a) 5 x (2137 + 2368)

=5 x4505

= 22525

b) 10236 - 1028 x 6

= 10236 - 6168

=4068

\(\frac{\sqrt{x^2y^2}}{xy}+\frac{\sqrt{\left(x-y\right)^2x^2}}{x\left(x-y\right)}-\frac{\sqrt{\left(x-y\right)y^2}}{y\left(x-y\right)}\)

\(=\sqrt{\frac{x^2y^2}{x^2y^2}}+\sqrt{\frac{\left(x-y\right)^2x^2}{x^2\left(x-y\right)^2}}-\sqrt{\frac{\left(x-y\right)y^2}{y^2\left(x-y\right)^2}}\)

\(=1+1-\frac{1}{\sqrt{x-y}}\)

\(=2-\frac{1}{\sqrt{x-y}}\)

x-y<0 thì sao ạ

\(a+b+c=2008;\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2008\)

\(\Rightarrow a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

\(\Leftrightarrow\frac{ab+bc+ac}{abc}=\frac{1}{a+b+c}\)

\(\Leftrightarrow\left(ab+bc+ac\right)\left(a+b+c\right)=abc\)

\(\Leftrightarrow\left(ab+bc+ac\right)\left(a+b+c\right)-abc=0\)

\(\Leftrightarrow\left(ab+bc+ac\right)\left(b+c\right)+a\left(ab+ac\right)+abc-abc=0\)

\(\Leftrightarrow\left(ab+bc+ac\right)\left(b+c\right)+a^2\left(b+c\right)=0\)

\(\Leftrightarrow\left(ab+bc+ac+a^2\right)\left(b+c\right)=0\)

\(\Leftrightarrow\left[b\left(a+c\right)+a\left(a+c\right)\right]\left(b+c\right)=0\)

\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)

\(\Leftrightarrow\)Hoặc a + b = 0 hoặc b + c = 0 hoặc a + c = 0 

Vậy 1 trong 3 số bằng 2008 (đpcm)

Cảm ơn bạn đã giúp mình nhưng bạn bị nhầm 2 dòng đầu . Mình sửa lại cho bạn 2 dòng đầu như sau:

      \(a+b+c=2008;\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2008}\) ;

    \(\Rightarrow\frac{1}{a+b+c}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

\(5\left(2x-1\right)^3=40\)

\(\Leftrightarrow\left(2x-1\right)^3=8\)

\(\Leftrightarrow2x-1=2\)

\(\Leftrightarrow2x=3\)

\(\Leftrightarrow x=\frac{3}{2}\)

\(5\cdot\left(2x-1\right)^3=40\)

\(\Rightarrow\left(2x-1\right)^3=8\)

\(\Rightarrow\left(2x-1\right)^3=2^3\)

\(\Rightarrow2x-1=2\)

\(\Rightarrow2x=3\)

\(\Rightarrow x=\frac{3}{2}\)

ti ck nha

3ab=11.ab +10

=>300+ab=11.ab +10

=>300=10.(ab+1)

=>ab+1=30

=>ab=29

A)

\(x^3=-125\)

\(x=-5\)

Vậy x = -5 

B) 

| x | = x + 1 

\(\Leftrightarrow\orbr{\begin{cases}x=x+1\\x=-\left(x+1\right)\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x-x=1\\x=-x-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}0=1\left(VN\right)\\2x=-1\end{cases}}\)

\(\Leftrightarrow x=-\frac{1}{2}\)

Vậy x = -1/2

C) \(x^2=x\)

\(\Leftrightarrow x^2-x=0\)

\(\Leftrightarrow x\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

Vậy x1 = 0  ;   x2 = 1

A) x³ = -125

x       = -125 : 3

x      \(\approx\)-41,5

B) | x | = x + 1

\(\orbr{\begin{cases}x=x\\x=-x\end{cases}}\)

=>\(\orbr{\begin{cases}x>0\\x< 0\end{cases}}\)

C) x² = x

x \(\varepsilon\)N

~ Hok tốt ~

\(\sqrt{5+2\sqrt{5}}\)

\(=\sqrt{5+2\sqrt{5}+1-1}\)

\(=\sqrt{\left(\sqrt{5}+1\right)^2-1}\)

\(=\sqrt{5}+1-1\)

\(=\sqrt{5}\)

A=\(2^{2011}\left(2^0+2^1+2^2+2^3+2^4+2^5\right)\)

A=\(2^{2011}\left(1+2+4+8+16+32\right)=2^{2011}.63\)

A=\(2^{2011}.3.21=>A⋮21\)

Study well