Question

Hộ mik bài hình mik cần gấp

Answer 1

Bài 3: 

a: \(\sqrt{3x-5}=7\)

\(\Leftrightarrow3x-5=49\)

\(\Leftrightarrow3x=54\)

hay x=18

Answer 2

Bài 3: 

b. \(\sqrt{100-200x}-\dfrac{3}{4}\sqrt{16-32x}+2\sqrt{1-2x}=18\)

<=> \(\sqrt{100\left(1-2x\right)}-\dfrac{3}{4}\sqrt{16\left(1-2x\right)}+2\sqrt{1-2x}=18\)

<=> \(10\sqrt{1-2x}-3\sqrt{1-2x}+2\sqrt{1-2x}=18\)

<=> \(\left(10-3+2\right)\sqrt{1-2x}=18\)

<=> \(11\sqrt{1-2x}=18\)

<=> \(\sqrt{1-2x}=\dfrac{18}{11}\)

<=> 1 - 2x = \(\left(\dfrac{18}{11}\right)^2\)

<=> 1 - 2x = \(\dfrac{324}{121}\)

<=> \(1-\dfrac{324}{121}=2x\)

<=> \(2x=\dfrac{-203}{121}\)

<=> \(x=\dfrac{-203}{242}\)

Answer 3

Bài 4:

Áp dụng hệ thức lượng trong tam giác vuông:

$AH^2=BH.CH$

$3^2=y.5\Rightarrow y=\frac{9}{5}=1,8$ 

Áp dụng định lý Pitago:

$x=\sqrt{3^2+y^2}=\sqrt{9+1,8^2}=\frac{3\sqrt{34}}{5}$ 

$\tan \alpha = \frac{BH}{AH}=\frac{y}{3}=\frac{1,8}{3}=\frac{3}{5}$

$\Rightarrow \alpha \approx 31^0$

b.

$AH\perp BC, CD\perp BC$ nên $AH\parallel CD$

Áp dụng định lý Talet:

$\frac{AH}{CD}=\frac{BH}{BC}$

$\Rightarrow DC=\frac{AH.BC}{BH}=\frac{3(y+5)}{y}=\frac{3(1,8+5)}{1,8}=\frac{34}{3}$ 

$\tan \widehat{HDC}=\frac{HC}{CD}=\frac{5}{\frac{34}{3}}=\frac{15}{34}$

$\Rightarrow \widehat{HDC}=23,8^0$

$\widehat{AHD}=\widehat{HDC}=23,8^0$ (hai góc so le trong)

Answer 4

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