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The problem of quadratic equation with one variable in mathematics
①5√8-2√32+√50

=5*3√2-2*4√2+5√2

=√2( 15-8+5)

= 12√2

②√6-√3/2-√2/3

=√6-√6/2-√6/3

=√6/6

③(√45+√27)-(√4/3+√ 125)

=(3√5+3√3)-(2√3/3+5√5)

=-2√5+7√5/3

④(√4a-√50b)-2(√b/2+√9a)

=(2√a-5√2b)-2(√2b/2+3√a)

=-4√a-6√2b

⑤√4x*(√3x/2-√x/6)

=2√x(√6x/2-√6x/6)

=2√x*(√6x/3)

=2/3*|x|*√6

⑥(x√y-y√x)÷√xy

=x√y÷√xy-y√x÷√xy

=√x-√y

⑦(3√7+2√3)(2√3-3√7)

=(2√3)^2-(3√7)^2

= 12-63

=-5 1

⑧(√32-3√3)(4√2+√27)

=(4√2-3√3)(4√2+3√3)

=(4√2)^2-(3√3)^2

=32-27

=5

⑨(3√6-√4)2

=(3√6)^2-2*3√6*√4+(√4)^2

=54- 12√6+4

=58- 12√6

⑩( 1+√2-√3)( 1-√2+√3)

=[ 1+(√2-√3)][ 1-(√2-√3)]

= 1-(√2-√3)^2

= 1-(2+3+2√6)

=-4-2√6

√3 X- 1=√2 X to find X {√5 X-3√ Y= 1} {√3 X-√5 Y=2} Note: X is not in the root sign √ (1/2x) 2+10. 0, if x = 7/(6x)

It's too difficult! ! ! ! ! ! ! !