giải giúp mình câu pt vô tỉ nay với.

A

a012

pt nay ban nham dc no la x=2. vay nen bay gio minh phai them bot hang tu thui!
pt <=> x2-4-(x+2)can(x-1)+4=x-2
<=> (x-2)(x+1)- (x+2)can(x-1)+4+x-x=0
<=>(x-2) (x+1) - (x+2)can(x-1) +2+x+2-x=0
<=>x(x-2) - (x+2)[can(x-1) +1]=0
nhan lien hop:
pt <=>x(x-2) - (x+2)(x-2)/[can(x-1)-1] =0
<=> x-2=0 hoac la x-(x+2)/can(x-1)-1]=0
pt thu 2 ban quy dong,khu mau roi binh phuong cung co no la x=2.
 
P

pe_lun_hp

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

đkxđ: $x ≥ 1$

Đặt $\sqrt{x-1} = t ≥ 0$

$(1) \leftrightarrow x^2 - (x+2)t = 2x - x + 2$

$\leftrightarrow x^2 - (x+2)t = 2(t^2 + 1) - x + 2$

$\leftrightarrow 2t^2 + (x+2)t - x^2 - x = 0$

Coi như pt bậc 2 ẩn t.

$\Delta = (3x + 2)^2$

$\left[\begin{matrix}t_1 = \dfrac{-(x+2) + (3x +2)}{4} = \dfrac{x}{2}\\ t_2 = \dfrac{-(x+2) - (3x + 2)}{4} = -x - 1 \end{matrix}\right.$

$\rightarrow \left[\begin{matrix}\sqrt{x-1} = \dfrac{x}{2} \\ \sqrt{x-1} = -x - 1 \end{matrix}\right.$

Đơn giản rồi ạ
 
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