giải phương trình ai giúp e vs ạ T_T
\(\dfrac{25x+\left(3x-4\right)}{\dfrac{5}{15}}=\dfrac{\left(3-x\right)+105x}{\dfrac{15}{5}}+\dfrac{5\left(1-x\right)}{5}\)
giải phương trình
a.\(\left(2x-3\right)^2=\left(2x-3\right)\left(x+1\right)\)
b.\(x\left(2x-9\right)=3x\left(x-5\right)\)
c.\(3x-15=2x\left(x-5\right)\)
d.\(\dfrac{5-x}{2}=\dfrac{3x-4}{6}\)
e.\(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
a) Ta có: \(\left(2x-3\right)^2=\left(2x-3\right)\left(x+1\right)\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x-3-x-1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=4\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{3}{2};4\right\}\)
b) Ta có: \(x\left(2x-9\right)=3x\left(x-5\right)\)
\(\Leftrightarrow x\left(2x-9\right)-3x\left(x-5\right)=0\)
\(\Leftrightarrow x\left(2x-9\right)-x\left(3x-15\right)=0\)
\(\Leftrightarrow x\left(2x-9-3x+15\right)=0\)
\(\Leftrightarrow x\left(6-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
Vậy: S={0;6}
c) Ta có: \(3x-15=2x\left(x-5\right)\)
\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{5;\dfrac{3}{2}\right\}\)
d) Ta có: \(\dfrac{5-x}{2}=\dfrac{3x-4}{6}\)
\(\Leftrightarrow6\left(5-x\right)=2\left(3x-4\right)\)
\(\Leftrightarrow30-6x=6x-8\)
\(\Leftrightarrow30-6x-6x+8=0\)
\(\Leftrightarrow-12x+38=0\)
\(\Leftrightarrow-12x=-38\)
\(\Leftrightarrow x=\dfrac{19}{6}\)
Vậy: \(S=\left\{\dfrac{19}{6}\right\}\)
e) Ta có: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)
\(\Leftrightarrow6x+4-3x-1=12x+10\)
\(\Leftrightarrow3x+3-12x-10=0\)
\(\Leftrightarrow-9x-7=0\)
\(\Leftrightarrow-9x=7\)
\(\Leftrightarrow x=-\dfrac{7}{9}\)
Vậy: \(S=\left\{-\dfrac{7}{9}\right\}\)
Giải các phương trình sau:
\(h.\dfrac{3\left(2x-1\right)}{4}-\dfrac{3x+1}{10}+1=\dfrac{2\left(3x+2\right)}{5}\)
\(i.\dfrac{\left(2x+1\right)^2}{5}-\dfrac{\left(x-1\right)^2}{3}=\dfrac{7x^2-14x-5}{15}\)
\(k.x+\dfrac{2x+\dfrac{x-1}{5}}{3}=1-\dfrac{3x-\dfrac{1-2x}{3}}{5}\)
\(i.\dfrac{\left(2x+1\right)^2}{5}-\dfrac{\left(x-1\right)^2}{3}=\dfrac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\dfrac{4x^2+4x+1}{5}-\dfrac{x^2-2x+1}{3}=\dfrac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\dfrac{12x^2+12x+3}{15}-\dfrac{5x^2-10x+5}{15}=\dfrac{7x^2-14x-5}{15}\)
\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5=7x^2-14x-5\)
\(\Leftrightarrow36x=-3\)
\(\Leftrightarrow x=-\dfrac{1}{12}\)
\(k.x+\dfrac{2x+\dfrac{x-1}{5}}{3}=1-\dfrac{3x-\dfrac{1-2x}{3}}{5}\)
\(\Leftrightarrow\dfrac{15x}{15}+\dfrac{10x+x-1}{15}=\dfrac{15}{15}-\dfrac{9x-1+2x}{15}\)
\(\Leftrightarrow15x+9x-1=14-7x\)
\(\Leftrightarrow31x=15\)
\(\Leftrightarrow x=\dfrac{15}{31}\)
giải các phương trình sau
1/ \(\dfrac{x-1}{2}-\dfrac{7x+3}{15}=\dfrac{2x+1}{3}+\dfrac{3-2x}{5}\)
2/ \(\dfrac{2\left(3x+1\right)+1}{4}-5=\dfrac{2\left(3x-1\right)}{5}-\dfrac{3x+2}{10}\)
Giải các phương trình sau:
1. \(a,\dfrac{6}{x-1}-\dfrac{4}{x-3}=\dfrac{8}{2x-6}\)
\(b,\dfrac{1}{x-2}+\dfrac{5}{x+1}=\dfrac{3}{2-x}\)
\(c,\dfrac{3x}{x-2}-\dfrac{x}{x-5}=\dfrac{3x}{\left(x-2\right)\left(5-x\right)}\)
2. \(a,\left(x+2\right)\left(3-4x\right)=x^2+4x+4\)
\(b,2x^2-6x+1\)
1a.
