căn (3x+1) = x+2 căn ( x^2 -4x+4)=2x-5 căn((x-1+2 nhân căn (x-2)) + căn((x-1-2 nhân căn (x-2))=4 31/07/2021 Bởi Madelyn căn (3x+1) = x+2 căn ( x^2 -4x+4)=2x-5 căn((x-1+2 nhân căn (x-2)) + căn((x-1-2 nhân căn (x-2))=4
1) √3x+1=x+23x+1=x+2 (1)(1) ĐK:x≥−13ĐK:x≥−13 (1)⇔3x+1=(x+2)2(1)⇔3x+1=(x+2)2 ⇔3x+1=x2+4x+4⇔3x+1=x2+4x+4 ⇔x2+x+3=0⇔x2+x+3=0 (vô nghiệm) 2) √x2−4x+4=2x−5x2−4x+4=2x−5 (2)(2) ĐK:x≥52ĐK:x≥52 (2)⇔|x−2|=2x−5(2)⇔|x−2|=2x−5 ⇔[x−2=2x−5x−2=5−2x⇔[x−2=2x−5x−2=5−2x ⇔⎡⎢⎣x=3(nhận)x=73(loại)⇔[x=3(nhận)x=73(loại) Vậy phương trình có nghiệm x=3x=3 3) √x−1+2√x−2+√x−1−2√x−2=4x−1+2x−2+x−1−2x−2=4 (3)(3) ĐK:x≥2ĐK:x≥2 (3)⇔√x−2+2√x−2+1+√x−2−2√x−2+1=4(3)⇔x−2+2x−2+1+x−2−2x−2+1=4 ⇔|√x−2+1|+|√x−2−1|=4⇔|x−2+1|+|x−2−1|=4 ⇔√x−2+1+|√x−2−1|=4⇔x−2+1+|x−2−1|=4 ⇔√x−2+|√x−2−1|=3⇔x−2+|x−2−1|=3 tuananhgh@ Bình luận
1) $\sqrt{3x + 1} = x + 2$ $(1)$ $ĐK: \, x \ge -\dfrac{1}{3}$ $(1) \Leftrightarrow 3x + 1 = (x + 2)^2$ $\Leftrightarrow 3x + 1 = x^2 + 4x + 4$ $\Leftrightarrow x^2 + x + 3 = 0$ (vô nghiệm) 2) $\sqrt{x^2 – 4x + 4} = 2x – 5$ $(2)$ $ĐK: \, x \ge \dfrac{5}{2}$ $(2)\Leftrightarrow |x -2| = 2x – 5$ $\Leftrightarrow \left[\begin{array}{l}x – 2 = 2x – 5\\x – 2 = 5 – 2x\end{array}\right.$ $\Leftrightarrow \left[\begin{array}{l}x = 3 \quad (nhận)\\x = \dfrac{7}{3}\quad (loại)\end{array}\right.$ Vậy phương trình có nghiệm $x = 3$ 3) $\sqrt{x – 1 + 2\sqrt{x -2}} + \sqrt{x – 1 – 2\sqrt{x – 2}} = 4$ $(3)$ $ĐK: \, x \geq 2$ $(3) \Leftrightarrow \sqrt{x – 2 + 2\sqrt{x -2} + 1} + \sqrt{x – 2 – 2\sqrt{x – 2} +1} = 4$ $\Leftrightarrow |\sqrt{x -2} +1| + |\sqrt{x -2} -1| = 4$ $\Leftrightarrow \sqrt{x -2} +1 + |\sqrt{x -2} -1| = 4$ $\Leftrightarrow \sqrt{x -2}+ |\sqrt{x -2} -1| = 3$ $\Leftrightarrow \left[\begin{array}{l}\sqrt{x -2}+ \sqrt{x -2} -1 = 3\\\sqrt{x -2}+ 1 – \sqrt{x -2} = 3\end{array}\right.$ $\Leftrightarrow \sqrt{x – 2} = 2$ $\Leftrightarrow x = 6$ Vậy phương trình có nghiệm $x = 6$ Bình luận
1) √3x+1=x+23x+1=x+2 (1)(1)
ĐK:x≥−13ĐK:x≥−13
(1)⇔3x+1=(x+2)2(1)⇔3x+1=(x+2)2
⇔3x+1=x2+4x+4⇔3x+1=x2+4x+4
⇔x2+x+3=0⇔x2+x+3=0 (vô nghiệm)
2) √x2−4x+4=2x−5x2−4x+4=2x−5 (2)(2)
ĐK:x≥52ĐK:x≥52
(2)⇔|x−2|=2x−5(2)⇔|x−2|=2x−5
⇔[x−2=2x−5x−2=5−2x⇔[x−2=2x−5x−2=5−2x
⇔⎡⎢⎣x=3(nhận)x=73(loại)⇔[x=3(nhận)x=73(loại)
Vậy phương trình có nghiệm x=3x=3
3) √x−1+2√x−2+√x−1−2√x−2=4x−1+2x−2+x−1−2x−2=4 (3)(3)
ĐK:x≥2ĐK:x≥2
(3)⇔√x−2+2√x−2+1+√x−2−2√x−2+1=4(3)⇔x−2+2x−2+1+x−2−2x−2+1=4
⇔|√x−2+1|+|√x−2−1|=4⇔|x−2+1|+|x−2−1|=4
⇔√x−2+1+|√x−2−1|=4⇔x−2+1+|x−2−1|=4
⇔√x−2+|√x−2−1|=3⇔x−2+|x−2−1|=3
tuananhgh@
1) $\sqrt{3x + 1} = x + 2$ $(1)$
$ĐK: \, x \ge -\dfrac{1}{3}$
$(1) \Leftrightarrow 3x + 1 = (x + 2)^2$
$\Leftrightarrow 3x + 1 = x^2 + 4x + 4$
$\Leftrightarrow x^2 + x + 3 = 0$ (vô nghiệm)
2) $\sqrt{x^2 – 4x + 4} = 2x – 5$ $(2)$
$ĐK: \, x \ge \dfrac{5}{2}$
$(2)\Leftrightarrow |x -2| = 2x – 5$
$\Leftrightarrow \left[\begin{array}{l}x – 2 = 2x – 5\\x – 2 = 5 – 2x\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = 3 \quad (nhận)\\x = \dfrac{7}{3}\quad (loại)\end{array}\right.$
Vậy phương trình có nghiệm $x = 3$
3) $\sqrt{x – 1 + 2\sqrt{x -2}} + \sqrt{x – 1 – 2\sqrt{x – 2}} = 4$ $(3)$
$ĐK: \, x \geq 2$
$(3) \Leftrightarrow \sqrt{x – 2 + 2\sqrt{x -2} + 1} + \sqrt{x – 2 – 2\sqrt{x – 2} +1} = 4$
$\Leftrightarrow |\sqrt{x -2} +1| + |\sqrt{x -2} -1| = 4$
$\Leftrightarrow \sqrt{x -2} +1 + |\sqrt{x -2} -1| = 4$
$\Leftrightarrow \sqrt{x -2}+ |\sqrt{x -2} -1| = 3$
$\Leftrightarrow \left[\begin{array}{l}\sqrt{x -2}+ \sqrt{x -2} -1 = 3\\\sqrt{x -2}+ 1 – \sqrt{x -2} = 3\end{array}\right.$
$\Leftrightarrow \sqrt{x – 2} = 2$
$\Leftrightarrow x = 6$
Vậy phương trình có nghiệm $x = 6$