2Fe(OH)3+3H2SO4---> Fe2(SO4)3+3H2O

\(\dfrac{2a}{3c}\)

Câu 2

b, \(\sqrt{\left(1-x\right)^2}=2x+1\) (x\(\ne1\) ; x\(\ge\dfrac{1}{2}\))

<=> \(|1-x|=2x+1\)

<=> \(\left[{}\begin{matrix}1-x=2x+1\\1-x=-1-2x\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=0\left(tmđk\right)\\x=-2\left(loai\right)\end{matrix}\right.\)

Vậy S=\(\left\{0\right\}\)

a(a+b)+a(a+c)+a(b+c)=a(2a+2b+2c)=2a(a+b+c)=10+12+14=36

Suy ra a(a+b+c)=18. Ta co: 

a(a+b+c)-a(a+b)=18-10=8=ac.

a(a+c)=a^2+ac=a^2+8=12. Suy ra a=-2 hoac a=2.Suy ra b=-3;3 ; c=-4;4

Câu 3

a, ĐKXĐ: x>0, x\(\ne\)4

M=( \(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\)). \(\dfrac{\sqrt{x}+2}{\sqrt{4x}}\)

M= \(\left(\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\). \(\dfrac{\sqrt{x}+2}{\sqrt{4x}}\)

M= \(\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\). \(\dfrac{\sqrt{x}+2}{\sqrt{4x}}\)

M= \(\dfrac{2x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+2}{\sqrt{4x}}\)

M= \(\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

b, Thay x= \(6+4\sqrt{2}\) ( x>0, x\(\ne\)4) ta có:

M= \(\dfrac{\sqrt{6+4\sqrt{2}}}{\sqrt{6+4\sqrt{2}}-2}\)

= \(\dfrac{\sqrt{\left(\sqrt{2}+2\right)^2}}{\sqrt{\left(\sqrt{2}+2\right)^2-2}}\) = \(\dfrac{\sqrt{2}+2}{\sqrt{2}+2-2}\)

= \(\dfrac{\sqrt{2}\left(1+\sqrt{2}\right)}{\sqrt{2}}\) = \(1+\sqrt{2}\)

Vậy khi x= \(6+4\sqrt{2}\) thì M= \(1+\sqrt{2}\)

c, Để M<1 <=> \(\dfrac{\sqrt{x}}{\sqrt{x}-2}< 1\)

<=> \(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-2}< 0\)

<=> \(\dfrac{2}{\sqrt{x}-2}< 0\)

Vì 2>0 <=> \(\sqrt{x}-2< 0\)

<=> \(\sqrt{x}< 2\)

<=> x<4

Vậy để M<1 thì 0<x<4

<=>

(NH4)2CO3+2NaOH--->Na2CO3+2NH3+2H2O

Na2SO4 chứ k phải là Ba2SO4 đâu Nguyễn Đức Tiến

A=5(1+5)+5^3(1+5)+........+5^99(1+5)

A=5.6+5^3.6+............+5^99.6

A=6(5+5^3+......+5^99)

=>A chia het cho 6

tick cho minh nha!!!

olm thay em lam dung ko?Kiem tra ho em voi!!

2K+2H2O--->2KOH+H2

2Fe(OH)3--->Fe2O3+3H2O