Câu 4:

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ Na+H_2O\to NaOH+\dfrac{1}{2}H_2\\ Na_2O+H_2O\to 2NaOH\\ \Rightarrow n_{Na}=2n_{H_2}=0,2(mol)\\ a,\%_{Na}=\dfrac{0,2.23}{10,8}.100\%=42,59\%\\ \%_{Na_2O}=100\%-42,59\%=57,41\%\\ b,n_{Na_2O}=\dfrac{10,8-0,2.23}{62}=0,1(mol)\\ \Rightarrow \Sigma n_{NaOH}=0,2+0,2=0,4(mol)\\ \Rightarrow m_{NaOH}=0,4.40=16(g)\)

Câu 5:

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ R+H_2O\to ROH+\dfrac{1}{2}H_2\\ R_2O+H_2O\to 2ROH\\ \Rightarrow n_{R}=2n_{H_2}=0,2(mol)\\ \Rightarrow n_{R_2O}=0,1(mol)\\ \Rightarrow M_R.0,2+(2M_R+16).0,1=10,8\\ \Rightarrow M_R=23(g/mol)\)

Vậy R là Na

\(\text{Δ}=\left(-2m\right)^2-4\left(m^2-m-2\right)\)

\(=4m^2-4m^2+4m+8\)

=4m+8

Để phương trình có hai nghiệm thì 4m+8>=0

hay m>=-2

Theo đề, ta có: \(\left(x_1+x_2\right)^2-2x_1x_2=4\)

\(\Leftrightarrow\left(-2m\right)^2-2\left(m^2-m-2\right)=4\)

\(\Leftrightarrow4m^2-2m^2+2m=0\)

=>2m(m+1)=0

=>m=0 hoặc m=-1

\(a,A=0,2\left(5x-1\right)-\dfrac{1}{2}\left(\dfrac{2}{3}x+4\right)+\dfrac{2}{3}\left(3-x\right)\)

\(=x-0,2-\dfrac{1}{3}x-2+2-\dfrac{2}{3}x\)

\(=\left(-0,2-2+2\right)+\left(x-\dfrac{1}{3}x-\dfrac{2}{3}x\right)\)

\(=-0,2\)

\(b,B=\left(x-2y\right)\left(x^2+2xy+4y^2\right)-\left(x^3-8y^3+10\right)\)

\(=x^3-8y^3-x^3+8y^3-10\)

\(=-10\)

\(c,C=4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)-4x\)

\(=4\left(x^2+2x+1\right)+\left(4x^2-4x+1\right)-8\left(x^2-1\right)-4x\)

\(=4x^2+8x+4+4x^2-4x+1-8x^2+8-4x\)

\(=13\)

a) \(A=0,2\left(5x-1\right)-\dfrac{1}{2}\left(\dfrac{2}{3}x+4\right)+\dfrac{2}{3}\left(3-x\right)\)

\(A=x-\dfrac{1}{5}-\dfrac{1}{3}x-2+2-\dfrac{2}{3}x\)

\(A=\left(x-\dfrac{1}{3}x-\dfrac{2}{3}x\right)-\left(\dfrac{1}{5}+2-2\right)\)

\(A=-\dfrac{1}{5}\)

Vậy: ...

b) \(B=\left(x-2y\right)\left(x^2+2xy+4y^2\right)-\left(x^3-8y^3+10\right)\)

\(B=\left[x^3-\left(2y\right)^3\right]-\left[x^3-\left(2y\right)^3\right]-10\)

\(B=-10\)

Vậy: ...

c) \(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x+1\right)\left(x-1\right)-4x\)

\(=4\left(x^2+2x+4\right)+\left(4x^2-4x+1\right)-8\left(x^2-1\right)-4x\)

\(=4x^2+8x+4+4x^2-4x+1-8x^2+8-4x\)

\(=\left(4x^2+4x^2-8x^2\right)+\left(8x-4x-4x\right)+\left(4+1+8\right)\)

\(=13\)

Vậy:...

=x−0,2−13x−2+2−23x=�−0,2−13�−2+2−23�

Bài 5: 

d: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y-z}{2+3-4}=\dfrac{-20}{1}=-20\)

Do đó: x=-40; y=-60; z=-80

Trên ADN:

A = 40%, T = 20%, G = 27%, X = 13%

Trên ADN:

Amạch gốc  =mU =  40%

Tmạch gốc =mA =  20%

Gmạch gốc = mX = 27%

 Xmạch gốc =mG=  13%

Bài 2.8
Cửa hàng lãi:\(\left(1000000:800000\right)-100\%=25\%\left(giavon\right)\)
Bài 2.9
Số tiền phải bán để lãi 25% giá vốn:\(740000+\left(740000\times25\%\right)==925000\left(đồng\right)\)

mình làm những bài bn chưa lm nhé

9B

10A

bài 2

have repainted

bàii 3

ride - walikking

swimming

watch

2 have repainted

4 have done 

8 have stepped

9 have you ever played

7) 1 riding -walking

2 swimming

3 watching

8)

2 my mother doesn't like listening to music

5 doing more exercises makes you healthier

còn lại ok nha

chịu ???????????????????

C5:

TN: Xa xa bờ bên kia

CN1: Thiên Mụ

VN1: hiện ra mờ ảo

CN2: ngọn tháp Phước Duyên

VN2: dát ánh trăng vàng

Đây là câu trần thuật