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B1) Từ \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Rightarrow\frac{xy+yz+zx}{xyz}=0\)
\(\Rightarrow xy+yz+zx=0\)
Ta có \(\left(x+y+z\right)^2=x^2+y^2+z^2+2\left(xy+yz+zx\right)\)
\(=x^2+y^2+z^2+2.0\)
\(=x^2+y^2+z^2\left(đpcm\right)\)
B2) \(a^2+b^2+c^2=ab+bc+ca\)
\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Vì \(\hept{\begin{cases}\left(a-b\right)^2\ge0\forall a;b\\\left(b-c\right)^2\ge0\forall b;c\\\left(c-a\right)^2\ge0\forall c;a\end{cases}\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0}\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow a=b=c\left(đpcm\right)}\)
\(a^2+b^2+c^2=ab+bc+ca\)
\(\Leftrightarrow\left(a^2+b^2+c^2\right).2=\left(ab+bc+ca\right).2\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Ta có: \(\hept{\begin{cases}\left(a-b\right)^2\ge0\forall a,b\\\left(b-c\right)^2\ge0\forall b,c\\\left(c-a\right)^2\ge0\forall a,c\end{cases}}\)\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\forall a,b,c\)
Mà \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\Leftrightarrow a=b=c\)
Vậy \(a^2+b^2+c^2=ab+bc+ca\)thì \(a=b=c\)
a) \(\frac{x+1}{2x+6}\)+\(\frac{2x+3}{x\left(x+3\right)}\)
= \(\frac{x+1}{2\left(x+3\right)}\)+ \(\frac{2x+3}{x\left(x+3\right)}\)
= \(\frac{x\left(x+1\right)}{2x\left(x+3\right)}\)+ \(\frac{2\left(2x+3\right)}{2x\left(x+3\right)}\)
= \(\frac{x^2+x+4x+6}{2x\left(x+3\right)}\)
= \(\frac{x^2+5x+6}{2x\left(x+3\right)}\)
= \(\frac{\left(x+2\right)\left(x+3\right)}{2x\left(x+3\right)}\)
= \(\frac{x+2}{2x}\)
b) \(\frac{x-1}{x}\)+ \(\frac{x+2}{2}\)
= \(\frac{2\left(x-1\right)}{2x}\)+ \(\frac{x\left(x+2\right)}{2x}\)
= \(\frac{2x-2+x^2+2x}{2x}\)
= \(\frac{x^2+4x-2}{2x}\)
c) \(\frac{1}{x+y}\)+ \(\frac{-1}{x-y}\)+ \(\frac{2x}{x^2+y^2}\)
= \(\frac{\left(x-y\right)\left(x^2+y^2\right)}{\left(x^2+y^2\right)\left(x-y\right)\left(x+y\right)}\)+\(\frac{-\left(x+y\right)\left(x^2+y^2\right)}{\left(x^2+y^2\right)\left(x-y\right)\left(x+y\right)}\)+ \(\frac{2x\left(x-y\right)\left(x+y\right)}{\left(x^2+y^2\right)\left(x-y\right)\left(x+y\right)}\)
= \(\frac{x^3+xy^2-x^2y-y^3-x^3-xy^2-xy^2-y^3+2x^3+2x^2y-2x^2y+2xy^2}{\left(x^2+y^2\right)\left(x^2-y^2\right)}\)
= \(\frac{2x^3+xy^2-x^2y-2y^3}{\left(x^2+y^2\right)\left(x^2-y^2\right)}\)
= \(\frac{\left(2x^3-2y^3\right)-\left(x^2y-xy^2\right)}{\left(x^2+y^2\right)\left(x^2-y^2\right)}\)
= \(\frac{2\left(x-y\right)\left(x^2+xy+y^2\right)-xy\left(x-y\right)}{\left(x^2+y^2\right)\left(x^2-y^2\right)}\)
= \(\frac{\left(x-y\right)\left(2x^2+2xy+2y^2-xy\right)}{\left(x^2+y^2\right)\left(x^2-y^2\right)}\)
= \(\frac{2x^2+xy+2y^2}{\left(x+y\right)\left(x^2+y^2\right)}\)
e) = \(\frac{3x^2-6xy+3y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
= \(\frac{3\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
=\(\frac{3x-3y}{x^2+xy+y^2}\)
( Mình bận rồi, lát làm câu d nhé)
a, Chứng minh \(x^3+y^3+z^3=\left(x+y\right)^3-3xy.\left(x+y\right)+z^3\)
Biến đổi vế phải thì ta phải suy ra điều phải chứng minh
b, Ta có: \(a+b+c=0\)thì
\(a^3+b^3+c^3==\left(a+b\right)^3-3ab\left(a+b\right)+c^3=-c^3-3ab\left(-c\right)+c^3=3abc\)
( Vì \(a+b+c=0\)nên \(a+b=-c\))
Theo giả thuyết \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Rightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\)
Khi đó \(A=\frac{yz}{x^2}+\frac{xz}{y^2}+\frac{xy}{z^2}\)
\(=\frac{xyz}{x^3}+\frac{xyz}{y^3}+\frac{xyz}{z^3}\)
\(=xyz\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)\)
\(=xyz.\frac{3}{xyz}=3\)
Ta có: \(A=\frac{x^2-x-6}{x-2}\)(ĐKXĐ: \(x\ne2\))
\(\Rightarrow A=\frac{x^2-3x+2x-6}{x-2}\)
\(\Rightarrow A=\frac{\left(x-2\right)\left(x+3\right)}{x-2}\)
\(\Rightarrow A=x+3\)
Mà \(x\in Z\)
=> A là số nguyên
Câu 1 :
a, \(\frac{3}{x+3}-\frac{x-6}{x^2+3x}=\frac{3x-x+6}{x\left(x+3\right)}=\frac{2x+6}{x\left(x+3\right)}=\frac{2}{x}\)
b, \(\frac{2x^2-x}{x-1}+\frac{x+1}{1-x}+\frac{2-x^2}{x-1}=\frac{2x^2-x-x-1+2-x^2}{x-1}\)
\(=\frac{x^2-2x+1}{x-1}=\frac{\left(x-1\right)^2}{x-1}=x-1\)
Bài 2 :
a, Với \(x\ne\pm2\)
\(A=\left(\frac{x}{x^2-4}+\frac{1}{x+2}-\frac{2}{x-2}\right):\left(1-\frac{x}{x+2}\right)\)
\(=\left(\frac{x+x-2-2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\right):\left(\frac{x+2-x}{x+2}\right)\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{2}=\frac{-3}{x-2}\)
b, Thay x = -4 vào biểu thức trên ta được :
\(-\frac{3}{-4-2}=-\frac{3}{-6}=\frac{1}{2}\)
c, Để A \(\inℤ\Rightarrow x-2\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)
| x - 2 | 1 | -1 | 3 | -3 |
| x | 3 | 1 | 5 | -1 |