Tham khảo!

https://youtu.be/z8c3fae6IDA

Tham khảo!

3:

a:Các tia trên hình là Ax,Ay,Bx,By,Cx,Cy

=>Có 6 tia

b: AB<AC

=>B nằm giữa A và C

=>AB+BC=AC

=>BC=4cm

c: AI=3/2=1,5cm

CI=7-1,5=5,5cm

1) Áp dụng định lí Pytago vào ΔABC vuông tại A, ta được:

\(BC^2=AB^2+AC^2\)

\(\Leftrightarrow BC^2=6^2+8^2=100\)

hay BC=10(cm)

Áp dụng hệ thức lượng trong tam giác vuông vào ΔABC vuông tại A có AH là đường cao ứng với cạnh huyền BC, ta được:

\(AH\cdot BC=AB\cdot AC\)

\(\Leftrightarrow AH\cdot10=6\cdot8=48\)

hay AH=4,8(cm)

Hello, everybody! Allow me to introduce myself. My name is Hai, I am 13 years old and a student at Gia Thụy secondary school . My favorite subject is math. I was born and grew up in Ha Noi. At the moment, I am living with my parents and a younger brother.I am a quiet, a little bit childish but also factual, dexterous and roman lady. In the education views, I am diligent, clever and patient with my study or even work and I never give up even though I know a lot of hardships are waiting for me in the future. My hobbies are jewellery collections, taking a walk in the afternoon, listening to music and watching movies. Parularly, I really love to watch how to make a sweet cake videos from the internet. I also enjoy playing badminton and riding a bicycle.My dream is to be a successful business woman so that now I am trying to study as hard as I can to get into the right subject that I love. And I also hope that my dream will come true and surely I will do everything to make it come true.

ngày hôm qua làm việc của mình chi chi đó nà

Câu 81:

1: ĐKXĐ: x>=0

\(5\sqrt{12x}+4\cdot\sqrt{3x}+2\cdot\sqrt{48x}=14\)

=>\(5\cdot2\cdot\sqrt{3x}+4\cdot\sqrt{3x}+2\cdot4\sqrt{3x}=14\)

=>\(22\sqrt{3x}=14\)

=>\(\sqrt{3x}=\frac{14}{22}=\frac{7}{11}\)

=>\(3x=\frac{49}{121}\)

=>\(x=\frac{49}{121\cdot3}=\frac{49}{363}\) (nhận)

2: ĐKXĐ: x>=5

\(\sqrt{4x-20}+\sqrt{x-5}-\frac13\cdot\sqrt{9x-45}=4\)

=>\(2\sqrt{x-5}+\sqrt{x-5}-\frac13\cdot3\sqrt{x-5}=4\)

=>\(2\sqrt{x-5}=4\)

=>\(\sqrt{x-5}=2\)

=>x-5=4

=>x=9(nhận)

3: ĐKXĐ: x>=-1

Ta có: \(\sqrt{16x+16}-\sqrt{9x+9}=1\)

=>\(4\cdot\sqrt{x+1}-3\cdot\sqrt{x+1}=1\)

=>\(\sqrt{x+1}=1\)

=>x+1=1

=>x=0(nhận)

4: ĐKXĐ: x>=0

\(3\sqrt{2x}+5\cdot\sqrt{8x}-20-\sqrt{18x}=0\)

=>\(3\sqrt{2x}+5\cdot2\sqrt{2x}-3\sqrt{2x}=20\)

=>\(10\sqrt{2x}=20\)

=>\(\sqrt{2x}=2\)

=>2x=4

=>x=2(nhận)

5: ĐKXĐ: x>=2

\(\sqrt{49x-98}-14\cdot\sqrt{\frac{x-2}{49}}=3\sqrt{x-2}+8\)

=>\(7\sqrt{x-2}-14\cdot\frac{\sqrt{x-2}}{7}-3\sqrt{x-2}=8\)

=>\(2\sqrt{x-2}=8\)

=>\(\sqrt{x-2}=4\)

=>x-2=16

=>x=18(nhận)

6: ĐKXĐ: x>=-3

\(\sqrt{9x+27}+5\sqrt{x+3}-\frac34\cdot\sqrt{16x+48}=5\)

=>\(3\sqrt{x+3}+5\sqrt{x+3}-\frac34\cdot4\sqrt{x+3}=5\)

=>\(5\sqrt{x+3}=5\)

=>\(\sqrt{x+3}=1\)

=>x+3=1

=>x=-2(nhận)

7: ĐKXĐ: x>=2

\(\sqrt{x-2}+\sqrt{9x-18}+\sqrt{81x-162}=4\)

=>\(\sqrt{x-2}+3\sqrt{x-2}+9\sqrt{x-2}=4\)

=>\(13\sqrt{x-2}=4\)

=>\(\sqrt{x-2}=\frac{4}{13}\)

=>\(x-2=\frac{16}{169}\)

=>\(x=2+\frac{16}{169}=\frac{354}{169}\) (nhận)

8: ĐKXĐ: x>=2

\(\sqrt{x-2}+\sqrt{9x-18}+\sqrt{81x-162}=4\)

=>\(\sqrt{x-2}+3\sqrt{x-2}+9\sqrt{x-2}=4\)

=>\(13\sqrt{x-2}=4\)

=>\(\sqrt{x-2}=\frac{4}{13}\)

=>\(x-2=\frac{16}{169}\)

=>\(x=2+\frac{16}{169}=\frac{354}{169}\) (nhận)

9: ĐKXĐ: x>=1

\(\sqrt{16x-16}-\sqrt{9x-9}+\sqrt{4x-4}+\sqrt{x-1}=8\)

=>\(4\sqrt{x-1}-3\sqrt{x-1}+2\sqrt{x-1}+\sqrt{x-1}=8\)

=>\(4\sqrt{x-1}=8\)

=>\(\sqrt{x-1}=2\)

=>x-1=4

=>x=5(nhận)

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