Hình tự kẻ bạn nhé.

a)Xét \(\Delta CHE\)và \(\Delta CDF\),có :

\(\hept{\begin{cases}\widehat{CHE}=\widehat{CDF}\left(=90^0\right)\\\widehat{DCF}:chung\end{cases}}\)\(\Rightarrow\Delta CHE~\Delta CDF\left(g.g\right)\)

\(\Rightarrow\frac{CH}{CD}=\frac{CE}{CF}\)

\(\Rightarrow CH.CF=CD.CE\)

Bạn tử kẻ hình nhé .

a)\(\Delta ABD~\Delta ACE\left(g.g\right)\)

\(\Rightarrow\frac{AB}{AC}=\frac{AD}{AE}\)

\(\Rightarrow\Delta ADE~\Delta ABC\left(c.g.c\right)\)

\(\Rightarrow\frac{S_{ADE}}{S_{ABC}}=\left(\frac{AD}{AB}\right)^2=cos^2\widehat{BAC}\)

\(\Rightarrow S_{ADE}=S_{ABC}.cos^2\widehat{BAC}\)

b)Ta có : \(S_{BCDE}=S_{ABC}-S_{ADE}=S_{ABC}-S_{ABC}.cos^2\widehat{BAC}=S_{ABC}\left(1-cos^2\widehat{BAC}\right)=S_{ABC}.sin^2\widehat{BAC}\)

Áp dụng định lý Py - ta - Go vào tam giác ABC vuông tại A có :

AC² = BC² - AB²

AC² = $\sqrt{5^2-3^2}=3\left(AC>0\right)$

Ta có : $S_{ABC}=\frac{1}{2}AB.AC$

Mà : $S_{ABC}=\frac{1}{2}AH.BC$

⇒ $\frac{1}{2}AB.AC=\frac{1}{2}AH.BC$

⇔ AH = $\frac{AB.AC}{BC}=\frac{3.4}{5}=2,4\left(cm\right)$

a) Áp dụng pi ta go ta có : AB² = AH² + BH² = 16² + 25² = 881 

=> AB = $\sqrt{881}$

Lại có : BH.HC =  AH²

<=> HC.25 = 16²

<=> HC.25 = 256

<=> HC = 256 : 25 = 10,24

Ta có : BC = HC + BH = 10,24 + 25 = 35,24 

Áp dụng bi ta go : AC² = AH² + HC² = 16² + 10,24² = 360,8576

=> AC = $\sqrt{\text{360,8576}}$

đk: \(x\ge\frac{1}{2}\)

\(pt\Leftrightarrow\left[4x^2-4x\sqrt{x+3}+x+3\right]+\left[1-2\sqrt{2x-1}+2x-1\right]=0\)

\(\Leftrightarrow\left(2x-\sqrt{x+3}\right)^2+\left(1-\sqrt{2x-1}\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{x+3}=2x\\\sqrt{2x-1}=1\end{cases}\Leftrightarrow x=1}\)

đk: \(x>-3\)

pt \(\Leftrightarrow\left(2-\sqrt{\frac{1}{x+3}}\right)+\left(2-\sqrt{\frac{5}{x+4}}\right)=0\)

\(\Leftrightarrow\frac{4-\frac{1}{x+3}}{2+\sqrt{\frac{1}{x+3}}}+\frac{4-\frac{5}{x+4}}{2+\sqrt{\frac{5}{x+4}}}=0\)

\(\Leftrightarrow\frac{4x+11}{\left(x+3\right)\left(2+\sqrt{\frac{1}{x+3}}\right)}+\frac{4x+11}{\left(x+4\right)\left(2+\sqrt{\frac{5}{x+4}}\right)}=0\)

Vì x>-3 \(\Rightarrow\frac{1}{\left(x+3\right)\left(2+\sqrt{\frac{1}{x+3}}\right)}+\frac{1}{\left(x+4\right)\left(2+\sqrt{\frac{5}{x+4}}\right)}=0\)

=> 4x+11=0 => x=\(\frac{-11}{4}\left(tm\right)\)

Đk: x \(\ge\)-3/2

Ta có: \(\sqrt{2x+3}=\frac{8x^3+4x}{2x+5}\)

<=> \(\left(2x+5\right)\sqrt{2x+3}=8x^3+4x\)

<=> \(\left(2x+3\right)\sqrt{2x+3}+2\sqrt{2x+3}=8x^3+4x\)

Đặt \(\sqrt{2x+3}=a\)(a \(\ge\)0)

Do đó: \(a^3+2a=8x^3+4x\)

<=> \(\left(a-2x\right)\left(a^2+2ax+4x^2\right)+2\left(a-2x\right)=0\)

<=> \(\left(a-2x\right)\left(a^2+2ax+4x^2+2\right)=0\)

<=> \(a=2x\)(vì \(a^2+2ax+4x^2+2=\left(a+x\right)^2+3x^2+2>0\))

<=> \(\sqrt{2x+3}=2x\)(đk: x \(\ge\)0)

<=> \(4x^2=2x+3\)

<=> \(4x^2-2x-3=0\)

\(\Delta'=\left(-1\right)^2+4.3=13>0\)=> pt có 2 nghiệm pb

\(x_1=\frac{1+\sqrt{13}}{4}\); \(x_2=\frac{1-\sqrt{13}}{4}\)

đk: \(\frac{-3}{2}\le x\le12\)

pt \(\Leftrightarrow\left(x^2-2x\sqrt{2x+3}+2x+3\right)+\left(9-6\sqrt{12-x}+12-x\right)=0\)

\(\Leftrightarrow\left(x-\sqrt{2x+3}\right)^2+\left(3-\sqrt{12-x}\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}x-\sqrt{2x+3}=0\left(1\right)\\3-\sqrt{12-x}=0\left(2\right)\end{cases}}\)

pt(1)\(\Leftrightarrow x=\sqrt{2x+3}\Leftrightarrow\hept{\begin{cases}x\ge0\\x^2-2x-3=0\end{cases}}\) \(\Leftrightarrow x\ge0\) và \(\orbr{\begin{cases}x=-1\\x=3\end{cases}}\) \(\Leftrightarrow x=3\)

pt(2) \(\Leftrightarrow\sqrt{12-x}=3\Leftrightarrow12-x=9\Leftrightarrow x=3\)

Vậy pt có nghiệm: x=3 (tm)