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ĐK \(x\ne-2;-3;-5;-6\)
\(\Leftrightarrow\dfrac{x-1}{x+2}-1-\left(\dfrac{x-2}{x+3}-1\right)=\dfrac{x-4}{x+5}-1-\left(\dfrac{x-5}{x+6}-1\right)\)
\(\Leftrightarrow\dfrac{x-1-x-2}{x+2}-\dfrac{x-2-x-3}{x+3}=\dfrac{x-4-x-5}{x+5}-\dfrac{x-5-x-6}{x+6}\)
\(\Leftrightarrow\dfrac{-3}{x+2}-\dfrac{-5}{x+3}=\dfrac{-9}{x+5}-\dfrac{-11}{x+6}\)
\(\Leftrightarrow\dfrac{3}{x+2}-\dfrac{5}{x+3}=\dfrac{9}{x+5}-\dfrac{11}{x+6}\)
\(\Leftrightarrow\dfrac{3}{x+2}+\dfrac{11}{x+6}=\dfrac{9}{x+5}+\dfrac{5}{x+3}\)
\(\Leftrightarrow\dfrac{3\left(x+6\right)+11\left(x+2\right)}{\left(x+2\right)\left(x+6\right)}=\dfrac{9\left(x+3\right)+5\left(x+5\right)}{\left(x+3\right)\left(x+5\right)}\)
\(\Leftrightarrow\dfrac{14x+40}{\left(x+2\right)\left(x+6\right)}=\dfrac{14x+52}{\left(x+3\right)\left(x+5\right)}\)
\(\Leftrightarrow\left(x+2\right)\left(x+6\right)\left(14x+52\right)=\left(x+3\right)\left(x+5\right)\left(14x+40\right)\)
\(\Leftrightarrow\left(x^2+8x+12\right)\left(14x+52\right)=\left(x^2+8x+15\right)\left(14x+40\right)\)
\(\Leftrightarrow14x\left(x^2+8x+12\right)+52\left(x^2+8x+12\right)=14x\left(x^2+8x+15\right)+40\left(x^2+8x+15\right)\)
\(\Leftrightarrow14x\left(x^2+8x\right)+12.14x+52\left(x^2+8x\right)+52.12=14x\left(x^2+8x\right)+15.14x+40\left(x^2+8x\right)+15.40\)
\(\Leftrightarrow12\left(x^2+8x\right)-42x+24=0\)
\(\Leftrightarrow12x^2+54x+24=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-4\end{matrix}\right.\)
a) A = {\(\dfrac{1}{n\left(n+1\right)}\)| \(n\in\mathbb{N},1\le n\le5\)}
b) B = {\(\dfrac{1}{n^2-1}\)|\(n\in\mathbb{N},2\le n\le6\)\(\)}
1) \(A=1+2+2^2+2^3+......+2^{2015}\)
\(\Leftrightarrow2A=2+2^2+2^3+......+2^{2016}\)
\(\Leftrightarrow2A-A=\left(2+2^2+2^3+......+2^{2016}\right)-\left(1+2+2^2+2^3+......+2^{2015}\right)\)
\(\Leftrightarrow A=2^{2016}-1\)
Vậy \(A=2^{2016}-1\)
6)Ta có: \(13+23+33+43+.......+103=3025\)
\(\Leftrightarrow2.13+2.23+2.33+2.43+.......+2.103=2.3025\)
\(\Leftrightarrow26+46+66+86+.......+206=6050\)
\(\Leftrightarrow\left(23+3\right)+\left(43+3\right)+\left(63+3\right)+\left(83+3\right)+.......+\left(203+3\right)=6050\)
\(\Leftrightarrow23+43+63+83+.......+203+3.10=6050\)
\(\Leftrightarrow23+43+63+83+.......+203+=6050-30\)
\(\Leftrightarrow23+43+63+83+.......+203+=6020\)
Vậy S=6020
b, B có 19 thừa số
=> \(-B=(1-\frac{1}{4})(1-\frac{1}{9})(1-\frac{1}{16})...(1-\frac{1}{400}) \)
<=>\(-B=\frac{(2-1)(2+1)(3-1)(3+1)(4-1)(4+1)...(20-1)(20+1)}{4.9.16...400} \)
<=>\(-B=\frac{(1.2.3.4...19)(3.4.5...21)}{(2.3.4.5.6...20)(2.3.4.5...20)} \)
<=>\(-B=\frac{21}{20.2} =\frac{21}{40} \)
<=>\(B=\frac{-21}{40} \)
- Thay từng giá trị vào, ta thấy A. \(\dfrac{15}{4}\) thỏa mãn.
