Bài 1:
\(a,A=\sqrt{175}-5\sqrt{63}+2\sqrt{7}\\ =5\sqrt{7}-5\cdot3\sqrt{7}+2\sqrt{7}\\ =5\sqrt{7}-15\sqrt{7}+2\sqrt{7}\\ =-8\sqrt{7}\\ b,B=\left(2\sqrt{12}+6\sqrt{27}\right):\sqrt{3}\\ =\left(2\cdot2\sqrt{3}+6\cdot3\sqrt{3}\right):\sqrt{3}\\ =\left(4\sqrt{3}+18\sqrt{3}\right):\sqrt{3}\\ =22\sqrt{3}:\sqrt{3}\\ =22\\ c,C=\sqrt{\left(\sqrt{3}-2\right)^2}+\sqrt{5+2\sqrt{6}}\\ =\left|\sqrt{3}-2\right|+\sqrt{\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}\\ =2-\sqrt{3}+\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\\ =2-\sqrt{3}+\sqrt{3}+\sqrt{2}\\ =2+\sqrt{2}\\ d,D=\dfrac{\sqrt{6}-\sqrt{2}}{\sqrt{3}-1}-\sqrt{2}\\ =\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\sqrt{2}\\ =\sqrt{2}-\sqrt{2}\\ =0\)
2) a) \(\sqrt{2x+1}-2=1\)
\(\Leftrightarrow\sqrt{2x+1}=3\left(3\ge0\right)\)
\(\Leftrightarrow2x+1=9\)
\(\Leftrightarrow x=8\)
b) \(\sqrt{4x-20}+\sqrt{x-5}=6\)
\(\Leftrightarrow\sqrt{4\left(x-5\right)}+\sqrt{x-5}=6\)
\(\Leftrightarrow3\sqrt{x-5}=6\)
\(\Leftrightarrow\sqrt{x-5}=6\left(6\ge0\right)\)
\(\Leftrightarrow x-5=36\)
\(\Leftrightarrow x=41\)
c) \(\dfrac{3x-5}{\sqrt{x-4}}=\sqrt{x-4}\left(x>4\right)\)
\(\Leftrightarrow3x-5=x-4\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
d) \(\sqrt{x^2-2x+5}=x-2\)
\(\Leftrightarrow x^2-2x+5=\left(x-2\right)^2\left(x\ge2\right)\)
\(\Leftrightarrow x^2-2x+5=x^2-4x+4\)
\(\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\left(ktm\right)\Rightarrow x\in\varnothing\)
e) \(\sqrt{x^2-4x+4}+2=x\)
\(\Leftrightarrow\sqrt{\left(x-2\right)^2}=x-2\)
\(\Leftrightarrow\left|x-2\right|=x-2\)
\(\Leftrightarrow x-2=x-2\left(x\ge2\right)\)
\(\Leftrightarrow0x=x\Leftrightarrow x\in R\)
So với điều kiện ta được \(x\ge2\)
f) \(\sqrt{4x^2-9}=2\sqrt{2x+3}\)
\(\Leftrightarrow\sqrt{\left(2x+3\right)\left(2x-3\right)}-2\sqrt{2x+3}=0\)
\(\Leftrightarrow\sqrt{2x+3}\left(\sqrt{2x-3}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\left(x\ge-\dfrac{3}{2}\right)\\\sqrt{2x-3}=2\left(2\ge0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\left(tm\right)\\2x-3=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\left(tm\right)\\x=\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)
Bài 3:
a: BC=BH+CH=9+4=13(cm)
Xét ΔABC vuông tại A có AH là đường cao
nên \(\left\{{}\begin{matrix}AB^2=BH\cdot BC\\AC^2=CH\cdot CB\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}AB=\sqrt{9\cdot13}=3\sqrt{13}\left(cm\right)\\AC=\sqrt{4\cdot13}=2\sqrt{13}\left(cm\right)\end{matrix}\right.\)
Xét ΔABC vuông tại A có \(tanC=\dfrac{AB}{AC}=\dfrac{3\sqrt{13}}{2\sqrt{13}}=\dfrac{3}{2}\)
nên \(\widehat{C}\simeq56^0\)
ΔABC vuông tại A
=>\(\widehat{ABC}+\widehat{ACB}=90^0\)
=>\(\widehat{ABC}\simeq90^0-56^0=34^0\)
b: Ta có: \(\widehat{CAD}+\widehat{BAD}=\widehat{BAC}=90^0\)
\(\widehat{CDA}+\widehat{HAD}=90^0\)(ΔHAD vuông tại H)
mà \(\widehat{BAD}=\widehat{HAD}\)(AD là phân giác của góc BAH)
nên \(\widehat{CAD}=\widehat{CDA}\)
=>ΔCAD cân tại C
c: Xét ΔAHB có AD là phân giác
nên \(\dfrac{DH}{DB}=\dfrac{AH}{AB}\)
Xét ΔABC vuông tại A có AH là đường cao
nên \(AH\cdot BC=AB\cdot AC\)
=>\(\dfrac{AH}{AB}=\dfrac{AC}{BC}\)
=>\(\dfrac{AC}{BC}=\dfrac{DH}{DB}\)
=>\(AC\cdot DB=BC\cdot DH\)
Bài 1:
e: \(E=\left(2+\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}\right)\left(2+\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}\right)\)
\(=\left(2-\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}\right)\left(2+\dfrac{\sqrt{5}\left(3+\sqrt{5}\right)}{3+\sqrt{5}}\right)\)
\(=\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)\)
=4-5=-1
f: \(F=\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(11+\sqrt{6}\right)\)
\(=\left(\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{6-4}-\dfrac{12\left(3+\sqrt{6}\right)}{9-6}\right)\left(\sqrt{6}+11\right)\)
\(=\left[3\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)\right]\left(\sqrt{6}+11\right)\)
\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)=6-121=-115\)
