Ta có : x + y = xy

=> x = xy - y

=> x = y(x - 1)

=> x : y = \(\frac{y\left(x-1\right)}{y}=x-1\) ( do y \(\ne\)0)

Mặt khác ta có : x : y = x + y

=> \(x-1=x+y\)

=> \(x-1-x-y=0\)

=> \(x-x-1-y=0\)

=> \(-1-y=0\Rightarrow y=-1\)

Thay y = -1 vào x + y = xy ta có :

x + (-1) = x.(-1)

=> x - 1 = -x

=> x - 1 + x = 0

=> (x + x) - 1 = 0

=> 2x - 1 = 0 => x = 1/2

Vậy x = 1/2,y = -1

\(x\cdot y=x:y\) 

\(x:x=y\cdot y\) 

\(y^2=1\)

\(\Leftrightarrow y=\pm1\)  

TH 1 : 

\(y=1\) 

\(x+y=x\cdot y\) 

\(x+1=x\cdot1\) 

\(x+1=x\) 

\(0=1\left(sai\right)\) 

Suy ra loại \(y=1\) 

TH 2 : 

\(y=-1\) 

\(x+y=x\cdot y\) 

\(x+\left(-1\right)=x\cdot\left(-1\right)\) 

\(x-1=-x\) 

\(x+x=1\) 

\(2x=1\) 

\(x=\frac{1}{2}\) 

Vậy \(x=\frac{1}{2};y=-1\) là  nghiệm                                                                             

Xét tam giác ABC: \(\widehat{ABC}+\widehat{BAC}+\widehat{BCA}=180\Rightarrow\widehat{BAC}=180-\widehat{BCA}-\widehat{ABC}\)

\(=180-\left(\widehat{BCM}+\widehat{ACM}\right)-\left(\widehat{ABM}+\widehat{CBM}\right)\)

\(\Leftrightarrow\widehat{BAC}+\widehat{ABM}+\widehat{ACM}=180-\widehat{BCM}-\widehat{CBM}\)

Xét tam giác BMC: \(\widehat{BMC}+\widehat{CBM}+\widehat{BCM}=180\Leftrightarrow\widehat{BMC}=180-\widehat{BCM}-\widehat{CBM}\)

Vậy \(\widehat{BMC}=\widehat{BAC}+\widehat{ABM}+\widehat{ACM}\)

a) \(x^2=x\Leftrightarrow x^2-x=0\Leftrightarrow x\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

b) \(x^2=2x\Leftrightarrow x^2-2x=0\Leftrightarrow x\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

c) \(x^2=-1\)vì \(x^2\ge0,\forall x\)nên phương trình vô nghiệm.

d) \(x^2=1\Leftrightarrow x^2-1=0\Leftrightarrow\left(x-1\right)\left(x+1\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

a, x² = x

x² - x = 0

x (x - 1) = 0

=> x = 0 hoặc x - 1 = 0

=> x = 0 hoặc x = 1

Vậy x thuộc {0 ; 1}.

b, x² = 2x

x² - 2x = 0

x (x - 2) = 0

=> x = 0 hoặc x - 2 = 0

=> x = 0 hoặc x = 2

Vậy x thuộc {0 ; 2}.

c, x² = -1

Ta có: x² >= 0 với mọi x

=> x² = -1 (vô lí)

Vậy x thuộc tập hợp rỗng.

d, x² = 1

=> x² = 1² = (-1)²

=> x = 1 hoặc x = -1

Vậy x thuộc {-1 ; 1}.

a, x² = x

=> x² - x =0

=> x(x-1) =0

=> x = 0 hoặc x=1

b, x² = 2x

=> x² - 2x =0

=> x(x-2) = 0

=> x= 0 hoặc x=2

c, x² = -1

vì x² \(\ge\)0 với mọi x 

=> x² +1 >0

=> x² > -1

=> x² =-1 là vô lí

d, x² =1

=> x = 1 hoặc x =-1

Bài làm :

\(a,x^2=x\)

\(\Leftrightarrow x^2-x=0\)

\(\Leftrightarrow x\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

\(b,x^2=2x\)

\(\Leftrightarrow x^2-2x=0\)

\(\Leftrightarrow x\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

\(c,x^2=-1\)  ( sai )

Vì \(x^2\ge0\forall x\)