1.
\(x^4-6x^2-12x-8=0\)
\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)
\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)
\(\Leftrightarrow x=1\pm\sqrt{5}\)
3.
ĐK: \(x\ge-9\)
\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)
\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)
Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)
\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)
\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)
\(\Leftrightarrow...\)
a) \(\dfrac{5+x}{4-x}=\dfrac{1}{2}\)
\(\Leftrightarrow2\left(5+x\right)=4-x\)
\(\Leftrightarrow2\left(5+x\right)-\left(4-x\right)=0\)
\(\Leftrightarrow10+2x-4+x=0\)
\(\Leftrightarrow6+3x=0\)
\(\Leftrightarrow3x=-6\)
\(\Leftrightarrow x=-2\)
Vậy x=-2
b) \(\dfrac{25}{14}=\dfrac{x+7}{x-4}\)
\(\Leftrightarrow25\left(x-4\right)=14\left(x+7\right)\)
\(\Leftrightarrow25\left(x-4\right)-14\left(x+7\right)=0\)
\(\Leftrightarrow25x-100-14x-98=0\)
\(\Leftrightarrow11x-198=0\)
\(\Leftrightarrow11x=198\)
\(\Leftrightarrow x=18\)
Vậy x=18
c) \(\dfrac{3x-5}{x+4}=\dfrac{5}{2}\)
\(\Leftrightarrow2\left(3x-5\right)=5\left(x+4\right)\)
\(\Leftrightarrow2\left(3x-5\right)-5\left(x+4\right)=0\)
\(\Leftrightarrow6x-10-5x-20=0\)
\(\Leftrightarrow x-30=0\)
\(\Leftrightarrow x=30\)
Vậy x=30
d) \(\dfrac{3x-1}{2x+1}=\dfrac{3}{7}\)
\(\Leftrightarrow7\left(3x-1\right)=3\left(2x+1\right)\)
\(\Leftrightarrow7\left(3x-1\right)-3\left(2x+1\right)=0\)
\(\Leftrightarrow21x-7-6x-3=0\)
\(\Leftrightarrow15x-10=0\)
\(\Leftrightarrow15x=10\)
\(\Leftrightarrow x=\dfrac{10}{15}=\dfrac{2}{3}\)
Vậy \(x=\dfrac{2}{3}\)
a) \(\dfrac{3}{5}.\dfrac{1}{x}-\dfrac{1}{3}=\dfrac{4}{6}\)
\(\Leftrightarrow\dfrac{3}{5x}=1\)
\(\Leftrightarrow x=\dfrac{5}{3}\)
b) \(\dfrac{x}{2}=-\dfrac{2y}{8}=\dfrac{3z}{15}\)
áp dụng dãy tí số = nhau
\(\dfrac{x}{2}=-\dfrac{2y}{8}=\dfrac{3z}{15}=\dfrac{x-2y+3z}{2+8+15}=\dfrac{1200}{15}=80\)
\(\Leftrightarrow\dfrac{x}{2}=80\Rightarrow x=160\)
\(\Leftrightarrow-\dfrac{y}{4}=80\Rightarrow y=-320\)
\(\Leftrightarrow\dfrac{z}{5}=80\Rightarrow z=400\)
\(\Leftrightarrow\left(\dfrac{2}{3}+\dfrac{x}{5}\right)\cdot30=80\)
=>1/5x+2/3=8/3
=>1/5x=2
hay x=10