x}{2\left(x-1\right)}\) : \(\frac{4^2-4}{2.\left(4-1\right)}=\frac{12}{6}=2\)
\(\left(x+y\right)^2-\left(x-y\right)^2\)
\(= (x²+2xy+y²)-(x²-2xy+y²)\)
\(= x²+2xy+y²-x²+2xy-y²\)
\(= 4xy\)
\(4x^2+4x+1=\left(2x+1\right)^2=\left(2.2+1\right)^2=25\)
Câu 2:
\(x^2+x=0\)
\(\Rightarrow x\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
\(x^2.\left(x-1\right)+4-4x=0\)
\(\Rightarrow x^2.\left(x-1\right)+4\left(1-x\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x^2-4\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x-2\right)\left(x+2\right)=0\)
Trường hợp 1: \(x-1=0\Rightarrow x=1\)
Trường hợp 2: \(x-2=0\Rightarrow x=2\)
Trường hợp 3: \(x+2=0\Rightarrow x=-2\)
Câu 3: Bạn xem lại đề bài nhé.
\(a.\) Với \(a+b+c=0\) thì \(\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{\left(-c\right).\left(-a\right).\left(-b\right)}{abc}=\frac{-abc}{abc}=-1\)
\(b.\) Công thức tổng quát: \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
Ta có:
\(\frac{1}{x\left(x+1\right)}=\frac{1}{x}-\frac{1}{x+1}\)
\(\frac{1}{\left(x+1\right)\left(x+2\right)}=\frac{1}{x+1}-\frac{1}{x+2}\)
\(\frac{1}{\left(x+2\right)\left(x+3\right)}=\frac{1}{x+2}-\frac{1}{x+3}\)
\(\frac{1}{\left(x+3\right)\left(x+4\right)}=\frac{1}{x+3}-\frac{1}{x-4}\)
\(\frac{1}{\left(x+4\right)\left(x+5\right)}=\frac{1}{x+4}-\frac{1}{x+5}\)
Do đó, suy ra được: \(A=\frac{1}{x}-\frac{1}{x+5}=\frac{x+5-x}{x\left(x+5\right)}=\frac{5}{x\left(x+5\right)}\)
\(1.\)
\(a,\left(a+b\right)^2=a^2+2ab+b^2\)
\(\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab=a^2+2ab+b^2\)
\(\Rightarrow\left(a+b\right)^2=\left(a-b\right)^2+4ab\left(đpcm\right)\)
a) \(x^2+x+1=x^2+x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)(luôn dương)
b) \(x^2-x+\frac{1}{2}=x^2-x+\frac{1}{4}+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2+\frac{1}{4}>0\)(luôn dương)
a, \(I=s\left(s^2-t\right)+\left(t^2+s\right)=s^3-st+t^2+s\)
Thay t = -1 và s = 1 vào biểu thức trên ta được :
\(1+1+1+1=4\)
b, \(N=u^2\left(u-v\right)-v\left(v^2-u^2\right)=u^2\left(u-v\right)+v\left(u+v\right)\left(u-v\right)\)
\(=\left(u-v\right)\left(u^2+v\left(u+v\right)\right)\)
Thay \(u=0,5=\frac{1}{2};v=-\frac{1}{2}\)
\(=\left(\frac{1}{2}+\frac{1}{2}\right).\frac{1}{4}=\frac{1}{4}\)
a) 3 u 2 − 8 u + 3 ( u 2 + 1 ) ( u − 1 ) b) 1 − 4 u 4 ( 4 u + 1 )