Gấp thì giúp đây ^_^ !!

+) Ta có : AM = BM ; M thuộc cạnh huyền BC 

=> AM là đường trung tuyến ứng với cạnh huyền BC

=> AM = BM = MC 

+) \(\widehat{BAC}+\widehat{B}+\widehat{C}=180^o\)

\(\Leftrightarrow90^o+30^o+\widehat{C}=180^o\)

\(\Leftrightarrow\widehat{C}=60^o\)

Xét tam giác AMC có :

\(\hept{\begin{cases}\widehat{C}=60^o\\AM=MC\end{cases}}\)

=> AMC là tam giác đều ( đpcm )

13 giờ 10 phút = \(\frac{79}{6}\)giờ

13giờ 10phút=13,16giờ

học tốt

a) 2x² - 4x + 5

= 2( x² - 2x + 1 ) + 3

= 2( x - 1 )² + 3 ≥ 3 > 0 ∀ x ( đpcm )

b) 3x² + 2x + 1

= 3( x² + 2/3x + 1/9 ) + 2/3

= 3( x + 1/3 )² + 2/3 ≥ 2/3 > 0 ∀ x ( đpcm )

c) -x² + 6x - 10

= -x² + 6x - 9 - 1

= -( x² - 6x + 9 ) - 1

= -( x - 3 )² - 1 ≤ -1 < 0 ∀ x ( đpcm )

d) -x² + 3x - 3

= -x² + 3x - 9/4 - 3/4

= -( x² - 3x + 9/4 ) - 3/4

= -( x - 3/2 )² - 3/4 ≤ -3/4 < 0 ∀ x ( đpcm )

e) \(\frac{x^2+4x+5}{2}>0\)

Vì 2 > 0

=> x² + 4x + 5 > 0

=> x² + 4x + 4  + 1 > 0

=> ( x + 2 )² + 1 > 0 ( đúng )

=> \(\frac{x^2+4x+5}{2}>0\)∀ x ( đpcm )

f) \(\frac{-6+2x-x^2}{x^2+1}< 0\)

Vì x² + 1 ≥ 1 ∀ x

=> -6 + 2x - x² < 0

=> -x² + 2x - 1 - 5

= -( x² - 2x + 1 ) - 5

= -( x - 1 )² - 5 < 0 ( đúng )

=> \(\frac{-6+2x-x^2}{x^2+1}< 0\)∀ x ( đpcm )

a,Ta có :\(2x^2-4x+5=\left(x^2-2x+1\right)+\left(x^2-2x+1\right)+3\)

\(=\left(x-1\right)^2+\left(x-1\right)^2+3=2\left(x-1\right)^2+3\)

Do \(2\left(x-1\right)^2\ge0\Leftrightarrow2\left(x-1\right)^2+3\ge3\forall x\inℝ\)

Hay :\(2x^2-4x+5>0\)

Vậy nên BPT luôn đúng với mọi số thực x 

b,Ta có : \(3x^2+2x+1=x^2+2x+1+2x^2\)

\(=\left(x+1\right)^2+2x^2\)

Do \(\hept{\begin{cases}\left(x+1\right)^2\ge0\forall x\inℝ\\2x^2\ge0\forall x\inℝ\end{cases}}\Leftrightarrow\left(x+1\right)^2+2x^2\ge0\forall x\inℝ\)

Vậy nên BPT luôn đúng với mọi số thực x

c,Ta có : \(-x^2+6x-10=-\left(x^2-6x+10\right)\)

\(=-\left(x^2-6x+9\right)-1=-\left(x-3\right)^2-1\)

Do \(\left(x-3\right)^2\ge0\forall x\inℝ\Leftrightarrow-\left(x-3\right)^2-1\le-1\forall x\inℝ\)

Hay \(-x^2+6x-10\le-1\forall x\inℝ\)

Vậy nên BPT luôn đúng với mọi số thực x

d, Ta có :\(-x^2+3x-3=-\left(x^2-3x+3\right)\)

\(=-\left(x^2-2.\frac{3}{2}.x+\frac{9}{4}\right)-\frac{3}{4}=-\left(x-\frac{3}{2}\right)^2-\frac{3}{4}\)

Do \(\left(x-\frac{3}{2}\right)^2\ge0\forall x\inℝ\Leftrightarrow-\left(x-\frac{3}{2}\right)^2-\frac{3}{4}\le-\frac{3}{4}\forall x\inℝ\)

Hay \(-x^2+3x-3\le0\forall x\inℝ\)

Vậy nên BPT luôn đúng với mọi số thực x

2 câu còn lại bạn nào làm giúp mình nha

\(x^2+x+6=x.\left(x+1\right)+6\)

x^2 + x + 6
<=> x ( x + 1) +6

Cách chuyển đổi câu giữa thì hiện tại hoàn thành và thì quá khứ đơn (lớp 6)

1,It is two months since we last was?

2,How long ago did you start/begin to write this book?

3,When did you start/begin learning how to drive a car?

4,When did Mr.Ba start/begin buying this flat?

5,It is the first time I have ever talked to headmaster

6,When did she start/begin writing an e-mail?

7,I've never worst film before

8,Is this me go to school late?