\(d,x^2=1\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

Học tốt 

a, ( x - 3 ) . ( x - 4 )  = 0              

=> x - 3 = 0 hoặc x - 4 = 0 

Nếu x - 3 = 0 => x = 3 

Nếu x - 4 = 0 => x = 4 

b, (\(\frac{1}{2}\)x  - 4 ) . ( x - \(\frac{1}{4}\)) = 0 

=>(  \(\frac{1}{2}\)x - 4 ) = 0    Hoặc  ( x - \(\frac{1}{4}\)) = 0 

Nếu ( \(\frac{1}{2}\)x - 4 ) = 0  => x = \(\frac{8}{1}\)

Nếu ( x - \(\frac{1}{4}\)) = 0     => x = \(\frac{1}{4}\)

c, (\(\frac{1}{3}\)- x ) . ( \(\frac{1}{2}\)+ 1 : x ) = 0 

=> ( \(\frac{1}{3}\)- x ) = 0 Hoặc ( \(\frac{1}{2}\)+ 1 : x ) = 0

Nếu (\(\frac{1}{3}\)- x ) = 0 => x = \(\frac{1}{3}\)

Nếu ( \(\frac{1}{2}\)+ 1 : x ) = 0 => x = \(\frac{-2}{1}\)

d, ( x + 3 ) . (  x - 4 ) + 2.(x + 3 ) = 0

=> (X + 3 ) = 0 Hoặc  ( x - 4 ) = 0 Hoặc 2. ( x + 3 ) = 0

Nếu x + 3 = 0 => x = 0

Nếu ( x - 4 ) = 0 => x = 4 

Nếu 2.(x + 3) = 0  => x = 3 

# Cụ MAIZ 

a. ( x - 3 ) ( x - 4 ) = 0

<=> \(\orbr{\begin{cases}x-3=0\\x-4=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=3\\x=4\end{cases}}\)

b. \(\left(\frac{1}{2}x-4\right)\left(x-\frac{1}{4}\right)=0\)

<=> \(\orbr{\begin{cases}\frac{1}{2}x-4=0\\x-\frac{1}{4}=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=8\\x=\frac{1}{4}\end{cases}}\)

\(x-\frac{3}{8}=\frac{1}{6}-\frac{1}{5}\)

=> \(x-\frac{3}{8}=\frac{5}{30}-\frac{6}{30}=-\frac{1}{30}\)

=> \(x=-\frac{1}{30}+\frac{3}{8}\)

=> \(x=\frac{41}{120}\)

\(-\frac{7}{10}\left(x+\frac{1}{3}\right)=\frac{4}{5}\)

=> \(-\frac{7}{10}x-\frac{7}{30}=\frac{4}{5}\)

=> \(-\frac{7}{10}x=\frac{4}{5}+\frac{7}{30}=\frac{31}{30}\)

=> \(x=\frac{31}{30}:\left(-\frac{7}{10}\right)=\frac{31}{30}\cdot\left(-\frac{10}{7}\right)=-\frac{31}{21}\)

\(x-\frac{4}{3}=\frac{5}{6}\Rightarrow x=\frac{5}{6}+\frac{4}{3}=\frac{5}{6}+\frac{8}{6}=\frac{13}{6}\)

Thiếu đề 

\(\frac{6}{5}+\left(x-\frac{2}{3}\right)=\frac{4}{7}\)

=> \(\frac{6}{5}+x-\frac{2}{3}=\frac{4}{7}\)

=> \(\frac{6}{5}+x=\frac{4}{7}+\frac{2}{3}=\frac{26}{21}\)

=> \(x=\frac{26}{21}-\frac{6}{5}=\frac{4}{105}\)

Bài làm :

\(a,\left(x-3\right)\left(x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=4\end{cases}}\)

\(b,\left(\frac{1}{2}x-4\right)\left(x-\frac{1}{4}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\frac{1}{2}x-4=0\\x-\frac{1}{4}=0\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{1}{2}x=4\\x=\frac{1}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}x=8\\x=\frac{1}{4}\end{cases}}\)

\(c,\left(\frac{1}{3}-x\right).\left(\frac{1}{2}+1:x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\frac{1}{3}-x=0\\\frac{1}{2}+1:x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=-2\end{cases}}\)

\(d,\left(x+3\right)\left(x-4\right)+2\left(x+3\right)=0\)

\(\Rightarrow\left(x+3\right)\left(x-4+2\right)=0\)

\(\Rightarrow\left(x+3\right)\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+3=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)

Học tốt nhé