Rút gọn biểu thức : A=\(\frac{5a^2-33a+60}{a\left(a-1\right)\left(a-4\right)\left(a-5\right)}\)
Những câu hỏi liên quan
Rút gọn phân thức A= \(\frac{\left(a+2\right)^2\left(5a-15a^2\right)}{\left(a-3\right)\left(4a-a^3\right)}\)
\(A=\frac{\left(a+2\right)^2\left(5a-15a^2\right)}{\left(a-3\right)\left(4a-a^3\right)}=\frac{\left(a+2\right)^2.5a.\left(1-3a\right)}{\left(a-3\right).a.\left(2-a\right)\left(a+2\right)}\)
\(=\frac{\left(a+2\right).5.\left(1-3a\right)}{\left(a-3\right).\left(2-a\right)}\)
Đúng 0
Bình luận (0)
Rút gọn biểu thức A= \(\frac{\left(\frac{2}{3}\right)^3\cdot\left(-\frac{3}{4}\right)^2\cdot\left(-1\right)^{2017}}{\left(\frac{2}{5}\right)^2\cdot\left(-\frac{5}{12}\right)^3}-\frac{71}{5}\)
SGK Kết nối tri thức và cuộc sống
15 tháng 8 2023 lúc 19:00
Rút gọn biểu thức: \(A = \frac{{{{\left( {{a^{\sqrt 2 - 1}}} \right)}^{1 + \sqrt 2 }}}}{{{a^{\sqrt 5 - 1}}.{a^{3 - \sqrt 5 }}}}\,\,\,\left( {a > 0} \right).\)
\(=\dfrac{a^{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}}{a^{\left(\sqrt{5}-1\right)+\left(3-\sqrt{5}\right)}}=\dfrac{a}{a^{\sqrt{5}-1+3-\sqrt{5}}}=\dfrac{a}{a^2}=\dfrac{1}{a}\)
Đúng 1
Bình luận (0)
Rút gọn biểu thức
\(A=\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{n^2}\right)\)
\(A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{n^2}\right)\)
\(=\left(\frac{2^2-1}{2^2}\right)\left(\frac{3^2-1}{3^2}\right)\left(\frac{4^2-1}{4^2}\right)...\left(\frac{n^2-1}{n^2}\right)\)
\(=\text{[}\frac{\left(2-1\right)\left(2+1\right)}{2^2}\text{]}.\text{[}\frac{\left(3-1\right)\left(3+1\right)}{3^2}\text{]}.\text{[}\frac{\left(4-1\right)\left(4+1\right)}{4^2}\text{]}...\text{[}\frac{\left(n-1\right)\left(n+1\right)}{n^2}\text{]}\)
\(=\left(\frac{1.3}{2^2}\right).\left(\frac{2.4}{3^2}\right).\left(\frac{3.5}{4^2}\right)...\text{[}\frac{\left(n-1\right)\left(n+1\right)}{n^2}\text{]}\)
\(=\frac{\text{[}1.2.3...\left(n-1\right)\text{]}.\text{[}3.4.5...\left(n+1\right)\text{]}}{\text{[}2.3.4...n\text{]}.\text{[}2.3.4...n\text{]}}\)
\(=\frac{1}{n}.\frac{n+1}{2}\)
\(=\frac{n+1}{2n}\)
Đúng 0
Bình luận (0)
Cho hai biểu thức
\(A=\left(2x+3\right)\left(x-1\right)-\left(x+1\right)\left(2x-5\right)-2\)
\(B=\left(x-4\right)\left(x-2\right)-\left(3x+1\right)\left(\frac{1}{3}x-2\right)+2\frac{1}{3}x-10\)
a) Rút gọn các biểu thức đã cho
b) Tìm công thức liên hệ giữa hai biểu thức A và B đã cho
