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\(a,\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)
\(\left(x-1\right)\left(5x+3-3x+8\right)=0\)
\(\left(x-1\right)\left(2x+11\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\2x+11=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\2x=-11\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=-\frac{11}{2}\end{cases}}}\)
\(b,3x\left(25x+15\right)-35\left(5x+3\right)=0\)
\(15x\left(5x+3\right)-35\left(5x+3\right)=0\)
\(\left(5x+3\right).5\left(3x-7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x+3=0\\5\left(3x-7\right)=0\end{cases}\Rightarrow\orbr{\begin{cases}5x=-3\\3x-7=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{3}{5}\\3x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{3}{5}\\x=\frac{7}{3}\end{cases}}}\)
a. (x−1)(5x+3)=(3x−8)(x−1)(x−1)(5x+3)=(3x−8)(x−1)
⇔(x−1)(5x+3)−(3x−8)(x−1)=0⇔(x−1)[(5x+3)−(3x−8)]=0⇔(x−1)(5x+3−3x+8)=0⇔(x−1)(2x+11)=0⇔(x−1)(5x+3)−(3x−8)(x−1)=0⇔(x−1)[(5x+3)−(3x−8)]=0⇔(x−1)(5x+3−3x+8)=0⇔(x−1)(2x+11)=0
⇔x−1=0⇔x−1=0hoặc 2x+11=02x+11=0
+ x−1=0⇔x=1x−1=0⇔x=1
+ 2x+11=0⇔x=−5,52x+11=0⇔x=−5,5
Phương trình có nghiệm x = 1 hoặc x = -5,5
b. 3x(25x+15)−35(5x+3)=03x(25x+15)−35(5x+3)=0
⇔15x(5x+3)−35(5x+3)=0⇔(15x−35)(5x+3)=0⇔15x(5x+3)−35(5x+3)=0⇔(15x−35)(5x+3)=0
⇔15x−35=0⇔15x−35=0 hoặc 5x+3=05x+3=0
+ 15x−35=0⇔x=3515=7315x−35=0⇔x=3515=\(\frac{7}{3}\)
+ 5x+3=0⇔x=−355x+3=0⇔x=−\(\frac{3}{5}\)
Phương trình có nghiệm x=\(\frac{7}{3}\)x=\(\frac{7}{3}\) hoặc x=−\(\frac{3}{5}\)
a) (x - 1)(5x + 3) = (3x - 8)(x - 1)
\(\Leftrightarrow\left(x-1\right)\left(5x+3\right)-\left(3x-8\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x+3-3x+8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+11\right)=0\)
\(\Leftrightarrow x-1=0\Rightarrow x=1\)
và\(2x+11=0\Rightarrow x=\frac{-11}{2}\)
a) 3(25x+15)-35(5x+3)=0
\(\Leftrightarrow3\cdot5\left(5x+3\right)-35\left(5x+3\right)=0\)
\(\Leftrightarrow\left(5x+3\right)\left(3\cdot5-35\right)=0\)
\(\Leftrightarrow\left(5x+3\right)=0\Leftrightarrow x=\frac{-3}{5}\)
Vậy x=-3/5
a) 6x2 - 5x + 3 = 2x - 3x(2 - x)
<=> 6x2 - 5x + 3 = 2x - 6x + 3x2
<=> 6x2 - 5x + 3 = -4x + 3x2
<=> 6x2 - 5x + 3 + 4x - 3x2 = 0
<=> 3x2 - x + 3 = 0
=> Pt vô nghiệm
b) 25x2 - 9 = (5x + 3)(2x + 1)
<=> 25x2 - 9 = 10x2 + 5x + 6x + 3
<=> 25x2 - 9 = 10x2 + 11x + 3
<=> 25x2 - 9 - 10x2 - 11x - 3 = 0
<=> 15x2 - 12 - 11x = 0
<=> 15x2 + 9x - 20x - 12 = 0
<=> 3x(5x + 3) - 4(5x + 3) = 0
<=> (5x + 3)(3x - 4) = 0
<=> 5x + 3 = 0 hoặc 3x - 4 = 0
<=> x = -3/5 hoặc x = 4/3
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a) (x - 1)(5x + 3) = (3x - 8)(x - 1)
⇔ (x - 1)(5x + 3) - (3x - 8)(x - 1) = 0
⇔ (x - 1)(5x + 3 - 3x + 8) = 0
⇔ (x - 1)(2x + 11) = 0
⇔\(\left[{}\begin{matrix}x-1=0\\2x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{-11}{2}\end{matrix}\right.\)
Vậy S = {1; \(\frac{-11}{2}\)}
b) 3x(25x + 15) - 35(5x + 3) = 0
⇔ 15x(5x + 3) - 35(5x + 3) = 0
⇔ 5(3x - 7)(5x + 3) = 0
⇔ \(\left[{}\begin{matrix}3x-7=0\\5x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{3}\\x=\frac{-3}{5}\end{matrix}\right.\)
Vậy S = {\(\frac{7}{3};\frac{-3}{5}\)}
