Giải phương trình:
\(\dfrac{t +3} {t - 2}\) +\(\dfrac{t - 2} { t+3}\) = \(\dfrac{ 5t +15} {t^2 + t -6}\)
MAI HỌC RỒI !! HELP ME /?!/
a)(3x-1)(x2+2)=(3x-1)(7x-10)
b)\(\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{t^2+t-6}\)
giup mik voi mai minh di hoc rau help me
a, (3x-1)(x2+2)=(3x-1)(7x-10)
<=>(3x-1)(x2+2)-(3x-1)(7x-10)=0
<=>(3x-1)(x2+2-7x+10)=0
<=>(3x-1)(x2-7x+12)=0
<=>(3x-1)(x2-3x-4x+12)=0
<=>(3x-1)(x-3)(x-4)=0
<=>\(\left[{}\begin{matrix}3x-1=0\\x-3=0\\x-4=0\end{matrix}\right.\)<=>\(\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=3\\x=4\end{matrix}\right.\)
Vậy ft có tập nghiệm S=\(\left\{\dfrac{1}{3},3,4\right\}\)
b,\(\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{t^2+t-6}\) (ĐKXĐ:t\(\ne2;t\ne-3\))
<=>\(\dfrac{\left(t+3\right)^2+\left(t-2\right)^2}{\left(t-2\right)\left(t+3\right)}\)=\(\dfrac{5t+15}{t^2-2t+3t-6}\)
<=>\(\dfrac{t^2+6t+9+t^2-4t+4}{\left(t-2\right)\left(t+3\right)}\)=\(\dfrac{5t+15}{\left(t-2\right)\left(t+3\right)}\)
=>2t2+2t+13=5t+15
<=>2t2+2t-5t+13-15=0
<=>2t2-3t-2=0
<=>2t2-4t+t-2=0
<=>(t-2)(2t+1)=0
<=>\(\left[{}\begin{matrix}t-2=0\\2t+1=0\end{matrix}\right.< =>\left[{}\begin{matrix}t=2\left(loại\right)\\t=\dfrac{-1}{2}\left(tmđkxđ\right)\end{matrix}\right.\)
Vậy ft có nghiệm duy nhất x=\(\dfrac{-1}{2}\)
Giải:
a) \(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)
Chia cả hai vế cho 3x-1, ta được:
\(x^2+2=7x-10\)
\(\Leftrightarrow x^2-7x+10+2=0\)
\(\Leftrightarrow x^2-7x+12=0\)
\(\Leftrightarrow x^2-4x-3x+12=0\)
\(\Leftrightarrow x\left(x-4\right)-3\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
Vậy ...
b) \(\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{t^2+t-6}\) (1)
ĐKXĐ: \(t\ne2;t\ne-3\)
\(\left(1\right)\Leftrightarrow\dfrac{\left(t+3\right)\left(t+3\right)}{\left(t-2\right)\left(t+3\right)}+\dfrac{\left(t-2\right)\left(t-2\right)}{\left(t-2\right)\left(t+3\right)}=\dfrac{5t+15}{\left(t-2\right)\left(t+3\right)}\)
\(\Rightarrow\left(t+3\right)^2+\left(t-2\right)^2=5t+15\)
\(\Leftrightarrow t^2+6t+9+t^2-4t+4=5t+15\)
\(\Leftrightarrow2t^2+2t+13=5t+15\)
\(\Leftrightarrow2t^2+2t+13-5t-15=0\)
\(\Leftrightarrow2t^2-3t-2=0\)
\(\Leftrightarrow2t^2-4t+t-2=0\)
\(\Leftrightarrow2t\left(t-2\right)+\left(t-2\right)=0\)
\(\Leftrightarrow\left(2t+1\right)\left(t-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2t+1=0\\t-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=-\dfrac{1}{2}\left(tm\right)\\t=2\left(ktm\right)\end{matrix}\right.\)
Vậy ...
a,\(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2+2-7x+10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x^2-7x+12=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=1\\\left(x-4\right)\left(x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x-4=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\\x=3\end{matrix}\right.\)
Vậy...
b,\(\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{t^2+t-6}\)
\(\Leftrightarrow\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{\left(t+3\right)\left(t-2\right)}\)
\(\Leftrightarrow\dfrac{\left(t+3\right)^2}{\left(t-2\right)\left(t+3\right)}+\dfrac{\left(t-2\right)^2}{\left(t-2\right)\left(t+3\right)}=\dfrac{5t+15}{\left(t+3\right)\left(t-2\right)}\)
\(\Leftrightarrow t^2+6t+9+t^2-4t+4=5t+15\)
\(\Leftrightarrow t^2+t^2+6t-4t-5t=15-9-4\)
\(\Leftrightarrow2t^2-3t=2\)
\(\Leftrightarrow\left(2t+1\right)\left(t+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2t+1=0\\t+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}t=-\dfrac{1}{2}\\t=-2\end{matrix}\right.\)
Vậy...
