ta có tam giác BCD cân tại C

=>góc CDB bằng góc CBD

=>BC//AD(goc ADB = gocCBD) 

=>DPCM ABCD là hình thang

                     Học tốt

\(DB\)là phân giác \(\widehat{ADC}\)suy ra \(\widehat{ADB}=\widehat{CDB}\)(1)

\(BC=CD\)suy ra \(\Delta CBD\)cân tại \(C\)suy ra \(\widehat{CBD}=\widehat{CDB}\)(2)

(1)(2) suy ra \(\widehat{ADB}=\widehat{CBD}\)

mà hai góc này ở vị trí so le trong suy ra \(BC//AD\).

Suy ra \(ABCD\)là hình thang. 

a) \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left(x^2+5x+6\right)-\left(x^2+3x-10\right)=0\)

\(\Leftrightarrow2x+16=0\)

\(\Leftrightarrow x=-8\)

b) \(\left(x-5\right)\left(-x+4\right)-\left(x-1\right)\left(x+3\right)=-2x^2\)

\(\Leftrightarrow\left(-x^2+9x-20\right)-\left(x^2+2x-3\right)=-2x^2\)

\(\Leftrightarrow7x-17=0\)

\(\Leftrightarrow x=\frac{17}{7}\)

a 

\(\left(x-3\right)\left(x-2\right)-\left(x+1\right)\left(x-5\right)=0\)   

\(x^2-5x+6-\left(x^2-4x-5\right)=0\)   

\(x^2-5x+6-x^2+4x+5=0\)   

\(-x+11=0\)   

\(-x=-11\)   

\(x=11\)

b 

\(3\left(2\sqrt{x}-1\right)\left(3\sqrt{x}-1\right)-\left(2\sqrt{x}-3\right)\left(9\sqrt{x}-1\right)=0\)   

\(18x-15\sqrt{x}+3-\left(18x-29\sqrt{x}+3\right)=0\)   

\(18x-15\sqrt{x}+3-18x+29\sqrt{x}-3=0\)   

\(14\sqrt{x}=0\)   

\(\sqrt{x}=0\)   

\(x=0\)

a) ĐK: \(x\ne2\).

\(\frac{x-3}{x-2}-2=\frac{x+1}{x-2}\)

\(\Leftrightarrow\frac{x-3}{x-2}-\frac{2\left(x-2\right)}{x-2}=\frac{x+1}{x-2}\)

\(\Rightarrow x-3-2\left(x-2\right)=x+1\)

\(\Leftrightarrow2x=0\)

\(\Leftrightarrow x=0\)(thỏa mãn) 

2) ĐK: \(x\ne\pm1\).

\(\frac{x+2}{x^2-1}+\frac{1}{x-1}=0\)

\(\Leftrightarrow\frac{x+2}{x^2-1}+\frac{x+1}{\left(x-1\right)\left(x+1\right)}=0\)

\(\Rightarrow x+2+x+1=0\)

\(\Leftrightarrow x=-\frac{3}{2}\)(tm)

c) ĐK:\(x\ne0,x\ne-1\).

\(\frac{x}{x+1}-\frac{1}{x}=\frac{1}{x^2+x}\)

\(\Leftrightarrow\frac{x^2}{x\left(1+x\right)}-\frac{x+1}{x\left(x+1\right)}=\frac{1}{x\left(x+1\right)}\)

\(\Rightarrow x^2-\left(x+1\right)=1\)

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

\(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)

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

d) ĐK: \(x\ne1\).

\(\frac{3x-1}{x^3-1}=\frac{1}{x-1}-\frac{x}{x^2+x+1}\)

\(\Leftrightarrow\frac{3x-1}{x^3-1}=\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(\Rightarrow3x-1=x^2+x+1-x\left(x-1\right)\)

\(\Leftrightarrow3x-1=2x+1\)

\(\Leftrightarrow x=2\)(tm) 

Hello, kb ko?