2,6g dung dịch 

ez

k cho mình nha đúng 100 %

\(\orbr{\begin{cases}2x-3=0\\5x+10=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=3\\5x=-10\end{cases}\Rightarrow}}\orbr{\begin{cases}x=\frac{3}{2}\\x=-2\end{cases}}\)

(2x - 3).(5x + 10) = 0

<=> \(\orbr{\begin{cases}2x-3=0\\5x+10=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=1,5\\x=-2\end{cases}}\)

Vậy x = 1,5 hoặc x = -2

A=100

B=104

Vậy A<B

Áp dụng hằng đẳng thức số 3, ta có:

A=\(\left(26-24\right)\left(26+24\right)=2.50=100\)

B=\(\left(27-25\right)\left(27+25\right)=2.52=104\)

Vậy: A<B

\(a,\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}=\frac{2\left(\sqrt{3}+1\right)-2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}=\frac{2\sqrt{3}+2-2\sqrt{3}+2}{3-1}\)

\(=\frac{4}{2}=2\)

\(b,\frac{2+\sqrt{2}}{2-\sqrt{2}}+\frac{2-\sqrt{2}}{2+\sqrt{2}}=\frac{\left(2+\sqrt{2}\right)\left(2+\sqrt{2}\right)+\left(2-\sqrt{2}\right)\left(2-\sqrt{2}\right)}{\left(2-\sqrt{2}\right)\left(2+\sqrt{2}\right)}\)

\(=\frac{4+4\sqrt{2}+2+4-4\sqrt{2}+2}{4-2}\)

\(=\frac{8+4}{2}=\frac{12}{2}=6\)

Trả lời

Mk chưa hiểu đề cho lắm.

Bạn có thể giải thích giúp mk được ko?

Hok tốt !

hãy viết 4 phân số bằng phân số 5/6 VÀ 6/7

1) \(\widehat{ADB}\) là góc ngoài của t/giác ABC => \(\widehat{ADB}=\widehat{C}+\widehat{DAC}\)

\(\widehat{ADC}\)là góc ngoài của t/giác AD => \(\widehat{ADC}=B+\widehat{DAB}\)

Mà \(\widehat{B}=\widehat{C}\)(gt); \(\widehat{DAB}=\widehat{DAC}\) (gt)

=> \(\widehat{DAB}=\widehat{DAC}\)

2) Xét t/giác ABD và t/giác ADC

có: \(\widehat{BAD}=\widehat{CAD}\) (gt)

   AD : chung

  \(\widehat{ADB}=\widehat{ADC}\)(cmt)

=> t/giác ABD = t/giác ADC (g.c.g)

a) \(\frac{x}{x+1}=\frac{1}{2}\)

=> 2x = x + 1

=> 2x - x = 1

=> x = 1

b) \(\frac{x}{2}=\frac{x}{3}\)

=> 3x = 2x

=> 3x - 2x = 0

=> x = 0

c) \(\frac{x+1}{2}=\frac{x+1}{2017}\)

=> \(2017\left(x+1\right)=2\left(x+1\right)\)

=> 2017x + 2017 = 2x + 2

=> 2017x - 2x = 2 - 2017

=> 2015x = -2015

=> x = -2015 : 2015

=> x = -1

i) \(\frac{3}{x}=\frac{x}{2017}\)

=> x² = 2017.3

=> x² = 6051

=> \(\orbr{\begin{cases}x=\sqrt{6051}\\x=-\sqrt{6051}\end{cases}}\)

còn lại tự lm

\(a,\frac{x}{x+1}=\frac{1}{2}\)

\(\Rightarrow x=\frac{1}{2}.\left(x+1\right)\)

\(\Rightarrow x=\frac{1}{2}x+\frac{1}{2}\)

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

\(\Rightarrow\frac{1}{2}x=\frac{1}{2}\)

\(\Rightarrow x=1\)

\(b,\frac{x}{2}=\frac{x}{3}\)

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

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

\(\Rightarrow3x=2x\)

\(\Rightarrow x=0\)

\(c,\frac{x+1}{2}=\frac{x+1}{2017}\)

\(\Rightarrow x+1=\frac{x+1}{2017}.2\)

\(\Rightarrow x+1=\frac{2x+2}{2017}\)

\(\Rightarrow2017x+2017=2x+2\)

\(\Rightarrow2017x-2x=2-2017\)

\(\Rightarrow2015x=-2015\)

\(\Rightarrow x=-1\)

\(i,\frac{3}{x}=\frac{x}{2017}\)

\(\Rightarrow x=3:\frac{x}{2017}\)

\(\Rightarrow x=\frac{6051}{x}\)

\(\Rightarrow x^2=6051\)

\(\Rightarrow x=\sqrt{6051}\)

\(o,\frac{x}{3}=\frac{x+1}{2}\)

\(\Rightarrow x=\frac{x+1}{2}.3\)

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

\(\Rightarrow2x=3x+3\)

\(\Rightarrow-x=3\)

\(\Rightarrow x=-3\)

\(m,\frac{x+1}{2}=\frac{x+2}{3}\)

\(\Rightarrow x+1=\frac{x+2}{3}.2\)

\(\Rightarrow x+1=\frac{2x+4}{3}\)

\(\Rightarrow3x+3=2x+4\)

\(\Rightarrow x=1\)

\(p,\frac{x+1}{2}=x\)

\(\Rightarrow2x=x+1\)

\(\Rightarrow x=1\)

\(m,\frac{2}{x}=\frac{x}{8}\)

\(\Rightarrow x=2:\frac{x}{8}\)

\(\Rightarrow x=\frac{16}{x}\)

\(\Rightarrow x^2=16\)

\(\Rightarrow x=4\)

\(Q,\frac{x^2}{2}=\frac{8}{x^2}\)

\(\Rightarrow x^2=\frac{8}{x^2}.2\)

\(\Rightarrow x^2=\frac{16}{x^2}\)

\(\Rightarrow x^4=16\)

\(\Rightarrow x=2\)

\(r,\frac{x^3}{2}=\frac{32}{x}\)

\(\Rightarrow x^3=\frac{32}{x}.2\)

\(\Rightarrow x^3=\frac{64}{x}\)

\(\Rightarrow x^4=64\)

\(\Rightarrow x=\sqrt[4]{64}\)

Theo ( 1 ), tính theo mod p, ta có 

\(-1\equiv\left(p-1\right)!\equiv\left(n-1\right)!n\left(n+1\right)...\left(p-1\right)\)

\(\equiv\left(n-1\right)!\left(p-\left(n-p\right)\right)\left(p-\left(p-n-1\right)\right)...\left(p-1\right)\)

\(\equiv\left(n-1\right)!\left(-1\right)^{p-n}\left(p-n\right)\left(p-n-1\right)\) )...1

\(\equiv\left(n-1\right)!\left(-1\right)^{p-n}\left(p-n\right)!\)

\(\equiv\left(n-1\right)!\left(-1\right)^{n-1}\left(p-n\right)!\) ( vì p lẻ )

Cbht