Tam giác ABC vuông tại A , có AB=5/7BC. Giá trị của cosC bằng:
A. cosC=\(\frac{\sqrt{6}}{7}\) B. cosC=\(\frac{2\sqrt{6}}{7}\) C. cos C=\(\frac{3\sqrt{6}}{7}\)D. cos C =\(\frac{4\sqrt{6}}{7}\)
1. Tính:
a) \(\frac{\sqrt{7}-5}{2}-\frac{6-2\sqrt{7}}{4}+\frac{6}{\sqrt{7}-2}-\frac{5}{4+\sqrt{7}}\)
b) \(\frac{2}{\sqrt{6}-2}+\frac{2}{\sqrt{6}+2}+\frac{5}{\sqrt{6}}\)
c) \(\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}\)
d) \(\frac{2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\)
1) Rút gọn biểu thức:
a) \(\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}\)
b) \(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}-\sqrt{6}\)
c) \(\frac{5}{4-\sqrt{11}}+\frac{1}{3+\sqrt{7}}-\frac{6}{\sqrt{7}-2}-\frac{\sqrt{7}-5}{2}\)
d) \(\frac{4}{\sqrt{5}-\sqrt{2}}+\frac{3}{\sqrt{5}-2}-\frac{2}{\sqrt{3}-2}+\frac{\sqrt{3}-1}{6}\)
a) \(=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\frac{14}{49-48}=14\)
b) \(=\frac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}-\frac{5\sqrt{6}}{5}+\frac{4\sqrt{3}-12\sqrt{2}}{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}\)
cho a, b, c thỏa mãn \(\sqrt{a}+\sqrt{b}+\sqrt{c}=7;a+b+c=23;\sqrt{abc}=3\)
tính giá trị biểu thức \(A=\frac{1}{\sqrt{ab}+\sqrt{c}-6}+\frac{1}{\sqrt{bc}+\sqrt{a}-6}+\frac{1}{\sqrt{ca}+\sqrt{b}-6}\)
Câu hỏi của hoàng thị huyền trang - Toán lớp 9 - Học toán với OnlineMath
Em tham khảo nhé!
cho a,b,c thỏa mãn \(\sqrt{a}+\sqrt{b}+\sqrt{c}=7;a+b+c=23;\sqrt{abc}=3\)Tính giá trị biểu thức\(N=\frac{1}{\sqrt{ab}+\sqrt{c}-6}+\frac{1}{\sqrt{bc}+\sqrt{a}-6}+\frac{1}{\sqrt{ac}+\sqrt{b}-6}\)
Từ giả thiết: \(\sqrt{a}+\sqrt{b}+\sqrt{c}=7\Leftrightarrow\sqrt{c}=7-\sqrt{a}-\sqrt{b}\)
Xét hạng tử: \(\frac{1}{\sqrt{ab}+\sqrt{c}-6}=\frac{1}{\sqrt{ab}+7-\sqrt{a}-\sqrt{b}-6}=\frac{1}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)}\)
Từ đó: \(N=\frac{1}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)}+\frac{1}{\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)}+\frac{1}{\left(\sqrt{c}-1\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}-3}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)}=\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}-3}{\sqrt{abc}-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)+\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)-1}\)
\(=\frac{7-3}{3-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)+7-1}=\frac{4}{9-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)}\)
Mặt khác: \(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2-\left(a+b+c\right)}{2}=13\)
Suy ra: \(N=\frac{4}{9-13}=-1\). Kết luận: N = -1.
