Rút gọn:
\(C=\left[\left(\dfrac{1}{a^2+1}\right)\cdot\dfrac{1}{a^2+2a+1}+\dfrac{2}{\left(a+1\right)^3}\cdot\left(\dfrac{1}{a}+1\right)\right]:\dfrac{a-1}{a^3}\)
Những câu hỏi liên quan
Rút gọn:
\(A=\left[\left(\dfrac{3}{1+x}-\dfrac{x}{x^2+x+1}\right):\dfrac{2x^2+3x}{x+1}+\dfrac{3}{x+1}\right]\cdot\dfrac{x^2+x}{1+3x}\)
\(B=\left[\dfrac{a}{2a-6}-\dfrac{a^2}{a^2-9}+\dfrac{a}{2a-9}\cdot\left(\dfrac{3}{a}+\dfrac{1}{3-a}\right)\right]:\dfrac{a^2-5a-6}{18-2a^2}\)
Câu 1:Cdfrac{1}{x+2}-dfrac{x^3-4x}{x^2+4}cdotleft(dfrac{1}{x^2+4x+4}-dfrac{1}{4-x^2}right)a) Rút gọn Cb) x bằng mấy để C 1?Câu 2:Bleft(dfrac{1}{sqrt{x}-1}-dfrac{1}{sqrt{x}}right):left(dfrac{sqrt{x}+1}{sqrt{x}-2}-dfrac{sqrt{x}+2}{sqrt{x}-1}right)a) Rút gọn Bb) x bằng mấy để left|Bright|BCâu 3: Rút gọn:Aleft[dfrac{left(1-aright)^2}{3a+left(a-1right)^2}+dfrac{2a^2-4a-1}{a^3-1}-dfrac{1}{1-a}right]:dfrac{2a}{a^3+a}
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Câu 1:
\(C=\dfrac{1}{x+2}-\dfrac{x^3-4x}{x^2+4}\cdot\left(\dfrac{1}{x^2+4x+4}-\dfrac{1}{4-x^2}\right)\)
a) Rút gọn C
b) x bằng mấy để C = 1?
Câu 2:
\(B=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
a) Rút gọn B
b) x bằng mấy để \(\left|B\right|=B\)
Câu 3: Rút gọn:
\(A=\left[\dfrac{\left(1-a\right)^2}{3a+\left(a-1\right)^2}+\dfrac{2a^2-4a-1}{a^3-1}-\dfrac{1}{1-a}\right]:\dfrac{2a}{a^3+a}\)
Rút gọn:
\(A=\left(\dfrac{1-2a}{1+2a}-\dfrac{1-4a+4a^2}{1+2a}\cdot\dfrac{1}{1-4a^2}\right)\cdot\left(\dfrac{a+1}{a}+\dfrac{1}{4a^2}\right)-\dfrac{1}{2a}\)
\(A=\left(\dfrac{-\left(2a-1\right)}{2a+1}+\dfrac{\left(2a-1\right)^2}{2a+1}\cdot\dfrac{1}{\left(2a-1\right)\left(2a+1\right)}\right)\cdot\left(\dfrac{4a\left(a+1\right)+1}{4a^2}\right)-\dfrac{1}{2a}\)
\(=\left(\dfrac{-\left(2a-1\right)}{2a+1}+\dfrac{2a-1}{\left(2a+1\right)^2}\right)\cdot\dfrac{4a^2+4a+1}{4a^2}-\dfrac{1}{2a}\)
\(=\dfrac{-\left(2a-1\right)\left(2a+1\right)}{\left(2a+1\right)^2}\cdot\dfrac{\left(2a+1\right)^2}{4a^2}-\dfrac{1}{2a}\)
\(=\dfrac{-\left(4a^2-1\right)}{4a^2}-\dfrac{2a}{4a^2}\)
\(=\dfrac{-4a^2-2a+1}{4a^2}\)
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P=\(\left(\dfrac{3\left(x+2\right)}{2x^2+8}-\dfrac{2x^2-x-10}{\left(x+1\right)\left[\left(x+1\right)^2-2x\right]}\right):\left(\dfrac{5}{x^2+1}+\dfrac{3}{2\left(x+1\right)}-\dfrac{3}{x-1}\right)\cdot\dfrac{2}{x-1}\)
a) rút gọn P
b)tìm tất cả các giá trị nguyên của x để P có giá trị là bội của 4
a: \(P=\left(\dfrac{3x+6}{2\left(x^2+4\right)}-\dfrac{2x^2-x-10}{\left(x+1\right)\left(x^2+1\right)}\right):\left(\dfrac{10\left(x^2-1\right)+3\left(x^2+1\right)\left(x-1\right)-6\left(x+1\right)\left(x^2+1\right)}{\left(x^2+1\right)\left(x+1\right)\left(x-1\right)\cdot2}\right)\cdot\dfrac{2}{x-1}\)
\(=\left(\dfrac{\left(3x+6\right)\left(x^3+x^2+x+1\right)-\left(2x^2+8\right)\left(2x^2-x-10\right)}{2\left(x^2+4\right)\left(x+1\right)\left(x^2+1\right)}\right)\cdot\dfrac{\left(x^2+1\right)\left(x-1\right)\left(x+1\right)\cdot2}{-3x^3+x^2-3x-13}\cdot\dfrac{2}{x-1}\)
\(=\dfrac{-x^4+11x^3+13x^2+17x+16}{\left(x^2+4\right)}\cdot\dfrac{2}{-3x^3+x^2-3x-13}\)
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Tính giá trị các biểu thức sau theo cách hợp lí nhất.
