Bài II
Câu 1:
1: ĐKXĐ: x>=-2
\(\sqrt{9x+18}+\sqrt{x+2}-\dfrac{1}{2}\cdot\sqrt{4x+8}=6\)
=>\(3\sqrt{x+2}+\sqrt{x+2}-\dfrac{1}{2}\cdot2\sqrt{x+2}=6\)
=>\(3\sqrt{x+2}=6\)
=>\(\sqrt{x+2}=2\)
=>x+2=4
=>x=2(nhận)
2: ĐKXĐ: x>=0
\(\dfrac{x+9\sqrt{x}}{\sqrt{x}+1}=5\)
=>\(x+9\sqrt{x}=5\left(\sqrt{x}+1\right)=5\sqrt{x}+5\)
=>\(x+9\sqrt{x}-5\sqrt{x}-5=0\)
=>\(x+4\sqrt{x}-5=0\)
=>\(\left(\sqrt{x}+5\right)\left(\sqrt{x}-1\right)=0\)
mà \(\sqrt{x}+5>=5>0\forall x\) thỏa mãn ĐKXĐ
nên \(\sqrt{x}-1=0\)
=>\(\sqrt{x}=1\)
=>x=1(nhận)
Câu 2:
Gọi CA là chiều cao của thang, CB là chiều dài của thang
Theo đề, ta có: AB\(\perp\)AC tại A; \(\widehat{B}=53^0\); BC=6,5(m)
Xét ΔABC vuông tại A có \(sinB=\dfrac{AC}{BC}\)
=>\(AC=6,5\cdot sin53\simeq5,19\left(m\right)\)
Bài I:
1: Khi x=4 thì \(B=\dfrac{\sqrt{4}}{\sqrt{4}-4}=\dfrac{2}{2-5}=\dfrac{2}{-3}=-\dfrac{2}{3}\)
2: \(A=\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{x+9\sqrt{x}}{x-9}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{x+9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}+3\right)-x-9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{2x+6\sqrt{x}-x-9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+3}\)
3: \(P=A:B=\dfrac{\sqrt{x}}{\sqrt{x}+3}:\dfrac{\sqrt{x}}{\sqrt{x}-5}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}-5}{\sqrt{x}}=\dfrac{\sqrt{x}-5}{\sqrt{x}+3}\)
\(P-1=\dfrac{\sqrt{x}-5-\sqrt{x}-3}{\sqrt{x}+3}=\dfrac{-8}{\sqrt{x}+3}< 0\)
=>P<1
