\(=\dfrac{1}{2\sqrt{2}}-3\sqrt{5}+2\sqrt{2}\\ =\dfrac{1-2\sqrt{2}.3\sqrt{5}+8}{2\sqrt{2}}=\dfrac{1-6\sqrt{10}+8}{2\sqrt{2}}=\dfrac{9-6\sqrt{10}}{2\sqrt{2}}=\dfrac{\sqrt{2}\left(9-6\sqrt{10}\right)}{4}\\ =\dfrac{9\sqrt{2}-12\sqrt{5}}{4}\)

\(=\dfrac{4\left(3-\sqrt{5}\right)}{9-5}-\dfrac{8\left(1-\sqrt{5}\right)}{1-5}+3\sqrt{5}\\ =3-\sqrt{5}-2\left(1-\sqrt{5}\right)+3\sqrt{5}\\ =3+2\sqrt{5}-2+2\sqrt{5}\\ =1-4\sqrt{5}\)

\(=x^2-2.4x+16-21\\ =\left(x-4\right)^2-21\)

\(a,=\left(2\sqrt{7}-2\sqrt{14}+\sqrt{7}\right).\sqrt{7}+14\sqrt{2}\\ =\left(3\sqrt{7}-2\sqrt{14}\right).\sqrt{7}+14\sqrt{2}\\ =21-14\sqrt{2}+14\sqrt{2}\\ =21\\ b,=\sqrt{9+2.3.\sqrt{2}+2}-\sqrt{9-2.3.\sqrt{2}+2}\\ =\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\\ =\left|3+\sqrt{2}\right|-\left|3-\sqrt{2}\right|\\ =3+\sqrt{2}-3+\sqrt{2}\\ =2\sqrt{2}\\ c,=\left(75\sqrt{2}+50\sqrt{2}-45\sqrt{2}\right):\sqrt{10}\\ =80\sqrt{2}:\sqrt{10}=16\sqrt{5}\)

\(=\dfrac{34}{5}+\dfrac{78\times14}{7\times39}\\ =\dfrac{34}{5}+\dfrac{1092}{273}\\ =\dfrac{34}{5}+4\\ =\dfrac{34+20}{5}\\ =\dfrac{54}{5}\)

Sửa Đề 

\(=\left(3x-1\right)\left(2x-1\right)\left(2x+1\right)-3\left(3x-1\right)\left(2x-1\right)^2\\ =\left(3x-1\right)\left(2x-1\right)\left(2x+1-6x+3\right)\\ =\left(3x-1\right)\left(2x-1\right)\left(-4x+4\right)\\ =4\left(3x-1\right)\left(2x-1\right)\left(-x+1\right)\)

Trắc nghiệm 

\(1-D\\ 2-B\\ 3-A\\ 4-C\\ 5-C\)

\(cotB=\dfrac{4}{5}=>tanB=\dfrac{5}{4}=>\widehat{B}=51^o\)

Ta có \(\widehat{B}+\widehat{C}=90^o=>\widehat{C}=39^o\\ AB=tanB.AC=>AB=12,5\left(cm\right)\\ \)

ÁP dụng Pytago vào tam giác ABC

\(AB^2+AC^2=BC^2=>BC=\sqrt{\left(12,5\right)^2+10^2}=16\left(cm\right)\)

\(a,=\left(98-2\right):2+1=49\left(phantu\right)\\ b,B=\left(70-6\right):4+1=17\left(phantu\right)\)

=>Cố tình khóa acc

=> nhưng iem ko chép mà chj iu ơi :((