Đề thi học kì 1 Toán 9 Kết nối tri thức - Đề số 3

a) Với \(x \ge 0,x \ne 1\) ta có:

\(A = \left( {\frac{1}{{\sqrt x {\rm{\;}} + 1}} - \frac{{2\sqrt x {\rm{\;}} - 2}}{{x\sqrt x {\rm{\;}} - \sqrt x {\rm{\;}} + x - 1}}} \right):\left( {\frac{1}{{\sqrt x {\rm{\;}} - 1}} - \frac{2}{{x - 1}}} \right)\)\(A = \left( {\frac{1}{{\sqrt x {\rm{\;}} + 1}} - \frac{{2\sqrt x {\rm{\;}} - 2}}{{\left( {x - 1} \right)\left( {\sqrt x {\rm{\;}} + 1} \right)}}} \right):\left( {\frac{1}{{\sqrt x {\rm{\;}} - 1}} - \frac{2}{{\left( {\sqrt x {\rm{\;}} - 1} \right)\left( {\sqrt x {\rm{\;}} + 1} \right)}}} \right)\)

\(A = \frac{{x - 1 - 2\sqrt x {\rm{\;}} + 2}}{{\left( {x - 1} \right)\left( {\sqrt x {\rm{\;}} + 1} \right)}}:\frac{{\sqrt x {\rm{\;}} + 1 - 2}}{{\left( {\sqrt x {\rm{\;}} - 1} \right)\left( {\sqrt x {\rm{\;}} + 1} \right)}}\)

\(A = \frac{{x - 2\sqrt x {\rm{\;}} + 1}}{{\left( {x - 1} \right)\left( {\sqrt x {\rm{\;}} + 1} \right)}}:\frac{{\sqrt x {\rm{\;}} - 1}}{{\left( {\sqrt x {\rm{ \;}} - 1} \right)\left( {\sqrt x {\rm{\;}} + 1} \right)}}\)

\(A = \frac{{{{\left( {\sqrt x {\rm{\;}} - 1} \right)}^2}}}{{\left( {x - 1} \right)\left( {\sqrt x {\rm{\;}} + 1} \right)}}.\left( {\sqrt x {\rm{\;}} + 1} \right)\)

\(A = \frac{{{{\left( {\sqrt x {\rm{\;}} - 1} \right)}^2}}}{{\left( {\sqrt x {\rm{\;}} - 1} \right)\left( {\sqrt x {\rm{\;}} + 1} \right)}}\)

\(A = \frac{{\sqrt x {\rm{\;}} - 1}}{{\sqrt x {\rm{\;}} + 1}}\).

b) Ta có \(A = \frac{{\sqrt x {\rm{\;}} - 1}}{{\sqrt x {\rm{\;}} + 1}} = \frac{{\sqrt x {\rm{\;}} + 1 - 2}}{{\sqrt x {\rm{\;}} + 1}} = 1 - \frac{2}{{\sqrt x {\rm{\;}} + 1}},{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \left( {x \ge 0} \right).\)

Đặt \(B = \sqrt x {\rm{\;}} + 1\), để A nguyên khi x nguyên thì B là ước nguyên của 2.

Vì \(x \ge 0\) nên \(B > 0{\mkern 1mu} {\mkern 1mu} \), suy ra B là ước nguyên dương của 2.

Ư\(\left( 2 \right) = \left\{ {1;2} \right\}\)

TH1: \(\sqrt x {\rm{\;}} + 1 = 1\) suy ra \(x = 0{\mkern 1mu} {\mkern 1mu} \left( {tm} \right)\)

TH2: \(\sqrt x {\rm{\;}} + 1 = 2\) suy ra \(x = 1{\mkern 1mu} {\mkern 1mu} \left( {ktm} \right)\)

Vậy \(x = 0\) thì A nguyên.

c)  Ta có \(A = \frac{{\sqrt x {\rm{\;}} - 1}}{{\sqrt x {\rm{\;}} + 1}} = 1 - \frac{2}{{\sqrt x {\rm{\;}} + 1}}\).

Vì \(\sqrt x {\rm{\;}} + 1 \ge 1{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \left( {do{\mkern 1mu} {\mkern 1mu} \sqrt x {\rm{\;}} \ge 0} \right)\) nên  \(\frac{2}{{\sqrt x {\rm{\;}} + 1}} \le \frac{2}{1}\)

Suy ra \( - \frac{2}{{\sqrt x {\rm{\;}} + 1}} \ge {\rm{\;}} - 2\)

Do đó \(1 - \frac{2}{{\sqrt x {\rm{\;}} + 1}} \ge {\rm{\;}} - 1\) hay \(A \ge {\rm{\;}} - 1\).

Dấu “=” xảy ra khi \(x = 0.\)

Vậy \(\min A = {\rm{\;}} - 1\) khi \(x = 0\).