Question

Va roog am nevoie...neaparata!!!

Answer 1

$\displaystyle 11).\left(2x+1\right)^2-\left(2x-1\right)^2=\\=\left(2x\right)^2+2 \cdot 2x \cdot 1+1^2-\left[\left(2x\right)^2-2 \cdot 2x \cdot 1+1^2\right]= \\ =4x^2+4x+1-\left(4x^2-4x+1\right)=4x^2+4x+1-4x^2+4x-1=8x \\ 12). \left( \sqrt{x+1} +x\right)^2+ \left( \sqrt{x+1} -x\right)^2= \\ =\left( \sqrt{x+1} \right)^2+2\cdot \sqrt{x+1} \cdot x+x^2+\left( \sqrt{x+1} \right)^2-2 \cdot \sqrt{x+1} \cdot x+x^2= \\ =x+1+2x \sqrt{x+1} +x^2+x+1-2x \sqrt{x+1} +x^2=2x^2+2x+2$
$\displaystyle 13).\left(x^2+x+1\right)^2-\left(x^2+x\right)^2= \\ =\left(x^2\right)^2+x^2+1^2+2 \cdot x^2 \cdot x+2 \cdot x \cdot 1+2 \cdot x^2 \cdot 1-\left(x^2+x\right)^2=\\=x^4+x^2+1+2x^3+2x+2x^2-\left[\left(x^2\right)^2+2 \cdot x^2 \cdot x+x^2\right]= \\ =x^4+3x^2+1+2x^3+2x-\left(x^4+2x^3+x^2\right)= \\ =x^4+3x^2+1+2x^3+2x-x^4-2x^3-x^2=2x^2+2x+1$
$\displaystyle 14).\left(x^2-x+2\right)^2+x^2 \left(-x^2+2x-5\right)= \\ =\left(x^2\right)^2+x^2+2^2-2 \cdot x^2 \cdot x+2 \cdot x^2 \cdot 2-2 \cdot x \cdot 2-x^4+2x^3-5x^2= \\ =x^4+x^2+4-2x^3+4x^2-4x-x^4+2x^3-5x^2=-4x+4$
$\displaystyle 15).E(x)=\left(x+1\right)^2-|x+1| \\ E(0)=\left(0+1\right)^2-|0+1| \\ E(0)=1^2-|1| \\ E(0)=1-1 \\ E(0)=0 \\ E(-1)=(-1+1)^2-|-1+1| \\ E(-1)=0^2-|0| \\ E(-1)=0$
$\displaystyle 16).E(x)=\left(1-x\right)^{1001}+\left(x-1\right)^{1001} \\ E(2)=\left(1-2\right)^{1001}+\left(2-1\right)^{1001} \\ E(2)=\left(-1\right)^{1001}+1^{1001} \\ E(2)=-1+1 \\ E(2)=0$
$\displystyle 17).E(x)=x^4+3x^2-5~~~~~~~~~~~~~~~~~~~~~x= \sqrt{5} \\ E\left( \sqrt{5} \right)=\left( \sqrt{5} \right)^4+3 \cdot \left( \sqrt{5} \right)^2-5 \\ E \left ( \sqrt{5} \right)=\left (5^{ \frac{1}{2} }\right)^4+3 \cdot \left(5^{ \frac{1}{2} }\right)^2-5 \\ E\left( \sqrt{5} \right)=5^{ \frac{1}{2} \cdot 4}+3 \cdot 5^{ \frac{1}{2} \cdot 2}-5 \\ E\left( \sqrt{5} \right)=5^2+3 \cdot 5^1-5 \\ E\left( \sqrt{5} \right)=25+3 \cdot 5-5 \\ E\left( \sqrt{5} \right)=25+15-5 \\ E\left( \sqrt{5} \right)=35$
$\displaystyle 18).E(x)=x^2+x+ \sqrt{2} \left(2 \sqrt{3} +1 \right)~~~~~~~~~~~~~~~~x= \sqrt{3} - \sqrt{2} \\ E \left( \sqrt{3} - \sqrt{2} \right)=\left( \sqrt{3} - \sqrt{2} \right)^2+ \sqrt{3} - \sqrt{2} + \sqrt{2} \left(2 \sqrt{3} +1 \right) \\ E \left( \sqrt{3} - \sqrt{2} \right)= \left ( \sqrt{3} \right)^2-2\cdot \sqrt{3} \cdot \sqrt{2}+\left(\sqrt{2} \right)^2+ \sqrt{3} - \sqrt{2}+2 \sqrt{6} + \sqrt{2}$
