$MI=(x_{M} - x_{I}) ; (y_{M}-y_{I}) ; (z_{M}-z_{I})$

$IM=(x_{M} - x_{I}) ; (y_{M}-y_{I}) ; (z_{M}-z_{I})$

$MI=\sqrt{(x_{M} - x_{I})^{2}+(y_{M}-y_{I})^{2}+(z_{M}-z_{I})^{2}}$

$MI=(x_{M} - x_{I}) ; (y_{M}-y_{I}) ; (z_{M}-z_{I})$

$\int\limits_{}^{} \frac{x\ln x}{(x^{2} + 1)^{2}}(1)$$Đặt \begin{cases}u=x\ln x \\ dv=\frac{1}{(x^{2}+1)^{2}} \end{cases} \Rightarrow \begin{cases}du=(\ln x+1)dx \\ v= ln(x^{2}+1)^{2}\end{cases}$$(1) = x\ln x.\ln (x^{2}+1)^{2} + c - \int\limits_{}^{} ln(x^{2}+1)^{2}.(lnx+1)dx$$= xlnx.ln(x^{2}+1)^{2}-c+H$$H=\int\limits_{}^{}2ln(x^{2}+1).(lnx+1)dx$$= \int\limits_{}^{}[2ln(x^{2}+1)lnx]dx+\int\limits_{}^{}2ln(x^{2}+1)dx$$=2\int\limits_{}^{}ln(x+x^{2}+1)dx + 2\int\limits_{}^{}ln(x^{2}+1)dx$Bạn dùng công thức $\int\limits_{}^{} ln(x)=xlnx-x+c$ tính tiếp

$\int\limits_{}^{} \frac{x\ln x}{(x^{2} + 1)^{2}}(1)$$Đặt \begin{cases}u=x\ln x \\ dv=\frac{1}{(x^{2}+1)^{2}} \end{cases} \Rightarrow \begin{cases}du=(\ln x+1)dx \\ v= ln(x^{2}+1)^{2}\end{cases}$$(1) = x\ln x.\ln (x^{2}+1)^{2} + c - \int\limits_{}^{} ln(x^{2}+1)^{2}.(lnx+1)dx$$= xlnx.ln(x^{2}+1)^{2}-c+H$$H=\int\limits_{}^{}2ln(x^{2}+1).(lnx+1)dx$$= \int\limits_{}^{}[2ln(x^{2}+1)lnx]dx+\int\limits_{}^{}2ln(x^{2}+1)dx$$=2\int\limits_{}^{}ln(x+x^{2}+1)dx + 2\int\limits_{}^{}ln(x^{2}+1)dx$$=2(x+x^{2}+1)ln(x+x^{2}+1)-(x+x^{2}+1)+2(x^{2}+1)ln(x^{2}+1)-(x^{2}+1)+c$" Công thức $\int\limits_{}^{} ln(x)=xlnx-x+c$ "$(1)=..................... $

$\int\limits_{}^{} \frac{x\ln x}{(x^{2} + 1)^{2}}(1)$$Đặt \begin{cases}u=x\ln x \\ dv=\frac{1}{(x^{2}+1)^{2}} \end{cases} \Rightarrow \begin{cases}du=(\ln x+1)dx \\ v= ln(x^{2}+1)^{2}\end{cases}$$(1) = x\ln x.\ln (x^{2}+1)^{2} + c - \int\limits_{}^{} ln(x^{2}+1)^{2}.(lnx+1)dx$$= xlnx.ln(x^{2}+1)^{2}-c+H$$H=\int\limits_{}^{}(lnx+1)2ln(x^{2}+1)dx$$=2\int\limits_{}^{}ln(x+x^{2}+1)dx + 2\int\limits_{}^{}ln(x^{2}+1)dx$Bạn dùng công thức $\int\limits_{}^{} ln(x)=xlnx-x+c$ tính tiếp

