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\(\dfrac{5}{x}+1+\dfrac{4}{x}+1=\dfrac{3}{-13}\\ \Rightarrow\dfrac{9}{x}+2=-\dfrac{3}{13}\\ \Rightarrow\dfrac{9}{x}=-\dfrac{59}{13}\\ \Rightarrow x=-\dfrac{207}{59}\)
a. \(\dfrac{5}{x+1}+\dfrac{4}{x+1}=\dfrac{-3}{13}\)
ĐKXĐ: x ≠ -1
⇔ \(\dfrac{65}{13\left(x+1\right)}+\dfrac{52}{13\left(x+1\right)}=\dfrac{-3\left(x+1\right)}{13\left(x+1\right)}\)
⇔ 65 + 52 = -3(x + 1)
⇔ 117 = -3x - 3
⇔ 117 + 3 = -3x
⇔ 120 = -3x
⇔ x = \(\dfrac{120}{-3}=-40\) (TM)
b. -x + 2 + 2x + 3 + x + \(\dfrac{1}{4}\) + 2x + \(\dfrac{1}{6}\) = \(\dfrac{8}{3}\)
⇔ -x + 2x + x + 2x = \(\dfrac{8}{3}-\dfrac{1}{6}-\dfrac{1}{4}-3-2\)
⇔ 4x = -2,75
⇔ x = \(\dfrac{-2,75}{4}=\dfrac{-11}{16}\)
c. \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+2}\) = \(\dfrac{12}{26}\)
⇔ \(\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{2\left(3x+1\right)}=\dfrac{12}{26}\)
⇔ \(\dfrac{312\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\) + \(\dfrac{520\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\) - \(\dfrac{312\left(2x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\)
= \(\dfrac{48\left(2x+1\right)\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\)
⇔ 312(3x +1) + 520(3x + 1) - 312(2x + 1) = 48(2x + 1)(3x + 1)
⇔ 936x + 312 + 1560x + 520 - 624x - 312 = (96x + 48)(3x + 1)
⇔ 936x + 312 + 1560x + 520 - 624x - 312 = 288x2 + 96x + 144x + 48
⇔ 936x + 1560x - 624x - 96x - 144x - 288x2 = 48 - 312 - 520 + 312
⇔ 1632x - 288x2 = -472
⇔ -288x2 + 1632x + 472 = 0 (Tự giải tiếp, dùng phương pháp tách hạng tử)
⇔ x = 5,942459684 \(\approx\) 6
\(\frac{6}{x+27}=-\frac{7}{x+1}\)
\(\Rightarrow6\left(x+1\right)=-7\left(x+27\right)\)
\(6x+6=-7x+\left(-189\right)\)
\(6x+7x=-189-6\)
\(13x=195\)
\(x=195:13\)
\(x=15\)
Vậy \(x=15\)
Ta có: \(\frac{6}{x+27}=\frac{-7}{x+1}\)
\(\Leftrightarrow6\cdot\left(x+1\right)=-7\cdot\left(x+27\right)\)
\(\Leftrightarrow6x+6=-7x-189\)
\(\Leftrightarrow6x+7x=-189-6\)
\(\Leftrightarrow13x=-195\)
\(\Leftrightarrow x=-15\)
Vậy \(x=-15\)
\(\approx GOOD\)\(LUCK\approx\)
Các bạn ơi mình thiếu ở chỗ là
4x5y chia cho 2 ; 5 và 9 đều dư 1
Nha các bạn
\(A=\frac{1\cdot2+2\cdot4+3\cdot6+4\cdot8+5\cdot10+6\cdot12}{3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20+18\cdot24}\)
\(A=\frac{2\cdot3\left[1\cdot2\right]+2\cdot3\left[2\cdot4\right]+2\cdot3\left[3\cdot6\right]+2\cdot3\left[4\cdot8\right]+2\cdot3\left[5\cdot10\right]}{3\cdot4\left[3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20\right]}\)
\(A=\frac{\left[3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20\right]}{2\cdot3\left[3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20\right]}=\frac{1}{2\cdot3}=\frac{1}{6}\)
a,Để \(|2x+1|+|x-2|=0\Leftrightarrow\hept{\begin{cases}2x+1=0\\x-2=0\end{cases}}\)(vô lý)
=> ko có x thỏa mãn
b,\(|x+5|=2x-1\Leftrightarrow1-2x< x+5< 2x-1\)
Bài 2:
5.A=5+5^2+5^3+...+5^40
5.A-A=(5+5^2+5^3+...+5^40)-(1+5+5^2+...+5^39)
4.A=5^40-1
A=5^40-1/4
chúc bạn học tốt nha, câu 1 mk đang tính, xong mk gửi qua tin nhắn cho bạn nha
\(D=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{1979.1982}\)
\(\Rightarrow3D=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{1979.1982}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{1979}-\frac{1}{1982}\)
\(=\frac{1}{2}-\frac{1}{1982}=\frac{495}{991}\)
\(\Rightarrow D=\frac{495}{991}\div3=\frac{165}{991}\)