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3, đk : x =< 3/5
TH1 : \(x-2=3-5x\Leftrightarrow6x=5\Leftrightarrow x=\dfrac{5}{6}\)(ktm)
TH2 : \(x-2=5x-3\Leftrightarrow4x=1\Leftrightarrow x=\dfrac{1}{4}\)(tm)
4, \(\Leftrightarrow8x-14=3x+21\Leftrightarrow5x=35\Leftrightarrow x=7\)
Bài 3:
\(\Leftrightarrow x-2=3-5x\\ \Leftrightarrow x+5x=3+2\\ \Leftrightarrow6x=5\\ \Leftrightarrow x=\dfrac{5}{6}\)
Vậy \(x=\dfrac{5}{6}\)
Bài 4:
\(\Leftrightarrow8x-14=3x+3+18\)
\(\Leftrightarrow8x-3x=3+18+14\\ \Leftrightarrow5x=35\\ \Leftrightarrow x=\dfrac{35}{5}=7\)
Vậy x = 7
\(3,x^3-4x=0\)
\(x\left(x^2-4\right)=0\)
\(\left(x-2\right)x\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=0\\x=2\end{matrix}\right.\)
\(4,4x-3\left(x-2\right)=7-x\)
\(4x-3x+6=7-x\)
\(x+6=7-x\)
\(2x=1\)
\(x=\dfrac{1}{2}\)
\(3\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
4 \(\Leftrightarrow4x-3x+6-7+x=0\Leftrightarrow x=\dfrac{1}{2}\)
giải phương trình sau:
a, (3x+1/4)-1/3*(6x+9/5)=1
b, (5/2x+1)-(2x/1-2x)=1-(6-4x/4x^2-1)
giải hộ mk vs ạ
a, \(\left(3x+\frac{1}{4}\right)-\frac{1}{3}\left(6x+\frac{9}{5}\right)=1\)
\(3x+\frac{1}{4}-\frac{6}{3}x-\frac{3}{5}=1\)
\(x-\frac{7}{20}=1\Leftrightarrow x=\frac{27}{20}\)
b,ĐKXĐ : x \(\ne\)-1/2 ; 1/2
\(\left(\frac{5}{2x+1}\right)-\left(\frac{2x}{1-2x}\right)=1-\left(\frac{6-4x}{4x^2-1}\right)\)
\(\frac{5}{2x+1}-\frac{2x}{1-2x}=1-\frac{6-4x}{4x^2-1}\)
\(\frac{5}{2x+1}-\frac{2x}{1-2x}=1-\frac{2\left(3-2x\right)}{\left(2x+1\right)\left(2x-1\right)}\)
\(\frac{5\left(1-2x\right)\left(2x-1\right)\left(2x+1\right)}{\left(2x+1\right)^2\left(1-2x\right)\left(2x-1\right)}-\frac{2x\left(2x+1\right)^2\left(2x-1\right)}{\left(1-2x\right)\left(2x+1\right)^2\left(2x-1\right)}=\frac{\left(2x+1\right)^2\left(1-2x\right)\left(2x-1\right)}{\left(2x+1\right)^2\left(1-2x\right)\left(2x-1\right)}-\frac{2\left(3-2x\right)\left(2x+1\right)\left(1-2x\right)}{\left(2x+1\right)\left(2x-1\right)^2\left(2x-1\right)\left(1-2x\right)}\)
\(22x-5-20x^2-8x^3=18x-7-8x^3-4x^2\)
lm nốt nha,bị troll rồi ko vt đc nữa.
\(a)PT\Leftrightarrow4x^2-9-4x^2+20x+3x=0.\\ \Leftrightarrow23x=9.\\ \Leftrightarrow x=\dfrac{9}{23}.\\ b)PT\Leftrightarrow\left(2x+1\right)\left(4x-3\right)-\left(2x+1\right)\left(2x-1\right)=0.\\\Leftrightarrow\left(2x+1\right)\left(4x-3-2x+1\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(2x-2\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(x-1\right)=0. \)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}.\\x=1.\end{matrix}\right.\)
\(\dfrac{2x-1}{5}-\dfrac{4x}{3}=2x-\dfrac{x}{10}\\ \Leftrightarrow\dfrac{6\left(2x-1\right)}{30}-\dfrac{40x}{30}=\dfrac{60x}{30}-\dfrac{3x}{30}\\ \Leftrightarrow12x-6-40x=60x-3x\\ \Leftrightarrow-28x-6=57x\\ \Leftrightarrow57x+28x+6=0\\ \Leftrightarrow85x=-6\\ \Leftrightarrow x=-\dfrac{6}{85}\)
Bài 1:
a) Ta có: \(2\left(3-4x\right)=10-\left(2x-5\right)\)
\(\Leftrightarrow6-8x-10+2x-5=0\)
\(\Leftrightarrow-6x+11=0\)
\(\Leftrightarrow-6x=-11\)
hay \(x=\dfrac{11}{6}\)
b) Ta có: \(3\left(2-4x\right)=11-\left(3x-1\right)\)
\(\Leftrightarrow6-12x-11+3x-1=0\)
\(\Leftrightarrow-9x-6=0\)
\(\Leftrightarrow-9x=6\)
hay \(x=-\dfrac{2}{3}\)
a: \(\Leftrightarrow x\left(2x+10\right)-x\left(x-2\right)=0\)
=>x(2x+10-x+2)=0
=>x(x+12)=0
=>x=0 hoặc x=-12
b: \(\Leftrightarrow\left(2x-5\right)\left(x+11\right)+\left(2x-5\right)\left(2x+1\right)=0\)
=>(2x-5)(3x+12)=0
=>x=5/2 hoặc x=-4
c: \(\Leftrightarrow\left(2x\right)^2-\left(x+3\right)^2=0\)
=>(x-3)(3x+3)=0
=>x=3 hoặc x=-1
d: \(\Leftrightarrow\left(x+2\right)\left(5-4x\right)-\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(x+2\right)\left(5-4x-x-2\right)=0\)
=>(x+2)(-5x+3)=0
=>x=-2 hoặc x=3/5
\(a,\left(x-2\right)x=2x\left(x+5\right)\)
\(\Leftrightarrow\left(x-2\right)x-2x\left(x+5\right)=0\)
\(\Leftrightarrow x.\left(x-2-2x-10\right)=0\)
\(\Leftrightarrow x\left(-x-12\right)=0\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+12=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=-12\end{matrix}\right.\)
Sửa đề: 2(x-1)^2+4x-19=(2x-1)(2x+5)
=>2(x^2-2x+1)+4x-19=4x^2+10x-2x-5
=>2x^2-4x+2+4x-19=4x^2+8x-5
=>4x^2+8x-5=2x^2-17
=>2x^2+8x+12=0
=>x^2+4x+6=0
=>(x+2)^2+2=0(loại)
\(6x-3+5=4x-1\)
<=> \(6x+2=4x-1\)
<=> \(6x=4x-3\)
<=> \(2x=-3\)
<=> \(x=-\dfrac{3}{2}\)
\(_{_{\text{chắc sai ha :)}}}\)