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Dau coroană vă rog repede!

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Salut! Pe ce clasa&@(@₩!)@¥(+^-#**×(×(#¥÷₩#(*=*$*=*$£=¥¥÷¥÷¥÷¥÷¥¥÷¥÷¥÷¥÷¥÷¥÷¥÷¥÷¥÷¥÷¥÷¥÷¥÷¥÷¥÷¥÷¥¥÷¥÷¥÷¥÷¥÷¥÷¥¥÷¥÷¥÷¥÷¥÷¥¥÷¥÷¥÷¥¥¥¥¥¥¥¥=¥=¥=¥=¥=¥_¥_¥_¥_¥¥______¥_¥_¥_¥_¥_____¥_¥%¥%_____¥_¥_¥______¥_¥_¥______¥_¥_¥¥___
a) [tex] (x+5)^{2} = x^{2} + 10x + 5^{2} = x^{2} + 10x + 25 [/tex]
b)[tex] (3x+7)^{2} = (3x)^{2} + 42x + 49 [/tex]
c)[tex] (3,2+5x)^{2} = 3,2^{2} + 32x + (5x)^{2} = 10,24 + 32x + 25 x^{2} [/tex]
d)[tex] (2x-3)^{2}= (2x)^{2} - 12x+ 3^{2} =4 x^{2} -12x+9 [/tex]
f)[tex] (2,5x-8)^{2}= (2,5x)^{2} - 40x+ 8^{2}= 6,25x-40x+64[/tex]
g)[tex](x-3)(x+3)= x^{2} - 3^{2} = x^{2} -9[/tex]
h)[tex]( \sqrt{2} x- \sqrt{3} )( \sqrt{2}x+ \sqrt{3} )= ( \sqrt{2}x )^{2}- \sqrt{3} ^{2} =2 x^{2} -3[/tex]
i)[tex] (x-1 / x)^{2} = x^{2} -2 + (1 / x)^{2}= x^{2} -2+1/ x^{2} [/tex]
j)[tex](x-1)(x+1)( x^{2} +1)( x^{4}+1 )- x^{8}+1=( x^{2} -1)( x^{2} +1)( x^{4}+1 )- x^{8}+1[/tex][tex]=( x^{4}-1 )( x^{4}+1 )- x^{8}+1= x^{8}-1- x^{8}+1=0 [/tex]
k)[tex] ( x^{2} +x+2)^{2}= x^{4}+ x^{2} +4+2 x^{3}+4x+4 x^{2} [/tex]
l)[tex] (x+2)^{2} + (x-3)^{2}-(x+1)(x-1)= x^{2} +4x+4+ x^{2} -6x+9-[tex]+1= x^{2} -2x+14[/tex] x^{2} [/tex]
m)[tex] (6a-5)^{2} -(5a+4)(5a-4)- (4a+3)^{2}= [/tex][tex]36 a^{2}-60a+25-25 a^{2} +16-16 a^{2}-24a-9=-5 a^{2}-84a +32[/tex]
n)[tex] (3a+b)^{2}- (2a-5b)^{2}-(a-2b)(a+2b)=[/tex][tex]9 a^{2}+6ab+ b^{2}-4 a^{2}+20ab-25 b^{2}- a^{2}+4 b^{2}= [/tex][tex]4a^{2} +26ab-20 b^{2} [/tex]