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Question

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a) $\sqrt{x + 1}$ + $\sqrt{9x + 9}$ = 4

b) $\sqrt{x - 2}$ - $\sqrt{x^{2} - 2x}$ = 0

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

d) $\sqrt{3x - 1}$ = $\sqrt{\sqrt{3}}$

e) ( 6 - 2$\sqrt{5}$ ) $\times$ $x^{}$ = $\sqrt{5}$ - 1

f) $b^{2}$$x^{2}$ - $b^{2}$ = $x^{2}$ ( b > 1 )

g) a$x^{}$ + 3 = $\sqrt{3}$ $\times$ $x^{}$ + $a^{2}$ (a khác $\sqrt{3}$)

h) $\sqrt{4x - 20}$ + $\sqrt{x - 5}$ - $\frac{1}{3}$ $\times$ $\sqrt{9x - 45}$ = 4

❄⊰๖ۣۜNgốc๖ۣۜ ⊱ ❄ · Answer 1 · 7/6/2017 12:14:52 AM

a) ĐK : $x\geqslant -1$

$\sqrt{x+1} +\sqrt{9(x+1)} =4$

$<=>4\sqrt{x+1}=4$

$<=>\sqrt{x+1} =1$

$<=>x=0$

Sửa 06-07-17 12:14 AM

❄⊰๖ۣۜNgốc๖ۣۜ ⊱ ❄
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score 0 · Answer 2

$b,\sqrt{x-2}(1-\sqrt{x})=0\Leftrightarrow x=1;x=2$

2 Đáp án