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cái gì đây?

của bạn nè.Mik lớp 5 nhưng vẫn phải học thuộc hết 

4.2:

a: x^2-x+1=x^2-x+1/4+3/4

=(x-1/2)^2+3/4>=3/4>0 với mọi x

=>x^2-x+1 ko có nghiệm

b: 3x-x^2-4

=-(x^2-3x+4)

=-(x^2-3x+9/4+7/4)

=-(x-3/2)^2-7/4<=-7/4<0 với mọi x

=>3x-x^2-4 ko có nghiệm

5:

a: x^2+y^2=25

x^2-y^2=7

=>x^2=(25+7)/2=16 và y^2=16-7=9

x^4+y^4=(x^2)^2+(y^2)^2

=16^2+9^2

=256+81

=337

b: x^2+y^2=(x+y)^2-2xy

=1^2-2*(-6)

=1+12=13

x^3+y^3=(x+y)^3-3xy(x+y)

=1^3-3*1*(-6)

=1+18=19

Câu 7:

a, \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

b, \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Theo PT: \(n_{Fe}=n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,1.56}{10}.100\%=56\%\\\%m_{CuO}=44\%\end{matrix}\right.\)

c, \(n_{CuO}=\dfrac{10-0,1.56}{80}=0,055\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=n_{Fe}+n_{CuO}=0,155\left(mol\right)\)

\(\Rightarrow C\%_{H_2SO_4}=\dfrac{0,155.98}{100}.100\%=15,19\%\)

d, Theo PT: \(\left\{{}\begin{matrix}n_{FeSO_4}=n_{Fe}=0,1\left(mol\right)\\n_{CuSO_4}=n_{CuO}=0,055\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{FeSO_4}=0,1.152=15,2\left(g\right)\\m_{CuSO_4}=0,055.160=8,8\left(g\right)\end{matrix}\right.\)

Câu 8:

a, \(CuCO_3+2HCl\rightarrow CuCl_2+CO_2+H_2O\)

b, \(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Theo PT: \(n_{CuCO_3}=n_{CO_2}=0,15\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{CuCO_3}=\dfrac{0,15.124}{20}.100\%=93\%\\\%m_{CuCl_2}=7\%\end{matrix}\right.\)

c, \(n_{HCl}=2n_{CO_2}=0,3\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)

Bài 6: 

a, \(SO_2+Ca\left(OH\right)_2\rightarrow CaSO_3+H_2O\)

b, \(n_{SO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

Theo PT: \(n_{Ca\left(OH\right)_2}=n_{CaSO_3}=n_{SO_2}=0,05\left(mol\right)\)

\(\Rightarrow C\%_{Ca\left(OH\right)_2}=\dfrac{0,05.74}{100}.100\%=3,7\%\)

c, \(m_{CaSO_3}=0,05.120=6\left(g\right)\)

Câu 2.

\(C=\dfrac{N}{20}=\dfrac{4080}{20}=204\)(chu kì)

\(L=\dfrac{N}{2}\cdot3,4=\dfrac{4080}{2}\cdot3,4=6936A^o\)

\(G=X=30\%\cdot4080=1224nu\)

\(A=T=\dfrac{4080-2\cdot1224}{2}=816nu\)

\(a.C=\dfrac{N}{20}=204\)

\(b.L=\dfrac{Nx3,4}{2}=6936A\)

\(c.A=T=20\%N=816;G=X=30\%N=1224\)

\(d.rN=\dfrac{N}{2}=2040\)

b.L=Nx3,4/2=6936A

1, 

2, 

3,

4,

1.2 với \(x\ge0,x\in Z\)

A=\(\dfrac{2\sqrt{x}+7}{\sqrt{x}+2}=2+\dfrac{3}{\sqrt{x}+2}\in Z< =>\sqrt{x}+2\inƯ\left(3\right)=\left(\pm1;\pm3\right)\)

*\(\sqrt{x}+2=1=>\sqrt{x}=-1\)(vô lí)

*\(\sqrt{x}+2=-1=>\sqrt{x}=-3\)(vô lí
*\(\sqrt{x}+2=3=>x=1\)(TM)

*\(\sqrt{x}+2=-3=\sqrt{x}=-5\)(vô lí)

vậy x=1 thì A\(\in Z\)

Bài 1: 

a: \(=-10x^3+20x^4-5x\)

b: \(=\dfrac{1}{3}a^2b+7a^5-1\)

c: \(=a^3+8+25-a^3=33\)

d: \(=x^2-16+8-x^3=-x^3+x^2-8\)

e: \(=a^3+1+8-a^3=9\)

f: \(=\dfrac{7-2x+4x-8}{2x+3}=\dfrac{2x-1}{2x+3}\)

g: \(=\dfrac{3}{2\left(x+3\right)}-\dfrac{2}{x\left(x+3\right)}\)

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

cần câu nào ?b

C5:

nC2H5OH = 8,2 / 46 = 0,2 (mol)

C2H5OH + 3O2-- (t^o)-- > 2CO2 + 3H2O

VCO2 = 0,4 . 22,4 = 8,96 (l)

VO2 = 0,6.22,4=13,44 (l)