1. Tìm x
a) 0,5x - 2 - ( 2,5x + 1) - (-x+2) = x
b) - 1/2 . (2x + 3) + 3/4 . (4x + 2/3) = 1/2 . ( 2- 3x)
c) 5^x + 5^x+2 = 650
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18 tháng 2 2020
1)\(\left(4x-10\right)\left(24+5x\right)=0\)
\(\Leftrightarrow2\left(2x-5\right)\left(24+5x\right)=0\)
Vì 2≠0
nên \(\left[{}\begin{matrix}2x-5=0\\24+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\5x=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{-24}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{5}{2};\frac{-24}{5}\right\}\)
2) \(0,5x\left(x-3\right)=\left(x-3\right)\left(2,5x-4\right)\)
\(\Leftrightarrow0,5x\left(x-3\right)-\left(x-3\right)\left(2,5x-4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left[0,5x-\left(2,5x-4\right)\right]=0\)
\(\Leftrightarrow\left(x-3\right)\left(0,5x-2,5x+4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(-2x+4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(4-2x\right)=0\)
\(\Leftrightarrow\left(x-3\right)\cdot2\cdot\left(2-x\right)=0\)
Vì 2≠0
nên \(\left[{}\begin{matrix}x-3=0\\2-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
Vậy: x∈{2;3}
3) \(4x^2-1=\left(2x+1\right)\left(3x-5\right)\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-1\right)-\left(2x+1\right)\left(3x-5\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left[2x-1-\left(3x-5\right)\right]=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-1-3x+5\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(4-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\4-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-1\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=4\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{-1}{2};4\right\}\)
4) \(\left(2-3x\right)\left(x+11\right)=\left(3x-2\right)\left(2-5x\right)\)
\(\Leftrightarrow\left(2-3x\right)\left(x+11\right)-\left(3x-2\right)\left(2-5x\right)=0\)
\(\Leftrightarrow\left(2-3x\right)\left(x+11\right)+\left(2-3x\right)\left(2-5x\right)=0\)
\(\Leftrightarrow\left(2-3x\right)\left(x+11+2-5x\right)=0\)
\(\Leftrightarrow\left(2-3x\right)\left(13-4x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2-3x=0\\13-4x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\4x=13\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=\frac{13}{4}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{2}{3};\frac{13}{4}\right\}\)
18 tháng 5 2021
a) Ta có: \(6x\left(x-5\right)+3x\left(7-2x\right)=18\)
\(\Leftrightarrow6x^2-30x+21x-6x^2=18\)
\(\Leftrightarrow-9x=18\)
hay x=-2
Vậy: S={-2}
b) Ta có: \(2x\left(3x+1\right)+\left(4-2x\right)\cdot3x=7\)
\(\Leftrightarrow6x^2+2x+12x-6x^2=7\)
\(\Leftrightarrow14x=7\)
hay \(x=\dfrac{1}{2}\)
Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)
c) Ta có: \(0.5x\left(0.4-4x\right)+\left(2x+5\right)\cdot x=-6.5\)
\(\Leftrightarrow0.2x-2x^2+2x^2+5x=-6.5\)
\(\Leftrightarrow5.2x=-6.5\)
hay \(x=-\dfrac{5}{4}\)
Vậy: \(S=\left\{-\dfrac{5}{4}\right\}\)
18 tháng 5 2021
d) Ta có: \(\left(x+3\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)=6\)
\(\Leftrightarrow x^2+5x+6-\left(x^2+3x-10\right)=6\)
\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)
\(\Leftrightarrow2x+16=6\)
\(\Leftrightarrow2x=-10\)
hay x=-5
Vậy: S={-5}
e) Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)
\(\Leftrightarrow14x=0\)
hay x=0
Vậy: S={0}
14 tháng 12 2021
\(a,\Leftrightarrow x^2+6x+9-x^2+3x+10=1\\ \Leftrightarrow9x=-18\Leftrightarrow x=-2\\ b,\Leftrightarrow4x^2-4x+1-4x^2+17x+15=3\\ \Leftrightarrow13x=-13\Leftrightarrow x=-1\\ c,\Leftrightarrow3x\left(x-2\right)+4\left(x-2\right)=0\\ \Leftrightarrow\left(3x+4\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=2\end{matrix}\right.\\ d,\Leftrightarrow2x\left(3x+5\right)-6\left(3x+5\right)=0\\ \Leftrightarrow\left(x-3\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{3}\end{matrix}\right.\)
