Răspuns :
[tex]a).|x|=3 \Rightarrow x=3~sau~x=-3 \\ b).|x|=4 \Rightarrow x=4 ~sau~x=-4 \\ c).|x|=0 \Rightarrow x=0 \\ d).|x|=-1 \Rightarrow ~Nu~are~solutii[/tex]
[tex]e).|x+1|=3 \\ x+1=3 \Rightarrow x=3-1 \Rightarrow x=2 \\ x+1=-3 \Rightarrow x=-3-1 \Rightarrow x=-4[/tex]
[tex]f).|x-2|=4 \\ x-2=4 \Rightarrow x=4+2 \Rightarrow x=6 \\ x-2=-4 \Rightarrow x=-4+2 \Rightarrow x=-2[/tex]
[tex]g).|x+2|=0 \\ x+2=0 \Rightarrow x=0-2 \Rightarrow x=-2 \\ h).|x-3|=-6 \Rightarrow Nu~are~solutii[/tex]
[tex]\displaystyle i).3x \leq 6 \Rightarrow x \leq \frac{6}{3} \Rightarrow x \leq 2 \Rightarrow x \in ( - \infty , 2] \\ j).2x\ \textgreater \ -8 \Rightarrow x\ \textgreater \ - \frac{8}{2} \Rightarrow x\ \textgreater \ -4 \Rightarrow x \in (-4 , \infty )[/tex]
[tex]\displaystyle k).-4x\ \textless \ 12 \Rightarrow x\ \textless \ - \frac{12}{4} \Rightarrow x\ \textgreater \ -3 \Rightarrow x \in (-3 , \infty ) \\ l).-3x \geq 15 \Rightarrow x \geq - \frac{15}{3} \Rightarrow x \leq -5 \Rightarrow x \in ( - \infty , -5] \\ m).2(x+1)\ \textless \ 8 \Rightarrow 2x+2\ \textless \ 8 \Rightarrow 2x\ \textless \ 8-2 \Rightarrow 2x\ \textless \ 6 \Rightarrow x\ \textless \ \frac{6}{2} \Rightarrow \\ \Rightarrow x\ \textless \ 3 \Rightarrow x \in ( - \infty , 3)[/tex]
[tex]\displaystyle n).-3(x-1) \leq 7 \Rightarrow -3x+3 \leq 7 \Rightarrow -3x \leq 7-3 \Rightarrow -3x \leq 4 \Rightarrow \\ \Rightarrow x \geq - \frac{4}{3} \Rightarrow x \in \left[- \frac{4}{3} ,\infty \right) \\ o).-2(x+1)\ \textgreater \ -4 \Rightarrow -2x-2\ \textgreater \ -4 \Rightarrow -2x\ \textgreater \ -4+2 \Rightarrow \\ \Rightarrow -2x\ \textgreater \ -2 \Rightarrow x\ \textgreater \ \frac{-2}{-2} \Rightarrow x\ \textless \ 1 \Rightarrow x \in ( \infty , 1) [/tex]
[tex]\displaystyle p).-5(3-x) \geq 10 \Rightarrow -15+5x \geq 10 \Rightarrow 5x \geq 10+15 \Rightarrow \\ \Rightarrow 5x \geq 25 \Rightarrow x \geq \frac{25}{5} \Rightarrow x \geq 5\Rightarrow x \in [5 , \infty)[/tex]
[tex]\displaystyle q).1+2+3+...+40= \frac{40(40+1)}{2} = \frac{40 \cdot 41}{2} = \frac{1640}{2}=820 \\ \\ r). 1+2+3+...+100= \frac{100(100+1)}{2} = \frac{100 \cdot 101}{2} = \frac{10100}{2} =5050 \\ \\ s).1+2+3+...+101= \frac{101(101+1)}{2} = \frac{101 \cdot 102}{2} = \frac{10302}{2}=5151 [/tex]
[tex]t\displaystyle ).2+4+6+...+2000=2(1+2+3+...+1000)= \\ \\ =2 \cdot \frac{1000(1000+1)}{2} =2 \cdot \frac{1000 \cdot 1001}{2} =\not 2 \cdot \frac{1001000}{\not 2} =1001000[/tex]
[tex]\displaystyle u).3+6+9+...+1980=3(1+2+3+...+660)= \\ \\ =3 \cdot \frac{660(660+1)}{2} =3 \cdot \frac{660 \cdot 661}{2} =3 \cdot \frac{436260}{2} =3 \cdot 218130=654390[/tex]
