Bài 1 : Tính
A) √50 – √18 + √200 -√162
Bài 2 : rút gọn
a) √6 + √10 / √21 + √35
b) √2 + √3 + √4 – √6 – √9 – √12 / √2 + √3 + √4
c) √6-2(√5) / √5 – 1
Bài 1 : Tính
A) √50 – √18 + √200 -√162
Bài 2 : rút gọn
a) √6 + √10 / √21 + √35
b) √2 + √3 + √4 – √6 – √9 – √12 / √2 + √3 + √4
c) √6-2(√5) / √5 – 1
Đáp án:
bài 1
a) = 5√2 – 3√2 + 10√2 – 9√2
= 3√2
bài 2
a) √6 + √10 = √2. ( √3 + √5 )
√21 +√35 = √7. ( √3 + √5 )
b) √2 + √3 + √4 – √6 – √9 -√12 = √2 – √3 -1 – √6
√2 + √3 + √4 = √2 + √3 + 2
c) √6 – 2 ( √5 ) ≈ -2, 02265
√5- 1 ≈ 1, 23607
Giải thích các bước giải:
các bước như trên nhó
Đáp án:
Bài 1: $3\sqrt{2}$
Bài 2:
a) $\dfrac{\sqrt{14}}{7}$
b) $1-\sqrt{3}$
c) $\dfrac{\sqrt{30}+\sqrt{6}-10-2\sqrt{5}}{4}$
Giải thích các bước giải:
Bài 1:
$\sqrt{50}-\sqrt{18}+\sqrt{200}-\sqrt{162}\\=5\sqrt{2}-3\sqrt{2}+10\sqrt{2}-9\sqrt{2}\\=3\sqrt{2}$
Bài 2:
$a) \dfrac{\sqrt{6}+\sqrt{10}}{\sqrt{21}+\sqrt{35}}\\=\dfrac{\sqrt{2}(\sqrt{3}+\sqrt{5})}{\sqrt{7}(\sqrt{3}+\sqrt{5})}\\=\dfrac{\sqrt{2}}{\sqrt{7}}\\=\dfrac{\sqrt{2}}{\sqrt{7}}.\dfrac{\sqrt{7}}{\sqrt{7}}\\=\dfrac{\sqrt{2}\sqrt{7}}{\sqrt{7}\sqrt{7}}\\=\dfrac{\sqrt{14}}{7}\\b) \dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}-\sqrt{6}-\sqrt{9}-\sqrt{12}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\\=\dfrac{\sqrt{2}+\sqrt{3}+2-\sqrt{6}-3-2\sqrt{3}}{\sqrt{2}+\sqrt{3}+2}\\=\dfrac{\sqrt{2}-\sqrt{3}-1-\sqrt{6}}{\sqrt{2}+\sqrt{3}+2}\\=\dfrac{\sqrt{2}-\sqrt{3}-1-\sqrt{6}}{\sqrt{2}+\sqrt{3}+2}.\dfrac{\sqrt{2}+\sqrt{3}-2}{\sqrt{2}+\sqrt{3}-2}\\=\dfrac{(\sqrt{2}-\sqrt{3}-1-\sqrt{6})(\sqrt{2}+\sqrt{3}-2)}{(\sqrt{2}+\sqrt{3}+2)(\sqrt{2}+\sqrt{3}-2)}\\=\dfrac{2+\sqrt{6}-2\sqrt{2}-\sqrt{6}-3+2\sqrt{3}-\sqrt{2}-\sqrt{3}+2-\sqrt{12}-\sqrt{18}+2\sqrt{6}}{(\sqrt{2}+\sqrt{3})^{2}-4}\\=\dfrac{2-2\sqrt{2}-3+2\sqrt{3}-\sqrt{2}-\sqrt{3}+2-2\sqrt{3}-\sqrt{18}+2\sqrt{6}}{2+2\sqrt{6}+3-4}\\=\dfrac{2-2\sqrt{2}-3-\sqrt{2}-\sqrt{3}+2-3\sqrt{2}+2\sqrt{6}}{1+2\sqrt{6}}\\=\dfrac{1-6\sqrt{2}-\sqrt{3}+2\sqrt{6}}{1+2\sqrt{6}}\\=\dfrac{1-6\sqrt{2}-\sqrt{3}+2\sqrt{6}}{1+2\sqrt{6}}.\frac{1-2\sqrt{6}}{1-2\sqrt{6}}\\=\dfrac{1-6\sqrt{2}+24\sqrt{3}-\sqrt{3}+6\sqrt{2}-24}{1-24}\\=\dfrac{-23+23\sqrt{3}}{-23}\\=-\dfrac{23(-1+\sqrt{3})}{23}\\=-(-1+\sqrt{3})=1-\sqrt{3}\\c) \dfrac{\sqrt{6}-2\sqrt{5}}{\sqrt{5}-1}\\=\dfrac{\sqrt{6}-2\sqrt{5}}{\sqrt{5}-1}.\dfrac{\sqrt{5}+1}{\sqrt{5}+1}\\=\dfrac{(\sqrt{6}-2\sqrt{5})(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)}\\=\dfrac{\sqrt{30}+\sqrt{6}-10-2\sqrt{5}}{5-1}\\=\dfrac{\sqrt{30}+\sqrt{6}-10-2\sqrt{5}}{4}$