\(\dfrac{1}{\left(x^2-x\right)}+\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}+\dfrac{1}{\left(x^2+x\right)}=-1\)
Giúp mình với. help me now
a) \(\dfrac{5x-2}{3}+x=1+\dfrac{5-3x}{2}\)
b) \(\dfrac{\left(3x-1\right)\left(x+2\right)}{3}-\dfrac{2x^2+1}{2}=\dfrac{11}{2}\)
c) \(x-\dfrac{5x+2}{6}=\dfrac{7-3x}{4}\)
d) \(\dfrac{x-1}{2}+\dfrac{x-1}{3}-\dfrac{x-1}{6}=2\)
a, \(\Rightarrow10x-4+6x=6+15-9x\Leftrightarrow7x=25\Leftrightarrow x=\dfrac{25}{7}\)
b, \(\Rightarrow2\left(3x^2+5x-2\right)-6x^2-3=33\Leftrightarrow10x-7=33\Leftrightarrow x=4\)
c, \(\Rightarrow12x-10x-4=21-9x\Leftrightarrow11x=25\Leftrightarrow x=\dfrac{25}{11}\)
d, \(\Rightarrow3x-3+2x-2-x+1=12\Leftrightarrow4x=16\Leftrightarrow x=4\)
\(\dfrac{5x-2}{3}+x=1+\dfrac{5-3x}{2}\)
\(\Leftrightarrow\dfrac{5x-2+3x}{3}=\dfrac{2+5-3x}{2}\)
\(\Leftrightarrow\dfrac{8x-2}{3}=\dfrac{7-3x}{2}\)
\(\Leftrightarrow16x-4=21-9x\)
\(\Leftrightarrow16x+9x=21+4\)
\(\Leftrightarrow25x=25\)
\(\Leftrightarrow x=1\)
\(a,\dfrac{5x-2}{3}+x=1+\dfrac{-3x+5}{2}\)
\(2\left(5x-2\right)+6x=-9x+21\)
\(16x+9x=21+4\)
\(25x=25\)
\(x=1\)
\(b,\dfrac{3x^2+5x-2}{3}-\dfrac{2x^2+1}{2}=\dfrac{11}{2}\)
\(\dfrac{6x^2=10x-4-6x^2-3}{6}=\dfrac{11}{2}\)
\(\dfrac{10x-4-3}{6}=\dfrac{11}{2}\)
\(\dfrac{10x-7}{6}=\dfrac{11}{2}\)
\(10x=33+7\)
\(x=4\)
câu 1 \(\dfrac{3\left(2x+1\right)}{4}\)-\(\dfrac{2\left(3x-1\right)}{5}\)=5
câu 2 \(\dfrac{5x-3}{6}\)+5-\(\dfrac{7x-1}{4}\)=\(\dfrac{x+2}{12}\)
câu 3 \(\dfrac{\left(x-1\right)\left(x+1\right)}{2}\)+\(\dfrac{3\left(x+1\right)}{4}\)=\(\dfrac{\left(x-2\right)^2}{2}\)
câu 4 \(\dfrac{\left(x-4\right)^3}{6}\)+1=\(\dfrac{x\left(x+1\right)}{2}\)-\(\dfrac{\left(x-5\right)\left(x+5\right)}{3}\)
câu 5 \(\dfrac{3\left(x+2\right)^3}{5}\)-\(\dfrac{\left(x-1\right)^2}{10}\)=\(\dfrac{\left(x-3\right)\left(x+3\right)}{2}\)
-Mình xin lỗi.Nhưng thầy mình chưa giảng về bài này và làm bài tập với lại thầy bắt phải làm nên ko cho điểm xấu nên các bạn giúp mình được ko.Mình cầu xin sự giúp đỡ của các bạn và cảm ơn rất nhiều ( các bạn biết câu nào làm câu đó nha <:{ )
Câu 1:
=>15(2x+1)-8(3x-1)=100
=>30x+15-24x+8=100
=>6x+23=100
hay x=77/6
Câu 2:
=>2(5x-3)+12-3(7x-1)=x+2
=>10x-6+12-21x+3-x-2=0
=>-12x=-7
hay x=7/12
Câu 3:
\(\Leftrightarrow2\left(x^2-1\right)+3\left(x+1\right)=2\left(x^2-4x+4\right)\)
\(\Leftrightarrow2x^2-2+3x+3-2x^2+8x-8=0\)
=>11x-7=0
hay x=-7/11
Câu 4:
(x - 4)^3/6 + 1 = x(x + 1)/2 - (x - 5)(x + 5)/3
<=> (x - 4)^3 + 6/6 = x^2 + x/2 - x^2 - 25/3
<=> (x - 4)^3 + 6/6 = 3x^2 + 3x - 2x^2 + 50/6
<=> (x - 4)^3 + 6 = 3x^2 + 3x - 2x^2 + 50
<=> x^3 - 12x^2 + 48x - 58 = x^2 + 3x + 50
<=> x^3 -13x^2 + 45x - 108 = 0
Đến đây bạn bấm máy nhẩm nghiệm là ra nhé
Câu 5:
3(x + 2)^3/5 - (x - 1)^2/10 = (x - 3)(x + 3)/2
<=> 6(x + 2)^3 - (x - 1)^2/10 = 5(x^2 - 9)/10
<=> 6(x + 2)^3 - (x - 1)^2 = 5(x^2 - 9)
<=> 6x^3 + 36x^2 + 72x + 48 - x^2 + 2x - 1 - 5x^2 + 45 = 0
