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Toán Lớp 9: Bài 2: Tìm x b) $\sqrt{x-3}$ $+2x – 7$ = $0$ c) $x-2.\sqrt{x-1}$ = $\sqrt{3-2 √2}$

Home/ Toán Lớp 9: Bài 2: Tìm x b) $\sqrt{x-3}$ $+2x – 7$ = $0$ c) $x-2.\sqrt{x-1}$ = $\sqrt{3-2 √2}$

Toán Lớp 9: Bài 2: Tìm x b) $\sqrt{x-3}$ $+2x – 7$ = $0$ c) $x-2.\sqrt{x-1}$ = $\sqrt{3-2 √2}$

Dương Tháng Mười Hai 27, 2022 Toán học 1 Comment 0 views

Toán Lớp 9: Bài 2: Tìm x
b) $\sqrt{x-3}$ $+2x – 7$ = $0$
c) $x-2.\sqrt{x-1}$ = $\sqrt{3-2 √2}$

Comments ( 1 )

catlantuong
27 Tháng Mười Hai, 2022 at 2:43 chiều

Reply

$a)\sqrt{x-3}+2x-7=0\\x\in[3;3,5]\\ \sqrt{x-3}+2x-7=0\\⇔\sqrt{x-3}=7-2x\\⇔x-3=(7-2x)^2\\⇔4x^2-29x+52=0$

$⇔$\(\left[ \begin{array}{l}x=4(ktm)\\x=3,25(tm)\end{array} \right.\)

$b)x-2\sqrt{x-1}=\sqrt{3-2\sqrt{2}}\\⇔2\sqrt{x-1}=x-\sqrt{3-2\sqrt{2}}\\⇔2\sqrt{x-1}=x-\sqrt{2}+1\\⇔4(x-1)=x^2+2+1-2\sqrt{2}+2x-2\sqrt{2}x\\⇔x^2-4.(2-2\sqrt{2})x+7-2\sqrt{2}\\∆’=4(2-2\sqrt{2})^2-7+2\sqrt{2}=48-32\sqrt{2}-7+2\sqrt{2}=41-30\sqrt{2}\\∆'<0\\\Rightarrow\mathbb{Phương-trình-vô-nghiệm}$

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