Giải các phương trình:
a). \(\dfrac{1+x}{1-x}+3=\dfrac{3-x}{1-x}\)
b). \(\dfrac{1}{2x-3}-\dfrac{3}{x\left(2x-3\right)}=\dfrac{5}{x}\)
c). \(\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{t^2+t-6}\)
a,
\(\dfrac{1+x+3-3x-3+x}{1-x}=0\\ \dfrac{1-x}{1-x}=0\\ =>1-x=0\\ =>x=1\\ \)
b =>x-3 =10x -15
=>x-10x=-15+3
=>-9x=-12
=>x=4/3
\(\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{t^2+t-6}\)
\(\left(2x+3\right)\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)
giúp mình vs. mai thày kiểm tra vở rùi
mơn các bạn nhiều
a, \(\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{t^2+t-6}\) ĐKXĐ: t\(\ne\)2,t\(\ne\)-3
\(\Leftrightarrow\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{\left(t+3\right)\left(t-2\right)}\)
\(\Rightarrow\left(t+3\right)\left(t+3\right)+\left(t-2\right)\left(t-2\right)=5t+15\)
\(\Leftrightarrow t^2+6t+9+t^2-4t+4-5t-15=0\)
\(\Leftrightarrow-3t-2=0\)
\(\Leftrightarrow-3t=2\)
\(\Leftrightarrow t=\dfrac{-2}{3}\) (tđk)
\(\Rightarrow S=\left\{\dfrac{-2}{3}\right\}\)
b, \(\left(2x+3\right)\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)ĐKXĐ: x\(\ne\)\(\dfrac{2}{7}\)
\(\Leftrightarrow\) \(\left(2x+3\right)\left(\dfrac{3x+8}{2-7x}+1\right)-\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)=0\)
\(\Rightarrow\left(\dfrac{3x+8}{2-7x}+1\right)\left(2x+3-x+5\right)=0\)
\(\Leftrightarrow\) \(\Rightarrow\left(\dfrac{3x+8}{2-7x}+1\right)\left(x+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{3x+8}{2-7x}+1=0\\x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x+8+2-7x=0\\x=-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-4x+10=0\\x=-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-8\end{matrix}\right.\)
\(\Rightarrow S=\left\{\dfrac{5}{2};-8\right\}\)
ĐKXĐ: x khác 2 và x khác -3
\(\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{t^2+t-6}\)
\(\Leftrightarrow\dfrac{\left(t+3\right)\left(t+3\right)}{\left(t+3\right)\left(t-2\right)}+\dfrac{\left(t-2\right)\left(t-2\right)}{\left(t+3\right)\left(t-2\right)}=\dfrac{5t+15}{t^2+t-6}\)
\(\Rightarrow t^2+6t+9+t^2-4=5t+15\)
\(\Leftrightarrow2t^2+t-10=0\)
\(\Leftrightarrow2t^2-4t+5t-10=0\)
\(\Leftrightarrow2t\left(t-2\right)+5\left(t-2\right)=0\)
\(\Leftrightarrow\left(2t+5\right)\left(t-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=2\left(loại\right)\\t=\dfrac{-5}{2}\end{matrix}\right.\)
Vậy..................
ĐKXĐ: x khác \(\dfrac{2}{7}\)\(\left(2x+3\right)\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)
\(\Leftrightarrow\left(2x+3\right)\left(\dfrac{3x+8}{2-7x}+1\right)-\left(x+5\right)\left(\dfrac{3x+8}{2-7x}+1\right)=0\)
\(\Leftrightarrow\left(\dfrac{3x+8}{2-7x}+1\right)\left(2x+3-x-5\right)=0\)
\(\Leftrightarrow\left(\dfrac{-4x+10}{2-7x}\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{-4x+10}{2-7x}=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=2\end{matrix}\right.\)
Vậy...............
Tính nhanh:
1) \(\dfrac{2}{15}.6\dfrac{5}{11}+\dfrac{5}{11}.\dfrac{-2}{15}-\dfrac{2}{15}.2015^0\)
2)\(\dfrac{5}{2.7}+\dfrac{3}{14.11}+ \dfrac{4}{11.7}+\dfrac{1}{14.15}+\dfrac{13}{15.16}\)
Các bạn giup mk với !!!Chiều nay mk đi học rồi!!Help me!!!!!!!!!!!!!!!!!!!!!!!
