Tính:
1) \(\sqrt{4-2\sqrt{3}}\)
2) \(\sqrt{5+2\sqrt{6}}\)
3) \(\sqrt{7-2\sqrt{10}}\)
4) \(\sqrt{14-6\sqrt{6}}\)
5) \(\sqrt{8+2\sqrt{15}}\)
6) \(\sqrt{10-2\sqrt{21}}\)
7) \(\sqrt{11+2\sqrt{18}}\)
LÀM CHI TIẾT GIÚP MK NHÉ!
Những câu hỏi liên quan
Tính:
1) \(\sqrt{14-2\sqrt{33}}\)
2) \(\sqrt{12-2\sqrt{35}}\)
3) \(\sqrt{16-2\sqrt{55}}\)
4) \(\sqrt{14-6\sqrt{5}}\)
5) \(\sqrt{17-12\sqrt{2}}\)
6) \(\sqrt{27-12\sqrt{5}}\)
7) \(\sqrt{4+\sqrt{15}}\)
LÀM CHI TIẾT GIÚP MK NHÉ!
1)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(\sqrt{11}-\sqrt{3}\right)^2}=\sqrt{11}-\sqrt{3}\)
2)
\(=\sqrt{\left(\sqrt{7}\right)^2-2.\sqrt{7}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}=\sqrt{7}-\sqrt{5}\)
3)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{11}-\sqrt{5}\right)}=\sqrt{11}-\sqrt{5}\)
4)
\(=\sqrt{3^2-2.3.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(3-\sqrt{5}\right)^2}=3-\sqrt{5}\)
5)
\(=\sqrt{3^2-2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}=\sqrt{\left(3-2\sqrt{2}\right)^2}=3-2\sqrt{2}\)
Đúng 0
Bình luận (0)
\(\sqrt{3-2\sqrt{2}}-\sqrt{11+6\sqrt{2}}\)
\(\sqrt{4-2\sqrt{3}}-\sqrt{7-4\sqrt{3}}+\sqrt{19+8\sqrt{3}}\)
\(\sqrt{6-2\sqrt{5}}+\sqrt{9+4\sqrt{5}}-\sqrt{14-6\sqrt{5}}\)
\(\sqrt{11-4\sqrt{7}}+\sqrt{23-8\sqrt{7}}+\sqrt{\left(-2^6\right)}\)
rút gọn:giải chi tiết hộ mình nha
a) Ta có: \(\sqrt{3-2\sqrt{2}}-\sqrt{11+6\sqrt{2}}\)
\(=\sqrt{2}-1-3-\sqrt{2}\)
=-4
b) Ta có: \(\sqrt{4-2\sqrt{3}}-\sqrt{7-4\sqrt{3}}+\sqrt{19+8\sqrt{3}}\)
\(=\sqrt{3}-1-2+\sqrt{3}+4+\sqrt{3}\)
\(=3\sqrt{3}+1\)
c) Ta có: \(\sqrt{6-2\sqrt{5}}+\sqrt{9+4\sqrt{5}}-\sqrt{14-6\sqrt{5}}\)
\(=\sqrt{5}-1+\sqrt{5}-2-3+\sqrt{5}\)
\(=3\sqrt{5}-6\)
d) Ta có: \(\sqrt{11-4\sqrt{7}}+\sqrt{23-8\sqrt{7}}+\sqrt{\left(-2\right)^6}\)
\(=\sqrt{7}-2+4-\sqrt{7}+8\)
=10
Đúng 1
Bình luận (0)
Tính
1) sqrt{18}.sqrt{2}
2) sqrt{15^2-9^2}
3) sqrt{46-6sqrt{5}}-sqrt{46+6sqrt{5}}
4)sqrt{21+6sqrt{6}}-sqrt{21-6sqrt{6}}
5) left(2+sqrt{5}right).sqrt{9-4sqrt{5}}
6)left(3-sqrt{2}right).sqrt{7+4sqrt{3}}
7)left(sqrt{3}+sqrt{5}right).sqrt{7-2sqrt{10}}
8)left(sqrt{6}+sqrt{10}right).sqrt{4-sqrt{15}}
9) sqrt{2}.left(sqrt{8}-sqrt{32}+3sqrt{18}right)
10) sqrt{2}left(sqrt{2}-sqrt{3-sqrt{5}}right)
11) sqrt{3}-sqrt{2}-sqrt{left(sqrt{3}+sqrt{2}right)^2}
12) left(sqrt{2}-sqrt{3+sqrt{5}}right).sqrt...
