a) \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\Rightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\Rightarrow\left(2x-3\right)\left(7x-2x+3\right)=0\Rightarrow\left[{}\begin{matrix}2x-3=0\\5x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{5}\end{matrix}\right.\)

b) \(\left(2x-7\right).\left(x-2\right)\left(x^2-4\right)=0\Rightarrow\left(2x-7\right)\left(x-2\right)^2\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}2x-7=0\\\left(x-2\right)^2=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\\x=-2\end{matrix}\right.\)

c)\(\left(9x^2-25\right)-\left(6x-10\right)=0\Rightarrow\left(3x-5\right)\left(3x+5\right)-2\left(3x-5\right)=0\Rightarrow\left(3x-5\right)\left(3x+5-2\right)=0\Rightarrow\left[{}\begin{matrix}3x-5=0\\3x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=1\end{matrix}\right.\)

a: Ta có: \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\)

\(\Leftrightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\)

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

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

b: Ta có: \(\left(2x-7\right)\left(x-2\right)\left(x^2-4\right)=0\)

\(\Leftrightarrow\left(2x-7\right)\left(x-2\right)^2\cdot\left(x+2\right)=0\)

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

c: Ta có: \(\left(9x^2-25\right)-\left(6x-10\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)

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: \(5x\left(2x-3\right)-6x+9=0\)

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

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

\(a,\Leftrightarrow\left(x-2\right)^3-3x\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-2-3x\right)=0\\ \Leftrightarrow\left(x-2\right)\left(-2x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\\ b,\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\\ \Leftrightarrow x\left(x-2\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)

e: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow2x=-7\)

hay \(x=-\dfrac{7}{2}\)

f: Ta có: \(x^3-6x^2+12x-19=0\)

\(\Leftrightarrow x^3-6x^2+12x-8-11=0\)

\(\Leftrightarrow\left(x-2\right)^3=11\)

hay \(x=\sqrt[3]{11}+2\)

a) (x-4)(x+4)-x(x+2)=0

     x²-16-x²-2x = 0

     -16 - 2x = 0

             2x = -16 

               x = -16/2

               x = -8

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

     (3x-1)(x-2)=0

=> x ∈ {1/3 ; 2 }

c) 6x - 12x² = 0

    6x(1-2x) = 0

=> x ∈ {0; 1/2 }

d) mình thấy có vẻ hơi sai đề nên mình ko giải được, bạn thông cảm nha

d/ 4x (3 - ¹/₄ x) + (x -2) ( x+ 2)

câu d bị  sai đề 

a) \(x^2+6x+9=144\)

\(\Leftrightarrow\left(x+3\right)^2=12^2\)

\(\Leftrightarrow x+3=12\)

\(\Leftrightarrow x=9\)

\(\text{a) }x^2+6x+9=144\\ \Leftrightarrow\left(x^2+6x+9\right)-144=0\\ \Leftrightarrow\left(x+3\right)^2-12^2=0\\ \Leftrightarrow\left(x+3+12\right)\left(x+3-12\right)=0\\ \Leftrightarrow\left(x+15\right)\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+15=0\\x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-15\\x=9\end{matrix}\right.\)

Vậy tập nghiệm phương trình là \(S=\left\{9;-15\right\}\)

\(\dfrac{x-19}{1999}+\dfrac{x-23}{1995}+\dfrac{x+82}{700}=5\\ \Leftrightarrow\left(\dfrac{x-19}{1999}-1\right)+\left(\dfrac{x-23}{1995}-1\right)+\left(\dfrac{x+82}{700}-3\right)=0\\ \Leftrightarrow\dfrac{x-2018}{1999}+\dfrac{x-2018}{1995}+\dfrac{x-2018}{700}=0\\ \Leftrightarrow\left(x-2018\right)\left(\dfrac{1}{1999}+\dfrac{1}{1995}+\dfrac{1}{700}\right)=0\\ \Leftrightarrow x-2018=0\left(\text{Vì }\dfrac{1}{1999}+\dfrac{1}{1995}+\dfrac{1}{700}\ne0\right)\\ \Leftrightarrow x=2018\)

Vậy nghiệm của phương trình là \(x=2018\)

\(\text{c) }x^3-3x^2+4=0\\ \Leftrightarrow x^3-2x^2-x^2+4=0\\ \Leftrightarrow\left(x^3-2x^2\right)-\left(x^2-4\right)=0\\ \Leftrightarrow x^2\left(x-2\right)-\left(x+2\right)\left(x-2\right)=0\\ \Leftrightarrow\left(x^2-x-2\right)\left(x-2\right)=0\\ \Leftrightarrow\left(x^2-2x+x-2\right)\left(x-2\right)=0\\ \Leftrightarrow\left[\left(x^2-2x\right)+\left(x-2\right)\right]\left(x-2\right)=0\\ \Leftrightarrow\left[x\left(x-2\right)+\left(x-2\right)\right]\left(x-2\right)=0\\\Leftrightarrow \left(x+2\right)\left(x-2\right)^2=0\\\Leftrightarrow\left[{}\begin{matrix}x+2=0\\\left(x-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=2\end{matrix}\right. \)

Vậy tập nghiệm phương trình là \(S=\left\{-2;2\right\}\)

Lời giải:

a. $(x.0,25+1999).2000=(53+1999).2000$

$x.0,25.2000+1999.2000=53.2000+1999.2000$

$x.0,25.2000=53.2000$

$x.0,25=53$

$x=53:0,25=212$

b. 