9,It is five months since he hasn't come back to his home village

10,The last time I have ever spoken to her

ks nhé!Học tốt!:))

1. We haven't been to this restaurant for two months.

=> It is two months since I last was to this restaurant.

2. When did you begin to write this book?

=> How long have you written this book?

3. How long have you learnt how to drive a car?

=> When did you start to learn how to drive a car?

4. How long is it since Mr Ba bought this flat?

=> When did Mr Ba start to buy this flat?

5. I've never talked to the headmaster before.

=> It's the first time I have ever talked to the headmaster.

6. How long is it since she wrote an e-mail?

=> When did she start to write an e-mail?

7. That's the worst film I've ever seen.

=>I've never seen such an bad film before.

8. Have you gone to school late before?

=> Is this the first time you have gone to school late?

9. He last came back to his home village five months ago.

=> It's five months since he last came back to his home village.

10. It's three years since I last spoke to her.

=>The last time I spoke to her is three years ago.

a) Để C đạt giá trị lớn nhất 

<=> (x-3)² + 1 đạt giá trị nhỏ nhất

Đặt \(A=\left(x-3\right)^2+1\ge1\)

Min A = 1 \(\Leftrightarrow x=3\)

Max C = 5 \(\Leftrightarrow x=3\)

b) Để D đạt giá trị lớn nhất

<=> |x-2|+2 đạt giá trị nhỏ nhất

Đặt B = | x-2 | + 2 

\(B=\left|x-2\right|+2\ge2\)

Min B = 2 \(\Leftrightarrow x=2\)

=> Max D = 2 \(\Leftrightarrow x=2\)

.a, \(\frac{x+1}{999}+\frac{x+2}{998}=\frac{x+3}{997}+\frac{x+4}{996}\)

.\(< =>\frac{x+1}{999}+1+\frac{x+2}{998}+1=\frac{x+3}{997}+1+\frac{x+4}{996}+1\)

.\(< =>\frac{x+1}{999}+\frac{999}{999}+\frac{x+2}{998}+\frac{998}{998}=\frac{x+3}{997}+\frac{997}{997}+\frac{x+4}{996}+\frac{996}{996}\)

.\(< =>\frac{x+1+999}{999}+\frac{x+2+998}{998}=\frac{x+3+997}{997}+\frac{x+4+996}{996}\)

.\(< =>\frac{x+1000}{999}+\frac{x+1000}{998}-\frac{x+1000}{997}-\frac{x+1000}{996}=0\)

.\(< =>\left(x+1000\right)\left(\frac{1}{999}+\frac{1}{998}-\frac{1}{997}-\frac{1}{996}\right)=0\)

.Do \(\frac{1}{999}+\frac{1}{998}-\frac{1}{997}-\frac{1}{996}\ne0\)

.Suy ra \(x+1000=0\Leftrightarrow x=-1000\)

.b, \(\frac{x+1}{1001}+\frac{x+2}{1002}=\frac{x+3}{1003}+\frac{x+4}{1004}\)

.\(< =>\frac{x+1}{1001}-1+\frac{x+2}{1002}-1=\frac{x+3}{1003}-1+\frac{x+4}{1004}-1\)

.\(< =>\frac{x+1}{1001}-\frac{1001}{1001}+\frac{x+2}{1002}-\frac{1002}{1002}=\frac{x+3}{1003}-\frac{1003}{1003}+\frac{x+4}{1004}-\frac{1004}{1004}\)

.\(< =>\frac{x+1-1001}{1001}+\frac{x+2-1002}{1002}=\frac{x+3-1003}{1003}+\frac{x+4-1004}{1004}\)

.\(< =>\frac{x-1000}{1001}+\frac{x+1000}{1002}-\frac{x+1000}{1003}-\frac{x+1000}{1004}=0\)

.\(< =>\left(x-1000\right)\left(\frac{1}{1001}+\frac{1}{1002}-\frac{1}{1003}-\frac{1}{1004}\right)=0\)

.Do \(\frac{1}{1001}+\frac{1}{1002}-\frac{1}{1003}-\frac{1}{1004}\ne0\)

.Suy ra \(x-1000=0\Leftrightarrow x=1000\)

cảm ơn

Gọi số cần tìm là abc. Theo đề bài

abc-ab=778 => 10xab+c-ab=778 => 9xab+c=778 => 9xab=778-c

9xab chia hết cho 9 => 778-c phải chia hết cho 9 và do c<=9 nên c=4

=> 9xab=778-4=774 => ab=86

Vậy abc=864

\(|3+x|\ge0\forall x\)

\(|5x-6|\ge0\forall x\)

\(\Rightarrow|3+x|=|5x-6|\)

\(\Rightarrow3+x=5x-6\)

\(3+6=5x-x\)

\(9=4x\)

\(x=9\div4\)

\(x=\frac{9}{4}\)

Giải:

a) Đồng dạng vì góc A = góc H = 90 độ

                           góc B chung

b) Vì AI là phân giác nên \(\frac{BI}{AB}=\frac{CI}{AC}\)

Suy ra \(\frac{AB}{AC}=\frac{BI}{CI}=\frac{2}{3}\)

Hay \(\frac{10}{CI}=\frac{2}{3}\)

Vậy CI = 15

Mik giải nhanh thôi còn bn tự trình bày lại sao cho đẹp