Bài 4: 

a: Xét ΔABI và ΔACI có 

AB=AC

AI chung

BI=CI

Do đó: ΔABI=ΔACI

b: Xét tứ giác ABDC có 

I là trung điểm của BC

I là trung điểm của AD

Do đó: ABDC là hình bình hành

Suy ra: AB=CD

0.125*37*8

=(0.125*8)*37

=1*37

0,125 x 37 x 8

= ( 0,125 x 8 ) x 37 

= 1 x 37 

= 37

0.125*37*8

=(0.125*8)*37

=1*37

Câu 4: 

a: Xét ΔABD và ΔAED có 

AB=AE

\(\widehat{BAD}=\widehat{EAD}\)

AD chung

Do đó: ΔABD=ΔAED

Câu 1:

\(a,=\dfrac{1}{2}+9\cdot\dfrac{1}{9}-18=\dfrac{1}{2}+1-18=-\dfrac{33}{2}\\ b,=2-1+4\cdot\dfrac{1}{4}+9\cdot\dfrac{1}{9}\cdot9=1+1+9=11\\ c,=-21,3\left(54,6+45,4\right)=-21,3\cdot100=-2130\\ d,B=\left(\dfrac{1}{16}+\dfrac{1}{2}-\dfrac{1}{16}\right):\left(\dfrac{1}{8}-\dfrac{1}{8}+1\right)=\dfrac{1}{2}:1=\dfrac{1}{2}\)

đề khó nhìn quá

I.

1A

2C

3A

4C

5B

6B

7C

8B

9B

10C

II. 

1T

2F

3F

4T

5F

6T

7F

8T

9F

10F

Bài 5

B=  \(\dfrac{2015}{2016+2017+2018}\)+\(\dfrac{2016}{2016+2017+2018}\)+\(\dfrac{2017}{2016+2017+2018}\)

Ta có:\(\dfrac{2015}{2016}\)>\(\dfrac{2015}{2016+2017+2018}\),\(\dfrac{2016}{2017}\)>\(\dfrac{2016}{2016+2017+2018}\),\(\dfrac{2017}{2018}\)>\(\dfrac{2017}{2016+2017+2018}\)

⇒A>B

Bài 5:

Ta có:

\(B=\dfrac{2015+2016+2017}{2016+2017+2018}\) 

\(B=\dfrac{2015}{2016+2017+2018}+\dfrac{2016}{2016+2017+2018}+\dfrac{2017}{2016+2017+2018}\) 

Vì \(\dfrac{2015}{2016}>\dfrac{2015}{2016+2017+2018}\) 

\(\dfrac{2016}{2017}>\dfrac{2016}{2016+2017+2018}\) 

\(\dfrac{2017}{2018}>\dfrac{2017}{2016+2017+2018}\) 

\(\Rightarrow A>B\)

bài 1 có ở phần dưới rồi nhé!

Kiên đi nhanh hơn

Vì quãng đường từ nhà bạn ấy đến trường mất 2,5 km nhưng thời gian bạn ấy đi lại bằng với thời gian Hùng đi từ nhà đến trường trên quãng đường 2km

làm ơn ghi luôn cách giải hộ mik với

7:

a: ĐKXĐ: x>=0; x<>1

\(D=\dfrac{1}{2\sqrt{x}-2}-\dfrac{1}{2\sqrt{x}+2}+\dfrac{\sqrt{x}}{1-x}\)

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

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

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

b: Khi x=4/9 thì \(D=\dfrac{-1}{\dfrac{2}{3}+1}=-1:\dfrac{5}{3}=-\dfrac{3}{5}\)

c: |D|=1/3

=>D=-1/3 hoặc D=1/3

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

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

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

=>x=4

6:

a: \(C=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\right):\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\)

\(=\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}:\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\dfrac{3\sqrt{x}-x+x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\cdot\dfrac{-\sqrt{x}\left(3-\sqrt{x}\right)}{2\sqrt{x}+4}\)

\(=\dfrac{3\left(\sqrt{x}+3\right)}{3+\sqrt{x}}\cdot\dfrac{-\sqrt{x}}{2\sqrt{x}+4}=\dfrac{-3\sqrt{x}}{2\sqrt{x}+4}\)

b: C<-1

=>C+1<0

=>\(\dfrac{-3\sqrt{x}+2\sqrt{x}+4}{2\sqrt{x}+4}< 0\)

=>\(-\sqrt{x}+4< 0\)

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

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

=>x>16

\(C=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\right):\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\\ =\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\dfrac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{1}{\sqrt{x}}\right)\\ =\left(\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}+\dfrac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\dfrac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\\ =\dfrac{3\sqrt{x}-x+x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\dfrac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\cdot\dfrac{-\sqrt{x}\left(3-\sqrt{x}\right)}{2\sqrt{x}+4}\\ =\dfrac{3\left(\sqrt{x}+3\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\cdot\dfrac{-\sqrt{x}\left(3-\sqrt{x}\right)}{2\sqrt{x}+4}\\ =\dfrac{-3\sqrt{x}}{2\sqrt{x}+4}\)

Để `C < -1` Ta có :

 \(\dfrac{-3}{2\sqrt{x}+4}< -1\\ \Leftrightarrow\dfrac{-3}{2\sqrt{x}+4}+1< 0\\ \Leftrightarrow\dfrac{-3}{2\sqrt{x}+4}+\dfrac{2\sqrt{x}+4}{2\sqrt{x}+4}< 0\\ \Leftrightarrow-3+2\sqrt{x}+4< 0\\ \Leftrightarrow2\sqrt{x}+1< 0\\ \Leftrightarrow2\sqrt{x}< -1\\ \Leftrightarrow\sqrt{x}< -\dfrac{1}{2}\\ \Leftrightarrow x< \dfrac{1}{4}\)