Casio Toán lớp 7: tính
a, \(A=\sqrt{1}+\sqrt{2}+\sqrt{3}+.....\sqrt{50}\)
b, \(B=\frac{1}{1}+\frac{1}{2}+...+\frac{1}{20}\)
c, \(C=2.5+5.8+8.11+...+62.65\)
Tính:
a) \(A=\sqrt{8-2\sqrt{15}}\left(\sqrt{3}+\sqrt{5}\right)-\left(\sqrt{45}-\sqrt{20}\right)\)
b) \(B=\left(\frac{\sqrt{21}-\sqrt{3}}{\sqrt{7}-1}-\frac{\sqrt{15}-\sqrt{3}}{1-\sqrt{5}}\right)\left(\frac{1}{2}\sqrt{6}-\sqrt{\frac{3}{2}}+3\sqrt{\frac{2}{3}}\right)\)
c) \(C=2\sqrt{3}+\sqrt{7-4\sqrt{3}}+\left(\sqrt{\frac{1}{3}}-\sqrt{\frac{4}{3}+}\sqrt{3}\right):\sqrt{3}\)
d) \(D=\left(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\right):\frac{1}{\sqrt{7-4\sqrt{3}}}\)
a) Ta có: \(A=\sqrt{8-2\sqrt{15}}\cdot\left(\sqrt{3}+\sqrt{5}\right)-\left(\sqrt{45}-\sqrt{20}\right)\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\left(\sqrt{9}-\sqrt{4}\right)\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)
\(=\left|\sqrt{5}-\sqrt{3}\right|\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)
\(=\left(\sqrt{5}-\sqrt{3}\right)\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)(Vì \(\sqrt{5}>\sqrt{3}\))
\(=5-3-\sqrt{5}\)
\(=2-\sqrt{5}\)
b) Ta có: \(B=\left(\frac{\sqrt{21}-\sqrt{3}}{\sqrt{7}-1}-\frac{\sqrt{15}-\sqrt{3}}{1-\sqrt{5}}\right)\left(\frac{1}{2}\sqrt{6}-\sqrt{\frac{3}{2}}+3\sqrt{\frac{2}{3}}\right)\)
\(=\left(\frac{\sqrt{3}\left(\sqrt{7}-1\right)}{\sqrt{7}-1}+\frac{\sqrt{3}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{3}{2}}+\sqrt{6}\right)\)
\(=\sqrt{3}+\sqrt{3}+\sqrt{6}\)
\(=2\sqrt{3}+\sqrt{6}\)
c) Ta có: \(C=2\sqrt{3}+\sqrt{7-4\sqrt{3}}+\left(\sqrt{\frac{1}{3}}-\sqrt{\frac{4}{3}}+\sqrt{3}\right):\sqrt{3}\)
\(=2\sqrt{3}+\sqrt{4-2\cdot2\cdot\sqrt{3}+3}+\sqrt{\frac{1}{3}:3}-\sqrt{\frac{4}{3}:3}+\sqrt{3:3}\)
\(=2\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\frac{1}{9}}-\sqrt{\frac{4}{9}}+\sqrt{1}\)
\(=2\sqrt{3}+\left|2-\sqrt{3}\right|+\frac{1}{3}-\frac{2}{3}+1\)
\(=2\sqrt{3}+2-\sqrt{3}+\frac{2}{3}\)(Vì \(2>\sqrt{3}\))
\(=\sqrt{3}+\frac{8}{3}\)
d) Ta có: \(D=\left(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\right):\frac{1}{\sqrt{7-4\sqrt{3}}}\)
\(=\left(\frac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\right)\cdot\sqrt{4-2\cdot2\cdot\sqrt{3}+3}\)
\(=\frac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\cdot\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\frac{60}{20}\cdot\left|2-\sqrt{3}\right|\)
\(=3\cdot\left(2-\sqrt{3}\right)\)(Vì \(2>\sqrt{3}\))
\(=6-3\sqrt{3}\)
J = \(\left(1+\frac{2+\sqrt{2}}{1+\sqrt{2}}\right)\) . \(\left(1-\frac{2-\sqrt{2}}{1-\sqrt{2}}\right)\)
M = \(\frac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}:\frac{1}{\sqrt{6}}\)
N = \(\frac{6}{1+\sqrt{7}}+\frac{1}{\sqrt{7}}\)
O = \(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{1+\sqrt{2}}-\frac{1}{2-\sqrt{3}}\)
Q = \(\left(\frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\frac{5}{\sqrt{5}}\right).\left(\sqrt{5}-\sqrt{2}\right)\)
S = \(\frac{2}{\sqrt{5}+1}-\sqrt{\frac{2}{3-\sqrt{5}}}\)
Các thầy cô, các bạn giúp em với ạ. Em cảm ơn !