$(5457+x:2):7=1075$

$5457+x:2=1075\times 7=7525$

$x:2=7525-5457=2068$

$x=2068\times 2=4136$

c. 

$1-(\frac{12}{5}+x-\frac{8}{9}): \frac{16}{9}=0$

$(\frac{12}{5}+x-\frac{8}{9}):\frac{16}{9}=1$

$\frac{12}{5}+x-\frac{8}{9}=1.\frac{16}{9}=\frac{16}{9}$

$\frac{68}{45}+x=\frac{16}{9}$

$x=\frac{16}{9}-\frac{68}{45}=\frac{4}{15}$

a) \(6x^2-72x=0\)

\(6x\left(x-12\right)=0\)

\(6x=0\) hoặc \(x-72=0\)

*) \(6x=0\)

\(x=0\)

*) \(x-12=0\)

\(x=12\)

Vậy \(x=0;x=12\)

b) \(-2x^4+16x=0\)

\(-2x\left(x^3-8\right)=0\)

\(-2x=0\) hoặc \(x^3-8=0\)

*) \(-2x=0\)

\(x=0\)

*) \(x^3-8=0\)

\(x^3=8\)

\(x=2\)

Vậy \(x=0;x=2\)

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

\(x^2-5x-x^2+6x-9=0\)

\(x-9=0\)

\(x=9\)

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

\(x^3-6x^2+12x-8-x^3+8=0\)

\(-6x^2+12x=0\)

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

\(-6x=0\) hoặc \(x-2=0\)

*) \(-6x=0\)

\(x=0\)

*) \(x-2=0\)

\(x=2\)

Vậy \(x=0;x=2\)

a. 2x\(^2\)-8=0

2x\(^2\)=8

x\(^2\)=4

x=2

b.3x\(^3\)-5x=0

x(3x\(^2\)-5)=0

\(\left[{}\begin{matrix}x=0\\x^2-5=0\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x=0\\x^2=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=^+_-\sqrt{5}\end{matrix}\right.\)

c.x+3x-4=0\(^{\left(\cdot\right)}\)

đặt t=x\(^2\) (t>0)

ta có pt: t\(^2\)+3t-4=0 \(^{\left(1\right)}\)

thấy có a+b+c=1+3+(-4)=0 nên pt\(^{\left(1\right)}\) có 2 nghiệm

t\(_1\)=1; t\(_2\)=\(\dfrac{c}{a}\)=-4

khi t\(_1\)=1 thì x\(^2\)=1 ⇒x=\(^+_-\)1

khi t\(_2\)=-4 thì x\(^2\)=-4 ⇒ x=\(^+_-\)2

vậy pt đã cho có 4 nghiệm x=\(^+_-\)1; x=\(^+_-\)2

d)3x\(^2\)+6x-9=0

thấy có a+b+c= 3+6+(-9)=0 nên pt có 2 nghiệm

x\(_1\)=1; x\(_2\)=\(\dfrac{c}{a}=\dfrac{-9}{3}=-3\)

e. \(\dfrac{x+2}{x-5}+3=\dfrac{6}{2-x}\)  (ĐK: x#5; x#2 )

⇔\(\dfrac{\left(x+2\right)\left(2-x\right)}{\left(x-5\right)\left(2-x\right)}+\dfrac{3\left(x+2\right)\left(2-x\right)}{\left(x-5\right)\left(2-x\right)}\)=\(\dfrac{6\left(x-5\right)}{\left(x-5\right)\left(2-x\right)}\)

⇒2x - x\(^2\) + 4 - 2x + 6x - 6x\(^2\) + 12 - 6x - 6x +30 = 0

⇔-7x\(^2\) - 6x + 46=0

Δ'=b'\(^2\)-ac = (-3)\(^2\) - (-7)\(\times\)46= 9+53 = 62>0

\(\sqrt{\Delta'}=\sqrt{62}\)

vậy pt có 2 nghiệm phân biệt

x\(_1\)=\(\dfrac{-b'+\sqrt{\Delta'}}{a}=\dfrac{3+\sqrt{62}}{-7}\)

x\(_2\)=\(\dfrac{-b'-\sqrt{\Delta'}}{a}=\dfrac{3-\sqrt{62}}{-7}\)

vậy pt đã cho có 2 nghiệm x\(_1\)=.....;x\(_2\)=......

câu g làm tương tự câu c