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

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

Câu 8:

a. Với $x,y$ là số nguyên thì $x, y-3$ cũng là số nguyên. Mà $x(y-3)=15$ nên ta có các TH:

TH1: $x=1, y-3=15\Rightarrow x=1; y=18$ (tm)

TH2: $x=-1, y-3=-15\Rightarrow x=-1; y=-12$ (tm)

TH3: $x=15; y-3=1\Rightarrow x=15; y=4$ (tm)

TH4: $x=-15; y-3=-1\Rightarrow x=-15; y=2$ (tm)

TH5: $x=3, y-3=5\Rightarrow x=3; y=8$ (tm)

TH6: $x=-3; y-3=-5\Rightarrow x=-3; y=-2$ (tm)

TH7: $x=5; y-3=3\Rightarrow x=5; y=6$ (tm)

TH8: $x=-5; y-3=-3\Rightarrow x=-5; y=0$ (tm)

Câu 8:

b. 

$xy-2y+3(x-2)=7$

$\Rightarrow y(x-2)+3(x-2)=7$

$\Rightarrow (x-2)(y+3)=7$

Do $x,y$ nguyên nên $x-2, y+3$ nguyên. Mà tích của chúng bằng $7$ nên ta có các TH sau:

TH1: $x-2=1, y+3=7\Rightarrow x=3; y=4$ (tm)

TH2: $x-2=-1; y+3=-7\Rightarrow x=1; y=-10$ (tm)

TH3: $x-2=7, y+3=1\Rightarrow x=9; y=-2$ (tm)

TH4: $x-2=-7; y+3=-1\Rightarrow x=-5; y=-4$ (tm)

Câu 8:

c. 

$xy-3x+y=15$

$\Rightarrow x(y-3)+(y-3)=12$

$\Rightarrow (x+1)(y-3)=12$

Do $x,y$ nguyên nên $x+1, y-3$ nguyên. Mà $(x+1)(y-3)=12$ nên ta xét các TH sau:

TH1: $x+1=1, y-3=12\Rightarrow x=0; y=15$ (tm)

TH2: $x+1=-1; y-3=-12\Rightarrow x=-2; y=-9$ (tm)

TH3: $x+1=12; y-3=1\Rightarrow x=11; y=4$ (tm)

TH4: $x+1=-12; y-3=-1\Rightarrow x=-13; y=2$ (tm)

TH5: $x+1=2; y-3=6\Rightarrow x=1; y=9$ (tm)

TH6: $x+1=-2; y-3=-6\Rightarrow x=-3; y=-3$ (tm)

TH7: $x+1=6; y-3=2\Rightarrow x=5; y=5$ (tm)

TH8: $x+1=-6; y-3=-2\Rightarrow x=-7; y=1$ (tm)

TH9: $x+1=3; y-3=4\Rightarrow x=2; y=7$ (tm)

TH10: $x+1=-3; y-3=-4\Rightarrow x=-4; y=-1$ (tm)

TH11: $x+1=4, y-3=3\Rightarrow x=3; y=6$ (tm)

TH12: $x+1=-4; y-3=-3\Rightarrow x=-5; y=0$ (tm)

a/ Ta có \(\widehat{B}=\widehat{C}\) => tam giác ABC cân tại A => AB=AC

AM là phân giác góc \(\widehat{A}\)

=> AM là đường cao của tg ABC (trong tg cân phân giác của góc ở đỉnh đồng thời là đường cao và đường trung trực)

\(\Rightarrow MB=MC\)

\(\Rightarrow AM\perp BC\Rightarrow\widehat{AMB}=\widehat{AMC}=90^o\)

b/ Xét tg BMI có

\(\widehat{AIB}=\widehat{AMB}+\widehat{IBM}\) (trong tg góc ngoài bằng tổng hai góc trong không kề với nó)

\(\Rightarrow\widehat{AIB}=90^o+\frac{\widehat{B}}{2}\) mà \(\widehat{B}=\widehat{C}\Rightarrow\widehat{AIB}=90^o+\frac{\widehat{C}}{2}\)

c/  Ta có

MN//AC;\(MB=MC\Rightarrow NA=NB=\frac{AB}{2}\) (Trong tg đường thẳng // với 1 cạnh và đi qua trung điểm 1 cạnh thì đi qua trung điểm cạnh còn lại)

=> MN là đường trung bình của tg ABC \(\Rightarrow MN=\frac{AC}{2}=\frac{AB}{2}\)

