Rút gọn căn bậc hai theo HẰNG ĐẲNG THỨC số 1 và 2:
1) \(\sqrt{28-10\sqrt{3}}\)
2) \(\sqrt{123+22\sqrt{2}}\)
3) \(\sqrt{7-3\sqrt{5}}\)
rút gọn biểu thức chưa căn thức bậc hai:
1,\(\sqrt{\left(1-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{2}+3\right)^2}\)
2, \(\sqrt{\left(\sqrt{3}-2\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\)
3,\(\sqrt{\left(\sqrt{5}-3\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}\)
4,\(\sqrt{\left(3+\sqrt{2}\right)^2}+\sqrt{\left(3-\sqrt{2}\right)^2}\)
5,\(\sqrt{\left(2-\sqrt{3}\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(1,=\left|1-\sqrt{2}\right|+\left|\sqrt{2}+3\right|\\ =1-\sqrt{2}+3+\sqrt{2}\\ =4\\ 2,=\left|\sqrt{3}-2\right|+\left|\sqrt{3}-1\right|\\ =\sqrt{3}-2+\sqrt{3}-1\\ =2\sqrt{3}-3\\ 3,=\left|\sqrt{5}-3\right|+\left|\sqrt{5}-2\right|\\ =\sqrt{5}-3+\sqrt{5}-2\\ =2\sqrt{5}-5\\ 4,=\left|3+\sqrt{2}\right|+\left|3-\sqrt{2}\right|\\ =3+\sqrt{2}+\sqrt{3}-\sqrt{2}\\ =3+\sqrt{3}\\ 5,=\left|2-\sqrt{3}\right|-\left|2+\sqrt{3}\right|\\ =2-\sqrt{3}-\left(2+\sqrt{3}\right)\\ =2-\sqrt{3}-2-\sqrt{3}\\ =-2\sqrt{3}\)
Bài 1
a. Tìm điều kiện để căn thức bậc hai có nghĩa \(\sqrt{\dfrac{2x+1}{x^2+1}}\)
b. \(\sqrt[3]{-27}+\sqrt[3]{64}-\dfrac{\sqrt[3]{-128}}{\sqrt[3]{2}}\)
* Rút gọn biểu thức
a. \(\sqrt{20}+2\sqrt{45}+\sqrt{125}-3\sqrt{80}\)
b. \(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{45}+\sqrt{\left(2-\sqrt{5}\right)^2}\)
c. \(\dfrac{5+\sqrt{5}}{5-\sqrt{5}}+\dfrac{5-\sqrt{5}}{5+\sqrt{5}}\)
Bài 1 :
a, ĐKXĐ : \(\dfrac{2x+1}{x^2+1}\ge0\)
Mà \(x^2+1\ge1>0\)
\(\Rightarrow2x+1\ge0\)
\(\Rightarrow x\ge-\dfrac{1}{2}\)
Vậy ...
b, Ta có : \(\sqrt[3]{-27}+\sqrt[3]{64}-\sqrt[3]{-\dfrac{128}{2}}\)
\(=-3+4-\left(-4\right)=-3+4+4=5\)
Bài 2 :
\(a,=2\sqrt{5}+6\sqrt{5}+5\sqrt{5}-12\sqrt{5}\)
\(=\sqrt{5}\left(2+6+5-12\right)=\sqrt{2}\)
\(b,=\sqrt{5}+\sqrt{5}+\left|\sqrt{5}-2\right|\)
\(=2\sqrt{5}+\sqrt{5}-2=3\sqrt{5}-2\)
\(c,=\dfrac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\)
\(=\dfrac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\)
\(=3\)
rút gọn căn bằng hằng đẳng thức
1.\(\sqrt{\frac{9}{4}-\sqrt{2}}\) 2.\(\sqrt{\frac{129}{16}-\sqrt{2}}\)
3.\(\sqrt{3+\sqrt{8}}\) 4.\(\sqrt{\frac{289+4\sqrt{72}}{16}}\)
5.\(\sqrt{28-10\sqrt{3}}\)
\(\sqrt{2-2.\frac{1}{2}\sqrt{2}+\frac{1}{4}}.\sqrt{8-2.2\sqrt{2}.\frac{1}{4}+\frac{1}{16}}=\sqrt{\left(\sqrt{2}-\frac{1}{2}\right)^2}\sqrt{\left(2\sqrt{2}-\frac{1}{4}\right)^2}\)