Đọc tiếp
Tính
1) \(\sqrt{18}.\sqrt{2}\)
2) \(\sqrt{15^2-9^2}\)
3) \(\sqrt{46-6\sqrt{5}}-\sqrt{46+6\sqrt{5}}\)
4)\(\sqrt{21+6\sqrt{6}}-\sqrt{21-6\sqrt{6}}\)
5) \(\left(2+\sqrt{5}\right).\sqrt{9-4\sqrt{5}}\)
6)\(\left(3-\sqrt{2}\right).\sqrt{7+4\sqrt{3}}\)
7)\(\left(\sqrt{3}+\sqrt{5}\right).\sqrt{7-2\sqrt{10}}\)
8)\(\left(\sqrt{6}+\sqrt{10}\right).\sqrt{4-\sqrt{15}}\)
9) \(\sqrt{2}.\left(\sqrt{8}-\sqrt{32}+3\sqrt{18}\right)\)
10) \(\sqrt{2}\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right)\)
11) \(\sqrt{3}-\sqrt{2}-\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)
12) \(\left(\sqrt{2}-\sqrt{3+\sqrt{5}}\right).\sqrt{2}+2\sqrt{5}\)
1: \(=\sqrt{36}=6\)
2: \(=\sqrt{\left(15-9\right)\left(15+9\right)}=\sqrt{24\cdot6}=12\)
3: \(=3\sqrt{5}-1-3\sqrt{5}-1=-2\)
4: \(=3\sqrt{2}+\sqrt{3}-3\sqrt{2}+\sqrt{3}=2\sqrt{3}\)
5: \(=\left(2+\sqrt{5}\right)\left(\sqrt{5}-2\right)=5-4=1\)
Đúng 0
Bình luận (0)
Tính
1, sqrt{3+2sqrt{2}} 7, sqrt{10-2sqrt{21}}
2, sqrt{4-2sqrt{3}} 8, sqrt{11+2sqrt{18}}
3, sqrt{5+2sqrt{6}} 9, sqrt{14-2sqrt{33}}
4, sqrt{7-2sqrt{10}}
5, sqrt{15-6sqrt{6}}
6, sqrt{8+2sqrt{15}}
các bạn ơi ! giúp mik với đi
Đọc tiếp
Tính
1, \(\sqrt{3+2\sqrt{2}}\) 7, \(\sqrt{10-2\sqrt{21}}\)
2, \(\sqrt{4-2\sqrt{3}}\) 8, \(\sqrt{11+2\sqrt{18}}\)
3, \(\sqrt{5+2\sqrt{6}}\) 9, \(\sqrt{14-2\sqrt{33}}\)
4, \(\sqrt{7-2\sqrt{10}}\)
5, \(\sqrt{15-6\sqrt{6}}\)
6, \(\sqrt{8+2\sqrt{15}}\)
các bạn ơi ! giúp mik với đi
Bạn chỉ cần tách chúng thành hằng đẳng thức sau đó áp dụng HĐT: \(\sqrt{A^2}=\left|A\right|\)
1, \(\sqrt{3+2\sqrt{2}}=\sqrt{2+2\sqrt{2}+1}=\sqrt{\left(\sqrt{2}+1\right)^2}=\left|\sqrt{2}+1\right|=\sqrt{2}+1\)
2, \(\sqrt{4-2\sqrt{3}}=\sqrt{3-2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}-1\right)^2}=\left|\sqrt{3}-1\right|=\sqrt{3}-1\)3, \(\sqrt{5+2\sqrt{6}}=\sqrt{3+2\sqrt{3}.\sqrt{2}+2}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}=\left|\sqrt{3}+\sqrt{2}\right|=\sqrt{3}+\sqrt{2}\)4, \(\sqrt{7-2\sqrt{10}}=\sqrt{5-2\sqrt{5}.\sqrt{2}+2}=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}=\left|\sqrt{5}-\sqrt{2}\right|=\sqrt{5}-\sqrt{2}\)5, \(\sqrt{15-6\sqrt{6}}=\sqrt{9-2.3.\sqrt{6}+6}=\sqrt{\left(3+\sqrt{6}\right)^2}=\left|3+\sqrt{6}\right|=3+\sqrt{6}\)Các câu còn lại tương tự nha!
