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Công thức tổng quát ''mở'' cho bài toán trên được hình thành trên cơ sở phân tích thành nhân tử và được phát biểu như sau:
\(a^4+4=\left(a^2-2a+2\right)\left(a^2+2a+2\right)\)
Khi đó, biểu thức \(A\) trở thành:
\(A=\frac{\left(1^2-2+2\right)\left(1^2+2+2\right)\left(5^2-2.5+2\right)\left(5^2+2.5+2\right)...\left(17^2-2.17+2\right)\left(17^2+2.17+2\right)}{\left(3^2-2.3+2\right)\left(3^2+2.3+2\right)\left(7^2-2.7+2\right)\left(7^2+2.7+2\right)...\left(19^2-2.19+2\right)\left(19^2+2.19+2\right)}\)
\(A=\frac{\left(1^2-2+2\right)}{\left(19^2+2.19+2\right)}=\frac{1}{401}\)
a) Ta có: \(\frac{3x-2}{6}-\frac{4-3x}{18}=\frac{4-x}{9}\)
\(\Leftrightarrow\frac{3\left(3x-2\right)}{18}-\frac{4-3x}{18}-\frac{2\left(4-x\right)}{18}=0\)
\(\Leftrightarrow9x-6-4+3x-\left(8-2x\right)=0\)
\(\Leftrightarrow12x-10-8+2x=0\)
\(\Leftrightarrow10x-18=0\)
\(\Leftrightarrow10x=18\)
hay \(x=\frac{9}{5}\)
Vậy: \(x=\frac{9}{5}\)
b) Ta có: \(\frac{2+3x}{6}-x+2=\frac{x-7}{9}\)
\(\Leftrightarrow\frac{3\left(2+3x\right)}{18}-\frac{18x}{18}+\frac{36}{18}-\frac{2\left(x-7\right)}{18}=0\)
\(\Leftrightarrow6+9x-18x+36-\left(2x-14\right)=0\)
\(\Leftrightarrow42-9x-2x+14=0\)
\(\Leftrightarrow56-11x=0\)
\(\Leftrightarrow11x=56\)
hay \(x=\frac{56}{11}\)
Vậy: \(x=\frac{56}{11}\)
c) ĐKXĐ: x∉{3;-3}
Ta có: \(\frac{6-x}{x^2-9}+\frac{2}{x+3}=\frac{-5}{x-3}\)
\(\Leftrightarrow\frac{6-x}{\left(x-3\right)\left(x+3\right)}+\frac{2\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\frac{-5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow6-x+2x-6=-5x-15\)
\(\Leftrightarrow x+5x+15=0\)
\(\Leftrightarrow6x=-15\)
hay \(x=\frac{-5}{2}\)(tm)
Vậy: \(x=\frac{-5}{2}\)
d) Ta có: \(\left(5x+2\right)\left(x^2-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+2=0\\x^2-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-2\\x^2=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-2}{5}\\x=\pm\sqrt{7}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{-2}{5};\sqrt{7};-\sqrt{7}\right\}\)
e) ĐKXĐ: x∉{4;-4}
Ta có: \(\frac{3}{x-4}+\frac{5x-2}{x^2-16}=\frac{4}{x+4}\)
\(\Leftrightarrow\frac{3\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}+\frac{5x-2}{\left(x-4\right)\left(x+4\right)}-\frac{4\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}=0\)
\(\Leftrightarrow3x+12+5x-2-\left(4x-16\right)=0\)
\(\Leftrightarrow8x+10-4x+16=0\)
\(\Leftrightarrow4x+26=0\)
\(\Leftrightarrow4x=-26\)
hay \(x=\frac{-13}{2}\)(tm)
Vậy: \(x=\frac{-13}{2}\)
Sửa đề: \(M=\frac{1^4+4}{3^4+4}\cdot\frac{5^4+4}{7^4+4}\cdot\frac{9^4+4}{11^4+4}\cdot...\cdot\frac{17^4+4}{19^4+4}\)
