\(\sqrt{25^2-24^2}=\sqrt{\left(25-24\right)\left(25+24\right)}=\sqrt{1.49}=\sqrt{49}=7\)
Đúng 5
Bình luận (0)
Chương I - Căn bậc hai. Căn bậc ba
Các câu hỏi tương tự
Tính giá trị biểu thức:
\(M=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{25\sqrt{24}+24\sqrt{25}}\)
Tính \(\dfrac{1}{\sqrt{25}+\sqrt{24}}+\dfrac{1}{\sqrt{24}+\sqrt{23}}+\dfrac{1}{\sqrt{23}+\sqrt{22}}+...+\dfrac{1}{\sqrt{2}+1}\)
Cho các biểu thức \(A=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+....\dfrac{1}{\sqrt{24}+\sqrt{25}}\)
a. Tính giá trị của A
Rút gọn :
M = \(\frac{1}{3.\left(\sqrt{1}+\sqrt{2}\right)}+\frac{1}{5.\left(\sqrt{2}+\sqrt{3}\right)}+\frac{1}{7.\left(\sqrt{3}+\sqrt{4}\right)}+....+\frac{1}{49.\left(\sqrt{24}+\sqrt{25}\right)}\)
Giải các phương trình sau:
\(a,\sqrt{1-4x+4x^2}=5\)
\(b,\sqrt{4-5x}=12\)
\(c,\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\)
\(d,\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
D sqrt{94-42sqrt{5}}-sqrt{94+42sqrt{5}}
A sqrt{sqrt{5}-sqrt{3-sqrt{29-12sqrt{5}}}}
C dfrac{2sqrt{4-sqrt{5+sqrt{21+sqrt{80}}}}}{sqrt{10}-sqrt{2}}
F sqrt{2+sqrt{3}}cdotsqrt{2+sqrt{2+sqrt{3}}cdotsqrt{2+sqrt{2+sqrt{2+sqrt{3}}}}}cdotsqrt{2-sqrt{2+sqrt{2+sqrt{3}}}}
B dfrac{1}{sqrt{1}+sqrt{2}}+dfrac{1}{sqrt{2}+sqrt{3}}+dfrac{1}{sqrt{3}+sqrt{4}}+...+dfrac{1}{sqrt{n-1}+sqrt{n}}
E dfrac{1}{sqrt{1}-sqrt{2}}-dfrac{1}{sqrt{2}-sqrt{3}}+dfrac{1}{sqrt{3}-sqrt{4}}-...-dfrac{1}{sqrt{24}-sqrt{25}}
Đọc tiếp
D = \(\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}\)
A = \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
C = \(\dfrac{2\sqrt{4-\sqrt{5+\sqrt{21+\sqrt{80}}}}}{\sqrt{10}-\sqrt{2}}\)
F = \(\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
B = \(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{n-1}+\sqrt{n}}\)
E = \(\dfrac{1}{\sqrt{1}-\sqrt{2}}-\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{1}{\sqrt{3}-\sqrt{4}}-...-\dfrac{1}{\sqrt{24}-\sqrt{25}}\)
\(S=\dfrac{\sqrt{2}-\sqrt{1}}{1+2}+\dfrac{\sqrt{3}-\sqrt{2}}{2+3}+\dfrac{\sqrt{4}-\sqrt{3}}{3+4}+...+\dfrac{\sqrt{25}-\sqrt{29}}{29+25}< \dfrac{2}{5}\)
Tính \(A=\sqrt{24-x^2}+\sqrt{8-x^2}\) biết \(\sqrt{24-x^2}-\sqrt{8-x^2}=2\)
Cho biểu thức:
A = \(\left(\frac{x+14\sqrt{x}-5}{x-24}+\frac{\sqrt{x}}{\sqrt{x}+5}\right):\frac{\sqrt{x}+2}{\sqrt{x}-5}\)
a) Rút gọn A vs x \(\ge\) 0; x \(\ne\) 25
b) Tìm x để A2 \(\ge\) A