a/ \(\Leftrightarrow\left(x-1\right)\left(5x+3\right)-\left(3x-8\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x+3-3x+8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+11\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\2x+11=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\frac{11}{2}\end{matrix}\right.\)
b/ \(3x.5\left(5x+3\right)-5.7\left(5x+3\right)=0\)
\(\Leftrightarrow5\left(3x-7\right)\left(5x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-7=0\\5x+3=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{7}{3}\\x=-\frac{3}{5}\end{matrix}\right.\)
\(3x\left(25x+15\right)-35\left(5x+3\right)=0\\ \Leftrightarrow75x^2+45x-175x-105=0\\\Leftrightarrow 75x^2-130x-105=0\\\Leftrightarrow 75\left(x^2-\frac{26}{15}x-\frac{7}{5}\right)=0\\\Leftrightarrow x^2-\frac{26}{15}x-\frac{7}{5}=0\\\Leftrightarrow x^2+\frac{3}{5}x-\frac{7}{3}x-\frac{7}{5}=0\\\Leftrightarrow \left(x+\frac{3}{5}\right)\left(x-\frac{7}{3}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+\frac{3}{5}=0\\x-\frac{7}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{3}{5}\\x=\frac{7}{3}\end{matrix}\right.\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{-\frac{3}{5};\frac{7}{3}\right\}\)
\(1.\left(5x+1\right)^2=\left(3x-2\right)^2\\ \Leftrightarrow\left(5x+1\right)^2-\left(3x-2\right)^2=0\\ \Leftrightarrow\left(5x+1-3x+2\right)\left(5x+1+3x-2\right)=0\\\Leftrightarrow \left(2x+3\right)\left(8x-1\right)=0\\\Leftrightarrow \left[{}\begin{matrix}2x+3=0\\8x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{2}{3}\\x=\frac{1}{8}\end{matrix}\right.\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{-\frac{2}{3};\frac{1}{8}\right\}\)
A. \(4\left(x+2\right)-7\left(2x-1\right)+9\left(3x-4\right)=30\)
\(\Leftrightarrow4x+8-14x+7+27x-36=30\)
\(\Leftrightarrow4x-14x+27x=30-8-7+36\)
\(\Leftrightarrow17x=51\)
\(\Leftrightarrow x=3\) . Vậy \(S=\left\{3\right\}\)
B. \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow10x-12x-12x=16-15-16+11\)
\(\Leftrightarrow10x=-4\)
\(\Leftrightarrow x=-\dfrac{2}{5}\) . Vậy \(S=\left\{-\dfrac{2}{5}\right\}\)
Câu C) bạn xem lại đề nha mik tính ko đc
D. \(\left(5x-3\right)4x-2x\left(10x-3\right)=15\)
\(\Leftrightarrow20x^2-12x-20x^2+6x=15\)
\(\Leftrightarrow-6x=15\)
\(\Leftrightarrow x=-\dfrac{5}{2}\) . Vậy \(S=\left\{-\dfrac{5}{2}\right\}\)
b) PT \(\Leftrightarrow15x\left(5x+3\right)-35\left(5x+3\right)=0\)
\(\Leftrightarrow\left(15x-35\right)\left(5x+3\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{3}{5};\dfrac{7}{3}\right\}\)
c) PT \(\Leftrightarrow\left(2-3x\right)\left(x-11\right)+\left(2-3x\right)\left(2-5x\right)=0\)
\(\Leftrightarrow\left(2-3x\right)\left(-9-4x\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{9}{4}\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{2}{3};-\dfrac{9}{4}\right\}\)
a)(x-1)(5x+3)=(3x-8)(x-1)
\(\Leftrightarrow\)(x-1)(5x+3)-(3x-8)(x-1)=0
\(\Leftrightarrow\left(x-1\right)\left(5x-3-3x+8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-5\right)=0\)
\(\left[{}\begin{matrix}x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{1;\dfrac{5}{2}\right\}\)