a) $mathrm{A}left(dfrac{2}{7} cdot dfrac{1}{4}-dfrac{1}{3} cdot dfrac{2}{7}right):left(dfrac{2}{7} cdot dfrac{3}{9}-dfrac{2}{7} cdot dfrac{2}{5}right)$;
b) $mathrm{B}dfrac{left(dfrac{1}{5}-dfrac{2}{7}right) cdot dfrac{3}{4}-dfrac{3}{4} cdotleft(dfrac{1}{3}-dfrac{2}{7}right)}{dfrac{1}{5} cdot dfrac{2}{7}-dfrac{1}{3} cdotleft(dfrac{2}{7}+dfrac{3}{9}right)+dfrac{3}{9} cdot dfrac{1}{5}} .$
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Tính giá trị các biểu thức sau theo cách hợp lí nhất.
a) $\mathrm{A}=\left(\dfrac{2}{7} \cdot \dfrac{1}{4}-\dfrac{1}{3} \cdot \dfrac{2}{7}\right):\left(\dfrac{2}{7} \cdot \dfrac{3}{9}-\dfrac{2}{7} \cdot \dfrac{2}{5}\right)$;
b) $\mathrm{B}=\dfrac{\left(\dfrac{1}{5}-\dfrac{2}{7}\right) \cdot \dfrac{3}{4}-\dfrac{3}{4} \cdot\left(\dfrac{1}{3}-\dfrac{2}{7}\right)}{\dfrac{1}{5} \cdot \dfrac{2}{7}-\dfrac{1}{3} \cdot\left(\dfrac{2}{7}+\dfrac{3}{9}\right)+\dfrac{3}{9} \cdot \dfrac{1}{5}} .$
Xem thêm câu trả lời
1, so sánh A;B biết: A=\(\left(\dfrac{\left(3\cdot\dfrac{2}{15}+\dfrac{1}{5}\right):2\cdot\dfrac{1}{2}}{\left(5\cdot\dfrac{3}{7}-2\cdot\dfrac{1}{4}\right):\dfrac{443}{56}}\right);B=\dfrac{1,2:\left(1\cdot\dfrac{1}{5}.1\cdot\dfrac{1}{4}\right)}{0,32+\dfrac{2}{25}}\)
1left(dfrac{sqrt{a}-2}{sqrt{a}+2}-dfrac{sqrt{a}+2}{sqrt{a}-2}right)cdotleft(sqrt{a}cdotdfrac{4}{sqrt{a}}right) (với a0,a≠4)
2left(dfrac{1}{1-sqrt{a}}-dfrac{1}{1+sqrt{a}}right)cdotleft(1-dfrac{1}{sqrt{a}}right) (với a0,a≠1)
3left(dfrac{sqrt{a}-1}{sqrt{a}+1}+dfrac{sqrt{a}+1}{sqrt{a}-1}right)cdotleft(1-dfrac{2}{a+1}right)^2 (với a0,a≠1)
Rút gọn các biểru thức sau
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1\(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right)\cdot\left(\sqrt{a}\cdot\dfrac{4}{\sqrt{a}}\right)\) (với a>0,a≠4)
2\(\left(\dfrac{1}{1-\sqrt{a}}-\dfrac{1}{1+\sqrt{a}}\right)\cdot\left(1-\dfrac{1}{\sqrt{a}}\right)\) (với a>0,a≠1)
3\(\left(\dfrac{\sqrt{a}-1}{\sqrt{a}+1}+\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\right)\cdot\left(1-\dfrac{2}{a+1}\right)^2\) (với a>0,a≠1)
Rút gọn các biểru thức sau
1. \(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right).\left(\sqrt{a}.\dfrac{4}{\sqrt{a}}\right)=\dfrac{\left(\sqrt{a}-2\right)^2-\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}.4=\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}.4=\dfrac{-64\sqrt{a}}{a-4}\)Nếu nhân tu thứ 2 của phép tính là \(\sqrt{a}-\dfrac{4}{\sqrt{a}}\) thì kết quả của phép tính là -16 nha bạn