Từ giả thiết: \sqrt{a}+\sqrt{b}+\sqrt{c}=7\Leftrightarrow\sqrt{c}=7-\sqrt{a}-\sqrt{b}a+b+c=7⇔c=7−a−b
Xét hạng tử: \frac{1}{\sqrt{ab}+\sqrt{c}-6}=\frac{1}{\sqrt{ab}+7-\sqrt{a}-\sqrt{b}-6}=\frac{1}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)}ab+c−61=ab+7−a−b−61=(a−1)(b−1)1
Từ đó: N=\frac{1}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)}+\frac{1}{\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)}+\frac{1}{\left(\sqrt{c}-1\right)\left(\sqrt{a}-1\right)}N=(a−1)(b−1)1+(b−1)(c−1)1+(c−1)(a−1)1
=\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}-3}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)}=\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}-3}{\sqrt{abc}-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)+\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)-1}=(a−1)(b−1)(c−1)a+b+c−3=abc−(ab+bc+ca)+(a+b+c)−1a+b+c−3
=\frac{7-3}{3-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)+7-1}=\frac{4}{9-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)}=3−(ab+bc+ca)+7−17−3=9−(ab+bc+ca)4
Mặt khác: \sqrt{ab}+\sqrt{bc}+\sqrt{ca}=\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2-\left(a+b+c\right)}{2}=13ab+bc+ca=2(a+b+c)2−(a+b+c)=13
Suy ra: N=\frac{4}{9-13}=-1N=9−134=−1. Kết luận: N = -1.
Ta có \(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=a+b+c+2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\)
mà \(\sqrt{a}+\sqrt{b}+\sqrt{c}=7;a+b+c=23\)nên \(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=13\)
Ta có: \(\sqrt{a}+\sqrt{b}+\sqrt{c}=7\Rightarrow\sqrt{c}-6=-\sqrt{a}-\sqrt{b}+1\)
nên \(\sqrt{ab}+\sqrt{c}-6=\sqrt{ab}-\sqrt{a}-\sqrt{b}+1=\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)\)
Tương tự \(\hept{\begin{cases}\sqrt{bc}+\sqrt{a}-6=\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)\\\sqrt{ac}+\sqrt{b}-6=\left(\sqrt{a}-1\right)\left(\sqrt{c}-1\right)\end{cases}}\)
Vậy \(H=\frac{1}{\sqrt{ab}+\sqrt{c}-6}+\frac{1}{\sqrt{bc}+\sqrt{a}-6}+\frac{1}{\sqrt{ac}+\sqrt{b}-6}\)
\(=\frac{1}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)}+\frac{1}{\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)}+\frac{1}{\left(\sqrt{c}-1\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{\sqrt{c}-1+\sqrt{a}-1+\sqrt{b}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)-3}{\sqrt{abc}+\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)}=\frac{7-3}{3+7-13-1}=-1\)
Thực hiện các phép tính sau
a, \(\frac{\sqrt{7}-5}{2}-\frac{6-2\sqrt{7}}{4}+\frac{6}{\sqrt{7}-2}-\frac{5}{4+\sqrt{7}}\)
b, \(\frac{1}{\sqrt{3}+\sqrt{2}-\sqrt{5}}-\frac{1}{\sqrt{3}+\sqrt{2}+\sqrt{5}}\)
c, \(\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}\)
a, = \(\frac{\sqrt{7}-5}{2}-\frac{2\left(3-\sqrt{7}\right)}{4}+\frac{6\left(\sqrt{7}+2\right)}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}-\frac{5\left(4-\sqrt{7}\right)}{\left(4-\sqrt{7}\right)\left(4+\sqrt{7}\right)}\)
a, = \(=\frac{\sqrt{7}-5}{2}-\frac{3-\sqrt{7}}{2}+\frac{6\sqrt{7}+12}{7-4}-\frac{20-5\sqrt{7}}{16-7}=\frac{\sqrt{7}-5-3+\sqrt{7}}{2}+\frac{6\sqrt{7}+12}{3}-\frac{20-5\sqrt{7}}{9}\)
b. = \(\frac{\sqrt{3}+\sqrt{2}+\sqrt{5}}{\left(\sqrt{3}+\sqrt{2}+\sqrt{5}\right)\left(\sqrt{3}+\sqrt{2}-\sqrt{5}\right)}-\frac{\sqrt{3}+\sqrt{2}-\sqrt{5}}{\left(\sqrt{3}+\sqrt{2}+\sqrt{5}\right)\left(\sqrt{3}+\sqrt{2}-\sqrt{5}\right)}=\frac{\sqrt{3}+\sqrt{2}+\sqrt{5}}{\left(\sqrt{3}+\sqrt{2}\right)^2-\left(\sqrt{5}\right)^2}-\frac{\sqrt{3}+\sqrt{2}-\sqrt{5}}{\left(\sqrt{3}+\sqrt{2}\right)^2-\left(\sqrt{5}\right)^2}\)
Thể tích của khối chóp cụt tam giác đều có cạnh đáy lớn bằng \(2a\), cạnh đáy nhỏ bằng \(a\) và chiều cao bằng \(\frac{{a\sqrt 6 }}{3}\) là
