Tìm x biết
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
Thà tách ra từng câu hỏi chả phải hay hơn à –  tranquynhat2002 06-07-17 06:34 AM
quá dài.Nhìn muốn nản –  tranquynhat2002 06-07-17 06:34 AM
a làm dùm e ạ –  linho0ocute 05-07-17 07:48 AM
em cứ bình phương là được mà cái nào cần chuyển vế thì chuyển –  Vũ Minh Hoàng 05-07-17 07:47 AM
a làm đc câu nào thì làm hộ e ạ –  linho0ocute 05-07-17 07:39 AM
dài quá nhìn nản –  Vũ Minh Hoàng 05-07-17 07:33 AM
$b,\sqrt{x-2}(1-\sqrt{x})=0\Leftrightarrow x=1;x=2$
a) ĐK : $x\geqslant -1$
$\sqrt{x+1} +\sqrt{9(x+1)} =4$
$<=>4\sqrt{x+1}=4$
$<=>\sqrt{x+1} =1$
$<=>x=0$


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