ụa, không có ai ạ?
em đang cần gấp lắm ;-;;
1. thực hiện phép tính
a.\(\frac{3+\sqrt{7}}{3-\sqrt{7}}-\frac{3-\sqrt{7}}{3+\sqrt{7}}\)
b,\(\left(\frac{\sqrt{2}+\sqrt{5}}{\sqrt{2}-5}-\frac{\sqrt{2}-5}{\sqrt{2}+5}\right):\frac{\sqrt{2}}{23}\)
c,\(5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{20}+\sqrt{5}\)
d,\(\sqrt{\frac{1}{2}}+\sqrt{4,5}+12,5\)
e, \(\frac{1}{2}\sqrt{48}-2\sqrt{75}-\sqrt{54}+5\sqrt{1\frac{1}{3}}\)
Bài 1: Thực hiện phép tính
a) Ta có: \(\frac{3+\sqrt{7}}{3-\sqrt{7}}-\frac{3-\sqrt{7}}{3+\sqrt{7}}\)
\(=\frac{\left(3+\sqrt{7}\right)^2}{\left(3-\sqrt{7}\right)\left(3+\sqrt{7}\right)}-\frac{\left(3-\sqrt{7}\right)^2}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}\)
\(=\frac{9+6\sqrt{7}+7-\left(9-6\sqrt{7}+7\right)}{9-7}\)
\(=\frac{16+6\sqrt{7}-16+6\sqrt{7}}{2}\)
\(=\frac{12\sqrt{7}}{2}=6\sqrt{7}\)
b)Sửa đề: \(\left(\frac{\sqrt{2}+5}{\sqrt{2}-5}-\frac{\sqrt{2}-5}{\sqrt{2}+5}\right):\frac{\sqrt{2}}{23}\)
Ta có: \(\left(\frac{\sqrt{2}+5}{\sqrt{2}-5}-\frac{\sqrt{2}-5}{\sqrt{2}+5}\right):\frac{\sqrt{2}}{23}\)
\(=\left(\frac{\left(\sqrt{2}+5\right)^2}{\left(\sqrt{2}-5\right)\left(\sqrt{2}+5\right)}-\frac{\left(\sqrt{2}-5\right)^2}{\left(\sqrt{2}+5\right)\left(\sqrt{2}-5\right)}\right)\cdot\frac{23}{\sqrt{2}}\)
\(=\left(\frac{27+10\sqrt{2}-\left(27-10\sqrt{2}\right)}{2-25}\right)\cdot\frac{23}{\sqrt{2}}\)
\(=\frac{27+10\sqrt{2}-27+10\sqrt{2}}{-23}\cdot\frac{23}{\sqrt{2}}\)
\(=\frac{20\sqrt{2}}{-\sqrt{2}}=-20\)
c) Ta có: \(5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{20}+\sqrt{5}\)
\(=\sqrt{25\cdot\frac{1}{5}}+\frac{1}{2}\cdot2\sqrt{5}+\sqrt{5}\)
\(=\sqrt{5}+\sqrt{5}+\sqrt{5}\)
\(=3\sqrt{5}\)
d) Ta có: \(\sqrt{\frac{1}{2}}+\sqrt{4.5}+12.5\)
\(=\frac{1}{\sqrt{2}}+\frac{3}{\sqrt{2}}+12.5\)
\(=2\sqrt{2}+12.5\)
e) Ta có: \(\frac{1}{2}\sqrt{48}-2\sqrt{75}-\sqrt{54}+5\sqrt{1\frac{1}{3}}\)
\(=\frac{1}{2}\cdot4\sqrt{3}-2\cdot5\sqrt{3}-3\sqrt{6}+5\cdot\sqrt{\frac{4}{3}}\)
\(=2\sqrt{3}-10\sqrt{3}-3\sqrt{6}+\frac{10}{\sqrt{3}}\)
\(=-8\sqrt{3}+\frac{10}{\sqrt{3}}-3\sqrt{6}\)
\(=\frac{-24+10}{\sqrt{3}}-\frac{9\sqrt{2}}{\sqrt{3}}\)
\(=\frac{-14-9\sqrt{2}}{\sqrt{3}}\)
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. A=(\(\frac{3x+16\sqrt{x}-7}{x+2\sqrt{x}-3}-\frac{\sqrt{x}+1}{\sqrt{x}+3}-\frac{\sqrt{x}+7}{\sqrt{x}-1}\)) : (\(2-\frac{\sqrt{x}}{\sqrt{x}-1}\))
b. B=(\(\frac{\sqrt{x}+1}{\sqrt{xy}+1}+\frac{\sqrt{xy}+\sqrt{x}}{1-\sqrt{xy}}+1\)) :( 1-\(\frac{\sqrt{xy}+\sqrt{x}}{\sqrt{xy}-1}-\frac{\sqrt{x}+1}{\sqrt{xy}+1}\))
