Question

[Kim Khánh Linh]

Bài 72 (trang 40 SGK Toán 9 Tập 1)

Phân tích thành nhân tử (với các số  $x, y, a, b$ không âm và $a \geq b$)

a) $x y-y \sqrt{x}+\sqrt{x}-1$ ;                                  b) $\sqrt{a x}-\sqrt{b y}+\sqrt{b x}-\sqrt{a y}$ ;

c) $\sqrt{a+b}+\sqrt{a^{2}-b^{2}}$ ;                                    d) $12-\sqrt{x}-x$

Answer 1

[ღᏠᎮღ🆃🆄ấ🅽ঔ 🅽🅰🅼ঌ❄๖ۣۜ]

\(a,\left(\sqrt{8}-3.\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)

\(=\sqrt{8}.\sqrt{2}-3\sqrt{2}.\sqrt{2}+\sqrt{10}.\sqrt{2}-\sqrt{5}\)

\(=\sqrt{16}-3.2+\sqrt{20}-\sqrt{5}\)

\(=\sqrt{4^2}-6+\sqrt{2^2.5}-\sqrt{5}\)

\(=2-6+2\sqrt{5}-\sqrt{5}\)

\(=-2+\sqrt{5}\)

\(b,\)

\(0,2\sqrt{\left(-10^2\right).3}+2\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}\)

\(=0,2.\left|-10\right|.\sqrt{3}+2\left|\sqrt{3}-\sqrt{5}\right|\)

\(=0,2.10.\sqrt{3}+2\left(\sqrt{5}-\sqrt{3}\right)\)

\(=2\sqrt{3}+2\sqrt{5}-2\sqrt{3}\)

\(=2\sqrt{5}\)

Answer 2

[Nguyễn Thị Như Hoa]

a) $xy-y\sqrt{x}+\sqrt{x}-1$

$=y\cdot \sqrt{x}\cdot \sqrt{x}-y\sqrt{x}+\sqrt{x}-1$

$=y\sqrt{x}(\sqrt{x}-1)+(\sqrt{x}-1)$

$=(\sqrt{x}-1)(y\sqrt{x}+1)$.

b) $\sqrt{ax}-\sqrt{by}+\sqrt{bx}-\sqrt{ay}$

$=(\sqrt{ax}+\sqrt{bx})-(\sqrt{ay}+\sqrt{by})$

$=(\sqrt{a}\cdot \sqrt{x}+\sqrt{b}\cdot \sqrt{x})-(\sqrt{a}\cdot \sqrt{y}+\sqrt{b}\cdot \sqrt{y})$

$=\sqrt{x}(\sqrt{a}+\sqrt{b})-\sqrt{y}(\sqrt{a}+\sqrt{b})$

$=(\sqrt{a}+\sqrt{b})(\sqrt{x}-\sqrt{y})$.

c) $\sqrt{a+b}+\sqrt{a^2-b^2}$

$=\sqrt{a+b}+\sqrt{(a+b)(a-b)}$

$=\sqrt{a+b}+\sqrt{a+b}\cdot \sqrt{a-b}$

$=\sqrt{a+b}(1+\sqrt{a-b})$.

d) $12-\sqrt{x}-x$

$=12-4\sqrt{x}+3\sqrt{x}-x$

$=4(3-\sqrt{x})+\sqrt{x}(3-\sqrt{x})$

$=(3-\sqrt{x})(4+\sqrt{x})$.