Tìm A,B,C cho biết:
\(\frac{\left(x^2-x+2\right)}{\left(x-1\right)^3}=\frac{A}{\left(x-1\right)^3}+\frac{B}{\left(x-1\right)^2}+\frac{C}{x-1}\)
1)tìm các số nguyên x và y thỏa mãn:\(y^2=x^2+x+1\)
2)cho các số thực x và y thỏa mãn \(\left(x+\sqrt{a+x^2}\right)\left(y+\sqrt{a+y^2}\right)\)=a
tìm giá trị biểu thức \(4\left(x^7+y^7\right)+2\left(x^5+y^5\right)+11\left(x^3+y^3\right)+2016\)
3)cho x;y là các số thực khác 0 thỏa mãn x+y khác 0
cmr \(\frac{1}{\left(x+y\right)^3}\left(\frac{1}{x^3}+\frac{1}{y^3}\right)+\frac{3}{\left(x+y\right)^4}\left(\frac{1}{x^2}+\frac{1}{y^2}\right)+\frac{6}{\left(x+y\right)^5}\left(\frac{1}{x}+\frac{1}{y}\right)\)\(=\frac{1}{x^3y^3}\)
4)cho a,b,c là các số dương.cmr\(\sqrt{\frac{a^3}{a^3+\left(b+c\right)^3}}+\sqrt{\frac{b^3}{b^3+\left(a+c\right)^3}}+\sqrt{\frac{c^3}{c^3+\left(a+b\right)^3}}\ge1\)
Tìm các số A , B , C để có
a) \(\frac{x^2-x+2}{\left(x-1\right)^3}=\frac{A}{\left(x-1\right)^3}+\frac{B}{\left(x-1\right)^2}+\frac{C}{x-1}\)
b) \(\frac{x^2+2x-1}{\left(x-1\right)\left(x^2+1\right)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}\)
Tìm các sô A; B; C để có :
a) \(\frac{x^2-x+2}{\left(x-1\right)^3}=\frac{A}{\left(x-1\right)^3}+\frac{B}{\left(x-1\right)^2}+\frac{C}{x-1}\)
b) \(\frac{x^2+2x-1}{\left(x-1\right)\left(x^2+1\right)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}\)
xác định các số hữu tỉ a,b,c,d sao cho:
a,\(\frac{1}{x\left(x+1\right)\left(x+2\right)}=\frac{a}{x\left(x+1\right)}+\frac{b}{\left(x+1\right)\left(x+2\right)}\)
b,\(\frac{x^3}{x^4-1}=\frac{a}{x-1}+\frac{b}{x+1}+\frac{cx+d}{x^2+1}\)
c,\(\frac{2x^2-x+1}{\left(x+1\right)\left(x-2\right)^2}=\frac{a}{x+1}+\frac{b}{x-2}+\frac{c}{x-2}\)
Cho :\(A=\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{x+3};B=\frac{a}{x\left(x+a\right)}+\frac{a}{\left(x+a\right)\left(x+2a\right)}+\frac{a}{\left(x+2a\right)\left(x+3a\right)}+\frac{1}{x+3a}\)CMR : A = B
tìm a,b,c biết :
\(\frac{21x^2+4x-41}{\left(x+1\right)\left(x+2\right)\left(x-3\right)}=\frac{a}{x+1}+\frac{b}{x+2}+\frac{c}{x-3}\)
\(VP=\frac{a\left(x^2-x-3\right)+b\left(x^2-2x-3\right)+c\left(x^2+3x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x-3\right)}\)
\(=\frac{\left(a+b+c\right)x^2+x\left(-a-2b+3c\right)+\left(-3a-3b+2c\right)}{\left(x+1\right)\left(x+2\right)\left(x-3\right)}\)
đồng nhất hệ số ta có
\(\hept{\begin{cases}a+b+c=21\\-a-2b+3c=4\\-3a-3b+2c=-41\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a=24\\b=\frac{-37}{5}\\c=\frac{22}{5}\end{cases}}\)
Tính A+B+C biết A=\(\frac{1}{\left(x+y\right)^3}.\left(\frac{1}{x^4}-\frac{1}{y^4}\right)\) , B=\(\frac{2}{\left(x+y\right)^4}.\left(\frac{1}{x^3}-\frac{1}{y^3}\right)\) ,C=\(\frac{1}{\left(x+y\right)^5}.\left(\frac{1}{x^2}-\frac{1}{y^2}\right)\)
\(\frac{x^2-x+2}{\left(x-1\right)^3}=\frac{A}{\left(x-1\right)^3}+\frac{B}{\left(x-1\right)^2}+\frac{C}{x-1}\)
Tìm A,B,C?
Bài 1 : Tìm x biết :
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
b, \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
c,\(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
Bài 2 : Tìm x biết :
a, | 2x - 5 | = x +1
b, | 3x - 2 | -1 = x
c, | 3x - 7 | = 2x + 1
d, | 2x-1 | +1 = x
1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)
=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)
b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c) TT
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)
=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)
=> \(\left|50x-140\right|=\left|25x+24\right|\)
=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)
Bài 2 : a. |2x - 5| = x + 1
TH1 : 2x - 5 = x + 1
=> 2x - 5 - x = 1
=> 2x - x - 5 = 1
=> 2x - x = 6
=> x = 6
TH2 : -2x + 5 = x + 1
=> -2x + 5 - x = 1
=> -2x - x + 5 = 1
=> -3x = -4
=> x = 4/3
Ba bài còn lại tương tự