Đúng 0
Bình luận (1)
6, \(\sqrt{8+2\sqrt{15}}=\sqrt{5+2\sqrt{5}.\sqrt{3}+3}=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\left|\sqrt{5}+\sqrt{3}\right|=\sqrt{5}+\sqrt{3}\)7, \(\sqrt{10-2\sqrt{21}}=\sqrt{7-2\sqrt{7}.\sqrt{3}+3}=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}=\left|\sqrt{7}-\sqrt{3}\right|=\sqrt{7}-\sqrt{3}\)8, \(\sqrt{11+2\sqrt{18}}=\sqrt{9+2\sqrt{9}.\sqrt{2}+2}=\sqrt{\left(\sqrt{9}+\sqrt{2}\right)^2}=\left|3+\sqrt{2}\right|=3+\sqrt{2}\)9, \(\sqrt{14-2\sqrt{33}}=\sqrt{11-2\sqrt{11}.\sqrt{3}+3}=\sqrt{\left(\sqrt{11}-\sqrt{3}\right)^2}=\left|\sqrt{11}-\sqrt{3}\right|=\sqrt{11}-\sqrt{3}\)Thử tự làm những câu còn lại rồi kiểm tra xem đúng hay sai nha!!!
Chúc bạn học tốt!!!
Đúng 0
Bình luận (0)
1/ \(=\sqrt{2+2\sqrt{2}+1}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)
2/ \(=\sqrt{3-2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
3/ \(=\sqrt{3+2\sqrt{6}+2}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}\)
4/ \(=\sqrt{5-2\sqrt{10}+2}=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}=\sqrt{5}-\sqrt{2}\)
5/ \(=\sqrt{9-6\sqrt{6}+6}=\sqrt{\left(3-\sqrt{6}\right)^2}=3-\sqrt{6}\)
6/ \(=\sqrt{5+2\sqrt{15}+3}=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\sqrt{5}+\sqrt{3}\)
7/ \(=\sqrt{7-2\sqrt{21}+3}=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}=\sqrt{7}-\sqrt{3}\)
8/ \(=\sqrt{9+2\sqrt{18}+2}=\sqrt{\left(3+\sqrt{2}\right)^2}=3+\sqrt{2}\)
9/ \(=\sqrt{11-2\sqrt{33}+3}=\sqrt{\left(\sqrt{11}-\sqrt{3}\right)^2}=\sqrt{11}-\sqrt{3}\)
Đúng 0
Bình luận (0)
Bài 1: Tính
1, Aleft(1-frac{5+sqrt{5}}{1+sqrt{5}}right).left(frac{5-sqrt{5}}{1-sqrt{5}}-1right)
2, Bleft(frac{3sqrt{125}}{15}-frac{10-4sqrt{6}}{sqrt{5}-2}right).frac{1}{sqrt{5}}
3, Cleft(frac{sqrt{1000}}{100}-frac{5sqrt{2}-2sqrt{5}}{2sqrt{5}-8}right).frac{sqrt{10}}{10}
4, Dfrac{1}{sqrt{49+20sqrt{6}}}-frac{1}{sqrt{49-20sqrt{6}}}+frac{1}{sqrt{7-4sqrt{3}}}
5, Efrac{1}{sqrt{4-2sqrt{3}}}-frac{1}{sqrt{7-sqrt{48}}}+frac{3}{sqrt{14-6sqrt{5}}}
6, Ffrac{1}{sqrt{2}-sqrt{3}}sqrt{frac{3sqrt{2}-2sqrt{3}...
