d,
Áp dụng định lí Pitago, ta có:
[tex]BD^2 = AD^2 + AB^2 = 100cm[/tex]
=> BD = 10cm
Xét 2 tam giác BMC và CBD có:
[tex]\widehat{BCM} = \widehat{BDC}[/tex]
[tex]\widehat{BCD} = \widehat{CBM} = 90^o[/tex]
=> [tex]\Delta BDC \sim \Delta MCB[/tex]
=> [tex]\frac{BM}{BC} = \frac{BC}{CD}[/tex]
=> [tex]BM = \frac{BC^2}{CD} = \frac{32}{3}cm[/tex]
=> [tex]AM = \frac{50}{3}cm[/tex]
Vì [tex]\Delta ABD \sim \Delta ANM[/tex]
=> [tex]\frac{AD}{BD} =\frac{AM}{MN}[/tex]
=> [tex]MN = \frac{AM.BD}{AD} = \frac{\frac{32}{3}.10}{8} = \frac{125}{6}cm[/tex]
5.
ĐKXĐ: [tex]x \geq -8[/tex]
Ta có:
[tex]3\sqrt{3}(x^2 + 4x + 2) - \sqrt{x + 8} = 0[/tex]
<=> [tex]9(x^2 + 4x + 2) - \sqrt{3x + 24} = 0[/tex]
<=> [tex](9x^2 + 36x + 42) + (3x + 6)\sqrt{3x + 24} - (3x + 7)\sqrt{3x + 24} - (3x + 24) = 0[/tex]
<=> [tex](3x + 6)(3x + 7) + (3x + 6)\sqrt{3x + 24} - (3x + 7)\sqrt{3x + 24} - (3x + 24) = 0[/tex]
<=> [tex](3x + 6)(3x + 7 + \sqrt{3x + 24}) - \sqrt{3x + 24}(3x + 7 + \sqrt{3x + 24}) = 0[/tex]
<=> [tex](3x + 6 -\sqrt{3x + 24})(3x + 7 + \sqrt{3x + 24}) = 0[/tex]
TH1:
[tex]3x + 6 - \sqrt{3x + 24} = 0 \Leftrightarrow 3x + 6 = \sqrt{3x + 24}[/tex]
Bình phương 2 vế, ta có:
[tex]9x^2 + 36x + 36 = 3x + 24[/tex]
<=> [tex]9x^2 + 33x + 12 = 0[/tex]
=> [tex]x = \frac{- 11 + \sqrt{73}}{6}[/tex] hoặc [tex]x = \frac{- 11 - \sqrt{73}}{6}[/tex] (Thỏa mãn)
TH2:
[tex]3x + 7 + \sqrt{3x + 24} = 0 \Leftrightarrow 3x + 7 = - \sqrt{3x + 24} [/tex]
Bình phương 2 vế, ta có:
[tex]9x^2 + 42x + 49 = 3x + 24[/tex]
<=> [tex]9x^2 + 39x + 25 = 0[/tex]
<=> [tex]x = \frac{- 13 + \sqrt{69}}{6}[/tex] hoặc [tex]x = \frac{- 13 - \sqrt{69}}{6}[/tex] (Thỏa mãn)
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