<=> 6x^3 + 30x^2 + 74x + 92 = 0
Đến đây bạn bấm máy nhẩm nghiệm như câu 4 nhé
Tính
a)\(\left(\dfrac{\left(x-1\right)^2}{\left(3x+x-1\right)^2}-\dfrac{1-2x^2+4x}{x^3-1}+\dfrac{1}{x-1}\right):\dfrac{x^2+x}{x^2+1}\)
b)\(\left(\dfrac{3\left(x+2\right)}{2\left(x^3+x^2+x+1\right)}+\dfrac{2x^2-x+10}{2\left(x^3+x^2+x+1\right)}\right):\left(\dfrac{5}{x^2+1}+\dfrac{3}{2\left(x+1\right)}-\dfrac{3}{2\left(x-1\right)}\right).\dfrac{2}{x-1}\)
c)\(\left(\dfrac{x^2}{x^2-5x+6}+\dfrac{x^2}{x^2-3x+2}\right):\dfrac{\left(x-1\right)\left(x-3\right)}{x^4+x^2+1}\)
Rút gọn :
a)\(A=\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+....+\dfrac{1}{\left(x+10\right)\left(x+11\right)}B=\dfrac{1}{x^2+x}+\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}\)
a) ta có : \(A=\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+...+\dfrac{1}{\left(x+10\right)\left(x+11\right)}\)
\(\Leftrightarrow A=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+10}-\dfrac{1}{x+11}\)
\(\Leftrightarrow A=\dfrac{1}{x}-\dfrac{1}{x+11}=\dfrac{11}{x\left(x+11\right)}\)
b) ta có : \(B=\dfrac{1}{x^2+x}+\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}\)
\(\Leftrightarrow B=\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}\)
\(\Leftrightarrow B=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}\)\(\Leftrightarrow B=\dfrac{1}{x}-\dfrac{1}{x+5}=\dfrac{5}{x\left(x+5\right)}\)
Thực hiện phép tính:
\(a,\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}\)
\(b,\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}\)
a) 1/x(x + 1) + 1/(x + 1)(x + 2) + 1/(x + 2)(x + 3) + 1/(x + 3)(x + 4)
( 1/x - 1/x+1) + (1/x+1 - 1/x+2) + (1/x+2 - 1/ x+3) + 1/(x+3 - 1/x+4)
(1/x +1/x+4) - ( 1/x+2 - 1/x+2) - ( 1/x+3 - 1/x+3)
1/x +1/x+4
2x+4/x(x+4)
Câu b bạn tách các mẫu thành nhân tử rồi làm như câu a nhé
Giúp mình với ạ
1) lim\(\dfrac{x-5x^2+1}{x^2-1}\)(x-->-∞)
2) lim\(\dfrac{5x^3\left(2-x^2\right)^3\left(4x^2+1\right)^2}{4x^{13}+x^2-6}\)(x-->+∞)
3) lim\(\dfrac{4x-\sqrt{9x^2+x}}{3-x}\)(x-->+∞)
\(\lim\limits_{x\rightarrow-\infty}\dfrac{x-5x^2+1}{x^2-1}=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{1}{x}-5+\dfrac{1}{x^2}}{1-\dfrac{1}{x^2}}=\dfrac{-5}{1}=-5\)
\(\lim\limits_{x\rightarrow+\infty}\dfrac{5x^3\left(2-x^2\right)^3\left(4x^2+1\right)^2}{4x^{13}+x^2-6}=\lim\limits_{x\rightarrow+\infty}\dfrac{5\left(\dfrac{2}{x^2}-1\right)^3\left(4+\dfrac{1}{x^2}\right)^2}{4+\dfrac{1}{x^{11}}-\dfrac{6}{x^{13}}}=\dfrac{5.\left(-1\right)^3.4^2}{4}=-20\)
\(\lim\limits_{x\rightarrow+\infty}\dfrac{4x-\sqrt{9x^2+x}}{3-x}=\lim\limits_{x\rightarrow+\infty}\dfrac{4-\sqrt{9+\dfrac{1}{x}}}{\dfrac{3}{x}-1}=\dfrac{4-3}{-1}=-1\)
Giúp mk nha
Giải các phương trình
1/ \(\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}=\dfrac{1}{6}\)
2/ \(\dfrac{1}{x^2-6x+8}+\dfrac{1}{x^2-10x+24}+\dfrac{1}{x^2-14x+48}=\dfrac{1}{9}\)
