Giải thích các bước giải:
Bài 1:
ĐKXĐ: $x\ge 0$
Ta có:
$\sqrt{x}+\sqrt{3x+2}=2$
$\to \sqrt{3x+2}=2-\sqrt{x}$
$\to (\sqrt{3x+2})^2=(2-\sqrt{x})^2$
$\to 3x+2=4-4\sqrt{x}+x$
$\to 2x+4\sqrt{x}-2=0$
$\to x+2\sqrt{x}-1=0$
$\to (x+2\sqrt{x}+1)-2=0$
$\to (\sqrt{x}+1)^2-2=0$
$\to (\sqrt{x}+1)^2=2$
$\to \sqrt{x}+1=\sqrt{2}$ vì $\sqrt{x}+1>0$
$\to \sqrt{x}=-1+\sqrt{2}$
$\to x=(-1+\sqrt{2})^2$
Bài 2:
Ta có:
$\widehat{BAH}=90^o-\widehat{HAC}=\widehat{ACH}=\widehat{ACB}$
$\to\cos\widehat{BAH}=\cos\widehat{ACB}=\dfrac{AC}{BC}=\dfrac45$
$\sin\widehat{BAH}=\sin\widehat{ACB}=\dfrac{AB}{BC}=\dfrac35$
$\to \cos\widehat{BAH}+3\sin\widehat{BAH}=\dfrac{13}{5}$
