Đề thi tuyển sinh vào lớp 10 môn Toán chuyên - Bình Phước - Năm 2019-2020
View attachment 115800
Câu 1
a.
\[\begin{align}
& (\frac{\sqrt{a}+1}{\sqrt{ab}+1}+\frac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}-1): (\frac{\sqrt{a}+1}{\sqrt{ab}+1}-\frac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}+1) \\
& DKXD:a;b\ge 0 \\
& =\frac{(\sqrt{a}+1)(\sqrt{ab}-1)+(\sqrt{ab}+\sqrt{a})(\sqrt{ab}+1)-ab+1}{ab-1}: \frac{(\sqrt{a}+1)(\sqrt{ab}-1)-(\sqrt{ab}+\sqrt{a})(\sqrt{ab}+1)+ab-1}{ab-1} \\
& =\frac{a\sqrt{b}+\sqrt{ab}-\sqrt{a}-1+ab+a\sqrt{b}+\sqrt{a}+\sqrt{ab}-ab+1}{ab-1}.\frac{ab-1}{a\sqrt{b}-\sqrt{a}+\sqrt{ab}-1-ab-a\sqrt{b}-\sqrt{a}-\sqrt{ab}+ab-1} \\
& =\frac{2\sqrt{ab}+2a\sqrt{b}}{ab-1}.\frac{ab-1}{-2\sqrt{a}-2} \\
& =\frac{2\sqrt{ab}(\sqrt{a}+1)}{-2(\sqrt{a}+1)} \\
& =-\sqrt{ab} \\
\end{align}\]
b.
\[\begin{align}
& {{x}^{5}}-3{{x}^{4}}-3{{x}^{3}}+6{{x}^{2}}-20x+2023 \\
& =({{x}^{5}}-4{{x}^{4}}+{{x}^{3}})+({{x}^{4}}-4{{x}^{3}}+{{x}^{2}})+5({{x}^{2}}-4x+1)+2018 \\
& =({{x}^{2}}-4x+1)({{x}^{3}}+{{x}^{2}}+5)+2018 \\
& {{x}^{2}}-4x+1={{(2-\sqrt{3})}^{2}}-4(2-\sqrt{3})+1=0 \\
& \Rightarrow H=2018 \\
\end{align}\]
Câu 3
a.
\[\begin{align}
& \sqrt{x+1}+\sqrt{6x-14}={{x}^{2}}-5 \\
& DKXD:x\ge \frac{7}{3} \\
& Pt\Leftrightarrow \sqrt{x+1}-2+\sqrt{6x-14}-2={{x}^{2}}-9 \\
& \Leftrightarrow \frac{x-3}{\sqrt{x+1}+2}+\frac{6x-18}{\sqrt{6x-14}+2}=(x-3)(x+3) \\
& \Leftrightarrow (x-3)\left[ \frac{1}{\sqrt{x+1}+2}+\frac{6}{\sqrt{6x-14}+2}-x-3 \right]=0 \\
\end{align}\]
\[\Leftrightarrow x=3\vee \frac{1}{\sqrt{x+1}+2}+\frac{6}{\sqrt{6x-14}+2}=x+3\]
Nếu \[\frac{1}{\sqrt{x+1}+2}+\frac{6}{\sqrt{6x-14}+2}=x+3\]
Có
\[\begin{align}
& \frac{1}{\sqrt{x+1}+2}<1\forall x\ge \frac{7}{3} \\
& \frac{6}{\sqrt{6x-14}+2}<3\forall x\ge \frac{7}{3} \\
& \Rightarrow \frac{1}{\sqrt{x+1}+2}+\frac{6}{\sqrt{6x-14}+2}<4\forall x\ge \frac{7}{3} \\
\end{align}\]
\[x+3>4\forall x\ge \frac{7}{3}\]
PTVN
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