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a) Với x1 = x2 = 1
\(\Rightarrow f\left(1\right)=f\left(1.1\right)\)
\(\Rightarrow f\left(1\right)=f\left(1\right).f\left(1\right)\)
\(\Rightarrow f\left(1\right).f\left(1\right)-f\left(1\right)=0\)
\(\Rightarrow f\left(1\right).\left[f\left(1\right)-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}f\left(1\right)=0\\f\left(1\right)-1=0\end{cases}}\)
Mà \(f\left(x\right)\ne0\) ( với mọi \(x\in R\) \(;\) \(x\ne0\) )
\(\Rightarrow f\left(1\right)\ne0\)
\(\Rightarrow f\left(1\right)-1=0\)
\(\Rightarrow f\left(1\right)=1\)
b) Ta có : \(f\left(\frac{1}{x}\right).f\left(x\right)=f\left(\frac{1}{x}.x\right)\)
\(\Rightarrow f\left(\frac{1}{x}\right).f\left(x\right)=f\left(1\right)=1\)
\(\Rightarrow f\left(\frac{1}{x}\right).f\left(x\right)=1\)
\(\Rightarrow f\left(\frac{1}{x}\right)=\frac{1}{f\left(x\right)}\)
\(\Rightarrow f\left(x^{-1}\right)=\left[f\left(x\right)\right]^{-1}\)
a) A + x2 - 4xy2 + 2xz - 3y2 = 0
=> A = -x2 + 4xy2 - 2xz + 3y2
b) B + 5x2 - 2xy = 6x2 + 9xy - y2
=> B = 6x2 + 9xy - y2 - 5x2 + 2xy= x2 + 11xy - y2
c) 3xy - 4y2 - A = x2 - 7xy + 8y2
=> A = 3xy - 4y2 - x2 + 7xy - 8y2 = -12y2 + 10xy - x2
Trả lời:
a, A + ( x2 - 4xy2 + 2xz - 3y2 ) = 0
=> A = - ( x2 - 4xy2 + 2xz - 3y2 ) = - x2 + 4xy2 - 2xz + 3y2
b, B + ( 5x2 - 2xy ) = 6x2 + 9xy - y2
=> B = 6x2 + 9xy - y2 - ( 5x2 - 2xy ) = 6x2 + 9xy - y2 - 5x2 + 2xy = x2 + 11xy - y2
c, ( 3xy - 4y2 ) - A = x2 - 7xy + 8y2
=> A = 3xy - 4y2 - ( x2 - 7xy + 8y2 ) = 3xy - 4y2 - x2 + 7xy - 8y2 = 10xy - 12y2 - x2
d, B + ( 4x2y + 5y2 - 3xz + z2 ) = x2 + 11xy - y2 + 4x2y + 5y2 - 3xz + z2 = x2 + 11xy + 4y2 + 4x2y - 3xz + z2
HÌNH TỰ VẼ
TA CÓ :A1+B1=AOB
MÀ A1+600 B2=450
600+450=1050AOB
VẬY AOB=1050
\(\frac{5}{4}-x-\frac{1}{10}=\frac{2}{5}\)
\(\frac{5}{4}-x=\frac{2}{5}+\frac{1}{10}\)
\(\frac{5}{4}-x=\frac{1}{2}\)
\(x=\frac{5}{4}-\frac{1}{2}\)
\(x=\frac{3}{4}\)
\(\frac{4}{5}-x-\frac{1}{10}=\frac{2}{5}\)
\(\frac{4}{5}-x=\frac{1}{2}\)
\(x=\frac{3}{10}\)
A 70 o 120 o B D C d' d 1 3 2 4 1 1
Vì \(\widehat{B_1}=\widehat{B_4}\left(\text{2 góc đối đỉnh}\right)\)\(\Rightarrow\widehat{B_4}=120^o\)
Vì \(\widehat{B_4}=\widehat{C_2}\left(\text{ 2 góc so le trong }\right)\)\(\Rightarrow\widehat{C_2}=120^o\)
Vì \(\widehat{C_2}+\widehat{C_3}=180^o\)( 2 góc kề bù ) \(\Rightarrow120^o+\widehat{C_3}=180^o\)\(\Rightarrow\widehat{C_3}=180^o-120^o=60^o\)
Vì \(\widehat{A_1}=\widehat{D_1}\left(\text{ 2 góc so le trong }\right)\)\(\Rightarrow\widehat{D_1}=70^o\)