Đọc tiếp
Bài 1: Tính
1, \(A=\left(1-\frac{5+\sqrt{5}}{1+\sqrt{5}}\right).\left(\frac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
2, \(B=\left(\frac{3\sqrt{125}}{15}-\frac{10-4\sqrt{6}}{\sqrt{5}-2}\right).\frac{1}{\sqrt{5}}\)
3, \(C=\left(\frac{\sqrt{1000}}{100}-\frac{5\sqrt{2}-2\sqrt{5}}{2\sqrt{5}-8}\right).\frac{\sqrt{10}}{10}\)
4, \(D=\frac{1}{\sqrt{49+20\sqrt{6}}}-\frac{1}{\sqrt{49-20\sqrt{6}}}+\frac{1}{\sqrt{7-4\sqrt{3}}}\)
5, \(E=\frac{1}{\sqrt{4-2\sqrt{3}}}-\frac{1}{\sqrt{7-\sqrt{48}}}+\frac{3}{\sqrt{14-6\sqrt{5}}}\)
6, \(F=\frac{1}{\sqrt{2}-\sqrt{3}}\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)
7, \(G=\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}-\sqrt{11-2\sqrt{10}}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}+\sqrt{12+8\sqrt{2}}}}\)
a. P (3+sqrt{2}+sqrt{6})(sqrt{6-3sqrt{3}})
b. A(frac{15}{sqrt{6}+1}+frac{4}{sqrt{6}-2}-frac{12}{3-sqrt{6}}): (sqrt{6}+11)
c. B frac{sqrt{8-2sqrt{12}}}{sqrt{3}-1}-sqrt{8}
d. C sqrt{4+sqrt{7}}-sqrt{4-sqrt{7}}-sqrt{2}
đ. Dfrac{1}{sqrt{2}-sqrt{3}}sqrt{frac{3sqrt{2}-2sqrt{3}}{3sqrt{2}+2sqrt{3}}}
e. E sqrt{8+2sqrt{10+2sqrt{5}}}+sqrt{8-2sqrt{10+2sqrt{5}}}
ê. G sqrt{4+5sqrt{3}+5sqrt{48-10sqrt{7+4sqrt{3}}}}
g. Hfrac{2sqrt{4+sqrt{5+21+sqrt{80}}}}{sqrt{10}-sqrt{2}}
i. Isqrt{frac{4-sqrt{7}}{4+sqrt{7...
Đọc tiếp
a. P= (\(3+\sqrt{2}+\sqrt{6}\))(\(\sqrt{6-3\sqrt{3}}\))
b. A=(\(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\)): (\(\sqrt{6}+11\))
c. B= \(\frac{\sqrt{8-2\sqrt{12}}}{\sqrt{3}-1}\)-\(\sqrt{8}\)
d. C= \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)
đ. D=\(\frac{1}{\sqrt{2}-\sqrt{3}}\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)
e. E= \(\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
ê. G= \(\sqrt{4+5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
g. H=\(\frac{2\sqrt{4+\sqrt{5+21+\sqrt{80}}}}{\sqrt{10}-\sqrt{2}}\)
i. I=\(\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}+\sqrt{\frac{4+\sqrt{7}}{4-\sqrt{7}}}\)
k. K=\(\frac{3+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{3-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
3 tháng 10 2021 lúc 13:53
a,\(\sqrt{8+2\sqrt{15}}\) -\(\sqrt{6+2\sqrt{15}}\)
b, \(\sqrt{17-2\sqrt{72}}-\sqrt{19+2\sqrt{18}}\)
c, \(\sqrt{8-2\sqrt{7}}+\sqrt{8+2\sqrt{7}}\)
d, \(\sqrt{12+2\sqrt{11}}-\sqrt{12-2\sqrt{11}}\)
e, \(\sqrt{10-2\sqrt{21}}-\sqrt{9-2\sqrt{14}}\)
