\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\Rightarrow\frac{x^2}{4}=\frac{y^2}{9}=\frac{2z^2}{32}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta được :

\(\frac{x^2}{4}=\frac{y^2}{9}=\frac{2z^2}{32}=\frac{x^2-y^2+2z^2}{4-9+32}=\frac{108}{27}=4\)

\(\Rightarrow\hept{\begin{cases}x^2=16\\y^2=36\\z^2=64\end{cases}}\) \(\Rightarrow\hept{\begin{cases}x=\pm4\\y=\pm6\\z=\pm8\end{cases}}\)

\(1^2+2^2+3^2+...+10^2=385\)

Mà \(1^2.2=2^2\), \(2^2.2=4^2\)

\(\Rightarrow\left(1+2^2+3^2+...+10^2\right).2=S\)

\(\Rightarrow S=385.2=770\)

S=(1².2²)+(2².2²)+(2².3²)+..................+(10².2²)

S=2².(1²+2²+3²+...................+10²)

S=4.385=1540

Vậy S=1540

\(\frac{2^{13}\cdot3^7}{4^7\cdot9^3}=\frac{2^{13}\cdot3^7}{\left[2^2\right]^7\cdot\left[3^2\right]^3}=\frac{2^{13}\cdot3^7}{2^{14}\cdot3^6}=2^{-1}\cdot3=\frac{1}{2}\cdot3=\frac{3}{2}\)

Bài làm

Ta có: \(\left(-\frac{1}{4}\right)^2=\left(\frac{1}{4}\right)^2\)

\(\left(\frac{1}{8}\right)^5=\left[\left(\frac{1}{4}\right)^2\right]^5=\left(\frac{1}{4}\right)^{10}\)

Mà \(2< 10\)

=> \(\left(\frac{1}{4}\right)^2< \left(\frac{1}{4}\right)^{10}\)

Hay \(\left(-\frac{1}{4}\right)^2< \left(\frac{1}{8}\right)^5\)

Vậy \(\left(-\frac{1}{4}\right)^2< \left(\frac{1}{8}\right)^5\)

# Học tốt #

Thank you very much!!!!!!

ta thấy /x+19/5/>=0

           /y+18/19/>=0

           /x-2004/>=0

Mà /x+19/5/+/y+18/19/+/z-2004/=0

=> x+19/5=0=>x=-19/5

    y+18/19=0=>y=-18/19

z-2004=0=>z=2004

Câu còn lại tương tự nha bạn

Tích mik nha 

b, \(\left|x+\frac{3}{4}\right|+\left|y-\frac{1}{5}\right|+\left|x+y+z\right|=0\)

vì \(\left|x+\frac{3}{4}\right|\ge0\forall x;\left|y-\frac{1}{5}\right|\ge0\forall y;\left|x+y+z\right|\ge0\forall z\)

Dâu ''='' xảy ra <=> x = -3/4 ; y = 1/5 ; \(-\frac{3}{4}+\frac{1}{5}+z=0\Leftrightarrow z=\frac{11}{20}\)

a) Đặt \(x-1=a\)

\(pt\Leftrightarrow\frac{13}{a}+\frac{5}{2a}=\frac{6}{3a}\)

\(\Leftrightarrow\frac{31}{2a}=\frac{6}{3a}\)

\(\Leftrightarrow\frac{31}{2}=2\)(vô lí)

Vậy pt vô nghiệm

a) \(\frac{13}{x-1}+\frac{5}{2x-2}=\frac{6}{3x-3}\)

\(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}=\frac{6}{3\left(x-1\right)}\)

\(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}=\frac{2}{x-1}\)

\(\frac{31}{2\left(x-1\right)}=\frac{2}{x-1}\)

\(\frac{31}{2}=2\)

=> không có x thỏa mãn đề bài.

b) \(\frac{1}{x-1}+\frac{-2}{3}\left(\frac{3}{4}-\frac{6}{5}\right)=\frac{5}{2-2x}\)

\(\frac{1}{x-1}+\frac{-2}{3}.\frac{-9}{20}=\frac{5}{2\left(1-x\right)}\)

\(\frac{1}{x-1}-\frac{-18}{60}=\frac{5}{2\left(1-x\right)}\)

\(\frac{1}{x-1}+\frac{3}{10}=\frac{5}{2\left(1-x\right)}\)

\(10\left(1-x\right)+3\left(x-1\right)\left(1-x\right)=25\left(x-1\right)\)

\(7-4x-3x^2=25x-25\)

\(7-4x-3x^2-25x+25=0\)

\(32-29x-3x^2=0\)

\(3x^2+29x-30=0\)

\(3x^2+32x-3x-32=0\)

\(x\left(3x+32\right)-\left(3x+32\right)=0\)

\(\left(3x+32\right)\left(x-1\right)=0\)

\(\orbr{\begin{cases}3x+32=0\\x-1=0\end{cases}}\)

\(\orbr{\begin{cases}x=-\frac{32}{3}\\x=1\end{cases}}\)