Question

putin ajutorr aicii vrg

Answer 1

II. 2

a)

$\sqrt{3-2x}=x\\3-2x=x^2\\-x^2-2x+3=0\\x=1 (adevarat),\:x=-3(fals)\\x=1$

b)

$\sqrt[3]{2x-1}=3\\2x-1=27\\2x=26\\x=14$

c)

$\sqrt{x+2}=3\\x+2=9\\x=7$

d)

$\sqrt[3]{x+2\sqrt{2}}=\sqrt{2}\\x+2\sqrt{2}=2\sqrt{2}\\x=0$

e)

$\sqrt{x+1}=x-1\\x+1=x^2-2x+1\\-x^2+3x=0\\x(-x+3)=0\\x=0 (fals),x=3(adevarat)\\x=3$

f)

$\sqrt[3]{x^3-7}=x-1\\x^3-7=x^3-3x^2+3x-1\\3x^2-3x-6=0\\x=-1(adevarat),\:x=2(adevarat)$

g)

$\sqrt{x+1}=5-x\\x+1=25-10x+x^2\\-x^2+11x-24=0\\x=8(fals),\:x=3(adevarat)\\x=3$

h)

$\sqrt[3]{x^3+7}=x+1\\x^3+7=x^3+3x^2+3x+1\\-3x^2-3x+6=0\\x=1(adevarat),\:x=-2(adevarat)$

i)

$3\sqrt{x-4}=x-8\\9x-36=x^2-16x+64\\-x^2+25x-100=0\\x_{1,\:2}=\frac{-\left(-25\right)\pm \:15}{2}\\x=20(adevarat),\:x=5(fals)\\x=20$

j)

$\sqrt{x^2-4}=\sqrt{5}\\x^2-4=5\\x^2=9\\x=3,\:x=-3$

k)

$\sqrt{x^2-4x+3}=2\sqrt{2}\\x^2-4x+3=8\\x^2-4x-5=0\\x=5,\:x=-1$

l)

$\sqrt{x^2-3x+2}=1-x\\x^2-3x+2=1-2x+x^2\\-3x+2=1-2x\\-x=-1\\x=1$

m)

$\sqrt{20+21x-x^2}=x+5\\20+21x-x^2=x^2+10x+25\\-2x^2+11x-5=0\\x_{1,\:2}=\frac{-11\pm \sqrt{11^2-4\left(-2\right)\left(-5\right)}}{2\left(-2\right)}\\x=\frac{1}{2},\:x=5$

n)

$3x-\sqrt{2x-1}=34\\2x-1=1156-204x+9x^2\\9x^2-206x+1157=0\\x_{1,\:2}=\frac{-\left(-206\right)\pm \sqrt{\left(-206\right)^2-4\cdot \:9\cdot \:1157}}{2\cdot \:9}\\x=13(adevarat),\:x=\frac{89}{9}(fals)\\x=13$

o)

$2x+\sqrt{x^2+\frac{9}{4}}=\frac{3}{2}\\4x+2\sqrt{x^2+\frac{9}{4}}=3\\4x^2+9=9-24x+16x^2\\16x^2-24x=4x^2\\12x^2-24x=0\\x_{1,\:2}=\frac{-\left(-24\right)\pm \sqrt{\left(-24\right)^2-4\cdot \:12\cdot \:0}}{2\cdot \:12}\\x=2(fals),\:x=0(adevarat)\\x=0$

p)

$\sqrt{2x^2-4x-4}=x-2\\2x^2-4x-4=x^2-4x+4\\2x^2=x^2+8\\x^2=8\\x=\sqrt{8},\:x=-\sqrt{8}\\x=2\sqrt{2}(adevarat),\:x=-2\sqrt{2}(fals)\\x=2\sqrt{2}$

q)

$\sqrt{7-\sqrt{x-3}}=2\\-\sqrt{x-3}=-3\\x-3=9\\x=12$