1) \(\dfrac{2}{15}\cdot6\dfrac{5}{11}+\dfrac{5}{11}\cdot\dfrac{-2}{15}-\dfrac{2}{15}\cdot2015^0\)
\(=\dfrac{2}{15}\cdot\dfrac{71}{11}-\dfrac{1}{11}\cdot\dfrac{2}{3}-\dfrac{2}{15}\cdot1\)
\(=\dfrac{142}{165}-\dfrac{2}{33}-\dfrac{2}{15}\)
\(=\dfrac{2}{3}\)
2) \(\dfrac{5}{2\cdot7}+\dfrac{3}{14\cdot11}+\dfrac{4}{11\cdot7}+\dfrac{1}{14\cdot15}+\dfrac{13}{15\cdot16}\)
\(=\dfrac{5}{14}+\dfrac{3}{154}+\dfrac{4}{77}+\dfrac{1}{210}+\dfrac{13}{240}\)
\(=\dfrac{39}{80}\)
\(\dfrac{2}{15}.6\dfrac{5}{11}+\dfrac{5}{11}.\dfrac{-2}{15}-\dfrac{2}{15}.2015^0\)
\(=\dfrac{2}{15}.\dfrac{71}{11}+\dfrac{-5}{11}.\dfrac{2}{5}-\dfrac{2}{15}.1\)
\(=\dfrac{2}{15}\left(\dfrac{71}{11}+\dfrac{-5}{11}-1\right)=\dfrac{2}{15}.5=\dfrac{2}{3}\)
\(\dfrac{5}{2.7}+\dfrac{3}{14.11}+\dfrac{4}{11.7}+\dfrac{1}{14.15}+\dfrac{13}{15.16}\)
\(=\dfrac{5}{14}+\dfrac{3}{154}+\dfrac{4}{77}+\dfrac{1}{210}+\dfrac{13}{240}\)
\(=\dfrac{39}{80}\)
Tính :
1, \(2016-1-\dfrac{1}{3}-\dfrac{1}{6}-\dfrac{1}{10}-\dfrac{1}{15}-...-\dfrac{1}{1225}\)
2, \(\dfrac{3}{2}+\dfrac{5}{4}+\dfrac{9}{8}+\dfrac{17}{16}+\dfrac{33}{32}+\dfrac{65}{64}-7\)
3, A= \(\dfrac{1}{3\cdot4}-\dfrac{1}{4\cdot5}-\dfrac{1}{5\cdot6}-...-\dfrac{1}{9\cdot10}\)
help me ~ mai mik phải nộp rồi
Giải phương trình:
\(\dfrac{3t}{t^2+3t+2}+\dfrac{2t}{t^2+t+2}=1\)
Nhận thấy \(t=0\) ko phải nghiệm
Với \(t\ne0\) pt tương đương:
\(\dfrac{3}{t+3+\dfrac{2}{t}}+\dfrac{2}{t+1+\dfrac{2}{t}}=1\)
Đặt \(t+\dfrac{1}{t}+1=x\Rightarrow t+\dfrac{2}{t}+3=x+2\)
Pt trở thành:
\(\dfrac{3}{x+2}+\dfrac{2}{x}=1\)
\(\Rightarrow3x+2\left(x+2\right)=x\left(x+2\right)\)
\(\Leftrightarrow x^2-3x-4=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}t+\dfrac{2}{t}+1=-1\\t+\dfrac{2}{t}+1=4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}t^2+2t+2=0\left(vn\right)\\t^2-3t+2=0\end{matrix}\right.\)
\(\Rightarrow t=\left\{1;2\right\}\)
Tìm x
a,\(\left(\dfrac{3}{4}-5x\right)\)-\(\left(2x-4\dfrac{1}{4}\right)\)=\(6\dfrac{2}{7}\)
Help me, mai mình đi học
a) \(\left(\dfrac{3}{4}-5x\right)-\left(2x-\dfrac{17}{4}\right)\)=\(\dfrac{44}{7}\)
\(\left(\dfrac{3}{4}-\dfrac{17}{4}\right)-\left(5x-2x\right)\) = \(\dfrac{44}{7}\)
\(\dfrac{-14}{4}-3x\) = \(\dfrac{44}{7}\)
3x= \(\dfrac{-14}{4}-\dfrac{44}{7}\)=\(\dfrac{-98}{28}-\dfrac{176}{28}\)= \(\dfrac{-274}{28}=\dfrac{-137}{14}\)
x= \(\dfrac{-137}{14}:3\)
x= \(\dfrac{-137}{42}\)
1.Tính giá trị biểu thức : \(\dfrac{5.2010-1996}{14+4.2010}\).\(\dfrac{15+\dfrac{15}{7}-\dfrac{15}{11}+\dfrac{15}{2009}-\dfrac{15}{13}}{\dfrac{4}{2009}-\dfrac{4}{13}+\dfrac{4}{7}-\dfrac{4}{11}+4}\)
2. Tìm tất cả SNT p sao cho p + 2 và p + 28 là SNT
3. Cho A = 1 - 3 + 32 - 33 + ... - 32009 + 32010 . Chứng minh 4A - 1 là một luỹ thừa của 3.
Phép tính lằng nhằng ***** nhỉ ?????
