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Question Number 152532 by mondli66 last updated on 29/Aug/21

Answered by amin96 last updated on 29/Aug/21

$$\{\begin{array}{l}0+y+3z=-4\\ x+2y+z=7\\ x-2y+0=1\end{array}\Rightarrow {\left(\begin{array}{cccc}1& 2& 1& 7\\ 1& -2& 0& 1\\ 0& 1& 3& -4\end{array}\right)}^{\left(1\right)-\left(1\right)}=$$$$={\left(\begin{array}{cccc}1& 2& 1& 7\\ 0& 4& 1& 6\\ 0& 1& 3& -4\end{array}\right)}^{\left(3\right)\times 4-\left(2\right)}=\left(\begin{array}{cccc}1& 2& 1& 7\\ 0& 4& 1& 6\\ 0& 0& 11& -22\end{array}\right)$$$$\{\begin{array}{l}x+2y+z=7\\ 4y+z=6\\ 11z=-22\end{array}\Rightarrow z=-2\Rightarrow y=2\Rightarrow x=5$$$$answer(5;2;-2)$$$$$$

Commented by amin96 last updated on 29/Aug/21

$$Gaussmethod$$

Commented by SANOGO last updated on 29/Aug/21

Commented by KONE last updated on 29/Aug/21

$$K={\int }_{1/2}^2(1+\frac{1}{x^2})arctan\left(x\right)dx$$$$posontt=\frac{1}{x}\iff dx=-\frac{1}{t^2}dt$$$$x=2\iff t=\frac{1}{2};x=\frac{1}{2}\iff t=2$$$$K={\int }_2^{\frac{1}{2}}(1+t^2)arctan\left(\frac{1}{t}\right)\times (-\frac{1}{t^2})dt$$$$={\int }_{\frac{1}{2}}^2(1+\frac{1}{t^2})arctan\left(\frac{1}{t}\right)dt$$$$2K={\int }_{\frac{1}{2}}^2(1+\frac{1}{x^2})arctan\left(x\right)dx+{\int }_{\frac{1}{2}}^2(1+\frac{1}{t^2})arctan\left(\frac{1}{t}\right)dt$$$$={\int }_{\frac{1}{2}}^2(1+\frac{1}{x^2})(arctan\left(x\right)+arctan\left(\frac{1}{x}\right))dxcarxettsonmuette$$$$onsaitquex>0arctan\left(x\right)+arctan\left(\frac{1}{x}\right)=\frac{\mathrm{\Pi }}{2}$$$$d^{\prime }ou2K=\frac{\mathrm{\Pi }}{2}{\int }_{\frac{1}{2}}^2(1+\frac{1}{x^2})dx$$$$K=\frac{\mathrm{\Pi }}{4}{[x-\frac{1}{x}]}_{\frac{1}{2}}^2=\frac{\mathrm{\Pi }}{4}[2-\frac{1}{2}-\frac{1}{2}+2]$$$$K=\frac{3\mathrm{\Pi }}{4}\underset{―}{\mathcal{KAB}}$$

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