$\frac{x-11}{89}+\frac{x-13}{87}+\frac{x-15}{85}+\frac{x-17}{83}=4$$=>\left(\frac{x-11}{89}-1\right)+\left(\frac{x-13}{87}-1\right)+\left(\frac{x-15}{85}-1\right)+\left(\frac{x-17}{83}-1\right)=0$$=>\frac{x-100}{89}+\frac{x-100}{87}+\frac{x-100}{85}+\frac{x-100}{83}=0$$=>\left(x-100\right)\left(\frac{1}{89}+\frac{1}{87}+\frac{1}{85}+\frac{1}{83}\right)=0$=> x-100 =0 => x=100Vậy nghiệm là 100 

ta có 0=\(\frac{x-11}{89}-1+\frac{x-13}{87}-1+\frac{x-15}{85}-1+\frac{x-17}{83}-1=0\)

0=\(\frac{x-11-89}{89}+\frac{x-13-87}{87}+\frac{x-15-85}{85}+\frac{x-17-83}{83}\)

0=\(\frac{x-100}{89}+\frac{x-100}{87}+\frac{x-100}{85}+\frac{x-100}{83}\)

0=(x-100)(\(\frac{1}{89}+\frac{1}{87}+\frac{1}{85}+\frac{1}{83}\))

vì \(\frac{1}{89}+\frac{1}{87}+\frac{1}{85}+\frac{1}{83}\)khác 0      suy ra x-100=0 

                                                                                    x=100

x−11/89+x−13/87+x−15/85+x−17/83=4

=>(x−11/89−1)+(x−13/87−1)+(x−15/85−1)+(x−17/83−1)=0

=>x−100/89+x−100/87+x−100/85+x−100/83=0

=>(x−100)(1/89+1/87+1/85+1/83)=0

=> x-100 =0 

Vậy nghiệm là 100

Với \(x\ge\dfrac{5}{2}\)có: \(A=x+\sqrt{2x-5}\ge\dfrac{5}{2}+0=\dfrac{5}{2}\)

Dấu '=' xảy ra \(\Leftrightarrow x=\dfrac{5}{2}\)

\(\Rightarrow A_{min}=\dfrac{5}{2}\)

\(\left(x+2\right)^2-9=0\)

\(\Rightarrow\left(x+2\right)^2=9\)

\(\Rightarrow\left[{}\begin{matrix}x+2=3\\x+2=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)

Vậy...

 a) \(m_O=\dfrac{20.20}{100}=4\left(g\right)\)

=> \(n_{CaO}=n_O=\dfrac{4}{16}=0,25\left(mol\right)\)

\(\left\{{}\begin{matrix}\%m_{CaO}=\dfrac{0,25.56}{20}.100\%=70\%\\\%m_{Ca}=100\%-70\%=30\%\end{matrix}\right.\)

b) \(n_{Ca}=\dfrac{20.30\%}{40}=0,15\left(mol\right)\)

PTHH: Ca+ 2H₂O --> Ca(OH)₂ + H₂

          0,15-------------------->0,15

=> V = 0,15.22,4 = 3,36 (l)

\(n_{Fe_3O_4}=\dfrac{23,2}{232}=0,1\left(mol\right)\)

=> n_Fe = 0,3 (mol)

=> m_Fe = 0,3.56 = 16,8 (g)

=> \(m=\dfrac{16,8.100}{78,9474}=21,28\left(g\right)\)

c) Giả sử Fe₃O₄ bị khử thành Fe

Gọi số mol Fe₃O₄ pư là a (mol)

PTHH: Fe₃O₄ + 4H₂ --> 3Fe + 4H₂O

                 a--->4a----->3a

Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,15}{4}\) => Hiệu suất tính theo H₂

m = 23,2 - 232a + 168a = 21,28

=> a = 0,03 (mol)

=> \(\left\{{}\begin{matrix}n_{Fe_3O_4\left(pư\right)}=0,03\left(mol\right)\\n_{H_2\left(pư\right)}=0,12\left(mol\right)\end{matrix}\right.\)

\(H=\dfrac{n_{H_2\left(pư\right)}}{n_{H_2\left(bđ\right)}}=\dfrac{0,12}{0,15}.100\%=80\%\)

X x 62 + X x 38 = 4200

X x (62 + 38) = 4200

X x 100 = 4200

X = 4200 : 100

X = 42.

cả hai cái mũ 2 đều \(\ge\)0 với mọi x, y

Mà tổng của chúng = 0

=> (x-11+y)²=(x-y-4)²=0

=> x-11+y = 0 => x+y = 11 (1)

x-y-4 = 0 => x-y = 4 (2)

(1), (2) => (tổng hiệu) x = 7,5 ; y = 3,5

Does your father get to work by bicycle?

Does your father get to work by bicycle?

3. Does your father cycle to work?

Does your father get to work by bike?

B C C A  B D C

1 B

2 C

3 C

4 A

5 B

6 D

7 C

Part 6 : 

1. especially

2. with

3. goods 

4. perform

Part 7: 

1. T

2. F

3. F

4.T

-2/7,-5/6

-2/7,-5/6

-2/7

-5/6