a)
\(4a^2cos^260^o+2ab.cos^2180^o+\dfrac{4}{3}cos^230^o\)
\(=4a^2.\left(\dfrac{1}{2}\right)^2+2ab.\left(-1\right)^2+\dfrac{4}{3}.\left(\dfrac{\sqrt{3}}{2}\right)^2\)
\(=4a^2.\dfrac{1}{4}+2ab+\dfrac{4}{3}.\dfrac{3}{4}\)
\(=a^2+2ab+1\).
b)
\(\left(asin90^o+btan45^o\right)\left(acos0^o+bcos180^o\right)\)
\(=\left(a+b\right)\left(a-b\right)=a^2-b^2\).

b) \(\dfrac{6\cdot9-2\cdot17}{63\cdot3-119}\)

\(=\dfrac{2\left(3\cdot9-17\right)}{7\cdot\left(3\cdot9-17\right)}\)

\(=\dfrac{2}{7}\)

Ta có: \(\dfrac{3\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}+1\right)}\)

\(=\dfrac{3\left(\sqrt{5}-1\right)}{6+2\sqrt{5}}\)

\(=\dfrac{3\left(\sqrt{5}-1\right)\left(6-2\sqrt{5}\right)}{\left(6-2\sqrt{5}\right)\left(6+2\sqrt{5}\right)}\)

\(=\dfrac{3\left(6\sqrt{5}-10-6+2\sqrt{5}\right)}{16}\)

\(=\dfrac{3\left(8\sqrt{5}-16\right)}{16}\)

\(=\dfrac{3\cdot\left(\sqrt{5}-2\right)}{2}\)

1. = \(\dfrac{x+y}{x-y}\)
2. = \(\dfrac{x}{x+3}\)

\(\left(5-3x\right)\left(5+3x\right)-\left(x+1\right)^3\)

\(=25-9x^2-x^3-3x^2-3x-1\)

\(=-x^3-12x^2-3x+24\)

\(\frac{2^{15}.5^8+2^{14}.5^9}{2^{16}.5^7+2^{16}.5^8}=\frac{2^{14}.5^8\left(2+5\right)}{2^{16}.5^7\left(1+5\right)}=\frac{2^{14}.5^8.7}{2^{16}.5^7.6}=\frac{5.7}{4.6}=\frac{35}{24}\)

BÀi 1:

\(\sqrt{\frac43}+\sqrt{12}-\frac43\cdot\sqrt{\frac34}\)

\(=\sqrt{\frac{12}{9}}+\sqrt{12}-\frac43\cdot\frac{\sqrt{12}}{4}\)

\(=\frac{2\sqrt3}{3}+2\sqrt3-\frac{2\sqrt3}{3}=2\sqrt3\)

Bài 2:

a: \(\left(2-a\right)\cdot\sqrt{\frac{2a}{a-2}}\)

\(=-\left(a-2\right)\cdot\sqrt{\frac{2a}{a-2}}\)

\(=-\sqrt{\frac{2a\left(a-2\right)^2}{a-2}}=-\sqrt{2a\left(a-2\right)}\)

b: \(\left(x-5\right)\cdot\sqrt{\frac{x}{25-x^2}}=-\left(5-x\right)\cdot\sqrt{\frac{x}{25-x^2}}=-\sqrt{\frac{x}{\left(5-x\right)\left(5+x\right)}\cdot\left(5-x\right)^2}\)

\(=-\sqrt{\frac{x\left(5-x\right)}{5+x}}\)

c: \(\left(a-b\right)\cdot\sqrt{\frac{3a}{b^2-a^2}}=-\left(b-a\right)\cdot\sqrt{\frac{3a}{\left(b-a\right)\left(b+a\right)}}\)

\(=-\sqrt{\frac{3a\left(b-a\right)^2}{\left(b-a\right)\left(b+a\right)}}=-\sqrt{\frac{3a\left(b-a\right)}{b+a}}\)

\(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)

\(=\left[\left(3x+1\right)-\left(3x+5\right)\right]^2\)

\(=\left(3x+1-3x-5\right)^2\)

\(=\left(-4\right)^2=16\)