3/ \(\dfrac{1}{x^2-3x+3}+\dfrac{2}{x^2-3x+4}=\dfrac{6}{x^2-3x+5}\)
4/ \(\dfrac{6}{\left(x+1\right)\left(x+2\right)}+\dfrac{8}{\left(x-1\right)\left(x+4\right)}=1\)
5/ \(4\left(x^3+\dfrac{1}{x^3}\right)=13\left(x+\dfrac{1}{x}\right)\)
6/ \(\dfrac{4x}{4x^2-8x+7}+\dfrac{3x}{4x^2-10x+7}=1\)
7/ \(\dfrac{x^2-3x+5}{x^2-4x+5}-\dfrac{x^2-5x+5}{x^2-6x+5}=-\dfrac{1}{4}\)
8/ \(x.\dfrac{8-x}{x-1}\left(x-\dfrac{8-x}{x-1}\right)=15\)
1) điều kiện xác định : \(x\notin\left\{-1;-2;-3;-4\right\}\)
ta có : \(\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\) \(\Leftrightarrow\dfrac{\left(x+3\right)\left(x+4\right)+\left(x+1\right)\left(x+4\right)+\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)\(\Leftrightarrow\dfrac{x^2+7x+12+x^2+5x+4+x^2+3x+2}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{3x^2+15x+18}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)
\(\Leftrightarrow6\left(3x^2+15x+18\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18\left(x^2+5x+6\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18\left(x+2\right)\left(x+3\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18=\left(x+1\right)\left(x+4\right)\) ( vì điều kiện xác định )
\(\Leftrightarrow18=x^2+5x+4\Leftrightarrow x^2+5x-14=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+7=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-7\end{matrix}\right.\left(tmđk\right)\)
vậy \(x=2\) hoặc \(x=-7\) mấy câu kia lm tương tự nha bn
Thực hiện phép tính:
\(a,\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}\)
\(b,\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}\)
a ) \(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+4}\)
\(=\dfrac{4}{x\left(x+4\right)}\)
b ) \(\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}\)
\(=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}\)
\(=\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}\)
\(=\dfrac{1}{x+1}-\dfrac{1}{x+5}\)
\(=\dfrac{4}{\left(x+1\right)\left(x+5\right)}\)
:D
Giải phương trình:
\(\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{3}{130}\)
Help me now !!!!
\(\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{3}{130}\)
ĐK: \(\left\{{}\begin{matrix}x\ne-1\\x\ne-2\\x\ne-3\\x\ne-4\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{130\left(x+3\right)\left(x+4\right)+130\left(x+1\right)\left(x+4\right)+130\left(x+1\right)\left(x+2\right)}{130\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{3\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}{130\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}\)
\(\Leftrightarrow3x^2+15x-378=0\)
\(\Leftrightarrow\left(x-9\right)\left(x+14\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-14\end{matrix}\right.\)
@ngonhuminh @Nguyễn Huy Thắng @Đức Minh@Hoang Hung Quan@Nguyễn Huy Tú@Hoàng Thị Ngọc Anh.... và mb khác giúp mik đi mà, cần gấp lắm T_T