Câu 3, 4 bài nào em?

9:

1: Khi m=1 thì phương trình sẽ là:

\(x^2-2x\cdot1+1^2-1+1=0\)

=>\(x^2-2x+1=0\)

=>\(\left(x-1\right)^2=0\)

=>x-1=0

=>x=1

2: Thay x=-1 vào phương trình, ta được:

\(\left(-1\right)^2-2m\cdot\left(-1\right)+m^2-m+1=0\)

=>\(1+2m+m^2-m+1=0\)

=>\(m^2+m+2=0\)

=>\(\left(m+\dfrac{1}{2}\right)^2+\dfrac{7}{4}=0\)(vô lý)

3: \(\text{Δ}=\left(-2m\right)^2-4\left(m^2-m+1\right)\)

\(=4m^2-4m^2+4m-4=4m-4\)

Để phương trình có 2 nghiệm thì 4m-4>=0

=>4m>=4

=>m>=1

4:

\(A=x_1x_2-x_1-x_2\)

\(=x_1x_2-\left(x_1+x_2\right)\)

\(=m^2-m+1-2m=m^2-3m+1\)

\(=m^2-3m+\dfrac{9}{4}-\dfrac{5}{4}=\left(m-\dfrac{3}{2}\right)^2-\dfrac{5}{4}>=-\dfrac{5}{4}\)

Dấu = xảy ra khi m=3/2

a)\(\dfrac{-5}{21}:\dfrac{25}{14}=\dfrac{-5}{21}.\dfrac{14}{25}=\dfrac{-2}{15}\)

b)\(3,4:\dfrac{-17}{14}=\dfrac{17}{15}:\dfrac{-17}{14}=\dfrac{17}{15}.\dfrac{14}{-17}=\dfrac{-14}{15}\)

c) \(\left(-1,7\right):1\dfrac{2}{15}=\dfrac{-17}{10}:\dfrac{17}{15}=\dfrac{-17}{10}.\dfrac{15}{17}=\dfrac{-3}{2}\)

d) \(-8,4:\left(-2,8\right)=\dfrac{-42}{5}:\dfrac{-14}{5}=\dfrac{-42}{5}.\dfrac{5}{-14}=3\)

tick cho mình nhé! 😊

III)

36:A

37:B

38:B

39:A(mk ko chắc lắm)

40:B

IV)

1:It is about 100 km from Hanoi to Thanh hoa

2:How long is it from Hai duong to Hue

3:My father used to live in a small village when he was a child

4:Although he is young,he performs excellently

5:There are five rooms in Hoa's house

6:The red car is cheaper than the black car

7:Mr hung has fewer days off than Mr Minh

8:Let's play chess tonight

9:The children enjoy reading comics

a) Cl₂ + 2NaOH --> NaClO + NaCl + H₂O

Chất oxh: Cl₂, chất khử: Cl₂

b) \(n_{Cl_2}=\dfrac{17,92}{22,4}=0,8\left(mol\right);n_{NaOH}=0,5.4=2\left(mol\right)\)

PTHH: Cl₂ + 2NaOH --> NaClO + NaCl + H₂O

_____0,8---->1,6--------->0,8---->0,8

=> \(\left\{{}\begin{matrix}C_{M\left(NaCl\right)}=\dfrac{0,8}{0,5}=1,6M\\C_{M\left(NaClO\right)}=\dfrac{0,8}{0,5}=1,6M\\C_{M\left(NaOH\right)}=\dfrac{2-1,6}{0,5}=0,8M\end{matrix}\right.\)

Bài 1.

a)Gia tốc vật: \(a=\dfrac{F}{m}=\dfrac{0,2}{0,1}=2\)m/s²

Vận tốc vật: \(v^2-v^2_0=2aS\)

\(\Rightarrow v=\sqrt{2aS}=\sqrt{2\cdot2\cdot10}=\sqrt{40}\approx6,3\)m/s

b)Với \(v'=10\)m/s thì gia tốc vật là:

\(a'=\dfrac{v'^2-v^2_0}{2S}=\dfrac{10^2-0}{2\cdot10}=5\)m/s²

Lực kéo lúc này: \(F'=m\cdot a'=0,1\cdot5=0,5N\)

Công của lực kéo F là:

\(A=F'\cdot s=0,5\cdot10=5J\)

https://hoidap247.com/cau-hoi/558074

tham khảo ạ

nH2SO4=0,045mol    nKOH=0,03

PT: X + H2SO4 ==> XSO4 + H2 (1)

      H2SO4 + 2KOH ==> K2SO4 + 2H2O (2)

Từ 1 --> nH2SO4 dư = 0,015

--> nX=nH2SO4 pứ= 0,045-0,015=0,03

MX=1,2:0,03=40 --> X là Ca