a)x-5 √x-6 ( vs x ≥0)
b) x- √x -12 vs x ≥0
c) 2x-7 √x+5 vs x ≥0
d) x- √xy + √y-1 vs x, y ≥0
e) x √x+1 vs x ≥0
h) x √x-8
a)x-5 √x-6 ( vs x ≥0) b) x- √x -12 vs x ≥0 c) 2x-7 √x+5 vs x ≥0 d) x- √xy + √y-1 vs x, y ≥0 e) x √x+1 vs x ≥0 h) x √x-8
By Kennedy
Đáp án:
Giải thích các bước giải:
Đáp án:
\(\begin{array}{l}
a.\left( {\sqrt x + 1} \right)\left( {\sqrt x – 6} \right)\\
b.\left( {\sqrt x – 12} \right)\left( {\sqrt x + 3} \right)
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.x – 5\sqrt x – 6 = x + \sqrt x – 6\sqrt x – 6\\
= \sqrt x \left( {\sqrt x + 1} \right) – 6\left( {\sqrt x + 1} \right)\\
= \left( {\sqrt x + 1} \right)\left( {\sqrt x – 6} \right)\\
b.x – \sqrt x – 12 = x – 4\sqrt x + 3\sqrt x – 12\\
= \sqrt x \left( {\sqrt x – 4} \right) + 3\left( {\sqrt x – 12} \right)\\
= \left( {\sqrt x – 12} \right)\left( {\sqrt x + 3} \right)\\
c.2x – 7\sqrt x + 5 = 2x – 2\sqrt x – 5\sqrt x + 5\\
= 2\sqrt x \left( {\sqrt x – 1} \right) – 5\left( {\sqrt x – 1} \right)\\
= \left( {\sqrt x – 1} \right)\left( {2\sqrt x – 5} \right)\\
e.x\sqrt x + 1 = \left( {\sqrt x + 1} \right)\left( {x – \sqrt x + 1} \right)\\
h.x\sqrt x – {2^3} = \left( {\sqrt x – 2} \right)\left( {x + 2\sqrt x + 4} \right)
\end{array}\)