\(=\left(\sqrt{2}-\frac{1}{2}\right)\left(2\sqrt{2}-\frac{1}{4}\right)=\frac{33-10\sqrt{2}}{8}\)
\(\sqrt{2+2\sqrt{2}+1}.4\sqrt{\frac{288+2\sqrt{288}+1}{16}}=\sqrt{\left(\sqrt{2}+1\right)^2}.4\sqrt{\frac{\left(12\sqrt{2}+1\right)^2}{4^2}}\)
\(=\left(\sqrt{2}+1\right)\left(12\sqrt{2}+1\right)=25+13\sqrt{2}\)
\(\sqrt{28-10\sqrt{3}}=\sqrt{25-2.5\sqrt{3}+3}=\sqrt{\left(5-\sqrt{3}\right)^2}=5-\sqrt{3}\)
Rút gọn căn bậc hai theo hằng đẳng thức:
a)\(\left(4\sqrt{2}+\sqrt{30}\right).\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4-\sqrt{15}}\)
b)\(2.\left(\sqrt{10}-\sqrt{2}\right).\left(4+\sqrt{6-2\sqrt{5}}\right)\)
c)\(\left(7+\sqrt{14}\right).\sqrt{9-2\sqrt{14}}\)
d)\(\sqrt{\dfrac{289+4\sqrt{72}}{16}}\)
e) \(\left(\sqrt{21}+7\right).\sqrt{10-2\sqrt{21}}\)
f)\(\sqrt{2-\sqrt{3}.\left(\sqrt{6}+\sqrt{2}\right)}\)
g) \(\sqrt{2}\sqrt{8+3\sqrt{7}}\)
h) \(\sqrt{11+6\sqrt{2}}\)
a)\(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\sqrt{10\left(4-\sqrt{15}\right)}+\sqrt{6\left(4-\sqrt{15}\right)}\)
\(=\sqrt{40-10\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)
\(=\sqrt{\left(5-\sqrt{15}\right)^2}+\sqrt{\left(3-\sqrt{15}\right)^2}\)
\(=5-\sqrt{15}+\sqrt{15}-3\)
\(=2\)
b) \(2\left(\sqrt{10}-\sqrt{2}\right)\left(4+\sqrt{6-2\sqrt{5}}\right)\)
\(=\left(2\sqrt{10}-2\sqrt{2}\right)\left(4+\sqrt{\left(1-\sqrt{5}\right)^2}\right)\)
\(=\left(2\sqrt{10}-2\sqrt{2}\right)\left(4+\sqrt{5}-1\right)\)
\(=\left(2\sqrt{10}-2\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
\(=6\sqrt{10}+2\sqrt{50}-6\sqrt{2}-2\sqrt{10}\)
\(=6\sqrt{10}+10\sqrt{2}-6\sqrt{2}-2\sqrt{10}\)
\(=4\sqrt{10}+4\sqrt{2}\)
c) \(\left(\sqrt{7}+\sqrt{14}\right)\sqrt{9-2\sqrt{14}}\)
\(=\left(\sqrt{7}+\sqrt{14}\right)\sqrt{\left(\sqrt{2}-\sqrt{7}\right)^2}\)
\(=\left(\sqrt{7}+\sqrt{14}\right)\left(\sqrt{7}-\sqrt{2}\right)\)
\(=7\sqrt{7}-7\sqrt{2}+\sqrt{98}-\sqrt{28}\)
\(=7\sqrt{7}-7\sqrt{2}+7\sqrt{2}-2\sqrt{7}\)
\(=5\sqrt{7}\)
d) \(\sqrt{\dfrac{289+4\sqrt{72}}{16}}\)
\(=\sqrt{\dfrac{289+42\sqrt{2}}{16}}\)
\(=\dfrac{\sqrt{289+42\sqrt{2}}}{\sqrt{4^2}}\)
\(=\dfrac{\sqrt{\left(1+12\sqrt{2}\right)^2}}{4}\)
\(=\dfrac{1+12\sqrt{2}}{4}\)
e) \(\left(\sqrt{21}+7\right)\sqrt{10-2\sqrt{21}}\)
\(=\left(\sqrt{21}+\sqrt{7}\right)\sqrt{\left(\sqrt{3}-\sqrt{7}\right)^2}\)
\(=\left(\sqrt{21}+\sqrt{7}\right)\left(\sqrt{7}-\sqrt{3}\right)\)
\(=\sqrt{147}-\sqrt{63}+7-\sqrt{21}\)
\(=7\sqrt{3}-\sqrt{63}+7-\sqrt{21}\)
