giải bài trên Bàii11.. Rút gọn các biểu thức sau,$a)~A=\frac2{a-2}.\sqrt{2a^8(a^2-4a+4)}(a
e2)$ $b)~B=4\sqrt{25u}-\frac{15}2\sqrt{\frac{16u}9}-\frac2u\sqrt{\frac{169u^3}4}(u0)$,$c)~C=5\sqrt{4x}-3\sqrt{\frac{100x}9}-\frac4x\sqrt{\frac{x^3}4}(x0)$ $d)~D=-\sqrt{36b}-\frac13\sqrt{54b}+\frac15\sqrt{150b}(b\geq0)$,$e)~E=\frac13\sqrt{9+6v+v^2}+\frac{4v}3+5(v\leq-3)$ $f)~F=\frac t2+\frac32\sqrt{4-4t+t^2}-2(t\leq2)$,Lời giải

Question

giải bài trên Bàii11.. Rút gọn các biểu thức sau,$a)~A=\frac2{a-2}.\sqrt{2a^8(a^2-4a+4)}(a
e2)$ $b)~B=4\sqrt{25u}-\frac{15}2\sqrt{\frac{16u}9}-\frac2u\sqrt{\frac{169u^3}4}(u>0)$,$c)~C=5\sqrt{4x}-3\sqrt{\frac{100x}9}-\frac4x\sqrt{\frac{x^3}4}(x>0)$ $d)~D=-\sqrt{36b}-\frac13\sqrt{54b}+\frac15\sqrt{150b}(b\geq0)$,$e)~E=\frac13\sqrt{9+6v+v^2}+\frac{4v}3+5(v\leq-3)$ $f)~F=\frac t2+\frac32\sqrt{4-4t+t^2}-2(t\leq2)$,Lời giải

Accepted Answer

Bàii11.
a) Ta có:
$$ A = \frac{2}{a-2} \cdot \sqrt{2a^8(a^2 - 4a + 4)} $$
$$ = \frac{2}{a-2} \cdot \sqrt{2a^8(a-2)^2} $$
$$ = \frac{2}{a-2} \cdot |a-2| \cdot \sqrt{2a^8} $$
$$ = \frac{2}{a-2} \cdot (a-2) \cdot a^4 \sqrt{2} $$
$$ = 2a^4 \sqrt{2} $$

b) Ta có:
$$ B = 4\sqrt{25u} - \frac{15}{2}\sqrt{\frac{16u}{9}} - \frac{2}{u}\sqrt{\frac{169u^3}{4}} $$
$$ = 4 \cdot 5\sqrt{u} - \frac{15}{2} \cdot \frac{4\sqrt{u}}{3} - \frac{2}{u} \cdot \frac{13u\sqrt{u}}{2} $$
$$ = 20\sqrt{u} - 10\sqrt{u} - 13\sqrt{u} $$
$$ = -3\sqrt{u} $$

c) Ta có:
$$ C = 5\sqrt{4x} - 3\sqrt{\frac{100x}{9}} - \frac{4}{x}\sqrt{\frac{x^3}{4}} $$
$$ = 5 \cdot 2\sqrt{x} - 3 \cdot \frac{10\sqrt{x}}{3} - \frac{4}{x} \cdot \frac{x\sqrt{x}}{2} $$
$$ = 10\sqrt{x} - 10\sqrt{x} - 2\sqrt{x} $$
$$ = -2\sqrt{x} $$

d) Ta có:
$$ D = -\sqrt{36b} - \frac{1}{3}\sqrt{54b} + \frac{1}{5}\sqrt{150b} $$
$$ = -6\sqrt{b} - \frac{1}{3} \cdot 3\sqrt{6b} + \frac{1}{5} \cdot 5\sqrt{6b} $$
$$ = -6\sqrt{b} - \sqrt{6b} + \sqrt{6b} $$
$$ = -6\sqrt{b} $$

e) Ta có:
$$ E = \frac{1}{3}\sqrt{9 + 6v + v^2} + \frac{4v}{3} + 5 $$
$$ = \frac{1}{3}\sqrt{(3 + v)^2} + \frac{4v}{3} + 5 $$
$$ = \frac{1}{3}|3 + v| + \frac{4v}{3} + 5 $$
$$ = \frac{1}{3}(-3 - v) + \frac{4v}{3} + 5 $$
$$ = -1 - \frac{v}{3} + \frac{4v}{3} + 5 $$
$$ = 4 + v $$

f) Ta có:
$$ F = \frac{t}{2} + \frac{3}{2}\sqrt{4 - 4t + t^2} - 2 $$
$$ = \frac{t}{2} + \frac{3}{2}\sqrt{(2 - t)^2} - 2 $$
$$ = \frac{t}{2} + \frac{3}{2}|2 - t| - 2 $$
$$ = \frac{t}{2} + \frac{3}{2}(2 - t) - 2 $$
$$ = \frac{t}{2} + 3 - \frac{3t}{2} - 2 $$
$$ = 1 - t $$