=\(\frac{\left(1^4+4\right)\cdot\left(5^4+4\right)\cdot\left(9^4+4\right)\cdot...\cdot\left(17^4+4\right)}{\left(3^4+4\right)\cdot\left(7^4+4\right)\cdot\left(11^4+4\right)\cdot...\cdot\left(19^4+4\right)}\)
\(=\frac{1\cdot17\cdot13\cdot145\cdot257}{17\cdot65\cdot29\cdot257\cdot401}=1\cdot\frac{1}{5}\cdot5\cdot\frac{1}{401}=\frac{1}{401}\)
Vậy: \(M=\frac{1}{401}\)
\(a,\frac{-x}{4}+6=8\)\(\Leftrightarrow\frac{x}{-4}=2\Leftrightarrow x=-8\)
b,\(\frac{-4}{x}-7=-5\Leftrightarrow\frac{-4}{x}=2\Leftrightarrow x=-2\)
c,\(12+\frac{-6}{5x}=17\Leftrightarrow-\frac{6}{5x}=5\Leftrightarrow x=-\frac{6}{25}\)
d,\(\frac{3-x}{7}=\frac{x+5}{4}\Leftrightarrow12-4x=7x+35\Leftrightarrow-11x=23\Leftrightarrow x=-\frac{23}{11}\)
e,\(7-2x=-\frac{3}{3x}=-\frac{1}{x}\Leftrightarrow7x-2x^2+1=0\)
\(\Leftrightarrow-2\left(x^2+\frac{7}{2}x+\frac{49}{16}\right)+\frac{57}{8}=0\Leftrightarrow\left(x+\frac{7}{4}\right)^2=\frac{57}{16}\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{7}{4}=\frac{\sqrt{57}}{16}\\x+\frac{7}{4}=-\frac{\sqrt{57}}{16}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\sqrt{57}-28}{16}\\x=\frac{-\sqrt{57}-28}{16}\end{matrix}\right.\)
a, \(-\frac{x}{4}+6=8\)
=> \(-\frac{x}{4}=8-6=2\)
=> \(x=2.-4=-8\)
Vậy \(x\in\left\{-8\right\}\)
\(b,\frac{4}{-x}-7=-5\)
=> \(\frac{4}{-x}=-5+\left(-7\right)=-12\)
=> \(x=4:12=\frac{1}{3}\)
Vậy \(x\in\left\{\frac{1}{3}\right\}\)
\(c,12+\frac{-6}{5x}=17\)
=> \(-\frac{6}{5x}=17-12=5\)
=> \(5x=-6:5=-\frac{6}{5}\)
=> \(x=-\frac{6}{5}:5=\frac{6}{25}\)
Vậy \(x\in\left\{\frac{6}{25}\right\}\)
\(d,\frac{3-x}{7}=\frac{x+5}{4}\)
=>\(4\left(3-x\right)=7\left(x+5\right)\)
=> \(12-4x=7x+35\)
=> \(-4x-7x=35-12\)
=> \(-11x=23\)
=> \(x=23:\left(-11\right)=-\frac{23}{11}\)
Vậy \(x\in\left\{-\frac{23}{11}\right\}\)
e, \(7-2x=-\frac{3}{3x}\)
=> \(7-2x=-\frac{1}{x}\)
=> \(7=2x+\left(-\frac{1}{x}\right)\)
=> \(7=2x-\frac{1}{x}\)
=> \(7=\frac{2x^2}{x}-\frac{1}{x}\)
=> \(7=\frac{2x^2-1}{x}\)
=> :))
a) \(\frac{x+5}{4}\)-\(\frac{2x-5}{3}\)=\(\frac{6x-1}{3}\)+\(\frac{2x-3}{12}\)
⇔\(\frac{3\left(x+5\right)}{12}\)-\(\frac{4\left(2x-5\right)}{12}\)=\(\frac{4\left(6x-1\right)}{12}\)+\(\frac{2x-3}{12}\)
⇒ 3x+15-8x+20=24x-4+2x-3
⇔3x+15-8x+20-24x+4-2x+3=0
⇔-31x+42=0
⇔x=\(\frac{42}{31}\)
Vậy tập nghiệm của phương trình đã cho là:S={\(\frac{42}{31}\)}
Ta có : \(n^4+4=\left[\left(n-1\right)^2+1\right]\left[\left(n+1\right)^2+1\right]\)
Do đó :
\(M=\frac{1\left(2^2+1\right)}{\left(2^2+1\right)\left(4^2+1\right)}.\frac{\left(4^2+1\right)\left(6^2+1\right)}{\left(6^2+1\right)\left(8^2+1\right)}.\frac{\left(8^2+1\right)\left(10^2+1\right)}{\left(10^2+1\right)\left(12^2+1\right)}...\frac{\left(16^2+1\right)\left(18^2+1\right)}{\left(18^2+1\right)\left(20^2+1\right)}\)
\(M=\frac{1}{20^2+1}=\frac{1}{401}\)