[tex]e).|x+1|=3 \\ x+1=3 \Rightarrow x=3-1 \Rightarrow x=2 \\ x+1=-3 \Rightarrow x=-3-1 \Rightarrow x=-4[/tex]
[tex]f).|x-2|=4 \\ x-2=4 \Rightarrow x=4+2 \Rightarrow x=6 \\ x-2=-4 \Rightarrow x=-4+2 \Rightarrow x=-2[/tex]
[tex]g).|x+2|=0 \\ x+2=0 \Rightarrow x=0-2 \Rightarrow x=-2 \\ h).|x-3|=-6 \Rightarrow Nu~are~solutii[/tex]
[tex]\displaystyle i).3x \leq 6 \Rightarrow x \leq \frac{6}{3} \Rightarrow x \leq 2 \Rightarrow x \in ( - \infty , 2] \\ j).2x\ \textgreater \ -8 \Rightarrow x\ \textgreater \ - \frac{8}{2} \Rightarrow x\ \textgreater \ -4 \Rightarrow x \in (-4 , \infty )[/tex]
[tex]\displaystyle k).-4x\ \textless \ 12 \Rightarrow x\ \textless \ - \frac{12}{4} \Rightarrow x\ \textgreater \ -3 \Rightarrow x \in (-3 , \infty ) \\ l).-3x \geq 15 \Rightarrow x \geq - \frac{15}{3} \Rightarrow x \leq -5 \Rightarrow x \in ( - \infty , -5] \\ m).2(x+1)\ \textless \ 8 \Rightarrow 2x+2\ \textless \ 8 \Rightarrow 2x\ \textless \ 8-2 \Rightarrow 2x\ \textless \ 6 \Rightarrow x\ \textless \ \frac{6}{2} \Rightarrow \\ \Rightarrow x\ \textless \ 3 \Rightarrow x \in ( - \infty , 3)[/tex]
[tex]\displaystyle n).-3(x-1) \leq 7 \Rightarrow -3x+3 \leq 7 \Rightarrow -3x \leq 7-3 \Rightarrow -3x \leq 4 \Rightarrow \\ \Rightarrow x \geq - \frac{4}{3} \Rightarrow x \in \left[- \frac{4}{3} ,\infty \right) \\ o).-2(x+1)\ \textgreater \ -4 \Rightarrow -2x-2\ \textgreater \ -4 \Rightarrow -2x\ \textgreater \ -4+2 \Rightarrow \\ \Rightarrow -2x\ \textgreater \ -2 \Rightarrow x\ \textgreater \ \frac{-2}{-2} \Rightarrow x\ \textless \ 1 \Rightarrow x \in ( \infty , 1) [/tex]
[tex]\displaystyle p).-5(3-x) \geq 10 \Rightarrow -15+5x \geq 10 \Rightarrow 5x \geq 10+15 \Rightarrow \\ \Rightarrow 5x \geq 25 \Rightarrow x \geq \frac{25}{5} \Rightarrow x \geq 5\Rightarrow x \in [5 , \infty)[/tex]
[tex]\displaystyle q).1+2+3+...+40= \frac{40(40+1)}{2} = \frac{40 \cdot 41}{2} = \frac{1640}{2}=820 \\ \\ r). 1+2+3+...+100= \frac{100(100+1)}{2} = \frac{100 \cdot 101}{2} = \frac{10100}{2} =5050 \\ \\ s).1+2+3+...+101= \frac{101(101+1)}{2} = \frac{101 \cdot 102}{2} = \frac{10302}{2}=5151 [/tex]
[tex]t\displaystyle ).2+4+6+...+2000=2(1+2+3+...+1000)= \\ \\ =2 \cdot \frac{1000(1000+1)}{2} =2 \cdot \frac{1000 \cdot 1001}{2} =\not 2 \cdot \frac{1001000}{\not 2} =1001000[/tex]
[tex]\displaystyle u).3+6+9+...+1980=3(1+2+3+...+660)= \\ \\ =3 \cdot \frac{660(660+1)}{2} =3 \cdot \frac{660 \cdot 661}{2} =3 \cdot \frac{436260}{2} =3 \cdot 218130=654390[/tex]
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