C nằm giữa A và B

=>CA+CB=AB

=>CB+2=7

=>CB=5(cm)

D là trung điểm của AC

=>\(AD=DC=\dfrac{AC}{2}=\dfrac{2}{2}=1\left(cm\right)\)

E là trung điểm của CB

=>\(EC=EB=\dfrac{BC}{2}=2,5\left(cm\right)\)

CA và CB là hai tia đối nhau

=>CD và CE là hai tia đối nhau

=>C nằm giữa D và E

=>DE=DC+CE=2,5+1=3,5(cm)

F là trung điểm của DE

=>\(DF=\dfrac{DE}{2}=1,75\left(cm\right)\)

Vì DC<DF

nên C nằm giữa D và F

=>DC+CF=DF

=>CF+1=1,75

=>CF=0,75(cm)

a: \(3x\left(x-2\right)=x^2-4\)

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

=>\(\left(x-2\right)\left(3x-x-2\right)=0\)

=>(x-2)(2x-2)=0

=>2(x-2)(x-1)=0

=>(x-1)(x-2)=0

=>\(\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

b:

ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

 \(\dfrac{x+3}{x-1}+\dfrac{x}{x+1}=\dfrac{x^2+4x+5}{x^2-1}\)

=>\(\dfrac{\left(x+3\right)\left(x+1\right)+x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2+4x+5}{\left(x-1\right)\left(x+1\right)}\)

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

=>\(x^2+4x+3+x^2-x-x^2-4x-5=0\)

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

=>(x-2)(x+1)=0

=>\(\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-1\left(loại\right)\end{matrix}\right.\)

c: \(3\left(x-1\right)< 5\left(x+1\right)-2\)

=>\(3x-3< 5x+5-2\)

=>3x-3<5x+3

=>-2x<6

=>x>-3

d: \(x^3>-2x\)

=>\(x^3+2x>0\)

=>\(x\left(x^2+2\right)>0\)

mà \(x^2+2>0\forall x\)

nên x>0

a: Thay m=-2 vào (1), ta được:

\(x^2+x\cdot\left(-2\right)+\left(-2\right)-2=0\)

=>\(x^2-2x-4=0\)

\(\Leftrightarrow x^2-2x+1-5=0\)

=>\(\left(x-1\right)^2=5\)

=>\(x-1=\pm\sqrt{5}\)

=>\(x=1\pm\sqrt{5}\)

b: \(\text{Δ}=\left(-m\right)^2-4\cdot1\cdot\left(m-2\right)\)

\(=m^2-4m+8=m^2-4m+4+4=\left(m-2\right)^2+4>0\forall m\)

=>Phương trình luôn có hai nghiệm phân biệt

c: Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=m\\x_1x_2=\dfrac{c}{a}=m-2\end{matrix}\right.\)

\(\dfrac{x_1^2-2}{x_1-1}\cdot\dfrac{x_2^2-2}{x_2-1}=4\)

=>\(\dfrac{\left(x_1^2-2\right)\left(x_2^2-2\right)}{\left(x_1-1\right)\left(x_2-1\right)}=4\)

=>\(\dfrac{\left(x_1x_2\right)^2-2\left(x_1^2+x_2^2\right)+4}{x_1x_2-\left(x_1+x_2\right)+1}=4\)

=>\(\dfrac{\left(m-2\right)^2-2\left[\left(x_1+x_2\right)^2-2x_1x_2\right]+4}{m-2-m+1}=4\)

=>\(\dfrac{\left(m-2\right)^2-2\left[m^2-2\left(m-2\right)\right]+4}{-1}=4\)

=>\(\left(m-2\right)^2-2\left(m^2-2m+4\right)+4=-4\)

=>\(m^2-4m+4-2m^2+4m-8+8=0\)

=>\(-m^2+4=0\)

=>\(\left[{}\begin{matrix}m=2\left(nhận\right)\\m=-2\left(nhận\right)\end{matrix}\right.\)

a: a: Xét ΔABC và ΔAED có

\(\dfrac{AB}{AE}=\dfrac{AC}{AD}\left(\dfrac{15}{5}=\dfrac{21}{7}=3\right)\)

\(\widehat{BAC}\) chung

Do đó: ΔABC~ΔAED

Vì \(\dfrac{AB}{AE}=\dfrac{AC}{AD}\)

nên \(AB\cdot AD=AE\cdot AC\)

b: \(\dfrac{AB}{AE}=\dfrac{AC}{AD}\)

=>\(\dfrac{AB}{AC}=\dfrac{AE}{AD}\)

Xét ΔABE và ΔACD có

\(\dfrac{AB}{AC}=\dfrac{AE}{AD}\)