\(\Rightarrow MN=NA\) => tg AMN cân tại N \(\Rightarrow\widehat{NAM}=\widehat{NMA}\)

d/ Ta có \(\widehat{AIB}=90^o+\frac{\widehat{C}}{2}\Rightarrow\widehat{B}=\widehat{C}=2\widehat{AIB}-180^o=40^o\)

\(\Rightarrow\widehat{A}=180^o-\left(\widehat{B}+\widehat{C}\right)=180^o-80^o=100^o\)

thank bạn

Hello, my name is Phuong. I am 13 years old. I live with my family in Vinh city, Nghe An province, Vietnam. In my country, spring is the best season of the year because it has Tet holiday. In Tet holiday, the streets are very beautiful with green trees, flowers, and funny noisy. In every house, there are a big tree with many love flowers and plants. Any things else? Yes, there are a lot of delicous drinks, cakes and candies. In Tet holiday, children have money from adults. It's so greet. We enjoy a happy Tet holiday every years. Do you know out happy holiday in spring? It's Tet holiday. What about you? Can you tell me about your country 's best holiday? Thank you so much.

Tet is a traditional holiday of our country Vietnam. The season of Tet occurs around the biginning of February and end of Janaurỵ The meaning of Tet is ending old year and welcome a new year. Tet holiday is very special for Vietnamese families to reunite. At that time, all family members try to come back their homes and get together to prepare for Tet. Moreover they will buy new clothes and clean their house. A branch of Mai flower or a Kumquat tree can't be absent from every home. They bring a fresh atmosphere to the house. We decorate ancestral altar with a tray full of fruits and flowers. Some typical Tet's food such as sweets, coconut jam, candied fruits, lemon seeds, rice cake , "thit kho"( stew pork with eggs). A special food can not absent is sky rice cake.. It is our country's traditional food on Tet holiday. Tet meal is more dilicious than our every day.  On Tet holiday, people have many typical activities. Young persons takes part in the traditional games such as: tug; cooking rice, cock fighting, watching lion dance. Many people visit relatives to wish a happy new year. A dults will give lucky money for children. Woman going to pagoda to wish good things will come to their familỵ
In my oppinion, i cozy and fresh Tets atmosphere. All member of family talk together about a happy new year. I can hang out with my friend and don't worry about any thịng " Tet" holiday is always in Vietnamese's heart.
 

Cho mình hỏi bạn tự viết hay lấy ở đâu ạ ?

2:

a: =(1+căn 3)^2-5

=4+2căn 3-5

=2căn 3-1

b: \(=\sqrt{\dfrac{125}{7}\cdot\dfrac{35}{81}}=\sqrt{\dfrac{625}{81}}=\dfrac{25}{9}\)

c: \(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)-\sqrt{6}+\sqrt{2}\)

=2-căn 6+căn 2

3:

a: \(=\dfrac{2\sqrt{3}+3\sqrt{3}-\sqrt{3}}{\sqrt{3}}=2+3-1=5\)

b: \(=\dfrac{6\sqrt{2}+7\sqrt{2}-5\sqrt{2}}{\sqrt{2}}=13-5=8\)

c: \(=\dfrac{12-10+8}{2}=5\)

d: \(=\sqrt{\dfrac{1}{5}:5}-\sqrt{\dfrac{9}{5}:5}+\sqrt{5:5}\)

=1/5-3/5+1

=3/5

1: Ta có: \(\widehat{BAD}+\widehat{B}=90^0\)

\(\widehat{BCE}+\widehat{B}=90^0\)

Do đó: \(\widehat{BAD}=\widehat{BCE}\)

2: Ta có: \(\widehat{AHE}+\widehat{BAD}=90^0\)

\(\widehat{ABD}+\widehat{BAD}=90^0\)

Do đó: \(\widehat{AHE}=\widehat{ABD}\)

câu 3:

Xét tam giác AEH vuông tại E: góc AHE+ góc EAH= 90 độ

                                                  60 độ +góc EAH=90 độ

                                                           góc EAH=30 độ (1)

Ta có: góc A= góc EAH+ góc HAC= 30 độ +45 độ= 75 độ 

Xét tam giác ADB vuông tại D có: góc B + góc EAH= 90 độ

                                                     góc B= 90 độ - 30 độ= 60 độ

lại có: góc BAC+  góc B + góc ACB= 180 độ (đ/ lý tổng ba góc trong 1 tam giác)

=> góc ACB= 180 độ-( 75 độ + 60 độ )= 45 độ