c. C=( \(\frac{\sqrt{x}-4x}{1+4x}-1\)):(\(\frac{1+2x}{1-4x}-\frac{2\sqrt{x}}{2\sqrt{x}}-1\))
d. D=(\(\frac{\sqrt{a-b}}{\sqrt{a+b}+\sqrt{a+b}}+\frac{a-b}{\sqrt{a^2-b^2}-a+b}\))\(\frac{a^2+b^2}{\sqrt{a^2-b^2}}\)
e. E=\(\frac{\left(\sqrt{a}-\sqrt{b}\right)+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}-b\)
a) Cho \(A=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{25}}\)
Chứng minh : 7 < A < 8
b) Chứng minh : \(5\sqrt{2}< 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{50}}< 10\sqrt{2}\)
a.\(\frac{1}{\sqrt{n}}=\frac{2}{\sqrt{n}+\sqrt{n}}>\frac{2}{\sqrt{n}+\sqrt{n+1}}=\frac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{n+1-n}=2\left(\sqrt{n+1}+\sqrt{n}\right)\)
áp dụng công thức cho biểu thức A có A>\(2\left(-\sqrt{2}+\sqrt{26}\right)>7\left(1\right)\)
(so sánh bình phương 2 số sẽ ra nha)
\(\frac{1}{\sqrt{n}}=\frac{2}{\sqrt{n}+\sqrt{n}}< \frac{2}{\sqrt{n}+\sqrt{n-1}}=\frac{2\left(\sqrt{n}-\sqrt{n-1}\right)}{n-n+1}=2\left(\sqrt{n}-\sqrt{n-1}\right)\)
áp dụng công thức cho biểu thức A ta CM được
A<\(2\left(\sqrt{2}-\sqrt{2-1}+\sqrt{3}-\sqrt{3-1}+...+\sqrt{25}-\sqrt{25-1}\right)\)
=\(2\left(-\sqrt{1}+\sqrt{25}\right)=2\left(-1+5\right)=2\cdot4=8\left(2\right)\)
từ (1) và (2) => ĐPCM
b. tương tự câu a ta CM đc BT đã cho=B>\(2\sqrt{51}-2\)> \(5\sqrt{2}\left(1\right)\)
và B<\(2\sqrt{50}=\sqrt{2}\cdot\sqrt{2\cdot50}=10\sqrt{2}\left(2\right)\)
từ (1) và (2)=>ĐPCM
(bạn nhớ phải biến đổi 1 thành 1/\(\sqrt{1}\) trc khi áp dụng công thức nha)
MỜI BẠN THAM KHẢO
Bài 1 :Chứng minh các đẳng thức :
a ) \(2\sqrt{2}\left(\sqrt{3}-2\right)\) + \(\left(1+2\sqrt{2}\right)^2-2\sqrt{6}=9\)
b ) \(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\sqrt{6}\)
c ) \(\sqrt{11-6\sqrt{2}}+\sqrt{11+6\sqrt{2}}=6\)
Bài 2 : Rút gọn các biểu thức sau :
a ) \(\frac{1}{\sqrt{5}+\sqrt{3}}-\frac{1}{\sqrt{5}-\sqrt{3}}\)
b ) \(\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{6}-\sqrt{2}}\)
c ) \(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}\)
Bài 3 : Rút gọn các biểu thức sau :
a ) \(\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}\)
b ) \(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}\)
c ) \(\left(\sqrt{6}+\sqrt{5}\right)^2-\sqrt{120}\)
d ) \(\left(\frac{1}{2}\sqrt{\frac{1}{2}}-\frac{3}{2}\sqrt{2}+\frac{4}{5}\sqrt{200}\right):\frac{1}{8}\)
\(2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}=9\)
Bài 1: CMR:
a, (4+\(\sqrt{3}\)). (4-\(\sqrt{3}\))=13
b, \(\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}=2\)
