Tính
\(\frac{\text{ }\left[\left(e-m\right)^2-\left(e+m\right)^2\right]\text{ }\left[\left(y-1\right)^2-\left(y+1\right)^2\right]}{a.16.n.h}\)\(.\frac{e}{u^{-1}}\)
\(\frac{\left[\left(e-m\right)^2-\left(e+m^2\right)\right].\left[\left(y-1\right)^2-\left(y+1\right)^2\right]}{a.16.n.h}x\frac{ê}{u^{-1}}\)
\(\frac{\left[\left(e-m\right)^2-\left(e+m\right)^2\right]\left[\left(y-1\right)^2-\left(y+1\right)^2\right]}{a.16.n.h}\) . \(\frac{e}{u^{-1}}\)
=\(\frac{\left(e^2-2e.m+m^2-e^2-2e.m-m^2\right).\left(y^2-2y+1-y^2-2y-1\right)}{a.16.n.h}.\frac{e}{u^{-1}}\)
= \(\frac{-4e.m.\left(-4y\right)}{a.16.n.h}.\frac{e}{u^{-1}}\)
=\(\frac{16e.m.y}{16a.n.h}.\frac{e}{\frac{1}{4}}\)
=\(\frac{e.m.y}{a.n.h}.e.u=\frac{e.m.y.e.u}{a.n.h}\)
Rút gọn biểu thức:
\(\frac{\left[\left(e-m\right)^2-\left(e+m\right)^2\right].\left[\left(y-1\right)^2-\left(y+1\right)^2\right]}{a.16.n.h}.\frac{\text{ê}}{y^{-1}}\)
Đề đăng tạp ><
Giải hộ tớ vs để tớ gửi tặng người đó >.<
\(\frac{\left[\left(e-m\right)^2-\left(e+m\right)\right]\left[\left(y-1\right)^2\left(y+1\right)^2\right]}{a.16.nh}.\frac{ê}{u^{-1}}\)
\(=\frac{\left[\left(e-m\right)^2\left(e+m\right)^2\right]\left[\left(y-1\right)^2\left(y+1\right)^2\right]}{16.anh}.êu\)
\(=\frac{\left(e^2-2em+m^2-e^2-2em.m^2\right)\left(y^2-2y+1-y^2-2y-1\right)}{16anh}.êu\)
\(=-\frac{4em\left(-4y\right)}{16anh}.êu\)
\(=\frac{emy}{anh}.êu\)
\(=\frac{em.yêu}{anh}\)
Rút gọn biểu thức :
\(\frac{\left[\left(e-m\right)^2-\left(e+m\right)^2\right].\left[\left(y-1\right)^2-\left(y+1\right)^2\right]}{a.16.n.h}.\frac{ê}{u^{-1}}\)
bài này có phải là " Biểu thức tình yêu " không ?
\(\frac{\left[\left(e-m\right)^2-\left(e+m\right)^2\right]\left[\left(y-1\right)^2-\left(y+1\right)^2\right]}{a.16.n.h}\) \(\times\) \(\frac{ê}{u^{-1}}\)
Rút gọn biểu thức trên.
\(=\frac{\left(e^2-2em+m^2-e^2-2em-m^2\right)\left(y^2-2y+1-y^2-2y-1\right)}{a.16.n.h}\)\(\times\frac{ê}{u^{-1}}\)
= \(\frac{\left(-4\right)em.\left(-4\right)y}{a.16.n.h}\)\(\times\frac{ê}{u^{-1}}\)
= \(\frac{16.e.m.y}{16.a.n.h}\times\frac{ê}{u^{-1}}\)
= \(\frac{e.m.y}{a.n.h}\times\frac{ê}{\frac{1}{u}}\)
= \(\frac{e.m.y}{a.n.h}\timesê.u\)
= \(\frac{e.m.y.ê.u}{a.n.h}\)
Tính
\(\frac{\left[\left(e-m\right)^2-\left(e+m\right)^2\right]\left[\left(y-1\right)^2-\left(y+1\right)^2\right]}{a.16.nh}.\frac{ê}{\frac{1}{u}}\)
P/s đề này ms đúng nha
bài này hơi khó
\(\frac{\left[\left(e-m\right)^2-\left(e+m\right)^2\right]\left[\left(y-1\right)^2\left(y+1\right)^2\right]}{a.16.nh}.\frac{ê}{\frac{1}{u}}\)
\(=\frac{\left[\left(e-m\right)^2\left(e+m\right)^2\right]\left[\left(y-1\right)^2\left(y+1\right)^2\right]}{16.anh}.\frac{êu}{1}\)
\(=\frac{\left(e^2-2em+m^2-e^2-2em-m^2\right)\left(y^2-2y+1-y^2-2y-1\right)}{16.anh}\)
\(=\frac{-4em\left(-4y\right)}{16.anh}.êu\)
\(=\frac{emy}{anh}.êu\)
\(=\frac{em.yêu}{anh}\)
\(\frac{\left(\left(e-m\right)^2-\left(e+m\right)^2\right)\left(\left(y-1\right)^2-\left(y+1\right)^2\right).\frac{ê}{\frac{1}{u}}}{16.anh}\)
tui k bit lm nhưng tui bit kết quả là em yêu anh
\(\frac{\left[\left(e-m\right)^2-\left(e+m\right)^2\right].\left[\left(y-1\right)^2-\left(y+1\right)^2\right]}{a.16.h.n}.\frac{ê}{y^{-1}}\)
Tính
\(\left[\left(e-m\right)^2-\left(e+m\right)^2\right]\left[\left(y-1\right)^2-\left(y+1\right)^2\right].\frac{ê}{\frac{1}{u}}\)
\(\left[\left(e-m\right)^2-\left(e+m\right)^2\right]\left[\left(y-1\right)^2-\left(y+1\right)^2\right]\)
\(=\left[\left(e-m-e-m\right)\left(e-m+e+m\right)\right]\) \(\left[\left(y-1-y-1\right)\left(y-1+y+1\right)\right]\)
\(=\left[-2m.2e\right]\left[-2.2y\right]\)
\(=-4me.\left(-4y\right)\)
\(=16mey\)
\(\left(e-m\right)\left(e-m\right)-\left(e+m\right)\left(e+m\right)\)
\(e^2-2em+m^2-e^2+2em-m^2\)
\(\left(e^2-e^2\right)-\left(2em-2em\right)+\left(m^2-m^2\right)=0\)
vậy \(0.\left\{\left(y-1\right)^2-\left(y+1\right)^2\right\}.\frac{e}{\frac{1}{u}}=0\)