1)\(\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-c\right)\left(b-a\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\)
2)\(\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-c\right)\left(b-a\right)}\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)
3)\(\frac{1}{x^2+3x+2}+\frac{2x}{x^3+4x^2+4x}+\frac{1}{x^2+5x+6}\)
Giúp mik với
Tính nhanh:
a. A=\(\left(-1\right)^{2n}.\left(-1\right)^n.\left(-1\right)^{n+1}\left(n\in N\right)\)
b. B=\(\left(10000-1^2\right)\left(10000-2^2\right)\left(10000-3^2\right)..\left(10000-1000^2\right)\)
c. C=\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)\left(\frac{1}{125}-\frac{1}{3^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
d. D=\(1999^{\left(1000-1^3\right)\left(1000-2^3\right)\left(1000-3^3\right)...\left(1000-10^3\right)}\)
a) \(A=\left(-1\right)^{2n}.\left(-1\right)^n.\left(-1\right)^{n+1}=\left(-1\right)^{3n+1}\)
b) \(B=\left(10000-1^2\right)\left(10000-2^2\right).........\left(10000-1000^2\right)\)
\(=\left(10000-1^2\right)\left(10000-2^2\right)......\left(10000-100^2\right)....\left(10000-1000^2\right)\)
\(=\left(10000-1^2\right)\left(10000-2^2\right).....\left(10000-10000\right).....\left(10000-1000^2\right)=0\)
c) \(C=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)..........\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right).....\left(\frac{1}{125}-\frac{1}{5^3}\right)......\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)........\left(\frac{1}{125}-\frac{1}{125}\right).....\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)
d) \(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-10^3\right)}\)
\(=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-1000\right)}=1999^0=1\)
Bài 1:Giải phương trình
a)\(10x^2-5x\left(2x+3\right)=15\)
b)\(3x-7-\left(3-4x\right)\left(2x+1\right)=4x\left(2x-7\right)\)
c)\(\left(4x-5\right)^2-\left(7-2x\right)=4\left(2x-4\right)^2+6x\)
Bài 2:Giải phương trình
a)\(\frac{3\left(x-1\right)}{2}+4=\frac{2x}{3}+\frac{4-5x}{6}\)
b)\(\frac{4-x}{7}-\frac{1}{7}\left(\frac{7+3x}{9}+\frac{5-2x}{2}\right)=4-\frac{4x}{3}\)
c)\(\frac{2}{9}\left(2x-5\right)-\frac{5}{3}\left[\left(x-2\right)-\frac{7}{12}\right]=\frac{3}{4}\left(x-3\right)\)
Bài 3:Giải phương trình
a)\(\left(x-6\right)\left(2x-5\right)\left(3x+9\right)=0\)
b)\(2x\left(x-3\right)+5\left(x-3\right)=0\)
c)\(\left(x^2-4\right)-\left(x-2\right)\left(3-2x\right)=0\)
Bài 4:Tìm m để phương trình sau có nghiệm bằng 7:\(\left(2m-5\right)x-2m^2+8=43\)
Bài 5:Giải phương trình
a)\(\left(2x-1\right)^2-\left(2x+1\right)^2=0\)
b)\(\frac{1}{27}\left(x-3\right)^3-\frac{1}{125}\left(x-5\right)^3=0\)
Bài 3:
a) \(\left(x-6\right).\left(2x-5\right).\left(3x+9\right)=0\)
\(\Leftrightarrow\left(x-6\right).\left(2x-5\right).3.\left(x+3\right)=0\)
Vì \(3\ne0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\2x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\2x=5\\x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\frac{5}{2}\\x=-3\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{6;\frac{5}{2};-3\right\}.\)
b) \(2x.\left(x-3\right)+5.\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right).\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\frac{5}{2}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{3;-\frac{5}{2}\right\}.\)
c) \(\left(x^2-4\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x^2-2^2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2-3+2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2;\frac{1}{3}\right\}.\)
Chúc bạn học tốt!
Bài 4 xem lại đề nhé bác
a) \(\frac{5x-3}{50x^2-2}+\frac{5x-9}{12x-60x^2}+\frac{1}{12x}=\frac{8x-5}{80x^2+16x}\)
b) \(\frac{1}{x\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+9\right)}=\frac{1}{3}\left(27-\frac{1}{x+9}\right)\)
Bạn nào làm giúp mình với ạ!