$\int\limits_{}^{} \frac{x\ln x}{(x^{2} + 1)^{2}}(1)$$Đặt \begin{cases}u=x\ln x \\ dv=\frac{1}{(x^{2}+1)^{2}} \end{cases} \Rightarrow \begin{cases}du=(\ln x+1)dx \\ v= ln(x^{2}+1)^{2}\end{cases}$$(1) = x\ln x.\ln (x^{2}+1)^{2} + c - \int\limits_{}^{} ln(x^{2}+1)^{2}.(lnx+1)dx$$= xlnx.ln(x^{2}+1)^{2}-c+H$$H=\int\limits_{}^{}2ln(x^{2}+1).(lnx+1)dx$$= \int\limits_{}^{}[2ln(x^{2}+1)lnx]dx+\int\limits_{}^{}2ln(x^{2}+1)dx$$=2\int\limits_{}^{}ln(x+x^{2}+1)dx + 2\int\limits_{}^{}ln(x^{2}+1)dx$Bạn dùng công thức $\int\limits_{}^{} ln(x)=xlnx-x+c$ tính tiếp

$\int\limits_{}^{} \frac{x\ln x}{(x^{2} + 1)^{2}}(1)$$Đặt \begin{cases}u=x\ln x \\ dv=\frac{1}{(x^{2}+1)^{a}} \end{cases} \Rightarrow \begin{cases}du=(\ln x+1)dx \\ v= ln(x^{2}+1)^{2}\end{cases}$$(1) = x\ln x.\ln (x^{2}+1)^{2} + c - \int\limits_{}^{} ln(x^{2}+1)^{2}.(lnx+1)dx$$= xlnx.ln(x^{2}+1)^{2}-c+H$$H=\int\limits_{}^{}(lnx+1)2ln(x^{2}+1)dx$$=2\int\limits_{}^{}ln(x+x^{2}+1)dx + 2\int\limits_{}^{}ln(x^{2}+1)dx$Bạn dùng công thức $\int\limits_{}^{} ln(x)=xlnx-x+c$ tính tiếp

$\int\limits_{}^{} \frac{x\ln x}{(x^{2} + 1)^{2}}(1)$$Đặt \begin{cases}u=x\ln x \\ dv=\frac{1}{(x^{2}+1)^{2}} \end{cases} \Rightarrow \begin{cases}du=(\ln x+1)dx \\ v= ln(x^{2}+1)^{2}\end{cases}$$(1) = x\ln x.\ln (x^{2}+1)^{2} + c - \int\limits_{}^{} ln(x^{2}+1)^{2}.(lnx+1)dx$$= xlnx.ln(x^{2}+1)^{2}-c+H$$H=\int\limits_{}^{}(lnx+1)2ln(x^{2}+1)dx$$=2\int\limits_{}^{}ln(x+x^{2}+1)dx + 2\int\limits_{}^{}ln(x^{2}+1)dx$Bạn dùng công thức $\int\limits_{}^{} ln(x)=xlnx-x+c$ tính tiếp

cos2x + (1 + 2cosx)(sinx-cosx) = 0<=> (cos2x - sin2x) - (1 + 2cosx)(cosx - sinx) = 0<=> (cosx - sinx) (cosx + sinx - 1 - 2cosx) = 0* cosx - sinx = 0 <=> 1 - tanx = 0 (ĐK: cosx \neq 0)<=> tanx = 1 <=> x = \frac{\pi }{4} + k\pi* sinx - 1 - cosx = 0 <=> (sinx - cosx)2 = 1<=> (sin2x - 2sinxcosx + cos2x) = 1<=> 2sinxcosx = 0<=> sinx = 0 <=> x = k\pi cosx = 0 <=> x = \frac{\pi }{2} + k\pi

cos2x + (1 + 2cosx)(sinx-cosx) = 0<=> (cos2x - sin2x) - (1 + 2cosx)(cosx - sinx) = 0<=> (cosx - sinx) (cosx + sinx - 1 - 2cosx) = 0* cosx - sinx = 0 <=> 1 - tanx = 0 (ĐK: cosx khác 0)<=> tanx = 1 <=> x = pi/4 + k.pi* sinx - 1 - cosx = 0 <=> (sinx - cosx)2 = 1<=> (sin2x - 2sinxcosx + cos2x) = 1<=> 2sinxcosx = 0<=> sinx = 0 <=> x = k.pi cosx = 0 <=> x = pi/2 + k.pi