a: \(\Leftrightarrow3x\left(5x-1\right)=0\)

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

\(2x^2\left(x+5\right)-3x^2-15x=0\Leftrightarrow2x^2.\left(x+5\right)-3x\left(x+5\right)=0\)

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

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

\(\Leftrightarrow x=0\)hoặc \(x=-5\)hoặc \(x=\frac{3}{2}\)

tíck mình nha bn thanks nhìu <3 

( 2. l x l - 1) .(7 - 3x ) =0                                        ( x² + 1 ). ( 1/2 - 3x ) <0                                  ( l x l + 1 ) . ( 15x - 1 ) = 0

=> 2 . l x l - 1 = 0 hoặc 7 - 3x = 0                           => x²+1 hoặc 1/2 -3x < 0                              => l x l + 1 hoặc 15x - 1 =0

+  2 . l x l - 1 = 0 => 2 . l x l =1 => x = 1/2             + x² +1< 0 => x không tồn tại                       + l x l - 1 = 0 => l x l = 1 => thuộc 1 : -1

+ 7 - 3x = 0 => 3x = 7 => x = 7/3                           + 1/2 - 3x < 0 => 3x > 1/2 => x > 1                 + 15x - 1 = 0 => 15x =1 => x = 1/15

\(a.-3x^2+15x=0\)

\(\Leftrightarrow3x\left(-x+5\right)=0\)

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

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

\(b.2x^2-32=0\)

\(\Leftrightarrow2x^2=32\)

\(\Leftrightarrow x^2=16\)

\(\Leftrightarrow\left|x\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

\(c.2x^2-5x+1=0\)

\(a=2;b=-5;c=1\)

\(\Delta=\left(-5\right)^2-4.2.1=17>0\)

Do \(\Delta>0\) nên phương trình có 2 nghiệm phân biệt:

\(x_1=\dfrac{5+\sqrt{17}}{4}\)

\(x_2=\dfrac{5-\sqrt{17}}{4}\)

\(a,-3x^2+15x=0\\ -3x\left(x-5\right)=0\\ \Rightarrow\left\{{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=5\end{matrix}\right.\) 

\(b,\\ 2\left(x^2-16\right)=0\\ \Leftrightarrow x^2-16=0\\ \Leftrightarrow\left(x-4\right)\left(x+4\right)=0\\ \Rightarrow\left\{{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\) 

\(c,\\ \Delta=5^2-4.2=17\\ \Rightarrow x_1,x_2=\dfrac{\Delta\pm b}{2ac}\\ =\dfrac{5\pm\sqrt{17}}{4}\)

mk mới lớp 6 thôi ,lớp 9 mình .......mình.........chịu (I VERY SORRY YOU!!)

sorry, i cant do it

mình lớp 9 nhưng mình lười giải vì " QUÁ NHIỀU " lười viết

a, 15x³ - 15x = 0    

15x(x²-1)=0

15x=0 hoặc x²-1=0  (tự tính nhoa)

b,3x²-6x+3=0

3(x²-2x+1)=0

x² -2x+1=0:3=3

x²-2x=3-1=2

x(x-2)=0

x=0 hoặc x-2=0 (tự tính nhoa)

Bài làm

a) 15x³-15x=0

<=> 15x( x² - 1 ) = 0

<=> \(\orbr{\begin{cases}15x=0\\x^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}}\)

Vậy x = { 0; + 1 }

b) 3x² - 6x + 3 = 0

<=> 3( x² - 2x + 1 ) = 0

<=> x² - 2x + 1 = 0

<=> ( x - 1 )² = 0

<=> x - 1 = 0

<=> x = 1

Vậy x = 1

c) 5(x - 1) - 3x(1 - x) = 0

<=> 5(x - 1) + 3x(x - 1) = 0

<=> (5 + 3x)(x - 1) = 0

<=> \(\orbr{\begin{cases}5+3x=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{3}\\x=1\end{cases}}}\)

Vậy x = { -5/3; 1 }

e) -7(x + 2) = 2x(x + 2) 

<=> -7(x + 2 ) - 2x( x + 2 ) = 0

<=> (x + 2)(-7 - 2x) = 0

<=> \(\orbr{\begin{cases}x+2=0\\-7-2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{7}{2}\end{cases}}}\)

Vậy x = { -2; x = -7/2 }

f)(2x - 3)(3x + 5) = (x - 1)(3x + 5)

<=> (2x - 3)(3x + 5) - (x - 1)(3x + 5) = 0

<=> (3x + 5)(2x - 3 - x + 1) = 0

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

<=> \(\orbr{\begin{cases}3x+5=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{3}\\x=2\end{cases}}}\)

Vậy x = { -5/3; 2 }

không ai trả lời thì mình giúp vậy 

\(c,5\left(x-1\right)-3x\left(1-x\right)=0\)

\(< =>5x-5-3x+3x^2=0\)

\(< =>3x^2+2x-5=0< =>3x^2-3x+5x-5=0\)

\(< =>3x\left(x-1\right)+5\left(x-1\right)=\left(2x+5\right)\left(x-1\right)=0\)

\(< =>\orbr{\begin{cases}2x+5=0\\x-1=0\end{cases}}< =>\orbr{\begin{cases}2x=-5\\x=1\end{cases}}< =>\orbr{\begin{cases}x=-\frac{5}{2}\\x=1\end{cases}}\)

\(d,6x-\left(x+2\right)^2-5=\left(1-3x\right)^3-12x\)

\(< =>6x-x^2-4x-9=1-27x^3+27x^2-21x\)

\(< =>-x^2+2x-9-x+27x^3-27x^2+21x=0\)

\(< =>27x^3-28x^2+22x-9=0\)(vô nghiệm)

\(e,-7\left(x+2\right)=2x\left(x+2\right)\)

\(< =>-7x-14-2x^2-4x=0\)

\(< =>-2x^2-11x-14=0\)

\(< =>-2\left(x^2+\frac{11}{2}x+\frac{121}{16}\right)+\frac{9}{8}=0\)

\(< =>\left(x+\frac{11}{4}\right)^2=-\frac{9}{8}=\frac{9}{16}\)

\(< =>\orbr{\begin{cases}x+\frac{11}{4}=\sqrt{\frac{9}{16}}\\x+\frac{11}{4}=-\sqrt{\frac{9}{16}}\end{cases}}< =>\orbr{\begin{cases}x=\frac{3}{4}-\frac{11}{4}=-2\\x=-\frac{3}{4}-\frac{11}{4}=-\frac{14}{4}=-\frac{7}{2}\end{cases}}\)

\(f,\left(2x-3\right)\left(3x+5\right)=\left(x-1\right)\left(3x+5\right)\)

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

bài cuối mình chịu đấy 

Ý b) <0