cho A= (2x-1)^2 -5x.(x-1)+2.(x+1).(x-2) rut gon bieu thuc A
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24.
Ta có: \(3k^2+3k+1=k^3+3k^2+3k+1-k^3=\left(k+1\right)^3-k^3\)
Do đó \(a_k=\frac{\left(k+1\right)^3-k^3}{\left(k^2+k\right)^3}=\frac{\left(k+1\right)^3-k^3}{k^3.\left(k+1\right)^3}=\frac{1}{k^3}-\frac{1}{\left(k+1\right)^3}\)
Áp dụng ta được:
\(P=a_1+a_2+...+a_9\)
\(=\frac{1}{1^3}-\frac{1}{2^3}+\frac{1}{2^3}-\frac{1}{3^3}+...+\frac{1}{9^3}-\frac{1}{10^3}\)
\(=1-\frac{1}{10^3}=\frac{999}{1000}\)
23. Ta có:
\(B=\frac{1^2}{2^2-1}.\frac{3^2}{4^2-1}.\frac{5^2}{\left(6^2-1\right)}.....\frac{\left(2n+1\right)^2}{\left(2n+2\right)^2-1}\)
\(=\frac{1.1.3.3.5.5.....\left(2n+1\right)\left(2n+1\right)}{\left(1.3\right).\left(3.5\right).\left(5.7\right).....\left[\left(2n+1\right)\left(2n+3\right)\right]}\)
\(=\frac{\left[1.3.5.....\left(2n+1\right)\right].\left[1.3.5.....\left(2n+1\right)\right]}{\left[1.3.5.....\left(2n+1\right)\right].\left[3.5.7.....\left(2n+3\right)\right]}\)
\(=\frac{1}{2n+3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Trả lời:
a, \(5\left(2x-1\right)-4\left(8-3x\right)=-5\)
\(\Leftrightarrow10x-5-32+12x=-5\)
\(\Leftrightarrow22x-37=-5\)
\(\Leftrightarrow22x=32\)
\(\Leftrightarrow x=\frac{16}{11}\)
Vậy x = 16/11 là nghiệm của pt.
b, \(\left(x+1\right)\left(x+2\right)-\left(x-3\right)\left(x+4\right)=6\)
\(\Leftrightarrow x^2+3x+2-\left(x^2+x-12\right)=0\)
\(\Leftrightarrow x^2+3x+2-x^2-x+12=0\)
\(\Leftrightarrow2x+14=0\)
\(\Leftrightarrow2x=-14\)
\(\Leftrightarrow x=-7\)
Vậy x = - 7 là nghiệm của pt.
\(A=\left(2x-1\right)^2-5x\left(x-1\right)+2\left(x+1\right)\left(x-2\right)\)
\(=4x^2-4x+1-5x^2+5x+2\left(x^2-x-2\right)\)
\(=x^2-x-3\)