\(\left(a+b\right).\left(b+c\right).\left(c-a\right)+\left(b+c\right).\left(c+a\right).\left(a-b\right)+\left(c+a\right).\left(a+b\right).\left(b-c\right)\)

\(=\left(a+b\right).\left[\left(b+c\right).\left(c-a\right)+\left(c+a\right).\left(a-b\right)\right]+\left(c+a\right).\left(a+b\right).\left(b-c\right)\)

\(=\left(a+b\right).\left(ac-a^2+bc-ab+a^2-ab+ac-bc\right)+\left(c+a\right).\left(a+b\right).\left(b-c\right)\)

\(=-\left(a+b\right).2a.\left(b-c\right)+\left(c+a\right).\left(a+b\right).\left(b-c\right)\)

\(=\left(a+b\right).\left(b-c\right).\left(-2a+c+a\right)=\left(a+b\right).\left(b-c\right).\left(c-a\right)\)

giai lai:

\(\left(b+c\right).\left[\left(a+b\right).\left(c-a\right)+\left(c+a\right).\left(a-b\right)\right]+\left(c+a\right).\left(a+b\right).\left(b-c\right)\)

\(=-\left(b+c\right).2a.\left(b-c\right)+\left(b-c\right).\left(ac+bc+a^2+ab\right)\)

\(=\left(b-c\right).\left(-2ab-2ac+ac+bc+a^2+ab\right)\)

\(=\left(b-c\right).\left(-ab-ac+bc+a^2\right)\)

\(=\left(b-c\right).\left(a+b\right).\left(a-c\right)\)

\(5^x=25\div125\)

\(\Leftrightarrow5^x=\frac{1}{5}\)

\(\Leftrightarrow5^x=5^{-1}\)

Vậy x = -1

\(5^a=25:125\)

\(\Rightarrow5^a=\frac{1}{5}\)

\(\Rightarrow5^a=5^{-1}\)

\(\Rightarrow a=-1\)

\(x^2-y=0\)

\(\Leftrightarrow x^2-\left(\sqrt{y}\right)^2=0\)

\(\Leftrightarrow\left(x+\sqrt{y}\right)\left(x-\sqrt{y}\right)=0\)

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

ĐK với mọi x , y\(\ge\)0

\(PT\Leftrightarrow x^2=y\)

............................................

Ta co:\(\hept{\begin{cases}2a+b⋮13\\5a-4b⋮13\end{cases}\Rightarrow\hept{\begin{cases}-2.\left(2a+b\right)⋮13\\5a-4b⋮13\end{cases}}}\)

\(\Rightarrow\hept{\begin{cases}-4a-2b⋮13\\5a-4b⋮13\end{cases}}\Rightarrow-4a-2b+5a-4b=a-6b\)

DK: a,b thuoc N, a > 0

\(\overline{a0b}=100a+b⋮7\)

\(\Rightarrow4.\left(100a+b\right)⋮7\)

\(\Rightarrow400a+4b⋮7\)

\(\Rightarrow a+4b⋮7\text{ vi }399a⋮7\)

\(\)

\(x^4\left(x^2-y^3\right)+y^3\left(x^4-y\right)\)

\(=x^6-x^4y^3+x^4y^3-y^4\)

\(=x^6-y^4\)

\(=\left(x^3-y^2\right)\left(x^3+y^2\right)\)

Đặt bt là A

\(A\sqrt{2}=\left(2-\sqrt{3}\right)\sqrt{52+30\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{52-30\sqrt{3}}\)

\(=\left(2-\sqrt{3}\right)\sqrt{\left(3\sqrt{3}+5\right)^2}-\left(2+\sqrt{3}\right)\sqrt{\left(3\sqrt{3}-5\right)^2}\)\(=\left(2-\sqrt{3}\right)\left(3\sqrt{3}+5\right)-\left(2+\sqrt{3}\right)\left(3\sqrt{3}-5\right)\)

khai triển hết ra ta đc \(A\sqrt{2}=2\Rightarrow A=\sqrt{2}\)

= 57/40

học tốt

trả lời 

ko viết lại đề 

x-3/5=5/8+1/5

x-3/5=33/40

x=33/40+3/5

x=57/40 

x/13 + 15/26 = 23/26

x/13 = 23/26 - 15/26

x/13 = 8/26

x /13 = 4/13

x = 4

\(\frac{x}{13}+\frac{15}{26}=\frac{23}{26}\)

\(\Leftrightarrow\frac{x}{13}=\frac{8}{26}\)

\(\Leftrightarrow\frac{x}{13}=\frac{4}{13}\)

\(\Leftrightarrow x=4\)