<prompt> Pata thamani ya chini iwezekanavyo yasqrt58 - 42x sqrt149 - 140 sqrt1 - x2ambapo -1 le x le 1.  <response>: Kwa kuwa -1 le x le 1, kuna pembe theta, 0circ le theta le 180circ, kiasi kwamba cos theta x. Kisha sqrt1 - x2 sin theta.asyukubwa wa kitengo 1 cm;jozi O, X, Y, Z;O 0,0;X 10,0;Y 0,3;Z 7dir40;kuchoraO--X--Z--Y--mzunguko;kuchoraO--Z;leboO, O, SW;leboX, X, E;leboY, Y, NW;leboZ, Z, N;lebo10, O X2, S;lebo3, O Y2, W;lebo7,O Z2, SE;lebo90circ - theta, 1.5,0.4;lebotheta, 0.3,0.6;asyTengeneza pembetatu OXZ na OYZ ili OX 10, OY 3, OZ 7, angle YOZ theta na angle XOZ 90circ - theta . Kisha angle XOY 90circ.Pia, kwa Sheria ya Cosines kwenye pembetatu YOZ,YZ sqrt32 72 - 2 cdot 3 cdot 7 cos theta sqrt58 - 42x.Kwa Sheria ya Cosines kwenye pembetatu XOZ,anzaalignXZ sqrt72 102 - 2 cdot 7 cdot 10 cos 90circ - theta sqrt149 - 140 sin theta sqrt149 - 140 sqrt1 - x2.endalignKwa hivyo, usemi tunaotaka kupunguza ni YZ XZ. By the Triangle Inequality, YZ XZ ge XY sqrt109. Usawa hutokea wakati Z ni hatua. kwenye overlineXY kiasi kwamba OZ 7, kwa hivyo thamani ya chini ni sqrt109. 