#Cảm ơn nhiều! :)
a/ ĐKXĐ: ...
\(\frac{5x-3}{2\left(5x-1\right)\left(5x+1\right)}-\frac{5x-9}{12x\left(5x-1\right)}+\frac{1}{12x}=\frac{8x-5}{16x\left(5x+1\right)}\)
\(\Leftrightarrow\frac{5x-3}{\left(5x-1\right)\left(5x+1\right)}+\frac{1}{6x}\left(1-\frac{5x-9}{5x-1}\right)=\frac{8x-5}{8x\left(5x+1\right)}\)
\(\Leftrightarrow\frac{5x-3}{\left(5x-1\right)\left(5x+1\right)}+\frac{4}{3x\left(5x-1\right)}-\frac{8x-5}{8x\left(5x+1\right)}=0\)
\(\Leftrightarrow24x\left(5x-3\right)+32\left(5x+1\right)-3\left(5x-1\right)\left(8x-5\right)=0\)
\(\Leftrightarrow-120x^2+379x-55=0\)
Bạn có nhầm đề chỗ nào ko nhỉ? Con số thật khủng khiếp (nghiệm ko hề đẹp)
b/ ĐKXĐ:...
\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{x}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+9}\right)=\frac{1}{3}\left(27-\frac{1}{x+9}\right)\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+9}=27-\frac{1}{x+9}\)
\(\Leftrightarrow\frac{1}{x}=27\Rightarrow x=\frac{1}{27}\)
a, \(\frac{5x-3}{50x^2-2}+\frac{5x-9}{12x-60x^2}+\frac{1}{12x}=\frac{8x-5}{80x^2+16x}\) (ĐKXĐ: x \(\ne\) \(\pm\)\(\frac{1}{5}\); x \(\ne\) 0)
\(\Leftrightarrow\) \(\frac{5x-3}{2\left(5x-1\right)\left(5x+1\right)}+\frac{-5x+9}{12x\left(5x-1\right)}+\frac{1}{12x}=\frac{8x-5}{16x\left(5x+1\right)}\)
\(\Leftrightarrow\) \(\frac{24x\left(5x-3\right)\left(5x+1\right)}{48x\left(5x-1\right)\left(5x+1\right)}+\frac{-4\left(5x+1\right)\left(5x-9\right)}{48x\left(5-1x\right)\left(5x+1\right)}+\frac{4\left(5x-1\right)\left(5x+1\right)}{48x\left(5x-1\right)\left(5x+1\right)}=\frac{3\left(8x-5\right)\left(5x-1\right)}{48x\left(5x-1\right)\left(5x+1\right)}\)
\(\Leftrightarrow\) 24x(5x - 3) - 4(5x + 1)(5x - 9) + 4(5x - 1)(5x + 1) = 3(8x - 5)(5x - 1)
\(\Leftrightarrow\) 120x2 - 72x - 100x2 + 160x + 36 + 100x2 - 4 = 120x2 - 99x + 15
\(\Leftrightarrow\) 120x2 - 120x2 - 100x2 + 100x2 - 72x + 160x + 99x = 15 - 36 + 4
\(\Leftrightarrow\) 187x = -17
\(\Leftrightarrow\) x = \(\frac{-1}{11}\) (TM ĐKXĐ)
Vậy S = {\(\frac{-1}{11}\)}
Chúc bn học tốt!! (Đã được kiểm chứng không sai :)
b, \(\frac{1}{x\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+9\right)}=\frac{1}{3}\left(27-\frac{1}{x+9}\right)\) (ĐKXĐ: x \(\ne\) 0; x \(\ne\) -3; x \(\ne\) -6; x \(\ne\) -9)
\(\Leftrightarrow\) \(\frac{1}{3}\)(\(\frac{1}{x}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+9}\)) = \(\frac{1}{3}\)(27 - \(\frac{1}{x+9}\))
\(\Leftrightarrow\) \(\frac{1}{3}\)(\(\frac{1}{x}-\frac{1}{x+9}\)) = \(\frac{1}{3}\)(27 - \(\frac{1}{x+9}\))
\(\Leftrightarrow\) \(\frac{1}{3}\)(\(\frac{1}{x}-\frac{1}{x+9}\)) - \(\frac{1}{3}\)(27 - \(\frac{1}{x+9}\))
\(\Leftrightarrow\) \(\frac{1}{3}\)(\(\frac{1}{x}-\frac{1}{x+9}-27+\frac{1}{x+9}\)) = 0
\(\Leftrightarrow\) \(\frac{1}{3}\)(\(\frac{1}{x}-27\)) = 0
\(\Leftrightarrow\) \(\frac{1}{x}-27\) = 0
\(\Leftrightarrow\) x = \(\frac{1}{27}\) (TM ĐKXĐ)
Vậy S = {\(\frac{1}{27}\)}
Chúc bn học tốt!!
