a: \(x\left(1-2x\right)+2x^2=14\)

=>\(x-2x^2+2x^2=14\)

=>x=14

b: \(x\left(x-5\right)+3x-15=0\)

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

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

Viết dưới dạng latex để mik dễ hỗ trợ bn nhé

\(\sqrt{3x+15}=\sqrt{10} \)

\( \sqrt{4(1-x)^{2}}-6=0\)

a: ta có: \(\sqrt{3x+15}=\sqrt{10}\)

\(\Leftrightarrow3x+15=10\)

hay \(x=-\dfrac{5}{3}\)

\(a,\Rightarrow4x\left(x^2-9\right)=0\\ \Rightarrow4x\left(x-3\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ b,\Rightarrow\left(3x-5-x-1\right)\left(3x-5+x+1\right)=0\\ \Rightarrow\left(2x-6\right)\left(4x-4\right)=0\\ \Rightarrow2\left(x-3\right)4\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

a) \(\Rightarrow4x\left(x^2-9\right)=0\)

\(\Rightarrow4x\left(x-3\right)\left(x+3\right)=0\)

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

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

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

\(\Rightarrow8\left(x-3\right)\left(x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

a: (x+2)(x-3)>0

nên x+2;x-3 cùng dấu

=>x>3 hoặc x<-2

b: (x-1)(x+4)<=0

nên x-1 và x+4 khác dấu

=>-4<=x<=1

a) x(x - 5) - 4x + 20 = 0

\(\Leftrightarrow\) x(x - 5) - (4x + 20)

\(\Leftrightarrow\) x(x - 5) - 4(x - 5) = 0

\(\Leftrightarrow\) (x - 5)(x - 4)

Khi x - 5 = 0 hoặc x - 4 = 0

 \(\Leftrightarrow\) x = 5           \(\Leftrightarrow\) x = 4

 Vậy S = \(\left\{5;4\right\}\)

b) x(x + 6) - 7x - 42 = 0

 \(\Leftrightarrow\) x(x + 6) - (7x - 42) = 0

 \(\Leftrightarrow\) x(x + 6) - 7(x + 6) = 0

 \(\Leftrightarrow\) (x + 6)(x - 7) = 0

Khi x - 6 = 0 hoặc x - 7 = 0

   \(\Leftrightarrow\) x = 6           \(\Leftrightarrow\) x = 7

 Vậy S = \(\left\{6;7\right\}\)

c) x³ - 5x² - x + 5 = 0

 \(\Leftrightarrow\) (x³ - 5x²) - (x + 5) = 0

 \(\Leftrightarrow\) x² (x - 5) - (x - 5) = 0

 \(\Leftrightarrow\) (x - 5)(x² - 1) = 0

 \(\Leftrightarrow\) (x - 5)(x - 1)(x + 1) = 0

 Khi x - 5 = 0 hoặc x - 1 = 0 hoặc x + 1 = 0

   \(\Leftrightarrow\) x = 5           \(\Leftrightarrow\) x = 1            \(\Leftrightarrow\) x = -1

 Vậy S = \(\left\{5;1;-1\right\}\)

d) 4x² - 25 - (2x - 5)(3x + 7) = 0

\(\Leftrightarrow\) (2x)² - 5² - (2x - 5)(3x + 7) = 0

\(\Leftrightarrow\) (2x - 5)(2x + 5) - (2x - 5)(3x + 7) = 0

\(\Leftrightarrow\) (2x - 5) \([\left(2x+5\right)-\left(3x+7\right)]\) = 0

\(\Leftrightarrow\) (2x - 5) ( 2x + 5 - 3x + 7) = 0

\(\Leftrightarrow\) (2x - 5)( -x + 12) = 0

Khi 2x - 5 = 0 hoặc -x + 12 = 0

  \(\Leftrightarrow\) 2x = 5             \(\Leftrightarrow\)   -x = -12

  \(\Leftrightarrow\) x = \(\dfrac{5}{2}\)              \(\Leftrightarrow\) x = 12

 Vậy S = \(\left\{\dfrac{5}{2};12\right\}\)

e) x³ + 27 + (x + 3)(x - 9) = 0

\(\Leftrightarrow\) x³ - 3³ + (x + 3)(x - 9) = 0

\(\Leftrightarrow\) (x - 3)(x² - 3x + 9) + (x + 3)(x - 9) = 0

\(\Leftrightarrow\) (x - 3) \(\left[\left(x^2-3x+9\right)+\left(x-9\right)\right]\) = 0

\(\Leftrightarrow\) (x - 3) ( x² - 3x + 9 + x - 9) = 0

\(\Leftrightarrow\) (x - 3)(x² - 2x) = 0

\(\Leftrightarrow\) (x - 3)x(x - 2)

 Khi x - 3 = 0 hoặc x = 0 hoặc x - 2 = 0

    \(\Leftrightarrow\) x = 3                            \(\Leftrightarrow\) x = 2

 Vậy S = \(\left\{3;0;2\right\}\)

 Chúc bạn học tốt

a) Ta có: \(x\left(x-5\right)-4x+20=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\)

b) Ta có: \(x\left(x+6\right)-7x-42=0\)

\(\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\)

\(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)

c) Ta có: \(x^3-5x^2-x+5=0\)

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

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

d) Ta có: \(4x^2-25-\left(2x-5\right)\left(3x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+5-3x-7\right)=0\)

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

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

\(\left(3x-2\right)^2-6x+4=0\\ =>\left(3x-2\right)^2+2\left(-3x+2\right)=0\\ =>\left(2-3x\right)^2+2\left(2-3x\right)=0\\ =>\left(2-3x\right)\left(2-3x+2\right)=0\\ =>\left(2-3x\right)\left(4-3x\right)=0\\ \)

=> 2-3x=0 hoặc 4-3x=0

Nếu 2-3x=0 thì 3x=2 => \(x=\dfrac{2}{3}\)

Nếu 4-3x=0 thì 3x=4 => \(x=\dfrac{4}{3}\)

Vậy \(x=\dfrac{2}{3},x=\dfrac{4}{3}\)

a) Ta có: \(x^2-x-12=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)

b) Ta có: \(x^2+3x-18=0\)

\(\Leftrightarrow\left(x+6\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=3\end{matrix}\right.\)

a) Ta có: \(\left(x-3\right)=\left(3-x\right)^2\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

b) Ta có: \(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}=\dfrac{1}{64}\)

\(\Leftrightarrow x^3+3\cdot x^2\cdot\dfrac{1}{2}+3\cdot x\cdot\dfrac{1}{4}+\left(\dfrac{1}{2}\right)^3=\dfrac{1}{64}\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^3=\left(\dfrac{1}{4}\right)^3\)

\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)

hay \(x=-\dfrac{1}{4}\)

c) Ta có: \(8x^3-50x=0\)

\(\Leftrightarrow2x\left(4x^2-25\right)=0\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)

f) Ta có: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

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

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

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

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

a, \(2x\left(x-3\right)-15+5x=0\\ \Rightarrow2x\left(x-3\right)-\left(15-5x\right)=0\\ \Rightarrow2x\left(x-3\right)-5\left(3-x\right)=0\\ \Rightarrow\left(2x+5\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x=3\end{matrix}\right.\)

b, \(x^3-7x=0\\ \Rightarrow x\left(x^2-7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\pm7\end{matrix}\right.\)

c, \(\left(2x-3\right)^2-\left(x+5\right)^2=0\\ \Rightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\\ \Rightarrow\left(x-8\right)\left(3x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)

Xem lại đề câu d