Căn thức bậc 2 và hằng đẳng thức \(\sqrt{A^2}=\left|A\right|\)
1, \(\sqrt{\left(2\sqrt{3}-3\right)^2}\)=
2. \(\sqrt{16}.\sqrt{25}+\sqrt{196}:\sqrt{49}\)
3. 36 :\(\sqrt{18^2}-\sqrt{169}\)
4. \(\sqrt{\sqrt{81}}\)
5 . \(\sqrt{3^2}+4^2\)
Tính giá trị biểu thức (Nhân thêm số căn vào biểu thức để làm xuất hiện hằng đẳng thức \(\left(a\pm\sqrt{b}\right)^2\) hoặc \(\left(\sqrt{a}\pm\sqrt{b}\right)^2\) rồi phá căn)
a. \(\left(4\sqrt{2}+\sqrt{30}\right).\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4-\sqrt{15}}\)
b. \(\dfrac{\sqrt{3}+1}{2}.\sqrt{8-2\sqrt{3}}\)
a) \(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right).\sqrt{\dfrac{8-2\sqrt{15}}{2}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{25.6}-\sqrt{9.10}\right).\sqrt{\dfrac{\left(\sqrt{5}\right)^2-2\sqrt{5}.\sqrt{3}+\left(\sqrt{3}\right)^2}{2}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right).\sqrt{\dfrac{\left(\sqrt{5}-\sqrt{3}\right)^2}{2}}\)
\(=\left(\sqrt{10}+\sqrt{6}\right).\dfrac{\left|\sqrt{5}-\sqrt{3}\right|}{\sqrt{2}}=\sqrt{2}.\left(\sqrt{5}+\sqrt{3}\right).\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=2\)
a) Ta có: \(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{4-\sqrt{15}}\)
\(=\sqrt{8-2\sqrt{15}}\cdot\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(\sqrt{5}-\sqrt{3}\right)^2\cdot\left(4+\sqrt{15}\right)\)
\(=\left(8-2\sqrt{15}\right)\left(4+\sqrt{15}\right)\)
\(=32+8\sqrt{15}-8\sqrt{15}-30\)
=2
Tính
a) \(2\sqrt{\dfrac{25}{16}}-3\sqrt{\dfrac{49}{36}}+4\sqrt{\dfrac{81}{64}}\)
b) \(\left(3\sqrt{2}\right)^2-\left(4\sqrt{\dfrac{1}{2}}\right)^2+\dfrac{1}{16}.\left(\sqrt{\dfrac{3}{4}}\right)^2\)
c) \(\dfrac{2}{3}\sqrt{\dfrac{81}{16}}-\dfrac{3}{4}\sqrt{\dfrac{64}{9}}+\dfrac{7}{5}.\sqrt{\dfrac{25}{196}}\)
a: \(=2\cdot\dfrac{5}{4}-3\cdot\dfrac{7}{6}+4\cdot\dfrac{9}{8}=\dfrac{5}{2}-\dfrac{7}{2}+\dfrac{9}{2}=\dfrac{7}{2}\)
b: \(=18-16\cdot\dfrac{1}{2}+\dfrac{1}{16}\cdot\dfrac{3}{4}\)
=10+3/64
=643/64
c: \(=\dfrac{2}{3}\cdot\dfrac{9}{4}-\dfrac{3}{4}\cdot\dfrac{8}{3}+\dfrac{7}{5}\cdot\dfrac{5}{14}=\dfrac{3}{2}-2+\dfrac{1}{2}=2-2=0\)
Tính
a) \(2\sqrt{\frac{25}{16}}-3\sqrt{\frac{49}{36}}+4\sqrt{\frac{81}{64}}\)
b) \(\left(3\sqrt{2}\right)^2-\left(4\sqrt{\frac{1}{2}}\right)^2+\frac{1}{16}.\left(\sqrt{\frac{3}{4}}\right)^2\)
c) \(\frac{2}{3}\sqrt{\frac{81}{16}}-\frac{3}{4}\sqrt{\frac{64}{9}}+\frac{7}{5}.\sqrt{\frac{25}{196}}\)
a) = \(\frac{7}{2}\)
b) = \(\frac{643}{64}\)
c) = 0
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,\(\sqrt{\left(2-\sqrt{3}\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(1,=\left|1-\sqrt{2}\right|+\left|\sqrt{2}+3\right|\\ =1-\sqrt{2}+3+\sqrt{2}\\ =4\\ 2,=\left|\sqrt{3}-2\right|+\left|\sqrt{3}-1\right|\\ =\sqrt{3}-2+\sqrt{3}-1\\ =2\sqrt{3}-3\\ 3,=\left|\sqrt{5}-3\right|+\left|\sqrt{5}-2\right|\\ =\sqrt{5}-3+\sqrt{5}-2\\ =2\sqrt{5}-5\\ 4,=\left|3+\sqrt{2}\right|+\left|3-\sqrt{2}\right|\\ =3+\sqrt{2}+\sqrt{3}-\sqrt{2}\\ =3+\sqrt{3}\\ 5,=\left|2-\sqrt{3}\right|-\left|2+\sqrt{3}\right|\\ =2-\sqrt{3}-\left(2+\sqrt{3}\right)\\ =2-\sqrt{3}-2-\sqrt{3}\\ =-2\sqrt{3}\)
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: \(\sqrt{\left(7-3\sqrt{3}\right)^2}\)
\(=\left|7-3\sqrt{3}\right|\)
\(=7-3\sqrt{3}\)
Tính
a) \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)
b) \(\sqrt{16}.\sqrt{25}+\sqrt{196}:\sqrt{49}\)
c) \(-0.3.\sqrt{\left(-0.3\right)^2}\)
d) \(36:\sqrt{2.3^2.18}-\sqrt{169}:13\)
Bạn nào giải từng bước giúp mình với, mình cảm ơn nhiều
CĂN BẬC HAI - CĂN THỨC BẬC HAI VÀ HẰNG ĐẲNG THỨC \(\sqrt{A^2}\)=|A|
1) Các biểu thức sau đây xác định (có nghĩa) với giá trị nào của x:
a)\(\sqrt{\frac{2x-1}{2-x}}\) b)\(\sqrt{5x^2-3x-8}\) c)\(\sqrt{5x^2+4x+7}\)
2) Tính:
a) \(-0,8\sqrt{\left(-0,125\right)^2};\) \(\sqrt{\left(-2\right)^6};\) \(\sqrt{\left(\sqrt{3}-2\right)^2};\)\(\sqrt{\left(2\sqrt{2}-3\right)^2}\).
b)\(\sqrt{\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right)^2};\) \(\sqrt{\left(0,1-\sqrt{0,1}\right)^2}.\)
c)\(\sqrt{4-2\sqrt{3}};\) \(\sqrt{3+2\sqrt{2}};\) \(\sqrt{9-4\sqrt{5}};\) \(\sqrt{16-6\sqrt{7}}.\)
Giải được bài nào hay bài đó, giúp mk nhen, cảm ơn nhìu
Căn Bậc Hai, Căn Thức Bậc Hai Và hằng Đẳng Thức \(\sqrt{A^2}=\left|A\right|\)
1. Với giá trị nào của x thì các biểu thức sau xác định (có nghĩa)
a. \(\sqrt{\frac{2x-1}{2-x}}\) b.\(\sqrt{5x^2+4x+7}\)
2.Tính
a.\(\sqrt{\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right)^2}\)
b.\(\sqrt{\left(0,1-\sqrt{0,1}\right)^2}\)
Giải câu nào cx đc nhen, thanks