c, \(\frac{\sqrt{1}}{2+\sqrt{3}}+\frac{\sqrt{1}}{2-\sqrt{3}}=4\)
d, \(\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\frac{1}{\sqrt{a}-\sqrt{b}}=a-b\)(a>0, b>0, a≠b)
Bài 2: CMR:
a, \(\sqrt{a}+\frac{\sqrt{1}}{\sqrt{a}}\ge2\left(a>0\right)\)
b, a+b+\(\frac{1}{2}\ge\sqrt{a}+\sqrt{b}\left(a,b>0\right)\)
c, \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{1}{\sqrt{xyz}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{zx}}\left(x,y,z>0\right)\)
d, \(\frac{\sqrt{3}+2}{\sqrt{3}-2}-\frac{\sqrt{3}-2}{\sqrt{3}+2}=-8\sqrt{3}\)
e, \(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}}:\frac{1}{\sqrt{a}-\sqrt{b}}\)=a-b(a>0, b>0, a≠b)
Bài 3: Tìm Min hoặc Max(nếu có):
a, \(\sqrt{x^2+9}\)
b, \(\frac{2}{\sqrt{x^2+1}}\)
c, 1-\(\sqrt{5+2x-x^2}\)
2/
a/ \(\sqrt{a}+\frac{1}{\sqrt{a}}\ge2\sqrt{\sqrt{a}.\frac{1}{\sqrt{a}}}=2\), dấu "=" khi \(a=1\)
b/ \(a+b+\frac{1}{2}=a+\frac{1}{4}+b+\frac{1}{4}\ge2\sqrt{a.\frac{1}{4}}+2\sqrt{b.\frac{1}{4}}=\sqrt{a}+\sqrt{b}\)
Dấu "=" khi \(a=b=\frac{1}{4}\)
c/ Có lẽ bạn viết đề nhầm, nếu đề đúng thế này thì mình ko biết làm
Còn đề như vậy: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\) thì làm như sau:
\(\frac{1}{x}+\frac{1}{y}\ge\frac{2}{\sqrt{xy}}\) ; \(\frac{1}{y}+\frac{1}{z}\ge\frac{2}{\sqrt{yz}}\); \(\frac{1}{x}+\frac{1}{z}\ge\frac{2}{\sqrt{yz}}\)
Cộng vế với vế ta được:
\(\frac{2}{x}+\frac{2}{y}+\frac{2}{z}\ge\frac{2}{\sqrt{xy}}+\frac{2}{\sqrt{yz}}+\frac{2}{\sqrt{xz}}\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\)
Dấu "=" khi \(x=y=z\)
d/ \(\frac{\sqrt{3}+2}{\sqrt{3}-2}-\frac{\sqrt{3}-2}{\sqrt{3}+2}=\frac{\left(\sqrt{3}+2\right)\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\frac{\left(\sqrt{3}-2\right)\left(\sqrt{3}-2\right)}{\left(\sqrt{3}+2\right)\left(\sqrt{3}-2\right)}\)
\(=\frac{7+4\sqrt{3}}{3-4}-\frac{7-4\sqrt{3}}{3-4}=-7-4\sqrt{3}+7-4\sqrt{3}=-8\sqrt{3}\)
e/ \(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}}:\frac{1}{\sqrt{a}-\sqrt{b}}=\frac{\left(\sqrt{a}\right)^3+\left(\sqrt{b}\right)^3}{\sqrt{ab}}.\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}=\frac{\left(a-b\right)\left(a+b-\sqrt{ab}\right)}{\sqrt{ab}}\)
\(=\frac{a^2-b^2}{\sqrt{ab}}-\left(a-b\right)\) (bạn chép đề sai)
Cho a,b,c là các số dương, Cm:
\(\frac{1}{\sqrt{a}+3\sqrt{b}}+\frac{1}{\sqrt{b}+3\sqrt{c}}+\frac{1}{\sqrt{c}+3\sqrt{a}}\ge\frac{1}{\sqrt{a}+2\sqrt{b}+\sqrt{c}}+\frac{1}{\sqrt{b}+2\sqrt{c}+\sqrt{a}}+\frac{1}{\sqrt{c}+2\sqrt{a}+\sqrt{b}}\)
Giúp Mình Với các bạn ơi !!!!!
Chứng minh gì vậy bạn