doesn't she
isn't it
has he
wasn't he
aren't they
is it
must he
isn't she
don't they
hasn't he
do they
didn't he
did she
isn't it
didn't he
was it
does she
doesn't he
didn't he
do they
didn't they
aren't there

Ta có: \(AB^2+HC^2=\left(AA'^2+A'B^2\right)+\left(A'H^2+A'C^2\right)\)

\(=\left(AA'^2+A'C^2\right)+\left(A'B^2+A'H^2\right)=AC^2+HB^2\)

Lại có: \(BC^2+HA^2=\left(BB'^2+B'C^2\right)+\left(B'H^2+B'A^2\right)\)

\(=\left(BB'^2+B'A^2\right)+\left(B'C^2+B'H^2\right)=AB^2+HC^2\)

\(\Rightarrow AB^2+HC^2=AC^2+HB^2=BC^2+HA^2\)

I like playing soccer, but I must do my homework before.

We are saving money because we want to buy a new house.

I like English and French books.

Lan saves energy, so she turns off the lights.

1 I like playing soccer but I must do my homework before

2 We are saving money because we want to buy a new house

3 I like English books and French books

4 Lan saves money so she turns off the lights

I like playing soccer but i must do my ......

We are saving money so we want to ........

I like English books and I like ..........

Lan saves energy because she turns off ...

b: Vì (d)//(d') nên a=3

Vậy: (d):y=3x+b

Thay x=4 và y=-5 vào (d), ta được:

b+12=-5

hay b=-17

a: Ta có: BC⊥BA tại B

nên BC là tiếp tuyến của (A;AB)

b: Xét (A) có 

CB là tiếp tuyến

CD là tiếp tuyến

Do đó: CB=CD
hay C nằm trên đường trung trực của BD(1)

Ta có: AB=AD

nên A nằm trên đường trung trực của BD(2)

Từ (1) và (2) suy ra AC là đường trung trực của BD

hay AC\(\perp\)BD

Bài 2: 

Để hai đồ thị song song thì \(\left\{{}\begin{matrix}m^2-2=-1\\m+2\ne3\end{matrix}\right.\Leftrightarrow m=-1\)

Bài 2:

\(\Leftrightarrow\left\{{}\begin{matrix}m^2-2=-1\\m+2\ne3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m^2=1\\m\ne1\end{matrix}\right.\Leftrightarrow m=-1\)

3.(⅓x - ¼)² = ⅓ 

=> (\(\dfrac{1}{3x}\)- \(\dfrac{1}{4}\) )² = \(\dfrac{1}{9}\)

=>\(\left[{}\begin{matrix}\dfrac{1}{3x}-\dfrac{1}{4}=\dfrac{-1}{3}\\\dfrac{1}{3x}-\dfrac{1}{4}=\dfrac{1}{3}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}\dfrac{1}{3x}=\dfrac{-1}{12}\\\dfrac{1}{3x}=\dfrac{7}{12}\end{matrix}\right.\)        => \(\left[{}\begin{matrix}x=-4\\x=\dfrac{12}{21}=\dfrac{4}{7}\end{matrix}\right.\)

Vậy, tập nghiệm x thỏa mãn là S=\(\left\{-4;\dfrac{4}{7}\right\}\)