f) bạn xem đề lại nhé
g) \(\sqrt{2}\sqrt{8+3\sqrt{7}}\)
\(=\sqrt{2\left(8+3\sqrt{7}\right)}\)
\(=\sqrt{16+6\sqrt{7}}\)
\(=\sqrt{\left(3+\sqrt{7}\right)^2}\)
\(=3+\sqrt{7}\)
g)
Bài 1: Rút gọn căn bậc 2 theo hằng đẳng thức 1:
a)\(\sqrt{\left(23-15\sqrt{3}\right)^2}\)
b) \(\sqrt{\left(2-2\sqrt{3}\right)^2}\)
c) \(\sqrt{\left(15-4\sqrt{3}\right)^2}\)
d)\(\sqrt{\left(16-6\sqrt{7}\right)^2}\)
f)\(\sqrt{\left(22-8\sqrt{3}\right)^2}\)
g) \(\sqrt{\left(9-4\sqrt{2}\right)^2}\)
h) \(\sqrt{\left(13-4\sqrt{3}\right)^2}\)
i)\(\sqrt{\left(7-3\sqrt{3}\right)^2}\)
(mink đag cần gấp)
Bài 1:
a) Ta có: \(\sqrt{\left(23-15\sqrt{3}\right)^2}\)
\(=\left|23-15\sqrt{3}\right|\)
\(=\left|\sqrt{529}-\sqrt{675}\right|\)
\(=\sqrt{675}-\sqrt{529}\)
\(=15\sqrt{3}-23\)
b) Ta có: \(\sqrt{\left(2-2\sqrt{3}\right)^2}\)
\(=\left|2-2\sqrt{3}\right|\)
\(=2\sqrt{3}-2\)
c) Ta có: \(\sqrt{\left(15-4\sqrt{3}\right)^2}\)
\(=\left|15-4\sqrt{3}\right|\)
\(=15-4\sqrt{3}\)
d) Ta có: \(\sqrt{\left(16-6\sqrt{7}\right)^2}\)
\(=\left|16-6\sqrt{7}\right|\)
\(=\left|\sqrt{256}-\sqrt{252}\right|\)
\(=16-6\sqrt{7}\)
f) Ta có: \(\sqrt{\left(22-8\sqrt{3}\right)^2}\)
\(=\left|22-8\sqrt{3}\right|\)
\(=\left|\sqrt{484}-\sqrt{192}\right|\)
\(=22-8\sqrt{3}\)
g) Ta có: \(\sqrt{\left(9-4\sqrt{2}\right)^2}\)
\(=\left|9-4\sqrt{2}\right|\)
\(=9-4\sqrt{2}\)
h) Ta có: \(\sqrt{\left(13-4\sqrt{3}\right)^2}\)
\(=\left|13-4\sqrt{3}\right|\)
\(=13-4\sqrt{3}\)
i) Ta có: \(\sqrt{\left(7-3\sqrt{3}\right)^2}\)
\(=\left|7-3\sqrt{3}\right|\)
\(=7-3\sqrt{3}\)
rút gọn biểu thức: A= \(\dfrac{\sqrt[3]{2}+\sqrt{7+2\sqrt{10}}+\sqrt[3]{3\sqrt[3]{4}-3\sqrt[3]{2}-1}}{\sqrt{5}+\sqrt{2}+1}\)
Lời giải:
$\sqrt{7+2\sqrt{10}}=\sqrt{2+5+2\sqrt{2.5}}=\sqrt{(\sqrt{2}+\sqrt{5})^2}=\sqrt{2}+\sqrt{5}$
\(\sqrt[3]{3\sqrt[3]{3}-3\sqrt[3]{2}-1}=\sqrt[3]{(1-\sqrt[3]{2})^3}=1-\sqrt[3]{2}\)
Do đó:
\(\text{TS}=\sqrt[3]{2}+\sqrt{2}+\sqrt{5}+1-\sqrt[3]{2}=\sqrt{2}+\sqrt{5}+1=\text{MS}\)
\(A=\frac{\text{TS}}{\text{MS}}=1\)
Trục căn thức ở mẫu và rút gọn
a,\(\dfrac{\sqrt{2}}{\sqrt{5}-\sqrt{3}}\) b,\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)
c,\(\dfrac{5+2\sqrt{5}}{\sqrt{5}+\sqrt{2}}\) d,\(\dfrac{2\sqrt{6}-\sqrt{10}}{4\sqrt{3}-2\sqrt{5}}\)
Bài 1: So sánh các căn bậc hai số học
a) 1 và\(\sqrt{3}-1\) b) 2 và \(\sqrt{2}+1\) c) 2\(\sqrt{31}\)và 10 d)\(\sqrt{2}+\sqrt{11}\)và \(\sqrt{3}+5\)
Rút gọn căn bậc hai bằng hằng đẳng thức:
\(\sqrt{11+4\sqrt{6}}\)