Tim x biet:
a.2|2x-3|=1/2
b.(x-1)^2=9/16
c.3/5(x-5/6)-1/2(3/2-1)=-1/4
d.1/3+1/6+1/10+......+2/x(x+1)=2015/2016
thực hiện phép tính :
a) 5x+10/10xy^2 nhân 12x/x+2
b) x-4/3x-1 nhân 9x-3/x^2-16
c)4x+2/(x+4)^2/ chia 3(x+3)/x+4
d)5x-5/3x+3 chia x-1/x+1
a: \(=\dfrac{5\left(x+2\right)}{10xy^2}\cdot\dfrac{12x}{x+2}=\dfrac{60x}{10xy^2}=\dfrac{6}{y^2}\)
b: \(=\dfrac{x-4}{3x-1}\cdot\dfrac{3\left(3x-1\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{3}{x+4}\)
c: \(=\dfrac{2\left(2x+1\right)}{\left(x+4\right)^2}\cdot\dfrac{\left(x+4\right)}{3\left(x+3\right)}=\dfrac{2\left(2x+1\right)}{3\left(x+3\right)\left(x+4\right)}\)
d: \(=\dfrac{5\left(x-1\right)}{3\left(x+1\right)}\cdot\dfrac{x+1}{x-1}=\dfrac{5}{3}\)
Tim x
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
2/ tim x
\(\frac{x+2015}{5}+\frac{x+2016}{6}=\frac{x+2017}{7} +\frac{x+2018}{8}\)
3/ tim x
\(\frac{1}{3}+\frac{1}{6}+\frac{99}{101}+\frac{1}{15}+... +\frac{1}{x\left(2x+1\right)}=\frac{1}{10}\)
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)
\(\Leftrightarrow x=-2020\)
Cảm ơn bạn rất nhiều mình đã hiểu rồi
Chúc bạn học tốt nhé
giải pt sau:a,x.(x-3)=(2-x).(x-3)
b,x-1/2+x-1/3+x-1/2016=0
c,2x/3+2x-1/6=4
d,7+2x=4.(5-x)
e,x+2/x-2-1/x=2/x.(x-2)
a,
x-2/3-2/15-2/35-2/63=1/9
b
(x+1
3)+(x+1/9)+(x+1/27)+(x+1/81)=56/81
c,
(1/3*5+1/5*7+1/7*9+.....+1/19*21)*x=9/7
d,3+5+7+...+2015/2+4+6+...+2016+x=1
e,2/6+2/12+2/20+...+2/x*(x+1)=2013/2015
a) x+2/x-2-1/x=2/x*(x-2)
b)2/2x-6+2/2x+2+2x/(x+1)*(3-x)=0
c) x+1/2017+x+2/2016=x+3/2015+x+4/2014
d) x-45/5+x-44/6+x-43/7+x-42/8=4
e) x-3/2011+x+2/2012=x-2012/2+x-2011/3
a) ĐKXĐ: \(x\notin\left\{0;2\right\}\)
Ta có: \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{x-2}{x\left(x-2\right)}=\dfrac{2}{x\left(x-2\right)}\)
Suy ra: \(x^2+2x-x+2-2=0\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)
Vậy: S={-1}
a) 2x-1/11+2x-2/12+2x-3/13=2x+5/5+2x+6/4+2x+7/3
b) x-1/2016+x-2/2015+x-3/2014+x-4/2013+x-5/2012 -5=0
c) x+2017/2+x+2015/3+x+2013/4+x+2011/5+8=0
1.giải các phương trình sau:
a, 3(2x+1)/4 - 5x+3/6 = 2x-1/3 - 3-x/4
b, 19/4 - 2(3x-5)/5 = 3-2x/10 - 3x-1/4
c, x-2*3/2+3 + x-3*5/3+5 + x-5*2/5+2 = 10
d, x-3/5*7 + x-5/3*7 + x-7/3*5 = 2(1/3 + 1/5 + 1/7)
2. giải các phương trình:
a, x-1/9 + x-2/8 = x-3/7 + x-4/6
b, (1/1*2 + 1/2*3 + 1/3*4 + ... + 1/9*10) (x-1) + 1/10x = x- 9/10
Câu 1 :
a, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}=\frac{2x-1}{3}-\frac{3-x}{4}\)
\(\Leftrightarrow\frac{6x+3}{4}+\frac{3-x}{4}=\frac{2x-1}{3}+\frac{5x+3}{6}\)
\(\Leftrightarrow\frac{5x+6}{4}=\frac{9x+1}{6}\Leftrightarrow\frac{30x+36}{24}=\frac{36x+4}{24}\)
Khử mẫu : \(30x+36=36x+4\Leftrightarrow-6x=-32\Leftrightarrow x=\frac{32}{6}=\frac{16}{3}\)
tương tự
\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)
\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)
\(< =>95-24x+40=6-4x-15x+5\)
\(< =>-24x+135=-19x+11\)
\(< =>5x=135-11=124\)
\(< =>x=\frac{124}{5}\)
\(\frac{\left(x-2\right).3}{2}+3+\frac{\left(x-3\right).5}{3}+5+\frac{\left(x-5\right).2}{5}+2=10\)
\(< =>\frac{\left(x-2\right).3.15}{30}+\frac{\left(x-3\right).5.10}{30}+\frac{\left(x-5\right).2.6}{30}=10-2-3-5\)
\(< =>\frac{\left(x-2\right).45+\left(x-3\right).50+\left(x-5\right).12}{30}=0\)
\(< =>45x-90+50x-150+12x-60=0\)
\(< =>107x-300=0< =>x=\frac{300}{107}\)
a) (2x+1)^2-2x-1=2b) (x^2-3x)^2+5(x^2-3x)+6=0c) (x^2-x-1)(x^2-x)-2=0d) (5-2x)^2+4x-10=8e) (x^2+2x+3)(x^2+2x+1)=3f) x(x-1)(x^2-x+1)-6=0
sửa lại chút: a) (2x+1)^2-2x-1=2 b) (x^2-3x)^2+5(x^2-3x)+6=0 c) (x^2-x-1)(x^2-x)-2=0 d) (5-2x)^2+4x-10=8 e) (x^2+2x+3)(x^2+2x+1)=3 f) x(x-1)(x^2-x+1)-6=0
a) Ta có: \(\left(2x+1\right)^2-2x-1=2\)
\(\Leftrightarrow\left(2x+1\right)^2-\left(2x+1\right)-2=0\)
\(\Leftrightarrow\left(2x+1\right)^2-2\left(2x+1\right)+\left(2x+1\right)-2=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x+1-2\right)+\left(2x+1-2\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-1\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{1}{2};-1\right\}\)
\(A=\left(6:\frac{3}{5}-1\frac{1}{6}x\frac{6}{7}\right):\left(4\frac{1}{5}x\frac{10}{11}+5\frac{2}{11}\right)\)\(B=\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{4}\right)x.......x\left(1-\frac{1}{2015}\right)x\left(1-\frac{1}{2016}\right)\)
\(C=5\frac{9}{10}:\frac{3}{2}-\left(2\frac{1}{3}x4\frac{1}{2}-2x2\frac{1}{3}\right):\frac{7}{4}\)