\(\widehat{BAE}\) chung

Do đó: ΔABE~ΔACD

=>\(\widehat{ABE}=\widehat{ACD};\widehat{AEB}=\widehat{ADC}\)

c: Xét ΔOBD và ΔOCE có

\(\widehat{OBD}=\widehat{OCE}\)

\(\widehat{BOD}=\widehat{COE}\)(hai góc đối đỉnh)

Do đó: ΔOBD~ΔOCE
=>\(\dfrac{OB}{OC}=\dfrac{OD}{OE}\)

=>\(OB\cdot OE=OD\cdot OC\)

a: Thay x=2/3 vào A, ta được:

\(A=\dfrac{\dfrac{2}{3}-2}{\dfrac{2}{3}}=\dfrac{-4}{3}:\dfrac{2}{3}=-\dfrac{4}{3}\cdot\dfrac{3}{2}=-2\)

b: \(B=\dfrac{4x}{x+1}+\dfrac{x}{1-x}+\dfrac{2}{x^2-1}\)

\(=\dfrac{4x}{x+1}-\dfrac{x}{x-1}+\dfrac{2}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{4x\left(x-1\right)-x\left(x+1\right)+2}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{4x^2-4x-x^2-x+2}{\left(x-1\right)\left(x+1\right)}=\dfrac{3x^2-5x+2}{\left(x-1\right)\left(x+1\right)}\)

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

a: Thay m=2 vào (1), ta được:

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

=>\(x^2-2x-3=0\)

=>(x-3)(x+1)=0

=>\(\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

b: \(\text{Δ}=\left[-2\left(m-1\right)\right]^2-4\cdot1\cdot\left(m-5\right)\)

\(=4\left(m^2-2m+1\right)-4\left(m-5\right)\)

\(=4m^2-8m+4-4m+20\)

\(=4m^2-12m+24=4m^2-12m+9+15\)

\(=\left(2m-3\right)^2+15>=15>0\forall m\)

=>Phương trình (1) luôn có hai nghiệm phân biệt

Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2\left(m-1\right)=2m-2\\x_1x_2=\dfrac{c}{a}=m-5\end{matrix}\right.\)

\(P=\left|x_1-x_2\right|=\sqrt{\left(x_1-x_2\right)^2}\)

\(=\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}\)

\(=\sqrt{\left(2m-2\right)^2-4\left(m-5\right)}\)

\(=\sqrt{4m^2-8m+4-4m+20}\)

\(=\sqrt{4m^2-12m+24}\)

\(=\sqrt{4m^2-12m+9+15}=\sqrt{\left(2m-3\right)^2+15}>=\sqrt{15}\forall m\)

Dấu '=' xảy ra khi 2m-3=0

=>\(m=\dfrac{3}{2}\)

a: Thay m=-1 vào (1), ta được:

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

=>\(x^2+2x=0\)

=>x(x+2)=0

=>\(\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

b: \(\text{Δ}=\left[-2m\right]^2-4\left(m^2-1\right)\)

\(=4m^2-4m^2+4=4>0\forall m\)

=>Phương trình luôn có hai nghiệm phân biệt

Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2m\\x_1x_2=\dfrac{c}{a}=m^2-1\end{matrix}\right.\)

\(P=x_1^2+x_2^2\)

\(=\left(x_1+x_2\right)^2-2x_1x_2\)

\(=4m^2-2\left(m^2-1\right)\)

\(=4m^2-2m^2+2=2m^2+2>=2\forall m\)

Dấu '=' xảy ra khi m=0

a: Thay m=-1 vào (1), ta được:

\(x^2+\left(-1-1\right)x-\left[\left(-1\right)^2-1\right]=0\)

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

=>x(x-2)=0

=>\(\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

b: \(\text{Δ}=\left(m-1\right)^2-4\cdot1\cdot\left(-m^2+1\right)\)

\(=m^2-2m+1+4m^2-4=5m^2-2m-3\)

Để phương trình có hai nghiệm phân biệt thì Δ>0

=>\(5m^2-2m-3>0\)

=>(m-1)(5m+3)>0

=>\(\left[{}\begin{matrix}m>1\\m< -\dfrac{3}{5}\end{matrix}\right.\)

Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=-m+1\\x_1x_2=\dfrac{c}{a}=-m^2+1\end{matrix}\right.\)

\(x_1=-2x_2\)

\(x_1+x_2=-m+1\)

Do đó: \(-x_2=-m+1\)

=>\(x_2=m-1\)

=>\(x_1=-2\left(m-1\right)=-2m+2\)

\(x_1x_2=-m^2+1\)

=>\(\left(m-1\right)\left(-2m+2\right)=-m^2+1\)

=>\(-2m^2+2m+2m-2+m^2-1=0\)

=>\(-m^2+4m-3=0\)

=>(m-1)(m-3)=0

=>\(\left[{}\begin{matrix}m=1\left(loại\right)\\m=3\left(nhận\right)\end{matrix}\right.\)