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[tex]\it a) |4-\sqrt15{| +|-{5}|+|\sqrt{15}-3| =[/tex]

b) [tex]|\sqrt{11}-\sqrt{12}|+|3-\sqrt{11}|-|\sqrt{12}-3|=[/tex]


Răspuns :

[tex]\displaystyle|x|=\left \{ {{x,\text{ dac\u a } x\geq0} \atop {-x,\text{ dac\u a }x<0}} \right.[/tex]

a)

[tex]\sqrt{15}=3,...\\4-\sqrt{15}>0\Rightarrow|4-\sqrt{15}|=4-\sqrt{15}\\\sqrt{15}-3>0\Rightarrow|\sqrt{15}-3|=\sqrt{15}-3\\|4-\sqrt{15}|+|-5|+|\sqrt{15}-3|=4-\sqrt{15}+5+\sqrt{15}-3=6[/tex]

b)

[tex]\sqrt{11}<\sqrt{12}\Rightarrow \sqrt{11}-\sqrt{12}<0\Rightarrow|\sqrt{11}-\sqrt{12}|=-(\sqrt{11}-\sqrt{12})=\sqrt{12}-\sqrt{11}\\\sqrt{11}=3,..\Rightarrow 3-\sqrt{11}<0\Rightarrow|3-\sqrt{11}|=-(3-\sqrt{11})=\sqrt{11}-3\\\sqrt{12}=3,..\Rightarrow\sqrt{12}-3>0\Rightarrow|\sqrt{12}-3|=\sqrt{12}-3\\|\sqrt{11}-\sqrt{12}|+|3-\sqrt{11}|-|\sqrt{12}-3|=\sqrt{12}-\sqrt{11}+\sqrt{11}-3-(\sqrt{12}-3)=0[/tex]

[tex]a) |4 - \sqrt{15} | + | - 5| + | \sqrt{15} - 3 | = \\ \\ 4 - \sqrt{15} + 5 + | \sqrt{15} - 3 | = \\ \\ (4 + 5 - 3) + ( - \sqrt{15} + \sqrt{15}) = \\ \\ 6 [/tex]

[tex]b) | \sqrt{11} - \sqrt{12} | + |3 - \sqrt{11} | - | \sqrt{12} - 3 | = \\ \\ | \sqrt{11} - 2 \sqrt{3} | + \sqrt{11} - 3 - |2 \sqrt{3} - 3 | = \\ \\ 2 \sqrt{3} - \sqrt{11} + \sqrt{11} - 3 - (2 \sqrt{3} - 3) = \\ \\ 2 \sqrt{3} - 3 - 2 \sqrt{3} + 3 = \\ \\ (2 \sqrt{3} - 2 \sqrt{3}) + ( - 3 + 3) = \\ \\ 0 [/tex]