Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(\frac{1}{6}\right)^5+\left(\frac{1}{8}\right)^3=\frac{1}{2^5}.\frac{1}{3^5}+\frac{1}{2^9}=\frac{1}{2^5}\left(\frac{1}{3^5}+\frac{1}{2^4}\right)=\frac{259}{124416}\)
Chúc bạn
học tốt!!!!!!!!!!!!!!!
\(\left(\frac{1}{6}\right)^5+\left(\frac{1}{8}\right)^3\)
= \(\frac{1}{7776}+\frac{1}{512}\)
= \(\frac{16}{124416}+\frac{243}{124416}\)
= \(\frac{259}{124416}\)
\(\frac{2.3+4.6+14.21}{3.5+6.10+21.35}=\frac{2.\left(3+2.6+7.21\right)}{5.\left(3+2.6+7.21\right)}=\frac{2}{5}\)
Ta có : \(\frac{5-x}{-25-y}=\frac{x}{y}\)
\(\Leftrightarrow\left(5-x\right)y=\left(-25-y\right)x\)
\(\Rightarrow5y-xy=-25x-xy\)
\(\Rightarrow5y=-25x\)
\(\Rightarrow\frac{x}{y}=\frac{5}{-25}=-\frac{1}{5}\)
\(\frac{5-x}{-25-y}=\frac{x}{y}\)
\(\Leftrightarrow y\left(5-x\right)=x\left(-25-y\right)\)
\(\Leftrightarrow5y-xy=-25y-xy\)
\(\Leftrightarrow5x=-25y\)
\(\Rightarrow\frac{y}{x}=\frac{5}{-25}=-\frac{1}{5}\)
Hùng sai òi :
Ta có ; \(\frac{5-x}{-25-y}=\frac{x}{y}\)
\(\Rightarrow\left(5-x\right)y=\left(-25-y\right)x\)
\(\Rightarrow5y-xy=-25x-xy\)
\(\Rightarrow5y=-25x\)
Vậy \(\frac{x}{y}=\frac{5}{-25}=\frac{-1}{5}\)
Có: Đề \(\Leftrightarrow\frac{abz-acy}{a^2}=\frac{bcx-abz}{b^2}=\frac{acy-bcx}{c^2}\)\(=\frac{\left(abz-abz\right)+\left(bcx-bcx\right)+\left(acy-acy\right)}{a^2+b^2+c^2}\)
\(=\frac{0}{a^2+b^2+c^2}=0\)\(\left(ĐKXĐ:a,b,c\ne0\right)\)
\(\Rightarrow\hept{\begin{cases}bz-cy=0\\cx-az=0\\ay-bx=0\end{cases}\Leftrightarrow\hept{\begin{cases}bz=cy\\cx=az\\ay=bx\end{cases}}}\)\(\Leftrightarrow\hept{\begin{cases}\frac{y}{b}=\frac{z}{c}\\\frac{z}{c}=\frac{x}{a}\\\frac{x}{a}=\frac{y}{b}\end{cases}}\RightarrowĐpcm\)
\(\frac{bz-cy}{a}\)=\(\frac{cx-az}{b}\)=\(\frac{ay-bx}{c}\)=>\(\frac{a\left(bz-cy\right)}{a^2}\)=\(\frac{b\left(cx-az\right)}{b^2}\)=\(\frac{c\left(ay-bx\right)}{c^2}\)
=>\(\frac{abz-acy}{a^2}\)=\(\frac{bcx-abz}{b^2}\)\(\frac{cay-bcx}{c^2}\)=\(\frac{abz-acy+bcx-abz+cay-bcx}{a^2+b^2+c^2}\)= 0
=>\(\frac{bz-cy}{a}\)=\(\frac{cx-az}{b}\)=\(\frac{ay-bx}{c}\)= 0
=> bz - cy = cx - az = ay - bx = 0
+) bz - cy = 0 => bz = cy => y/b = z/c
+) cx - az = 0 => cx = az => x/a = z/c
=> x/a = y/b = z/c
\(\frac{3\cdot13-13\cdot18}{15\cdot40-80}=\frac{13\left(3-18\right)}{15\cdot40-40\cdot2}=\frac{-15\cdot13}{40\cdot13}=-\frac{3}{8}\)
Sao ảnh hai bn Linh Le và Trần Việt Linh giống nhau thế hay là cùng nick
\(\frac{6.9-2.17}{63.3-119}=\frac{2.3.9-2.17}{7.9.3-7.17}=\frac{2.27-2.17}{7.27-7.17}=\frac{2.\left(27-17\right)}{7.\left(27-17\right)}=\frac{2}{7}\)
\(\frac{6.9-2.17}{63.3-119}=\frac{2.3.3^2-2.17}{3^2.7.3-7.17}\)
=2.3/.3^2/-2.17/ phần 3^2/.7.3/-7.17/=\(\frac{2.\left(-2\right)}{7.\left(-7\right)}=\frac{-4}{-14}=\frac{2}{7}\)
bn ơi / là bỏ số đó nha bn nếu mk ghi bn ko hiểu bn cứ hỏi mk nhé
Ta có: \(\frac{a}{b}\)=\(\frac{c}{d}\) => \(\frac{a}{c}\)=\(\frac{b}{d}\)
Ta có: \(\frac{a}{c}\)=\(\frac{b}{d}\)\(\Rightarrow\)\(\frac{5a+3b}{5c+3d}\)=\(\frac{5a-3b}{5c-bd}\)
\(\Rightarrow\)\(\frac{5a+3b}{5a-3b}\)=\(\frac{5c+3d}{5c-3d}\) (đpcm)
Giải:
Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{c}=\frac{b}{d}=\frac{5a}{5c}=\frac{3b}{3d}=\frac{5a+3b}{5c+3d}=\frac{5a-3b}{5c-3d}\)
\(\frac{5a+3b}{5c+3d}=\frac{5a-3b}{5c-3d}\Rightarrow\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\left(đpcm\right)\)
=> \(\frac{10101.39-10101}{10101\left(50-7\right)}\)
=> \(\frac{10101\left(39-1\right)}{10101\left(50-7\right)}\)
=> \(\frac{38}{43}\)