Question

daca a^2 -b^2 =40 si a-b=4 atunci a+b este egal cu...?
scris ca patratul unei sume x^2 +2√6x +6 devine...?
daca A=7+2x si B=2x-3 sa se calculeze A^2 + B^2
daca a^2 - b^2=21 si a+b=7. calculati :(a-b)+(a-b)^2 + (a-b)^3
daca x+ 1supra x =2 , culaculati x^2 + 1supra x^2

Answer 1

${a}^{2} - {b}^{2} = 40$

$a - b = 4$

${a}^{2} - {b}^{2} = (a - b)(a + b)$

$(a - b)(a + b) = 40$

$4(a + b) = 40 \: | \div 4$

$a + b = 10$

${x}^{2} + 2 \sqrt{6} x + 6 = {(x + \sqrt{6} )}^{2}$

$a = 7 + 2x$

$b = 2x - 3$

${a}^{2} + {b}^{2} = {(7 + 2x)}^{2} + {(2x - 3)}^{2}$

$= 49 + 28x + 4 {x}^{2} + 4 {x}^{2} - 12x + 9$

$= 8 {x}^{2} + 16x + 58$

${a}^{2} - {b}^{2} = 21$

$a + b = 7$

${a}^{2} - {b}^{2} = (a - b)(a + b)$

$(a - b) \times 7 = 21$

$7(a - b) = 21 \: | \div 7$

$a - b = 3$

$a - b + {(a - b)}^{2} + {(a - b)}^{3}$

$3 + {3}^{2} + {3}^{3} = 3 + 9 + 27 = 12 + 27 = 39$

$x+\frac{ 1}{x} = 2\:|\times\:x$

${x}^{2} + 1 = 2x$

${x}^{2} - 2x +1= 0$

${x}^{2}-x-x+1=0$

$x(x-1)-(x-1)=0$

$(x-1)(x-1)=0$

$x-1=0=>x=1$

${x}^{2} +\frac{ 1 }{ {x}^{2} } = {1}^{2} + \frac{1}{ {1}^{2} } = 1+\frac{1}{1}=1+1=2$