Rút gọn các biểu thức sau :a) Aleft(0,04right)^{-1,5}-left(0,125right)^{frac{-2}{3}}b) Bleft(6^{frac{-2}{7}}right)^{-7}-left[left(left(0,2right)^{0,75}right)^{-4}right]c) Cfrac{a^{sqrt{5}+3}.a^{sqrt{5}left(sqrt{5}-1right)}}{left(a^{2sqrt{2}-1}right)^{2sqrt{2}+1}}d) Dleft(a^{frac{1}{2}}-b^{frac{1}{2}}right)^2:left(b-2bsqrt{frac{b}{a}}+frac{b^2}{a}right)left(a,b0right)
Đọc tiếp
Rút gọn các biểu thức sau :
a) \(A=\left(0,04\right)^{-1,5}-\left(0,125\right)^{\frac{-2}{3}}\)
b) \(B=\left(6^{\frac{-2}{7}}\right)^{-7}-\left[\left(\left(0,2\right)^{0,75}\right)^{-4}\right]\)
c) \(C=\frac{a^{\sqrt{5}+3}.a^{\sqrt{5}\left(\sqrt{5}-1\right)}}{\left(a^{2\sqrt{2}-1}\right)^{2\sqrt{2}+1}}\)
d) \(D=\left(a^{\frac{1}{2}}-b^{\frac{1}{2}}\right)^2:\left(b-2b\sqrt{\frac{b}{a}}+\frac{b^2}{a}\right)\left(a,b>0\right)\)
a) \(A=\left[\left(\frac{1}{5}\right)^2\right]^{\frac{-3}{2}}-\left[2^{-3}\right]^{\frac{-2}{3}}=5^3-2^2=121\)
b) \(B=6^2+\left[\left(\frac{1}{5}\right)^{\frac{3}{4}}\right]^{-4}=6^2+5^3=161\)
c) \(C=\frac{a^{\sqrt{5}+3}.a^{\sqrt{5}\left(\sqrt{5}-1\right)}}{\left(a^{2\sqrt{2}-1}\right)^{2\sqrt{2}+1}}=\frac{a^{\sqrt{5}+3}.a^{5-\sqrt{5}}}{a^{\left(2\sqrt{2}\right)^2-1^2}}\)
\(=\frac{a^{\sqrt{5}+3+5-\sqrt{5}}}{a^{8-1}}=\frac{a^8}{a^7}=a\)
d) \(D=\left(a^{\frac{1}{2}}-b^{\frac{1}{2}}\right)^2:\left(b-2b\sqrt{\frac{b}{a}}+\frac{b^2}{a}\right)\)
\(=\left(\sqrt{a}-\sqrt{b}\right)^2:b\left[1-2\sqrt{\frac{b}{a}}+\left(\sqrt{\frac{b}{a}}\right)^2\right]\)
\(=\left(\sqrt{a}-\sqrt{b}\right)^2:b\left(1-\sqrt{b}a\right)^2\)
Đúng 0
Bình luận (0)
Bài 6: Rút gọn biểu thức:
\(A=\frac{a^3-3a+\left(a^2-1\right)\sqrt{a^2-4}-2}{a^3-3a+\left(a^2-1\right)\sqrt{a^2-4}+2}\left(a>2\right)\)
\(B=\sqrt{\frac{1}{a^2+b^2}+\frac{1}{\left(a+b\right)^2}+\sqrt{\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{\left(a^2+b^2\right)^2}}}\left(ab\ne0\right)\)
rút gọn biểu thức sau
\(log_2\left(2a^2\right)+\left(log_2^a\right)a^{log_a\left(log_2^a+1\right)}+\frac{1}{2}log^2_2a^4\)
ta có:
\(log^{\left(2a^2\right)}_2+\left(log_2^a\right)a^{log_a^{\left(log^a_1+1\right)}}+\frac{1}{2}log^2_2a^4=log_2^2+log_2^{a^2}+log_2^a\left(log^a_2+1\right)+\frac{1}{2}log^2_2a^4\)
\(=1+2log^a_2+log^a_2\left(1+log^a_2\right)+2log^2a_2\)
\(=3log^2_2a+3log^a_2+1\)
Đúng 0
Bình luận (0)
Rút gọn biểu thức
a)\(\frac{\left(\frac{2}{3}\right)^3.\left(-\frac{3}{4}\right)^2.\left(-1\right)^5}{\left(\frac{2}{5}\right)^2.\left(-\frac{5}{12}\right)^2}\)
b)\(6^6+6^3.3^3+3^6\)/-73