A. \(\frac{{7\sqrt 2 }}{8}{a^3}\).
B. \(\frac{{\sqrt 2 }}{4}{a^3}\).
C. \(\frac{{7\sqrt 2 }}{{12}}{a^3}\).
D. \(\frac{{7\sqrt 3 }}{4}{a^3}\).
Diện tích đáy lớn là: \(S = \frac{{{{\left( {2{\rm{a}}} \right)}^2}\sqrt 3 }}{4} = {a^2}\sqrt 3 \)
Diện tích đáy bé là: \(S' = \frac{{{a^2}\sqrt 3 }}{4}\)
Thể tích của bồn chứa là: \(V = \frac{1}{3}.\frac{{a\sqrt 6 }}{3}\left( {{a^2}\sqrt 3 + \sqrt {{a^2}\sqrt 3 .\frac{{{a^2}\sqrt 3 }}{4}} + \frac{{{a^2}\sqrt 3 }}{4}} \right) = \frac{{7\sqrt 2 }}{{12}}{a^3}\)
Chọn C.
Tính
A/\(\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right).\frac{1}{\sqrt{6}}\)
B/ \(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)
C/ \(\frac{5}{4-\sqrt{11}}+\frac{1}{3+\sqrt{7}}-\frac{6}{\sqrt{7}-2}-\frac{\sqrt{7}-5}{2}\)
D/ \(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}\)
Tính
a/\(\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right).\frac{1}{\sqrt{6}}\)
b/\(\left(\frac{5}{4-\sqrt{11}}+\frac{1}{3+\sqrt{7}}-\frac{6}{\sqrt{7}-2}-\frac{\sqrt{7}-5}{2}\right)\)
c/\(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)
d/\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}\)
Tính
a/\(\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right).\frac{1}{\sqrt{6}}\)
b/\(\left(\frac{5}{4-\sqrt{11}}+\frac{1}{3+\sqrt{7}}-\frac{6}{\sqrt{7}-2}-\frac{\sqrt{7}-5}{2}\right)\)
c/\(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)
d/\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}\)
\(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\left(\frac{\sqrt{7}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}+\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\left(\frac{-\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\frac{-\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\left(-\sqrt{7}-\sqrt{5}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\frac{\sqrt{5}-\sqrt{7}}{\sqrt{7}+\sqrt{5}}=\frac{\left(\sqrt{5}-\sqrt{7}\right)\left(\sqrt{5}+\sqrt{7}\right)}{\left(\sqrt{7}+\sqrt{5}\right)^2}=\frac{2}{12+2\sqrt{35}}\)
\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{\left(\sqrt{5}-\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}+\frac{\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+3\right)}-\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{8-2\sqrt{15}}{2}+\frac{8+2\sqrt{15}}{2}-\frac{\left(\sqrt{5}+1\right)^2}{4}=8-\frac{6+2\sqrt{5}}{4}=\frac{26-2\sqrt{5}}{4}\)