ĐKXĐ: \(x\ne\left\{1;3\right\}\)
\(\Leftrightarrow\dfrac{6}{x-1}=\dfrac{4}{x-3}+\dfrac{4}{x-3}\)
\(\Leftrightarrow\dfrac{3}{x-1}=\dfrac{4}{x-3}\Leftrightarrow3\left(x-3\right)=4\left(x-1\right)\)
\(\Leftrightarrow3x-9=4x-4\Rightarrow x=-5\)
b.
ĐKXĐ: \(x\ne\left\{-1;2\right\}\)
\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{3}{2-x}+\dfrac{1}{2-x}\)
\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{4}{2-x}\Leftrightarrow5\left(2-x\right)=4\left(x+1\right)\)
\(\Leftrightarrow10-2x=4x+4\Leftrightarrow6x=6\Rightarrow x=1\)
1c.
ĐKXĐ: \(x\ne\left\{2;5\right\}\)
\(\Leftrightarrow\dfrac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}=\dfrac{-3x}{\left(x-2\right)\left(x-5\right)}\)
\(\Leftrightarrow3x\left(x-5\right)-x\left(x-2\right)=-3x\)
\(\Leftrightarrow2x^2-10x=0\Leftrightarrow2x\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=5\left(loại\right)\end{matrix}\right.\)
2a.
\(\Leftrightarrow-4x^2-5x+6=x^2+4x+4\)
\(\Leftrightarrow5x^2+9x-2=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{5}\end{matrix}\right.\)
2b.
\(2x^2-6x+1=0\Rightarrow x=\dfrac{3\pm\sqrt{7}}{2}\)
Giải các phương trình sau :
a) \(\dfrac{9x-0,7}{4}-\dfrac{5x-1,5}{7}=\dfrac{7x-1,1}{3}-\dfrac{5\left(0,4-2x\right)}{6}\)
b) \(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x+3}=1-\dfrac{4}{\left(x-1\right)\left(x+3\right)}\)
c) \(\dfrac{3}{4\left(x-5\right)}+\dfrac{15}{50-2x^2}=-\dfrac{7}{6\left(x+5\right)}\)
d) \(\dfrac{8x^2}{3\left(1-4x^2\right)}=\dfrac{2x}{6x-3}-\dfrac{1+8x}{4+8x}\)
Giải các bất phương trình sau
a/ (x+1).(x-1).(3x-6)>0
b/ \(\dfrac{x+3}{x-2}\le0\)
c/ \(\dfrac{\left(2x-5\right).\left(x+2\right)}{-4x+3}\ge0\)
d/ \(\dfrac{2x-5}{3x+2}< \dfrac{3x+2}{2x-5}\)
e/ \(\dfrac{2x^2+x}{1-2x}\ge1-x\)
f/ \(\dfrac{\left(2+x\right)^5.\left(x+1\right).\left(3-x\right)^{11}}{\left(2-x\right).\left(1-x\right)^{20}}\le0\)
a) \(\left(x+1\right)\left(x-1\right)\left(3x-6\right)>0\)
Lập bảng xét dấu ta được kết quả :
\(Bpt\Leftrightarrow\left[{}\begin{matrix}-1< x< 1\\x>2\end{matrix}\right.\)
b) \(\dfrac{x+3}{x-2}\le0\)
Lập bảng xét dấu ta được kết quả :
\(Bpt\Leftrightarrow-3\le x< 2\)
d) \(\dfrac{2x-5}{3x+2}< \dfrac{3x+2}{2x-5}\)
\(\Leftrightarrow\dfrac{2x-5}{3x+2}-\dfrac{3x+2}{2x-5}< 0\)
\(\Leftrightarrow\dfrac{\left(2x-5\right)^2-\left(3x+2\right)^2}{\left(3x+2\right)\left(2x-5\right)}< 0\)
