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a: Ta có: \(2x\left(x-1\right)-2x^2=-6\)

\(\Leftrightarrow2x^2-2x-2x^2=-6\)

\(\Leftrightarrow x=3\)

b: Ta có: \(2x\left(x-3\right)+5\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)

a) (x – 45).27 = 0 ó x – 45 = 0 ó x = 45

b) 45.(2x – 4).13 = 0 ó 2x – 4 = 0 ó 2x – 4 = 0 ó 2x = 4 ó x = 2

c) (x – 3).(x – 5) = 0

Vậy tập nghiệm của phương trình là S = {3;5}

a x=45+0=45

b) 5x(x-2000)-x+2000=0

\(\Rightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\\ \Rightarrow\left(x-2000\right)\left(5x-1\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x-2000=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0+2000\\5x=0+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2000\\5x=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2000\\x=\dfrac{1}{5}\end{matrix}\right.\)

Ai giúp minh làm bài 5 phía trên với

c) Ta có: \(2x\left(x-2\right)+3\left(x-2\right)=0\)

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

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

d) Ta có: \(5x^2+x=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-1}{5}\end{matrix}\right.\)

1/a)Ta có -5=-1*5=-5*1

​Sau đó bn lập bảng rồi dùng ước bội để tính nhé​

​ Tk mình nha bn!

a: (2x+1)(3-x)(4-2x)=0

=>(2x+1)(x-3)(x-2)=0

hay \(x\in\left\{-\dfrac{1}{2};3;2\right\}\)

b: 2x(x-3)+5(x-3)=0

=>(x-3)(2x+5)=0

=>x=3 hoặc x=-5/2

c: =>(x-2)(x+2)+(x-2)(2x-3)=0

=>(x-2)(x+2+2x-3)=0

=>(x-2)(3x-1)=0

=>x=2 hoặc x=1/3

d: =>(x-2)(x-3)=0

=>x=2 hoặc x=3

e: =>(2x+5+x+2)(2x+5-x-2)=0

=>(3x+7)(x+3)=0

=>x=-7/3 hoặc x=-3

f: \(\Leftrightarrow2x^3+5x^2-3x=0\)

\(\Leftrightarrow x\left(2x^2+5x-3\right)=0\)

\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)

hay \(x\in\left\{0;-3;\dfrac{1}{2}\right\}\)

a)(2x-3)(x+5)=0

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-5\end{matrix}\right.\)

Vậy x=3/2 hoặc x=-5

a) \(\left(2x-3\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x+5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-5\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là: \(S=\left\{\dfrac{3}{2};-5\right\}\)

b) \(3x\left(x-2\right)-7\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(3x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-7=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{7}{2}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là: \(S=\left\{2;\dfrac{7}{2}\right\}\)

c) \(5x\left(2x-3\right)-6x+9=0\)

\(\Leftrightarrow5x\left(2x-3\right)-3\left(2x-3\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(5x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\5x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là: \(S=\left\{\dfrac{3}{2};\dfrac{3}{5}\right\}\)

a: Ta có: \(\left(2x-3\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-5\end{matrix}\right.\)

b: Ta có: \(3x\left(x-2\right)-7\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(3x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{7}{3}\end{matrix}\right.\)

c: Ta có: \(5x\left(2x-3\right)-6x+9=0\)

\(\Leftrightarrow\left(2x-3\right)\left(5x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{5}\end{matrix}\right.\)

a) ${}^{}-25=0$

${}^{}=0+25=25$

${}^{}={}^{}={}^{}$

$\Rightarrow [\begin{array}{c}2x-1=5\\ 2x-1=-5\end{array}\Rightarrow [\begin{array}{c}2x=6\\ 2x=-4\end{array}\Rightarrow [\begin{array}{c}x=3\\ x=-2\end{array}$

b) $8{}^{}-50x=0$

$2x(4{}^{}$

câu b thiếu bn ơi

ok

a) ${}^{}-25=0$

${}^{}=0+25=25$

${}^{}={}^{}={}^{}$

$\Rightarrow [\begin{array}{c}2x-1=5\\ 2x-1=-5\end{array}\Rightarrow [\begin{array}{c}2x=6\\ 2x=-4\end{array}\Rightarrow [\begin{array}{c}x=3\\ x=-2\end{array}$

b) $8{}^{}-50x=0$

$2x(4{}^{}$

a: Ta có: \(\left(2x-1\right)^2-25=0\)

\(\Leftrightarrow\left(2x-6\right)\left(2x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)