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22 tháng 4 2020

Cách làm luôn bn ơi

22 tháng 4 2020

Phương trình đưa được về dạng ax + b = 0

21 tháng 1 2016

a) (x - 1)(5x + 3) = (3x - 8)(x - 1)

\(\Leftrightarrow\left(x-1\right)\left(5x+3\right)-\left(3x-8\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(5x+3-3x+8\right)=0\)

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

\(\Leftrightarrow x-1=0\Rightarrow x=1\)

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

8 tháng 9 2017

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

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

 

5 tháng 2 2021

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

22 tháng 6 2017

\(3x\left(25x+15\right)-35\left(5x+3\right)=0\)

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

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

\(\Rightarrow\orbr{\begin{cases}15x-35=0\\5x+3=0\end{cases}}\)                              \(\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=\frac{-3}{5}\end{cases}}\)

Vậy  \(x\in\left\{\frac{7}{3};\frac{-3}{5}\right\}\)

22 tháng 6 2017

       3x(25x + 15) - 35(5x + 3) = 0
<=> 15x(5x + 3) - 35(5x + 3) = 0
<=> (5x + 3)(15x - 35) = 0
<=> 5(5x + 3)(3x - 7) = 0
<=> 5x + 3 = 0      hay      3x - 7 = 0 (vì 5 \(\ne\)0)
<=> 5x       = -3       I <=> 3x      = 7
<=>   x     =\(\frac{-3}{5}\)I <=>    x     = \(\frac{7}{3}\)

Vậy S = {\(\frac{-3}{5}\)\(\frac{7}{3}\)}

29 tháng 12 2015

-0,6

tick roi mih noi cach giai

 

30 tháng 12 2015

Mình tick rùi đó, cki cko mik đi

26 tháng 3 2018

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=−\(\frac{3}{5}\)

7 tháng 4 2018

a)  x + 30 % x = − 1 , 3

x 1 + 3 10 = − 13 10 13 10 x = − 13 10 x = − 1

b)  1 3 x + 2 5 x − 1 = 0

1 3 x + 2 5 x − 2 5 = 0 11 15 x = 2 5 x = 2 5 : 11 15 x = 6 11

c)  3 x − 1 2 − 5 x + 3 5 = − x + 1 5

3 x − 3 2 − 5 x − 3 = − x + 1 5 x = − 3 2 − 3 − 1 5 x = − 47 10

\(a,\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)

\(\left(x-1\right)\left(5x+3-3x+8\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}x-1=0\\2x+11=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\2x=-11\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=-\frac{11}{2}\end{cases}}}\)

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

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

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

\(\Rightarrow\orbr{\begin{cases}5x+3=0\\5\left(3x-7\right)=0\end{cases}\Rightarrow\orbr{\begin{cases}5x=-3\\3x-7=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{3}{5}\\3x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{3}{5}\\x=\frac{7}{3}\end{cases}}}\)

17 tháng 6 2020

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 = {\(\frac{7}{3};\frac{-3}{5}\)}

NV
17 tháng 6 2020

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

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

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