Bài 1:

Cường độ dòng điện qua điện trở: I = U : R = 12 : 60 = 0,2 (A)

Bài 2:

Điện trở tương đương: R_tđ = R₁ + R₂ = 3 + 5 = 8 (\(\Omega\))

Cường độ dòng điện qua mạch chính: I = U : R_tđ = 12 : 8 = 1,5 (A)

Bài 3:

Điện trửo tương đương: R_tđ = (R₁.R₂) : (R₁ + R₂) = (3.6) : (3 + 6) = 2 (\(\Omega\))

Có: U = U₁ = U₂ = 12V (Vì R₁//R₂)

Cường độ dòng điện qua mạch chính và các mạch rẽ:

I = U : R_tđ = 12 : 2 = 6 (A)

I₁ = U₁ : R₂ = 12 : 3 = 4(A)

I₂ = U₂ : R₂ = 12 : 6 = 2(A)

Bài 5:

a: 2x-(3-5x)=4(x+3)

=>2x-3+5x=4x+12

=>7x-3=4x+12

=>3x=15

=>x=5

b: =>5/3x-2/3+x=1+5/2-3/2x

=>25/6x=25/6

=>x=1

c: 3x-2=2x-3

=>3x-2x=-3+2

=>x=-1

d: =>2u+27=4u+27

=>u=0

e: =>5-x+6=12-8x

=>-x+11=12-8x

=>7x=1

=>x=1/7

f: =>-90+12x=-45+6x

=>12x-90=6x-45

=>6x-45=0

=>x=9/2

c)\(\left\{{}\begin{matrix}u_1+u_3=3\\u_1^2+u_3^2=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}u_1+u_3=3\\\left(u_1+u_3\right)^2-2u_1u_3=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}u_1+u_3=3\\u_1u_3=2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}u_1=2\\u_3=1\end{matrix}\right.\\\left\{{}\begin{matrix}u_1=1\\u_3=2\end{matrix}\right.\end{matrix}\right.\)

Làm nốt (sử dụng công thức: \(u_n=u_1+\left(n-1\right)d\) để tìm được công sai

\(S_n=nu_1+\dfrac{n\left(n-1\right)}{2}d\) để tính tổng 15 số hạng đầu)

d)\(\left\{{}\begin{matrix}u_1+u_2+u_3=14\\u_1u_2u_3=64\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}u_2-d+u_2+u_2+d=14\\\left(u_2-d\right)u_2\left(u_2+d\right)=64\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}u_2=\dfrac{14}{3}\\\left(u_2^2-d^2\right)u_2=64\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\dfrac{14}{3}=u_2=u_1+d\\d=\dfrac{2\sqrt{889}}{21}\end{matrix}\right.\\\left\{{}\begin{matrix}\dfrac{14}{3}=u_1+d\\d=\dfrac{-2\sqrt{889}}{21}\end{matrix}\right.\end{matrix}\right.\) 

(Làm nốt,số xấu quá)

e)\(\left\{{}\begin{matrix}u_1+u_2+u_3=7\\u_1^2+u_2^2+u_3^2=21\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}u_1+u_2+u_3=7\\u_1u_2u_3=\dfrac{21-\left(u_1+u_2+u_3\right)^2}{2}=-14\end{matrix}\right.\)

Làm như ý d)

a: Xét tứ giác AOBM có

góc OAM+góc OBM=180 độ

=>AOBM nội tiếp

b: \(cosAOM=\dfrac{OA}{OM}=\dfrac{1}{3}\)

nên \(\widehat{AOM}\simeq71^0\)

=>\(\widehat{AOB}\simeq142^0\)

=>sđ cung nhỏ AB là 142 độ; sđ cung lơn AB=360-142=218 độ

c:

Xét (O) có

ΔBAC nội tiếp

BC là đường kính

=>ΔBAC vuông tại A

=>BA vuông góc AC

Xét(O) có

MA,MB là tiêp tuyến

nên MA=MB

mà OA=OB

nên OM là trung trực của AB

=>OM//AC

góc ACB=góc OAC

góc OAC=góc AOM

=>góc ACB=góc AOM=góc BOM

d: góc DOM+góc BOM=90 độ

góc DMO+góc AOM=90 độ

mà góc BOM=góc AOM

nên góc DOM=góc DMO

=>DO=DM

XIV.

1. Her sister is more beautiful than her

2. English exercises is more difficult than Literature exercises

3. My new class is bigger tham my old class

4. Phi is more handsome than anybody in the class

5. Loan is taller than anyone in the family

IV.

3. French is not as difficult as English

4. That house is bigger than this house

5. A new house is more expensive than an old house

6. Yesterday is not as hot as today

7. My sister is taller than me

8. Girls are not as strong as boys

X.

1. Lesson 1 is not as long as Lesson 2

2. Ha Noi is smaller than HCM City

3. Old English is not as difficult as new French

4. Quang is not as thin as Quan

5. Your computer is more expensive than my computer

XI.

3. Minh is thinner than Hoang

4. Driving a motorbike is not as difficult as driving a car

XII.

1. Boys are stronger than girls

2. I am more intelligent than my sister

3. A new house is more expensive than an old house

4. English is more difficult than French

5. That river is longer than this river
6. New films are more interesting than old films 

XIII.

3. Quan is the most intelligent student in my school

4. HCM City is the biggest city in VN

Kẻ đường cao AH

Áp dụng PTG: \(BC=\sqrt{AB^2+AC^2}=10\left(cm\right)\)

Áp dụng HTL: \(BH=\dfrac{AB^2}{BC}=\dfrac{18}{5}\left(cm\right);AH=\dfrac{AB\cdot AC}{BC}=\dfrac{24}{5}\left(cm\right)\)

Vì AD là p/g nên \(\dfrac{BD}{DC}=\dfrac{AB}{AC}=\dfrac{3}{4}\Rightarrow BD=\dfrac{3}{4}DC\)

Mà \(BD+DC=BC=10\Rightarrow\dfrac{7}{4}DC=10\Rightarrow DC=\dfrac{40}{7}\left(cm\right)\)

\(\Rightarrow BD=\dfrac{30}{7}\left(cm\right)\)

\(\Rightarrow HD=BD-BH=\dfrac{30}{7}-\dfrac{18}{5}=\dfrac{24}{35}\)

Áp dụng PTG: \(AD=\sqrt{AH^2+HD^2}=\sqrt{\left(\dfrac{24}{35}\right)^2+\left(\dfrac{24}{5}\right)^2}=\dfrac{24\sqrt{2}}{7}\approx4,85\left(cm\right)\)