\(M=\frac{-1,2:\left(1\frac{3}{5}.1,25\right)}{0,64-\frac{1}{25}}+\frac{\left(1,08-\frac{2}{5}\right):\frac{4}{7}}{\left(5\frac{5}{9}-\frac{9}{4}\right).\frac{36}{17}}+0,6.0,5:\frac{2}{5}\)

\(M=\frac{-1,2:\left(\frac{8}{5}.1,25\right)}{0,64-\frac{1}{25}}+\frac{0,68:\frac{4}{7}}{\left(\frac{50}{9}-\frac{9}{4}\right).\frac{36}{17}}+0,6.0,5.\frac{5}{2}\)

\(M=\frac{-1.2:2}{0,6}+\frac{1,19}{\frac{119}{36}.\frac{36}{17}}+0.6.0,5.\frac{5}{2}\)

\(M=\frac{-0,6}{0,6}+\frac{1,19}{7}+0,75\)

\(M=-1+0,17+0,75=-0,08=\frac{-2}{25}\)

cậu có thể tham khảo bài làm trên đây ạ, chúc cậu học tốt :>

Sửa Ta có : 6ab6 : ab = 192

=> 6000 + ab x 10 + 6 = 192 x ab

=> 6006 = 192 x ab - 10 x ab

=> 6006 = ab x (192 - 10)

=> 6006 = ab x 182

=> ab = 33

Vậy ab = 33

Hỏi nhiều quá nhỉ ?

Bg

Ta có: 6ab6 : ab = 192   (ab \(\inℕ^∗\))

=> 6000 + ab.10 + 6 = 192.ab 

=> 6006 + ab.10 = 192.ab 

=> 6006 = 192.ab - 10.ab      (192.ab - 10.ab = (192 - 10).ab)

=> 6006 = 182.ab 

=> 6006 : 182 = 33 = ab 

Vậy ab = 33

1. Has just come

2. Have driven

3. Has just finished

4. translated

5. Saw

6. Has gone

7. Swam

8. rememberred

9. read

10.have never written

11. did you have

1. Tom ( just come)__ has just come__. Do you want to talk to him?

2. This is the first time I (drive)__drove__a BMW.

3. She (just finish)_has just finished__her homework. She can go out with friends.

4. My father (translate)__translated__two pages yesterday.

5. My brother (see)__saw__him last week at Jane"s birthday party.

6. Mr.Jones (go)_went___to Australia. He is not in this country

7. Last summer i (swim)__swam__in the Red Sea

8. I (remember)__remembered___to close the door

9. I (read)_read___twenty novels last year

10. I ( never write) _have never written___any poem

11. How many lessons (you have) __have you had___today?

\(x\left(x-1\right)=x^2-x=\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\)

Vì \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)\(\Rightarrow\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0\Leftrightarrow x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)

Vậy GTNN của bt trên = - 1/4 <=> x = 1/2

x( x - 1 )

= x² - x

= ( x² - 2.1/2.x + 1/4 ) - 1/4

= ( x - 1/2 )² - 1/4

( x - 1/2 )² ≥ 0 ∀ x => ( x - 1/2 )² - 1/4 ≥ -1/4

Đẳng thức xảy ra <=> x - 1/2 = 0 => x = 1/2

Vậy GTNN của biểu thức = -1/4 <=> x = 1/2

\(x^3+x^2+x=3\)

\(\Leftrightarrow x^3+x^2+x-3=0\)

\(\Leftrightarrow x^3+2x^2+3x-x^2-2x-3=0\)

\(\Leftrightarrow x\left(x^2+2x+3\right)-\left(x^2+2x+3\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+2x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x^2+2x+3=0\end{cases}}\)

+) x - 1 = 0 <=> x = 1

+) x² + 2x + 3 = 0

Mà \(x^2+2x+3=\left(x+1\right)^2+2\ge2\)

=> Không có x tm trong th này

Vậy pt có nghiệm là x = 1

x³ + x² + x = 3

<=> x³ + x² + x - 3 = 0

<=> x³ + 2x² - x² + 3x - 2x - 3 = 0

<=> ( x³ + 2x² + 3x ) - ( x² + 2x + 3 ) = 0

<=> x( x² + 2x + 3 ) - 1( x² + 2x + 3 ) = 0

<=> ( x - 1 )( x² + 2x + 3 ) = 0

<=> \(\orbr{\begin{cases}x-1=0\\x^2+2x+3=0\end{cases}}\)

+) x - 1 = 0 => x = 1

+) x² + 2x + 3 = ( x² + 2x + 1 ) + 2 = ( x + 1 )² + 2 ≥ 2 > 0 ∀ x 

Vậy phương trình có nghiệm duy nhất là x = 1

VÌ BN TÍNH SAI NHA!!!

VÌ BN TÍNH SAI NHA!!!