vì 1/9 > 1/40 ; 1/29 > 1/40 ; 1/31 > 1/40; 1/39 > 1/40

nên 1/9 + 1/ 29 + 1/31 + 1/39 > 1/40 + 1/40 + 1/40 + 1/40 mà 1/40 + 1/40 + 1/40 + 1/40 = 1/10 

=) M > 1/10

M > 1/20 + 1/30 + 1/40 + 1/40 

M> 2/15 > 2/20 = 1/10
=> M > 1/10

Bài 22:

1: \(\sqrt{3-\sqrt5}=\frac{\sqrt{6-2\sqrt5}}{\sqrt2}\)

\(=\frac{\sqrt{\left(\sqrt5-1\right)^2}}{\sqrt2}=\frac{\sqrt5-1}{\sqrt2}=\frac{\sqrt{10}-\sqrt2}{2}\)

2: \(\sqrt{7+3\sqrt5}\)

\(=\frac{\sqrt{14+6\sqrt5}}{\sqrt2}\)

\(=\frac{\sqrt{\left(3+\sqrt5\right)^2}}{\sqrt2}=\frac{3+\sqrt5}{\sqrt2}=\frac{3\sqrt2+\sqrt{10}}{2}\)

3: \(\sqrt{9+\sqrt{17}}-\sqrt{9-\sqrt{17}}-2\)

\(=\frac{1}{\sqrt2}\left(\sqrt{18+2\sqrt{17}}-\sqrt{18-2\sqrt{17}}\right)-2\)

\(=\frac{1}{\sqrt2}\left(\sqrt{\left(\sqrt{17}+1\right)^2}-\sqrt{\left(\sqrt{17}-1\right)^2}\right)-2\)

\(=\frac{1}{\sqrt2}\left(\sqrt{17}+1-\sqrt{17}+1\right)-2=\frac{2}{\sqrt2}-2=\sqrt2-2\)

Bài 26:

1: \(\left|3-2x\right|=2\sqrt5\)

=>\(\left[\begin{array}{l}2x-3=2\sqrt5\\ 2x-3=-2\sqrt5\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=3+2\sqrt5\\ 2x=3-2\sqrt5\end{array}\right.\Rightarrow x=\frac{3\pm2\sqrt5}{2}\)

2: \(\sqrt{x^2}=12\)

=>|x|=12

=>x=12 hoặc x=-12

3: \(\sqrt{x^2-2x+1}=7\)

=>\(\sqrt{\left(x-1\right)^2}=7\)

=>|x-1|=7

=>\(\left[\begin{array}{l}x-1=7\\ x-1=-7\end{array}\right.\Rightarrow\left[\begin{array}{l}x=8\\ x=-6\end{array}\right.\)

22,

1, Đặt √(3-√5) = A

=> √2A=√(6-2√5)

=> √2A=√(5-2√5+1)

=> √2A=|√5 -1|

=> A=\(\dfrac{\sqrt{5}-1}{\text{√2}}\)

=> A= \(\dfrac{\sqrt{10}-\sqrt{2}}{2}\)

2, Đặt √(7+3√5) = B

=> √2B=√(14+6√5)

 => √2B=√(9+2√45+5)

=> √2B=|3+√5|

=> B= \(\dfrac{3+\sqrt{5}}{\sqrt{2}}\)

=> B= \(\dfrac{3\sqrt{2}+\sqrt{10}}{2}\)

3, 

Đặt √(9+√17) - √(9-√17) -\(\sqrt{2}\)=C

=> √2C=√(18+2√17) - √(18-2√17) -\(2\)

=> √2C=√(17+2√17+1) - √(17-2√17+1) -\(2\)

=> √2C=√17+1- √17+1 -\(2\)

=> √2C=0

=> C=0

26,

|3-2x|=2\(\sqrt{5}\)

TH1: 3-2x ≥ 0 ⇔ x≤\(\dfrac{-3}{2}\)

3-2x=2\(\sqrt{5}\)

-2x=2\(\sqrt{5}\) -3

x=\(\dfrac{3-2\sqrt{5}}{2}\) (KTMĐK)

TH2: 3-2x < 0 ⇔ x>\(\dfrac{-3}{2}\)

3-2x=-2\(\sqrt{5}\)

-2x=-2√5 -3

x=\(\dfrac{3+2\sqrt{5}}{2}\) (TMĐK)

Vậy x=\(\dfrac{3+2\sqrt{5}}{2}\)

2, \(\sqrt{x^2}\)=12 ⇔ |x|=12 ⇔ x=12, -12

3, \(\sqrt{x^2-2x+1}\)=7

⇔ |x-1|=7 

TH1: x-1≥0 ⇔ x≥1

x-1=7 ⇔ x=8 (TMĐK)

TH2: x-1<0 ⇔ x<1

x-1=-7 ⇔ x=-6 (TMĐK)

Vậy x=8, -6

4, \(\sqrt{\left(x-1\right)^2}\)=x+3

⇔ |x-1|=x+3

TH1: x-1≥0 ⇔ x≥1

x-1=x+3 ⇔ 0x=4 (KTM)

TH2: x-1<0 ⇔ x<1

x-1=-x-3 ⇔ 2x=-2 ⇔x=-1 (TMĐK)

Vậy x=-1

Lời giải:
$22+23-25+27-29+31-33$

$=22+(23-25)+(27-29)+(31-33)$

$=22+(-2)+(-2)+(-2)=22+(-2).3=22-6=16$

          Ta gọi tử của phân số B là A ta có:

A=1+2+2^2+2^3+...+2^2008

2A=2 + 2^2 + 2^3 + 2^4 +... + 2^2009

=>A=2^2009 - 1

   A=-1 + 2^2009

          ta thấy tử là số đối của mẫu =>B=\(\dfrac{-1}{1}\)

cảm ơn bạn nhiều

Bài 3:

a. \(R=R1+R2=15+30=45\Omega\)

b. \(\left\{{}\begin{matrix}I=U:R=9:45=0,2A\\I=I1=I2=0,2A\left(R1ntR2\right)\end{matrix}\right.\)

c. \(\left\{{}\begin{matrix}U1=R1.I1=15.0,2=3V\\U2=R2.I2=30.0,2=6V\end{matrix}\right.\)

Bài 4:

\(I1=U1:R1=6:3=2A\)

\(\Rightarrow I=I1=I2=2A\left(R1ntR2\right)\)

\(U=R.I=\left(3+15\right).2=36V\)

\(U2=R2.I2=15.2=30V\)