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2.\(\left(\dfrac{1}{1-\sqrt{a}}-\dfrac{1}{1+\sqrt{a}}\right).\left(1-\dfrac{1}{\sqrt{a}}\right)=\dfrac{1+\sqrt{a}-1+\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}.\dfrac{-\left(1-\sqrt{a}\right)}{\sqrt{a}}=\dfrac{-2\sqrt{a}}{\left(1+\sqrt{a}\right)\sqrt{a}}=\dfrac{-2}{1+\sqrt{a}}\)\(\left(a>0,a\ne1\right)\)
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Bài 1:Tính
a, Adfrac{3}{4}cdotdfrac{8}{9}cdotdfrac{15}{16}cdot....cdotdfrac{9999}{10000}
b,Bleft(1-dfrac{1}{21}right)cdotleft(1-dfrac{1}{28}right)cdotleft(1-dfrac{1}{36}right)cdot....cdotleft(1-dfrac{1}{1326}right)
c,Cleft(1+dfrac{1}{1cdot3}right)cdotleft(1+dfrac{1}{2cdot4}right)cdotleft(1+dfrac{1}{3cdot5}right)cdot....cdotleft(1+dfrac{1}{99cdot101}right)
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Bài 1:Tính
a, A=\(\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot....\cdot\dfrac{9999}{10000}\)
b,B=\(\left(1-\dfrac{1}{21}\right)\cdot\left(1-\dfrac{1}{28}\right)\cdot\left(1-\dfrac{1}{36}\right)\cdot....\cdot\left(1-\dfrac{1}{1326}\right)\)
c,C=\(\left(1+\dfrac{1}{1\cdot3}\right)\cdot\left(1+\dfrac{1}{2\cdot4}\right)\cdot\left(1+\dfrac{1}{3\cdot5}\right)\cdot....\cdot\left(1+\dfrac{1}{99\cdot101}\right)\)
a)
\(A=\dfrac{3}{4}.\dfrac{8}{9}...\dfrac{9999}{10000}\)
\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}...\dfrac{99.101}{100.100}\)
\(=\dfrac{1.2...99}{2.3...100}.\dfrac{3.4...101}{2.3...100}\)
\(=\dfrac{1}{100}.\dfrac{101}{2}\)
\(=\dfrac{101}{200}\)
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26 tháng 8 2023 lúc 15:21
Rút gọn biểu thức
\(\left(1-\dfrac{1}{1+2}\right)\cdot\left(1-\dfrac{1}{1+2+3}\right)\cdot\cdot\cdot\cdot\left(1-\dfrac{1}{1+2+3+4+5+.....+2006}\right)\)
Giúp em với ạ
\(\left(1-\dfrac{1}{1+2}\right)\cdot\left(1-\dfrac{1}{1+2+3}\right)\cdot\left(\dfrac{1}{1+2+3+...+2006}\right)\)
\(=\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{6}\right)\cdot\left\{\dfrac{1}{\left(2006+1\right)\left[\left(2006-1\right):1+1\right]}\right\}\)
\(=\dfrac{2}{3}\cdot\dfrac{5}{6}\cdot\dfrac{1}{2007\cdot2006}\)
\(=\dfrac{10}{18}\cdot\dfrac{1}{4026042}\)
\(=\dfrac{5}{9}\cdot\dfrac{1}{4026042}\)
\(=\dfrac{5}{36234378}\)
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