Question

самостоятельная с математикой
1) √1+x√x2+24=x+1
2) √22-x-√10-x=2
3) √5x-1-√3x-2=√x-1
4) √x-1-2√x-2+√x+7-6√x-2=6
5) √2x2+5x+2-√x2+x-2=√3x+6

Answer 1

1) $\sqrt{1+x\sqrt{x^2+24} } =x+1$

$1+x\sqrt{x^2+24} =x^2+2x+1$

$x\sqrt{x^2+24} =x^2+2x$

$x^2*(x^2+24)=x^4+4x^3+4x^2$

$x^4+24x^2=x^4+4x^3+4x^2$

$24x^2=4x^3+4x^2$

$24x^2-4x^3-4x^2=0$

$20x^2-4x^3=0$

$4x^2*(5-x)=0$

$x^2*(5-x)=0$

$x^2=0$

$5-x=0$

$x=0$

$x=5$

$\sqrt{1+0\sqrt{0^2+24} } =0+1$

$\sqrt{1+5\sqrt{5^2+24} } =5+1$

$1=1$

$6=6$

x₁ = 0, x₂ =5

2) $\sqrt{22-x} - \sqrt{10-x} =2$

$\sqrt{22-x} =2+\sqrt{10-x}$

$22-x=4+4\sqrt{10-x} +10-x$

$22=4+4\sqrt{10-x} +10$

$22=14+4\sqrt{10-x}$

$-4\sqrt{10-x} =14-22$

$-4\sqrt{10-x} =-8$

$\sqrt{10-x} =2$

$10-x=4$

$-x=4-10$

$-x=-6$

$x=6$

$\sqrt{22-6} - \sqrt{10-6} =2$

$2=2$

x=6

3) $\sqrt{5x-1} -\sqrt{3x-2} = \sqrt{x-1}$

$5x-1-2\sqrt{(5x-1)*(3x-2)} + 3x-2=x-1$

$5x-2\sqrt{15^2-10x-3x+2} +3x-2=x$

$8x-2\sqrt{15x^2-13x+2} -2=x$

$-2\sqrt{15x^2-13x+2} =x-8x+2$

$-2\sqrt{15x^2-13x+2} =-7x+2$

$4(15x^2-13x+2)=4-28x+49x^2$

$60x^2-52x+8=4-28x+49x^2$

$60x^2-52x+8-4+28x-49x^2=0$

$11x^2-24x+4=0$

$11x^2-2x-22x+4=0$

$x*(11x-2)-2(11x-2)=0$

$(11x-2)*(x-2)=0$

$11x-2=0$

$x-2=0$

$x=\frac{2}{11}$

$x=2$

$\sqrt{5*\frac{2}{11}-1 } - \sqrt{3*\frac{2}{11}-2 } =\sqrt{\frac{2}{11}-1 }$

$\sqrt{5*2-1} -\sqrt{3*2-2} =\sqrt{2-1}$

$\sqrt{-\frac{1}{11} }-\sqrt{3*\frac{2}{11} -2} }=\sqrt{\frac{2}{11}-1 }$

$1=1$

$x\neq \frac{2}{11}$

$x=2$

x=2