$\displaystyle E \left( \sqrt{3}-\sqrt{2} \right)=3-2 \sqrt{6}+2+\sqrt{3}-\sqrt{2} +2 \sqrt{6}+ \sqrt{2} \\ E \left( \sqrt{3}-\sqrt{2} \right)=\sqrt{3} +5$
$\displaystyle 19).(x+1)^2=x^2+1-Fals \\ (x+1)^2=x^2+2x+1~~~~~~~~~~~~~~~~~(x+1)^2 \not = x^2+1 \\ 20).(x+1)^2=x^2+1 -Adevarat \\ x^2+2x+1=x^2+1 \\ x^2-x^2+2x=1-1 \\ 2x=0 \\ x=0$
$\displaystyle 21).a-b=4;~a^2-b^2=24;~a+b=? \\ a^2-b^2=24 \Rightarrow (a-b)(a+b)=24 \Rightarrow 4(a+b)=24 \Rightarrow a+b= \frac{24}{4} \Rightarrow \\ \Rightarrow a+b=6 \\ 22).a+b=4;~a \cdot b=2;~ \frac{1}{a} + \frac{1}{b} =? \\ \frac{1}{a} + \frac{1}{b} = \frac{a+b}{ab} = \frac{4}{2} =2 \Rightarrow \frac{1}{a} + \frac{1}{b} =2$
$\displaystyle 23).a+b=4;~a \cdot b=2;~a^2+b^2=? \\ a+b=4 \Rightarrow (a+b)^2=4^2\Rightarrow a^2+2ab+b^2=16 \Rightarrow \\\Rightarrow a^2+2 \cdot 2+b^2=16 \Rightarrow a^2+4+b^2=16 \Rightarrow a^2+b^2=16-4 \Rightarrow \\ \Rightarrow a^2+b^2=12$
$\displaystyle 24).x- \frac{1}{x} =2;~x^2+ \frac{1}{x^2} =? \\ x- \frac{1}{x} =2 \Rightarrow \left(x- \frac{1}{x} \right)^2=2^2 \Rightarrow x^2-2 \cdot x \cdot \frac{1}{x} + \left( \frac{1}{x} \right)^2=4 \Rightarrow \\ \Rightarrow x^2-2+ \frac{1}{x^2 } =4 \Rightarrow x^2+ \frac{1}{x^2} =4+2 \Rightarrow x^2+ \frac{1}{x^2} =6$
$\displaystyle 25).a-b=1;~a^2+b^2-2ab+2a-2b+1=? \\ a^2+b^2-2ab+2a-2b+1=(a-b)^2+2(a-b)+1= \\ =1^2+2 \cdot 1+1=1+2+1=4 \\ a^2+b^2-2ab+2a-2b+1=4 \\ 26).a+b=4;~ac+ad+bc+bd=4 \sqrt{3} ;~c+d=? \\ ac+ad+bc+bd=4 \sqrt{3} \Rightarrow a(c+d)+b(c+d)=4 \sqrt{3} \Rightarrow \\ \Rightarrow (c+d)(a+b)=4 \sqrt{3} \Rightarrow 4(c+d)=4 \sqrt{3} \Rightarrow c+d= \frac{4 \sqrt{3} }{4} \Rightarrow \\ \Rightarrow c+d= \sqrt{3}$
$\displaystyle 27).x^2-5x=x(x-5)\\28).3x^{10}+9x^8=3x^8(x^2+3)\\29).x(x+1)-x-1=x(x+1)-(x+1)=(x+1)(x-1) \\ 30).x^2(x+1)+x(x+1)^2=x(x+1)(2x+1) \\ 31).xy-x-y+1=x(y-1)-(y-1)=(y-1)(x-1) \\ 32).9xy+3x-3y-1=3x(3y+1)-(3y+1)=(3y+1)(3x-1) \\ 33).25x^2-49=(5x+7)(5x-7) \\ 34).(x-2)^2-4x^2=-(x+2)(3x-2)$
$\displaystyle 35).x^2+6x+9-y^2=(x+3)^2-y^2=(x-y+3)(x+y+3) \\ 36).x^2+7x+6=x^2+x+6x+6=x(x+1)+6(x+1)= \\ =(x+1)(x+6) \\ 37).x^2-5x+6=x^2-2x-3x+6=x(x-2)-3(x-2)= \\ =(x-2)(x-3) \\ 38).x^2+2x-3=x^2-x+3x-3=x(x-1)+3(x-1)= \\ =(x-1)(x+3) \\ 39).2x^2+x-1=2x^2+2x-x-1=2x(x+1)-(x+1)= \\ =(x+1)(2x-1) \\ 40).x^3+x^2-16x-16=x^2(x+1)-16(x-1)= \\ =(x+1)(x^2-16)=(x+1)(x+4)(x-4)$
$\displaystyle 41).F(x)= \frac{3-x^2}{x^2+2} \\ F \left( \sqrt{2} \right)= \frac{3-\left( \sqrt{2} \right)^2}{\left( \sqrt{2} \right)^2+2} \\ F \left( \sqrt{2} \right)= \frac{3-2}{2+2} \\ F \left( \sqrt{2} \right)= \frac{1}{4}$