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Determinati,in fiecare exercitiu,numarul natural X,stiind ca:
a)x supra 6 din 72 este 60;
b)15 supra X din 44 este 165;
c)11supra 15 din X este 88;
d)X supra X+1 din 12 este 8;
e)3*X+1 supra 2*X+5 din 252 este 20;
f)X supra 3 din X+1 este 4;
g)5 supra X+5 din 3*X+2 este 10
h)X+1 supra X+2 din 2*X+4


Răspuns :

1) [tex] \frac{x}{6}*72=60 [/tex]⇔[tex] \frac{72x}{6} =60[/tex]⇔72x=360⇔x=5
2)[tex] \frac{15}{x}*44=165 [/tex]⇔660=65x⇔x=4
c)[tex] \frac{11}{15}*x=88 [/tex]⇔11x=1320⇔x=120
d)[tex] \frac{x}{x+1}*12=8 [/tex]⇔12x=8x+8⇔4x=8⇔x=2
e)[tex] \frac{3x+1}{2x+5}*252=20 [/tex]⇔756x+252=40x+100⇔716x=-152⇔x=-0,21
f)[tex] \frac{x}{3}* x+1=4[/tex]⇔[tex] x^{2} [/tex]+x=12⇔[tex] x^{2} [/tex]+x-12=0⇔x1,2=⇔[tex] \frac{-1+/- \sqrt{1+48} }{2}= \frac{-1+/1 \sqrt{49} }{2} = \frac{-1+/-7}{2} [/tex]⇔x1=[tex] \frac{-1+7}{2}=3 [/tex] ⇔x2=[tex] \frac{-1-7}{2}=-4 [/tex]
g)[tex] \frac{5}{x+5} *(3x=2)=10[/tex]⇔[tex] \frac{15x+10}{x+5}=10 [/tex]⇒15x+10=10x+50⇒5x=40⇒x=8

Iar punctul "h)" nu am putut să-l fac deoarece nu mi-ai scris toate datele la acesta.

[tex]a). \frac{x}{6} \cdot 72=60 \Rightarrow \frac{\not72x }{\not6 } =60 \Rightarrow12x=60 \Rightarrow x= \frac{60}{12} \Rightarrowx=5 \\ \\ b). \frac{15}{x} \cdot 44=165 \Rightarrow 660=165x \Rightarrow x= \frac{660}{165} \Rightarrow x=4 \\ \\ c). \frac{11}{15} \cdot x=88 \Rightarrow \frac{11x}{15} =88 \Rightarrow 11x=88 \cdot15 \Rightarrow11x=1320 \Rightarrow x=120 \\ \\ d). \frac{x}{x+1} \cdot 12=8 \Rightarrow \frac{12x}{x+1} =8 \Rightarrow 12x=8x+8 \Rightarrow 4x=8 \Rightarrow x= 2[/tex]

[tex]e). \frac{3x+1}{2x+5} \cdot 252=20 \Rightarrow \frac{252(3x+1)}{2x+5} =20 \Rightarrow \frac{756x+252}{2x+5} =20 \Leftrightarrow \\ \\ \Leftrightarrow 756x+252=40x+100 \Rightarrow 756x-40x=100-252 \Leftrightarrow \\ \\ \Leftrightarrow 716x=-152 \Rightarrow x=- \frac{152}{716} \Rightarrow x=-0,21[/tex]

[tex]f). \frac{x}{3} \cdot x+1=4 \Rightarrow \frac{x^2+x}{3} =4 \Rightarrow x^2+x=12 \Rightarrow x^2+x-12=0 \\ \\ x^2+x-12=0 \\ \\ d=1^2-4 \cdot1 \cdot(-12)=1+48=49\ \textgreater \ 0 \\ \\ x_1= \frac{-1+ \sqrt{49} }{2} = \frac{-1+7}{2} = \frac{6}{2} =3 \\ \\ x_2= \frac{-1-7}{2} = \frac{-8}{2} =-4[/tex]

[tex]g). \frac{5}{x+5} \cdot 3x+2=10 \Rightarrow \frac{5(3x+2)}{x+5} =10 \Rightarrow \frac{15x+10}{x+5} =10 \Leftrightarrow \\ \\ \Leftrightarrow 15x+10=10x+50 \Rightarrow 15x-10x=50-10 \Leftrightarrow \\ \\ \Leftrightarrow5x=40 \Rightarrow x= \frac{40}{5} \Rightarrow x=8[/tex]