B1 :Giải phương trình
a,\(\frac{3\left(x-3\right)}{4}-1=\frac{2x+3\left(x+1\right)}{6}-\frac{7+12x}{12}\)
b,\(\left(x+2\right)\left(3-4x\right)=x^2+4x+4\)
c,\(\frac{x-2}{x+2}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{x^2-4}\)
d,I7-xI-5x=1
B2:Giải bất phương trình
a,\(\left(x-2\right)\left(x+2\right)\ge x\left(x-4\right)\)
b,\(\frac{x-1}{4}-1\ge\frac{x+1}{3}+8\)
bài 1:chứng minh rằng:
\(\frac{y-z}{\left(x-y\right)\left(x-z\right)}+\frac{z-x}{\left(y-z\right)\left(y-x\right)}+\frac{x-y}{\left(z-x\right)\left(z-y\right)}=\frac{2}{x-y}+\frac{2}{y-z}+\frac{2}{z-x}\)
bài 2:cho m+n=1;m*n khác 0 chứng minh:
\(\frac{m}{n^3-1}+\frac{n}{m^3-1}=\frac{2\left(m-n-2\right)}{m^2\cdot n^2+3}\)
bài 3 cho a,b,c thỏa a*b*c=2013 chứng minh:
\(\frac{2013a}{ab+2013a+2013}+\frac{b}{bc+b+2013}+\frac{c}{ac+c+1}=1\)
bài 4:Tìm A,B,C để
\(\frac{x^2+2x-1}{\left(x-1\right)\left(x^2+1\right)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}\)
mình đag cần gấp giải giúp mình nha!
THANK YOU ❤❤>_<
mik đag cần gấp các bn giải nhanh dùm mik nha
Giải các phương trình sau:
a) \(\left(x-3\right)\left(1-x\right)=\left(2x-3\right)\left(x+5\right)\)
b) \(\frac{3}{5x-1}+\frac{2}{3-5x}=\frac{4}{\left(1-5x\right)\left(x-3\right)}\)
c) \(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x-1}=\frac{3}{x\left(x^4+x^2+1\right)}\)
d) \(\frac{m-3}{m+3}-\frac{1-3m}{1+3m}-1\)
e)\(\frac{x}{3+x}-\frac{x}{x-3}=\frac{1}{9-x^2}\)
Tính
a, \(\frac{1}{\left(y-1\right)\left(y-2\right)}+\frac{2}{\left(2-y\right)\left(3y-y\right)}+\frac{3}{\left(1-y\right)\left(y-3\right)}\)
b, \(\frac{x^2}{x+1}+\frac{2x}{x^2-1}-\frac{1}{1-x}+1\)
c, \(\frac{1}{x^2+3x+2}-\frac{2x}{x^2+4x^2+4x}+\frac{1}{x^2+5x+6}\)
Tìm tập nghiệp xác định và giải các phương trình sau:
a)\(\frac{1-x}{x+1}+3=\frac{2x+3}{x+1}\)
b)\(\frac{\left(x+2\right)^2}{2x-3}=\frac{x^2-10}{2x-3}\)
c)\(\frac{5x-2}{2-2x}+\frac{2x-1}{2}=1-\frac{x^2+x-3}{1-x}\)
d)\(\frac{5x-2}{3}+\frac{\left(x+1\right)\left(x-1\right)}{3x-1}=\frac{\left(x+2\right)\left(1-3x\right)}{9x-3}\)
Bạn nào giúp mình với ạ :<
