3x(25x + 15) – 35(5x + 3) = 0

⇔ 15x(5x + 3) – 35(5x + 3) = 0

⇔ (15x – 35)(5x + 3) = 0 ⇔ 15x – 35 = 0 hoặc 5x + 3 = 0

       15x – 35 = 0 ⇔ x = 35/15 = 7/3

      5x + 3 = 0 ⇔ x = - 3/5

Vậy phương trình có nghiệm x = 7/3 hoặc x = -3/5

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) 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) (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/ \(3x.5\left(5x+3\right)-5.7\left(5x+3\right)=0\)

\(\Leftrightarrow5\left(3x-7\right)\left(5x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-7=0\\5x+3=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{7}{3}\\x=-\frac{3}{5}\end{matrix}\right.\)

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 \(x\in\left\{1;\dfrac{5}{2}\right\}\)

a) Ta có: \(\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)

\(\Leftrightarrow5x^2+3x-5x-3=3x^2-3x-8x+8\)

\(\Leftrightarrow5x^2-2x-3=3x^2-11x+8\)

\(\Leftrightarrow5x^2-2x-3-3x^2+11x-8=0\)

\(\Leftrightarrow2x^2+9x-11=0\)

\(\Leftrightarrow2x^2+11x-2x-11=0\)

\(\Leftrightarrow x\left(2x+11\right)-\left(2x+11\right)=0\)

\(\Leftrightarrow\left(2x+11\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+11=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-11\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{11}{2}\\x=1\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{11}{2};1\right\}\)

b) Ta có: \(3x\left(25x+15\right)-35\left(5x+3\right)=0\)

\(\Leftrightarrow3x\cdot5\cdot\left(5x+3\right)-35\left(5x+3\right)=0\)

\(\Leftrightarrow15x\left(5x+3\right)-35\left(5x+3\right)=0\)

\(\Leftrightarrow\left(5x+3\right)\left(15x-35\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+3=0\\15x-35=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-3\\15x=35\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=\dfrac{7}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{3}{5};\dfrac{7}{3}\right\}\)

c) Ta có: \(\left(2-3x\right)\left(x-11\right)=\left(3x-2\right)\left(2-5x\right)\)

\(\Leftrightarrow2x-22-3x^2+33x=6x-15x^2-4+10x\)

\(\Leftrightarrow-3x^2+35x-22=-15x^2+16x-4\)

\(\Leftrightarrow-3x^2+35x-22+15x^2-16x+4=0\)

\(\Leftrightarrow12x^2+19x-18=0\)

\(\Leftrightarrow12x^2+27x-8x-18=0\)

\(\Leftrightarrow3x\left(4x+9\right)-2\left(4x+9\right)=0\)

\(\Leftrightarrow\left(4x+9\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+9=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-9\\3x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{9}{4}\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{9}{4};\dfrac{2}{3}\right\}\)

Bạn đăng từng câu một thì sẽ có người giúp bạn đấy!

Tick cho mình nhé!

dài thế

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\(2x+11=0\Rightarrow x=\frac{-11}{2}\)

\(3x\left(25x+15\right)-35\left(5x+3\right)=0\\ \Leftrightarrow75x^2+45x-175x-105=0\\\Leftrightarrow 75x^2-130x-105=0\\\Leftrightarrow 75\left(x^2-\frac{26}{15}x-\frac{7}{5}\right)=0\\\Leftrightarrow x^2-\frac{26}{15}x-\frac{7}{5}=0\\\Leftrightarrow x^2+\frac{3}{5}x-\frac{7}{3}x-\frac{7}{5}=0\\\Leftrightarrow \left(x+\frac{3}{5}\right)\left(x-\frac{7}{3}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+\frac{3}{5}=0\\x-\frac{7}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{3}{5}\\x=\frac{7}{3}\end{matrix}\right.\)

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 \(S=\left\{-\frac{2}{3};\frac{1}{8}\right\}\)