Ôn thi vào 10

\(B=\dfrac{1}{\sqrt{x}-2}-\dfrac{1}{\sqrt{x}+2}+\dfrac{\sqrt{9x^2}}{\sqrt{3x}}\)

\(=\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{\left(3x\right)^2}}{3x}\)

\(=\dfrac{\sqrt{x}+2-\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{3x}{3x}\)

\(=\dfrac{4}{x-4}+1\)

\(=\dfrac{4+x-4}{x-4}\)

\(=\dfrac{x}{x-4}\)

\(=\dfrac{x-0}{x-4}\Rightarrow a=0;b=4\)

\(\Rightarrow a-b=0-4=-4\)

\(tan30^o=\dfrac{AB}{AC}\Leftrightarrow AC=\dfrac{AB}{tan30^o}=\dfrac{28}{\dfrac{\sqrt[]{3}}{3}}=28\sqrt[]{3}\left(cm\right)\)

Nên câu 1) sai

\(BC^2=AB^2+AC^2=28^2+28^2.3=28^2.4=28^2.2^2\)

\(\Leftrightarrow BC=28.2=56\left(cm\right)\)

Nên câu 2) sai

\(\widehat{B}+\widehat{C}=90^o\Leftrightarrow\widehat{B}=90^o-30^o=60^o\)

Nên câu 3) đúng

Rút gọn \(M=\dfrac{\sqrt[]{x}-5}{\sqrt[]{x}+2}\)

\(\Leftrightarrow M=1-\dfrac{7}{\sqrt[]{x}+2}\)

\(x\ge0\Leftrightarrow\sqrt[]{x}\ge0\)

\(\Leftrightarrow\sqrt[]{x}+2\ge2\)

\(\Leftrightarrow\dfrac{7}{\sqrt[]{x}+2}\le\dfrac{7}{2}\)

\(\Leftrightarrow-\dfrac{7}{\sqrt[]{x}+2}\ge-\dfrac{7}{2}\)

\(\Leftrightarrow M=1-\dfrac{7}{\sqrt[]{x}+2}\ge1-\dfrac{7}{2}=-\dfrac{5}{2}\)

Vậy \(Min\left(M\right)=-\dfrac{5}{2}\)

Vậy chọn câu a

\(A=\dfrac{x+5\sqrt[]{x}}{25-x}\left(x\ge0;x\ne25\right)\)

\(A\ge0\Leftrightarrow A=\dfrac{x+5\sqrt[]{x}}{25-x}\ge0\)

\(\Leftrightarrow\dfrac{\sqrt[]{x}\left(\sqrt[]{x}+5\right)}{\left(5-\sqrt[]{x}\right)\left(5+\sqrt[]{x}\right)}\ge0\)

\(\Leftrightarrow\dfrac{\sqrt[]{x}}{\left(5-\sqrt[]{x}\right)}\ge0\)

\(\Leftrightarrow0\le x< 25\)

Vậy \(A\ge0\Leftrightarrow0\le x< 25\)

\(A=\dfrac{y-5\sqrt{y}}{\sqrt{y}-5}+\dfrac{10+2\sqrt{x}-5\sqrt{y}-\sqrt{xy}}{5+\sqrt{y}}\)

\(=\dfrac{\sqrt{y}\left(\sqrt{y}-5\right)}{\sqrt{y}-5}+\dfrac{2\left(5+\sqrt{x}\right)-\sqrt{y}\left(5+\sqrt{x}\right)}{5+\sqrt{x}}\)

\(=\sqrt{y}+\dfrac{\left(2-\sqrt{y}\right)\left(5+\sqrt{x}\right)}{5+\sqrt{x}}\)

\(=\sqrt{y}+2-\sqrt{y}\)

\(=2\)

\(\Rightarrow A+x< 5\Leftrightarrow2+x< 5\Leftrightarrow x< 3\)

Nguyên lớn nhất thỏa mãn là: 2 

3: Đặt \(P=A:B\)

\(=\dfrac{x+2\sqrt{x}}{x-1}:\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+2}\)

\(=\dfrac{x}{\sqrt{x}-1}\)

Để P>1 thì P-1>0

=>\(\dfrac{x-\sqrt{x}+1}{\sqrt{x}-1}>0\)

=>\(\sqrt{x}-1>0\)

=>\(\sqrt{x}>1\)

=>x>1

3) \(\dfrac{A}{B}>1\)

\(\Leftrightarrow\dfrac{x+2\sqrt{x}}{x-1}:\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-1\right)}>1\)

\(\Leftrightarrow\dfrac{x+2\sqrt{x}}{x-1}.\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+2}>1\)

\(\Leftrightarrow\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+2}>1\)

\(\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}+1}.\dfrac{\sqrt{x}}{1}>1\)

\(\Leftrightarrow\dfrac{x}{\sqrt{x}+1}-1>0\)

\(\Leftrightarrow\dfrac{x}{\sqrt{x}+1}-\dfrac{\sqrt{x}-1}{\sqrt{x}-1}>0\)

\(\Leftrightarrow x-\sqrt{x}+1>0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)^2>0\)

\(\Leftrightarrow\sqrt{x}-1>0\)

\(\Leftrightarrow x>1\)

a.

Do AB, AC là các tiếp tuyến \(\Rightarrow\widehat{ABO}=\widehat{ACO}=90^0\)

\(\Rightarrow\) B và C cùng nhìn AO dưới 1 góc vuông nên ABOC nội tiếp đường tròn đường kính OA

b.

M là trung điểm dây DE \(\Rightarrow OM\perp DE\)

\(\Rightarrow\widehat{OMA}=90^0\)

\(\Rightarrow M\) thuộc đường tròn đường kính OA

Theo t/c hai tiếp tuyến cắt nhau: \(AB=AC\)

\(\Rightarrow\widehat{BMA}=\widehat{AMC}\) (hai góc nt chắn 2 cung bằng nhau của đường tròn đường kính OA)

\(\Rightarrow MA\) là phân giác của \(\widehat{BMC}\)