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cum descompun in factori? 
a) (3x+1)² - ( x+3 )² = 
b) (3x+5)²-(x+1)² = 
c) (x+√2)² - 8 =
d) (2x + √3)² =
e) (√2x-1)² -32x² = 
 f) A la a sasea - b la a sasea


Răspuns :

a)=3[tex] x^{2} [/tex]+1-[tex] x^{2} [/tex]+6x+9=2[tex] x^{2} +6x+10[/tex]
b)=(3[tex] x^{2} )+2*3x*5+5 ^{2} -x ^{2} +2*x*1+1 ^{2} [/tex]=9[tex] x^{2} [/tex]+30x+25-[tex] x^{2} +2x+1=8 x^{2} +32x+26[/tex]
c)=[tex] x^{2} +2*x* \sqrt{2}+ (\sqrt{2}) ^{2} -8= x^{2} +2 \sqrt{2}x+2-8= x^{2} +2 \sqrt{2}x-6[/tex]
d)=(2[tex] x)^{2} [/tex]+2*2x*[tex] \sqrt{3}+( \sqrt{3})^{2} =4x ^{2}+4 \sqrt{3}x+3[/tex]
e)=([tex] \sqrt{2} ) ^{2} -2* \sqrt{2} *1+1 ^{2} -32 x^{2} =[/tex]2-[tex]2 \sqrt{2} +1-32 x^{2} =3-2 \sqrt{2} -32 x^{2} [/tex]
a) [tex] (3x+1)^{2} - (x+3)^{2} = \\ =3* x^{2} +6x+1- x^{2} -6x-9= \\ 2* x^{2} -8[/tex]
b) [tex] (3x+5)^{2} - (x+1)^{2} = \\ =9 x^{2} +30x+25- x^{2} -2x-1= \\ =8 x^{2} +28x+24[/tex]
c)[tex]( x+ \sqrt{2} )^{2} -8= \\ = x^{2} +2 \sqrt{2} x+2-8= \\ = x^{2} +2 \sqrt{2}x-6 [/tex]
d)[tex](2 x+ \sqrt{3} )^{2} = \\= 4 x^{2} +4 \sqrt{3} x+3[/tex]
e)[tex] ( \sqrt{2} x+ \sqrt{3} )^{2} -32 x^{2} = \\ =2 x^{2} +2 \sqrt{6} x+3-32 x^{2} = \\ =-30 x^{2} +2 \sqrt{6} x+3[/tex]
f) [tex] a^{6} - b^{6} = (a^{2} )^{3} -( b^{2}) ^{3} = \\ =( a^{2} - b^{2})*( a^{4} +a^{2}*b^{2}+ b^{4} ) = \\ =(a+b)*(a-b)*(a^{4}+a^{2}*b^{2}+b^{4})[/tex]