Help Me vs ! Bn nào nhanh nhất , đúng nhất mik tặng tik !*(Lời giải chi tiết )* nội trong đêm nay nhé !
3.
\(A=1-3+3^2-3^3+...-3^{2009}-3^{2010}\)
\(\Rightarrow3A=3-3^2+3^3-3^4+...-3^{2010}+3^{2011}\)
\(\Rightarrow4A=3-3^2+3^3-3^4+...-3^{2010}+3^{2011}+\left(1-3+3^2-3^3+...-3^{2009}+3^{2010}\right)\)\(\Rightarrow4A=3^{2011}-1\)
\(\Rightarrow4A=3^{2011}\)
\(\Rightarrow\)ĐPCM
Bài 1: Thực hiện các phép tính sau ( Tính nhanh nếu có thể)
a) \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)
b) \(\dfrac{3}{1.3}+\dfrac{3}{3.5}+\dfrac{3}{5.7}+...+\dfrac{3}{49.51}\)
Bài 2: Tìm x biết:
a) \(\dfrac{2}{3}.x+\dfrac{1}{4}=\dfrac{7}{12}\)
b) \(\dfrac{1}{2}+\dfrac{1}{3}:2.x=-1\)
c) (3.x-2)\(^2=\dfrac{16}{25}\)
d) (2 . x - 3).(6 - 4 . x)=0
Bài 3 Một trường THCS có 3020 học sinh. Số học sinh khối 6 bằng 0,3 số học sinh toàn trường. Số học sinh khối 9 bằng 20% số học sinh toàn trường. Số học sinh khối 8 bằng \(\dfrac{1}{2}\) số học sinh khối 6 và khối 9. Tính số học sinh khối 7?
Bài 4: Tính nhanh: \(\dfrac{3}{3.5}+\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{97.99}\)
Bài 5: Đoạn đương bộ đi từ Đà Nẵng đến Huế dài 108 km. Một xe máy xuất phát từ Đà Nẵng đã đi được \(\dfrac{5}{6}\) quãng đường. Hỏi xe máy còn cách Huế bao nhiêu km?
Bài 6: Tìm số tự nhiên x biết rằng:
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x.\left(x\right)+1}=\dfrac{2007}{2009}\)
HELP ME! Mk đang cần gấp,ai nhanh, đầy đủ mk tick cho!
Bài 1:
a)=2.( \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+......+\dfrac{1}{97}-\dfrac{1}{99}\)
=2. (1/3-1/99)
=2. (33/99-1/99)
=2. 32/99
=64/99
b) tương tự như trên.
Bài 1 :
a) \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)
\(=2\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{97.99}\right)\)
\(=2\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(=2\left(\dfrac{1}{3}-\dfrac{1}{99}\right)\)
\(=2\left(\dfrac{33}{99}-\dfrac{1}{99}\right)\)
\(=2.\dfrac{32}{99}\)
\(=\dfrac{2.32}{99}\)
\(=\dfrac{64}{99}\)
b) \(\dfrac{3}{1.3}+\dfrac{3}{3.5}+\dfrac{3}{5.7}+...+\dfrac{3}{49.51}\)
\(=2\left(\dfrac{3}{1.3}+\dfrac{3}{3.5}+\dfrac{3}{5.7}+...+\dfrac{3}{49.51}\right)\)
\(=3\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{49.51}\right)\)
\(=3\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
\(=3\left(1-\dfrac{1}{51}\right)\)
\(=3.\dfrac{50}{51}\)
\(=\dfrac{3.50}{51}\)
\(=\dfrac{1.50}{17}\)
\(=\dfrac{50}{17}\)
Bài 2:
a) \(\dfrac{2}{3}.x+\dfrac{1}{4}=\dfrac{7}{12}\)
\(\dfrac{2}{3}.x=\dfrac{7}{12}-\dfrac{1}{4}\)
\(\dfrac{2}{3}.x=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}:\dfrac{2}{3}\)
\(x=\dfrac{1}{2}\)
Vậy \(x=\dfrac{1}{2}\)
b) \(\dfrac{1}{2}+\dfrac{1}{3}:2.x=-1\)
\(\dfrac{1}{3}:2.x=-1-\dfrac{1}{2}\)
\(\dfrac{1}{3}:2.x=\dfrac{-3}{2}\)
\(2.x=\dfrac{1}{3}:\dfrac{-3}{2}\)
\(2.x=\dfrac{-2}{9}\)
\(x=\dfrac{-2}{9}:2\)
\(x=\dfrac{-1}{9}\)
Vậy \(x=\dfrac{-1}{9}\)