\(\Leftrightarrow\dfrac{\left(2x-5+3x+2\right)\left(2x-5-3x-2\right)}{\left(3x+2\right)\left(2x-5\right)}< 0\)
\(\Leftrightarrow\dfrac{-\left(5x-3\right)\left(x+7\right)}{\left(3x+2\right)\left(2x-5\right)}< 0\)
Lập bảng xét dấu ta được kết quả :
\(Bpt\Leftrightarrow\left[{}\begin{matrix}-7< x< -\dfrac{2}{3}\\\dfrac{5}{3}< x< \dfrac{5}{2}\end{matrix}\right.\)
Giải các bất phương trình sau
a/ (x+1).(x-1).(3x-6)>0
b/ \(\dfrac{x+3}{x-2}\le0\)
c/ \(\dfrac{\left(2x-5\right).\left(x+2\right)}{-4x+3}\ge0\)
d/ \(\dfrac{2x-5}{3x+2}< \dfrac{3x+2}{2x-5}\)
e/ \(\dfrac{2x^2+x}{1-2x}\ge1-x\)
f/ \(\dfrac{\left(2+x\right)^5.\left(x+1\right).\left(3-x\right)^{11}}{\left(2-x\right).\left(1-x\right)^{20}}\le0\)
Giải các phương trình sau:
\(a.\dfrac{5x-2}{3}=\dfrac{5-3x}{2}\)
\(b.\dfrac{10x+3}{12}=1+\dfrac{6+8x}{9}\)
\(c.2\left(x+\dfrac{3}{5}\right)=5-\left(\dfrac{13}{5}+x\right)\)
\(d.\dfrac{7}{8}x-5\left(x-9\right)=\dfrac{20x+1,5}{6}\)
\(e.\dfrac{7x-1}{6}+2x=\dfrac{16-x}{5}\)
\(f.\dfrac{x+4}{5}-x+4=\dfrac{x}{3}-\dfrac{x-2}{2}\)
a: =>10x-4=15-9x
=>19x=19
hay x=1
b: \(\Leftrightarrow3\left(10x+3\right)=36+4\left(8x+6\right)\)
=>30x+9=36+32x+24
=>30x-32x=60-9
=>-2x=51
hay x=-51/2
c: \(\Leftrightarrow2x+\dfrac{6}{5}=5-\dfrac{13}{5}-x\)
=>3x=6/5
hay x=2/5
d: \(\Leftrightarrow\dfrac{7x}{8}-\dfrac{5\left(x-9\right)}{1}=\dfrac{20x+1.5}{6}\)
\(\Leftrightarrow21x-120\left(x-9\right)=4\left(20x+1.5\right)\)
=>21x-120x+1080=80x+60
=>-179x=-1020
hay x=1020/179
e: \(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)
=>35x-5+60x=96-6x
=>95x+6x=96+5
=>x=1
f: \(\Leftrightarrow6\left(x+4\right)+30\left(-x+4\right)=10x-15\left(x-2\right)\)
=>6x+24-30x+120=10x-15x+30
=>-24x+96=-5x+30
=>-19x=-66
hay x=66/19
Giải phương trình sau:
\(\dfrac{\left(2x+1\right)^2}{5}-\dfrac{\left(x-1\right)^2}{3}=\dfrac{7x^2-14x-5}{15}\)
\(\dfrac{\left(2x+1\right)^2}{5}-\dfrac{\left(x-1\right)^2}{3}=\dfrac{7x^2-14x-5}{15}\)
⇔ \(\dfrac{3\left(2x+1\right)^2}{15}-\dfrac{5\left(x-1\right)^2}{15}=\dfrac{7x^2-14x-5}{15}\)
⇔ \(3\left(2x+1\right)^2-5\left(x-1\right)^2=7x^2-14x-5\)
⇔ \(3\left(4x^2+4x+1\right)-5\left(x^2-2x+1\right)=7x^2-14x-5\)
⇔ \(12x^2+12x+3-5x^2+10x-5=7x^2-14x-5\)
⇔ \(7x^2+22x-2=7x^2-14x-5\) ⇔ \(36x+3=0\) ⇔ x=\(\dfrac{-1}{12}\)
\(\Leftrightarrow3\left(4x^2+4x+1\right)-5\left(x^2-2x+1\right)=7x^2-14x-5\)
\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5-7x^2+14x+5=0\)
\(\Leftrightarrow36x=-3\)
hay x=-1/12