\(a,\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\\ =\sqrt{3}+\sqrt{5}-\left(\sqrt{5}+1\right)=\sqrt{3}-1\\ b,=3-2\sqrt{2}-\left(3\sqrt{2}+1\right)=2-5\sqrt{2}\\ c,=\sqrt{7}-1+\sqrt{7}+1=2\sqrt{7}\\ d,=\sqrt{11}+1-\left(\sqrt{11}-1\right)=2\\ e,=\sqrt{7}-\sqrt{3}-\left(\sqrt{7}-\sqrt{2}\right)=\sqrt{2}-\sqrt{3}\)
Đúng 2
Bình luận (1)
Thực hiện phép tính:a,left(sqrt{28}-2sqrt{14}+sqrt{7}right)cdotsqrt{7}+7sqrt{8}b,left(sqrt{8}-3sqrt{2}+sqrt{10}right)cdotleft(sqrt{2}-3sqrt{0.4}right)c,left(15sqrt{50}+5sqrt{200}-3sqrt{450}right):sqrt{10}d,sqrt{6+2sqrt{5}}+sqrt{6-2sqrt{5}}e,sqrt{11+6sqrt{2}}-sqrt{11-6sqrt{2}}f,sqrt[3]{5sqrt{2}+7}-sqrt[3]{5sqrt{2}-7}g,sqrt[3]{20+14sqrt{2}}+sqrt[3]{20-14sqrt{2}}h,sqrt[3]{26+15sqrt{3}}-sqrt[3]{26-15sqrt{3}}
Đọc tiếp
Thực hiện phép tính:
\(a,\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{8}\)
\(b,\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\cdot\left(\sqrt{2}-3\sqrt{0.4}\right)\)
\(c,\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}\)
\(d,\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)
\(e,\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(f,\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)
\(g,\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\)
\(h,\sqrt[3]{26+15\sqrt{3}}-\sqrt[3]{26-15\sqrt{3}}\)
g, h. Câu hỏi của Nữ hoàng sến súa là ta - Toán lớp 9 - Học toán với OnlineMath
Dạng 3.Chứng minh đẳng thức
Bài 1: CM
a)\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}=2\)
b)\(\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}=8\)
Bài 2 :CM
\(\dfrac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{2}}=\sqrt{\sqrt{5}+1}\)
Bài 1
a) Đặt VT = A
<=> \(2\sqrt{2}A=\left(8+2\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{8-2\sqrt{15}}\)
<=> \(2\sqrt{2}A=\left(\sqrt{5}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right).\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
<=> \(2A=\left(\sqrt{5}+\sqrt{3}\right)^2.\left(\sqrt{5}-\sqrt{3}\right)^2\)
<=> 2A = \(\left(5-3\right)^2=4\)
<=> A = 2
b) Đặt VT = B
<=> \(2\sqrt{2}B=\left(10+2\sqrt{21}\right).\left(\sqrt{14}-\sqrt{6}\right)\sqrt{10-2\sqrt{21}}\)
<=> \(2\sqrt{2}B=\left(\sqrt{7}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{7}-\sqrt{3}\right).\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)
<=> \(2B=\left(\sqrt{7}+\sqrt{3}\right)^2.\left(\sqrt{7}-\sqrt{3}\right)^2=\left(7-3\right)^2=16\)
<=> B = 8
Bài 2
Đặt VT = A
<=> A2 = \(\dfrac{\sqrt{5}+2+\sqrt{5}-2+2\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}{2}\)
<=> A2 = \(\dfrac{2\sqrt{5}+2\sqrt{5-4}}{2}=\dfrac{2\sqrt{5}+2}{2}=\sqrt{5}+1\)
<=> \(A=\sqrt{\sqrt{5}+1}\)
Đúng 0
Bình luận (0)