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Question Number 197541 by a.lgnaoui last updated on 20/Sep/23

$$\mathbf{Montrer}\mathbf{que}$$$$\mathbf{x}=\frac{\mathbf{an}+\mathbf{bm}}{\mathbf{m}+\mathbf{n}}$$

Commented by a.lgnaoui last updated on 20/Sep/23

Commented by a.lgnaoui last updated on 21/Sep/23

Commented by a.lgnaoui last updated on 21/Sep/23

Answered by HeferH last updated on 20/Sep/23

$$\frac{x-a}{b-x}=\frac{m}{n}$$$$xn-an=bm-mx$$$$x(n+m)=an+bm$$$$x=\frac{an+bm}{m+n}$$

Answered by a.lgnaoui last updated on 21/Sep/23

$$\mathbf{propriete}\mathbf{des}\mathbf{triangles}\mathbf{semblables}:$$$$\left(\mathbf{meme}\mathbf{angle}\left(\mathit{\alpha }\right)\mathrm{et}2\mathbf{cotes}\mathbf{paraleles}\right)$$$$\frac{\mathrm{x}-\mathrm{a}}{\mathrm{m}}=\frac{\mathrm{b}-\mathrm{x}}{\mathrm{n}}\Rightarrow \mathrm{m}(\mathrm{b}-\mathrm{x})=\mathrm{n}(\mathrm{x}-\mathrm{a})$$$$(\mathrm{m}+\mathrm{n})\mathrm{x}=\mathrm{an}+\mathrm{bm}\mathbf{x}=\frac{\mathbf{an}+\mathbf{bm}}{\mathbf{m}+\mathbf{n}}$$

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