Rút gọn căn bậc hai theo hằng đẳng thức:
\(\sqrt{11+4\sqrt{6}}\)
Rút gọn căn bậc hai bằng hằng đẳng thức:
\(\sqrt{11+4\sqrt{6}}\)
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) \(\sqrt{7-3\sqrt{5}}\)
\(\sqrt{25-2.5.\sqrt{3}+3}=\sqrt{\left(5-\sqrt{3}\right)^2}=5-\sqrt{3}\)
\(\sqrt{121+2.11.\sqrt{2}+2}=\sqrt{\left(11+\sqrt{2}\right)^2}=11+\sqrt{2}\)
\(\sqrt{\frac{9}{2}-2.\frac{3}{\sqrt{2}}.\frac{\sqrt{5}}{\sqrt{2}}+\frac{5}{2}}=\sqrt{\left(\frac{3}{\sqrt{2}}-\frac{\sqrt{5}}{\sqrt{2}}\right)^2}=\frac{3}{\sqrt{2}}-\frac{\sqrt{5}}{\sqrt{2}}=\frac{3\sqrt{2}-\sqrt{10}}{2}\)
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)
Rút gọn căn thức bậc hai
b, \(\sqrt{8-2\sqrt{7}}\)
c, \(\sqrt{29-12\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(\sqrt{8-2\sqrt{7}}=\sqrt{\left(\sqrt{7}-1\right)^2}=\left|\sqrt{7}-1\right|=\sqrt{7}-1\)
\(\sqrt{29-12\sqrt{5}}-\sqrt{9-4\sqrt{5}}=\sqrt{\left(2\sqrt{5}-3\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(=\left|2\sqrt{5}-3\right|-\left|\sqrt{5}-2\right|=2\sqrt{5}-3-\sqrt{5}+2=\sqrt{5}-1\)
b)\(=\sqrt{\left(\sqrt{7}-1\right)^2}=\sqrt{7}-1\)
c)\(=\sqrt{\left(2\sqrt{5}-3\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(=2\sqrt{5}-3-\sqrt{5}+2\)
\(=\sqrt{5}-1\)
b,\(\sqrt{8-2\sqrt{7}}=\sqrt{\left(\sqrt{7}-1\right)^2}=\sqrt{7}-1\)
c,\(\sqrt{29-12\sqrt{5}}-\sqrt{9-4\sqrt{5}}=\sqrt{\left(3-2\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}=3-2\sqrt{5}-\sqrt{5}+2=5-3\sqrt{5}\)
Trục căn thức và rút gọn
\(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
Help me plssssssss
\(=\left(-3+3\sqrt{6}+4+2\sqrt{6}-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
=(căn 6-11)(căn 6+11)
=6-121=-115
\(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}\right)^2-1^2}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}\right)^2-2^2}-\dfrac{12\left(3+\sqrt{6}\right)}{3^2-\left(\sqrt{6}\right)^2}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right)\left(\sqrt{6}+11\right)\)
\(=\left[3\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)\right]\left(\sqrt{6}+11\right)\)
\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)
\(=\left(\sqrt{6}\right)^2-11^2\)
\(=6-121\)
\(=-115\)
Căn thức bậc hai và hằng đẳng thức |A|= căn A
Giải phương trình: \(\sqrt{x+2\sqrt{x-1}}=2\) Chứng minh: \(\sqrt{a+a\sqrt{a-2}+2}+\sqrt{a-4\sqrt{a-2}+2}=4\) (với \(2\le a\le6\)\(\sqrt{x+2\sqrt{x-1}}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}\right)^2+2\sqrt{x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}=2\)
\(\Leftrightarrow\left|\sqrt{x-1}+1\right|=2\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x-1}+1=2\\\sqrt{x-1}+1=-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x-1}=1\Leftrightarrow x-1=1\Leftrightarrow x=2\\\sqrt{x-1}=-3\left(vl\right)\end{cases}}\)
Vậy phương trình có tập nghiệm \(S=\left\{2\right\}\)
Hãy rút gọn :
\(Q=\sqrt{6+\sqrt{6+\sqrt{6+...}}}\) \(\left(\text{vô số căn bậc hai}\right)\)
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,\(\sqrt{\left(2-\sqrt{3}\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(1,=\left|1-\sqrt{2}\right|+\left|\sqrt{2}+3\right|\\ =1-\sqrt{2}+3+\sqrt{2}\\ =4\\ 2,=\left|\sqrt{3}-2\right|+\left|\sqrt{3}-1\right|\\ =\sqrt{3}-2+\sqrt{3}-1\\ =2\sqrt{3}-3\\ 3,=\left|\sqrt{5}-3\right|+\left|\sqrt{5}-2\right|\\ =\sqrt{5}-3+\sqrt{5}-2\\ =2\sqrt{5}-5\\ 4,=\left|3+\sqrt{2}\right|+\left|3-\sqrt{2}\right|\\ =3+\sqrt{2}+\sqrt{3}-\sqrt{2}\\ =3+\sqrt{3}\\ 5,=\left|2-\sqrt{3}\right|-\left|2+\sqrt{3}\right|\\ =2-\sqrt{3}-\left(2+\sqrt{3}\right)\\ =2-\sqrt{3}-2-\sqrt{3}\\ =-2\sqrt{3}\)
Mọi người ơi giúp mình với nha!! Mình cần rất gấp
Giải phương trình:
\(A=\sqrt{3x^2+6x+12}+\sqrt{5x^4-10x^2+9}=-2x^2-4x+3\)
*Lưu ý: Mình mới học tới bài Căn thức bậc hai và hằng đẳng thức \(\sqrt{A^2}=\left|A\right|\)thôi.
Ta có:
\(\sqrt{3x^2+6x+12}+\sqrt{5x^4-10x^2+9}\)
\(=\sqrt{\left(3x^2+6x+3\right)+9}+\sqrt{\left(5x^4-10x^2+5\right)+4}\)
\(=\sqrt{3\left(x+1\right)^2+9}+\sqrt{5\left(x^2-1\right)^2+4}\ge3+2=5\left(1\right)\)
Ta lại có:
\(-2x^2-4x+3=-2\left(x+1\right)^2+5\le5\left(2\right)\)
Từ (1) và (2) dấu = xảy ra khi \(x=-1\)