21 tháng 1 2018
\(a,2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)
Vậy nghiệm của pt là \(S=\left\{3;-\dfrac{5}{2}\right\}\)
\(b,\left(2-3x\right)\left(x+11\right)=\left(3x-2\right)\left(2-5x\right)\)
\(\Leftrightarrow-\left(3x-2\right)\left(x+11\right)-\left(3x-2\right)\left(2-5x\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(-x-11-2+5x\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(4x-13\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\4x-13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{13}{4}\end{matrix}\right.\)
Vậy nghiệm của pt là \(S=\left\{\dfrac{2}{3};\dfrac{13}{4}\right\}\)
\(c,\left(2x+1\right)\left(3x-2\right)=\left(5x-8\right)\left(2x+1\right)\)
\(\Leftrightarrow\left(2x+1\right)\left(3x-2\right)-\left(5x-8\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(3x-2-5x+8\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(-2x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\-2x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)
Vậy nghiệm của pt là \(S=\left\{-\dfrac{1}{2};3\right\}\)
\(d,\left(x-1\right)\left(2x-1\right)=x\left(1-x\right)\)
\(\Leftrightarrow\left(x-1\right)\left(2x-1\right)+x\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-1+x\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy nghiệm của pt là \(S=\left\{1;\dfrac{1}{3}\right\}\)
\(e,0,5x\left(x-3\right)=\left(x-3\right)\left(1,5x-1\right)\)
\(\Leftrightarrow0,5x\left(x-3\right)-\left(x-3\right)\left(1,5x-1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(0,5x-1,5x+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(-x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\-x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
Vậy nghiệm của pt là \(S=\left\{1;3\right\}\)
\(f,\left(x+2\right)\left(3-4x\right)=x^2+4x=4\)
\(\Leftrightarrow\left(x+2\right)\left(3-4x\right)-x^2-4x-4=0\)
\(\Leftrightarrow\left(x+2\right)\left(3-4x\right)-\left(x^2+4x+4\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3-4x\right)-\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(x+2\right)\left(3-4x-x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(-5x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\-5x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{5}\end{matrix}\right.\)
Vậy nghiệm của pt là \(S=\left\{-2;\dfrac{1}{5}\right\}\)
\(g,\left(2x^2+1\right)\left(4x-3\right)=\left(x-12\right)\left(2x^2+1\right)\)
\(\Leftrightarrow\left(2x^2+1\right)\left(4x-3\right)-\left(x-12\right)\left(2x^2+1\right)=0\)
\(\Leftrightarrow\left(2x^2+1\right)\left(4x-3-x+12\right)=0\)
\(\Leftrightarrow\left(2x^2+1\right)\left(3x+9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2+1>0\forall x\\3x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x^2+1>0\\x=-3\end{matrix}\right.\)
Vậy nghiệm của pt là \(S=\left\{-3\right\}\)
\(h,2x\left(x-1\right)=x^2-1\)
\(\Leftrightarrow2x\left(x-1\right)-\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
Vậy nghiệm của pt là \(S=\left\{1\right\}\)
10 tháng 2 2022
b: \(\Leftrightarrow\dfrac{-3x^2+36x+12}{3\left(x+4\right)\left(x-1\right)}=\dfrac{36\left(x-1\right)}{3\left(x+4\right)\left(x-1\right)}+\dfrac{12\left(x+4\right)}{3\left(x-1\right)\left(x+4\right)}\)
\(\Leftrightarrow-3x^2+36x+12=36x-36+12x+48\)
\(\Leftrightarrow-3x^2+36x+12-48x-12=0\)
\(\Leftrightarrow3x\left(x+4\right)=0\)
=>x=0(nhận) hoặc x=-4(loại)
K
30 tháng 6 2021
a) 3x(4x-3)-2x(5-6x)=0
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow24x^2-19x=0\)
\(\Leftrightarrow x\left(24x-19\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)
Vậy x=0 hoặc x=\(\dfrac{19}{24}\)