Bài 5: 

a: BC=10cm

b: HA=4,8cm

HB=3,6(cm)

HC=6,4(cm)

Bài 6:

\(x^3=6+3\sqrt[3]{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)\\ \Leftrightarrow x^3=6+3x\sqrt[3]{1}\\ \Leftrightarrow x^3-3x=6\\ y^3=34+3\sqrt[3]{\left(17+12\sqrt{2}\right)\left(17-12\sqrt{2}\right)}\left(\sqrt[3]{17+12\sqrt{2}}+\sqrt[3]{17-12\sqrt{2}}\right)\\ \Leftrightarrow y^3=34+3y\sqrt[3]{1}\\ \Leftrightarrow y^3-3y=34\\ \Leftrightarrow P=x^3-3x+y^3-3y+1980=6+34+1980=2020\)

gfrưerrrrrrrrrrr

\(\left\{{}\begin{matrix}2x+y=1\\x+y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x+y=-1\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x=2\\y=-3\end{matrix}\right.\)

\(\left\{{}\begin{matrix}2x+2y=18\\x-y=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=9\\x-y=-6\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}2x=3\\x-y=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=\dfrac{15}{2}\end{matrix}\right.\)\(\left\{{}\begin{matrix}2x+3y=6\\x-2y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+6y=12\\3x-6y=9\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}7x=21\\3x-6y=9\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x=3\\y=0\end{matrix}\right.\)

a) ĐKXĐ: \(x>0;x\ne\pm1.\)

\(A=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}}{1-x}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{1-\sqrt{x}}{\sqrt{x}+1}\right).\\ A=\dfrac{x+2\sqrt{x}+1+x-\sqrt{x}-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\)

\(A=\dfrac{2x+1}{4\sqrt{x}}.\)

b) \(A=\dfrac{3}{4}.\Rightarrow\dfrac{2x+1}{4\sqrt{x}}=\dfrac{3}{4}.\Rightarrow12\sqrt{x}-8x+4=0.\\ \Leftrightarrow8x-12\sqrt{x}-4=0.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{3+\sqrt{17}}{4}.\\\sqrt{x}=\dfrac{3-\sqrt{17}}{4}.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13+3\sqrt{17}}{8}.\\x=\dfrac{13-3\sqrt{17}}{8}.\end{matrix}\right.\) (TM). 

1) 43 . 78 - 43 . 48 + 30 . 80 - 30 . 23

= 43.(78 - 48) + 30.(80 - 23)

= 43.30 + 30.57

= 30.(43 + 57)

= 30.100

= 3000

2) 31.175 - 31.50 + 69.125

= 31.(175 - 50) + 69.125

= 31.125 + 69.125

= 125.(31 + 69)

= 125.100

= 12500

3) 2.[(7 - 3¹³ : 3¹²) : 2² + 99] - 10²

= 2.[(7 - 3) : 4 + 99] - 100

= 2.(4 : 4 + 99) - 100

= 2.(1 + 99) - 100

= 2.100 - 100

= 200 - 100

= 100

4) 2²⁰¹⁹.2² : 2²⁰¹⁶ - 125 : 5² + 2019⁰

= 2²⁰²¹ : 2²⁰¹⁶ - 125 : 25 + 1

= 2⁵ - 5 + 1

= 32 - 4

= 28

XXét tứ giác AMDN có ^AMD=^MAN=^AND=90∞

⇒AMDN là hình chữ nhật

hcn AMDN có AD là phân giác góc A

⇒AMDN là hình vuông(dấu hiệu 3)

-my friend is traveling abroad so much. 
-Tom is doing his job well.
-he has an art class for more than 120 minutes.
-what toys are their favorite?
-What is the distance from your house to Van Hanh mall?
-how old are you
-37. mình không thấy
-what is your father's job?
 

1:

a: =>(x-1)(x-7)=0

=>x=1 hoặc x=7

b: =>x(x^2-9x+8)=0

=>x(x-1)(x-8)=0

=>\(x\in\left\{0;1;8\right\}\)

c: Đặt 1/căn x-7=a; 1/căn y+6=b

Theo đề, ta có:

7a-4b=5/3 và 5a+3b=13/6

=>a=1/3 và b=1/6

=>x-7=9 và y+6=36

=>x=16 và y=30

Bài 3:

a: Δ=(2m+3)^2-4(m^2+3m+2)

=4m^2+12m+9-4m^2-12m-8=1>0

=>PT luôn có hai nghiệm pb

b: x1^2+x2^2=1

=>(x1+x2)^2-2x1x2=1

=>(2m+3)^2-2(m^2+3m+2)=1

=>4m^2+12m+9-2m^2-6m-4-1=0

=>2m^2+6m+4=0

=>m=-1 hoặc m=-2

Phần trắc nghiệm

Câu 1: D

Câu 2: C

Câu 3: B

Câu 4: D