a) Ta có: \(\left(2x-3\right)^2=\left(2x-3\right)\left(x+1\right)\)

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

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

\(\Leftrightarrow\left(2x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=4\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{3}{2};4\right\}\)

b) Ta có: \(x\left(2x-9\right)=3x\left(x-5\right)\)

\(\Leftrightarrow x\left(2x-9\right)-3x\left(x-5\right)=0\)

\(\Leftrightarrow x\left(2x-9\right)-x\left(3x-15\right)=0\)

\(\Leftrightarrow x\left(2x-9-3x+15\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

Vậy: S={0;6}

c) Ta có: \(3x-15=2x\left(x-5\right)\)

\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)

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

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

Vậy: \(S=\left\{5;\dfrac{3}{2}\right\}\)

d) Ta có: \(\dfrac{5-x}{2}=\dfrac{3x-4}{6}\)

\(\Leftrightarrow6\left(5-x\right)=2\left(3x-4\right)\)

\(\Leftrightarrow30-6x=6x-8\)

\(\Leftrightarrow30-6x-6x+8=0\)

\(\Leftrightarrow-12x+38=0\)

\(\Leftrightarrow-12x=-38\)

\(\Leftrightarrow x=\dfrac{19}{6}\)

Vậy: \(S=\left\{\dfrac{19}{6}\right\}\)

e) Ta có: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)

\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)

\(\Leftrightarrow6x+4-3x-1=12x+10\)

\(\Leftrightarrow3x+3-12x-10=0\)

\(\Leftrightarrow-9x-7=0\)

\(\Leftrightarrow-9x=7\)

\(\Leftrightarrow x=-\dfrac{7}{9}\)

Vậy: \(S=\left\{-\dfrac{7}{9}\right\}\)

Rảnh rỗi thật sự .-.

a: \(\Leftrightarrow x^3-27-x\left(x^2-4\right)=1\)

\(\Leftrightarrow x^3-27-x^3+4x=1\)

=>4x-27=1

hay x=7

b: \(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+6\left(x+1\right)^2+3x^2=15\)

\(\Leftrightarrow-9x^2+27x+6x^2+12x+6+3x^2=15\)

=>39x+6=15

hay x=3/13

c: \(\Leftrightarrow x^3-3x^2+3x-1-x^3-27+3x^2-12=2\)

\(\Leftrightarrow3x-40=2\)

hay x=14

Bài 1:

a) Ta có: \(\frac{4}{5}x-3=\frac{1}{5}x\left(4x-15\right)\)

\(\Leftrightarrow\frac{4x}{5}-3=\frac{4x^2}{5}-3x\)

\(\Leftrightarrow\frac{12x}{15}-\frac{45}{15}-\frac{12x^2}{15}+\frac{45x}{15}=0\)

Suy ra: \(12x-45-12x^2+45x=0\)

\(\Leftrightarrow-12x^2+57x-45=0\)

\(\Leftrightarrow-12x^2+12x+45x-45=0\)

\(\Leftrightarrow-12x\left(x-1\right)+45\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(-12x+45\right)=0\)

\(\Leftrightarrow-3\left(x-1\right)\left(4x-15\right)=0\)

mà \(-3\ne0\)

nên \(\left[{}\begin{matrix}x-1=0\\4x-15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\4x=15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{15}{4}\end{matrix}\right.\)

Vậy: Tập nghiệm \(S=\left\{1;\frac{15}{4}\right\}\)

b) Ta có: \(\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3-x\right)}{4}\)

\(\Leftrightarrow\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}+\frac{\left(x-3\right)^2}{4}=0\)

\(\Leftrightarrow\frac{12\left(x-3\right)}{12}-\frac{2\left(x-3\right)\left(2x-5\right)}{12}+\frac{3\left(x-3\right)^2}{12}=0\)

Suy ra: \(12\left(x-3\right)-2\left(2x^2-11x+15\right)+3\left(x^2-6x+9\right)=0\)

\(\Leftrightarrow12x-36-4x^2+22x-30+3x^2-18x+27=0\)

\(\Leftrightarrow-x^2+16x-39=0\)

\(\Leftrightarrow-\left(x^2-16x+39\right)=0\)

\(\Leftrightarrow x^2-13x-3x+39=0\)

\(\Leftrightarrow x\left(x-13\right)-3\left(x-13\right)=0\)

\(\Leftrightarrow\left(x-13\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-13=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\\x=3\end{matrix}\right.\)

Vậy: Tập nghiệm S={3;13}

c) Ta có: \(\frac{\left(3x+1\right)\left(3x-2\right)}{3}+5\left(3x+1\right)=\frac{2\left(2x+1\right)\left(3x+1\right)}{3}+2x\left(3x+1\right)\)

\(\Leftrightarrow\frac{9x^2-3x-2}{3}+5\left(3x+1\right)-\frac{12x^2+10x+2}{3}-2x\left(3x+1\right)=0\)

\(\Leftrightarrow\frac{9x^2-3x-2-12x^2-10x-2}{3}-6x^2+13x+5=0\)

\(\Leftrightarrow\frac{-3x^2-13x-4}{3}+\frac{3\left(-6x^2+13x+5\right)}{3}=0\)

Suy ra: \(-3x^2-13x-4-18x^2+39x+15=0\)

\(\Leftrightarrow-21x^2+26x+11=0\)

\(\Leftrightarrow-21x^2-7x+33x+11=0\)

\(\Leftrightarrow-7x\left(3x+1\right)+11\left(3x+1\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(-7x+11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\-7x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\-7x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{3}\\x=\frac{11}{7}\end{matrix}\right.\)

Vậy: Tập nghiệm \(S=\left\{-\frac{1}{3};\frac{11}{7}\right\}\)

\(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\)

\(\Leftrightarrow x^3-9x^2+27x-27-\left(x^3-3^3\right)+9\left(x+1\right)^2=15\)

\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+9.\left(x^2+2x+1\right)=15\)

\(\Leftrightarrow-9x^2+27x+9x^2+18x+9=15\)

\(\Leftrightarrow45x=6\Leftrightarrow x=\frac{2}{15}\)

Vậy: \(S=\left\{\frac{2}{15}\right\}\)

pt <=> \(\left(x-3\right)^3-\left(x^3-27\right)+9\left(x+1\right)^2=15\)

   <=> \(x^3-3x^2.3+3x.3^2-27-x^3+27+9x^2+18x+9=15\)

   <=>  \(45x=6\)

   <=>  \(x=\frac{6}{45}=\frac{2}{15}\)

= \(\frac{2}{15}\) nha bn

\(=\dfrac{4\left(x+3\right)^2}{\left(x+5\right)\left(5x+5\right)}-\dfrac{x^2-25}{\left(3x-2x-5\right)\left(3x+2x+5\right)}-\dfrac{\left(3x-3-x\right)\left(3x-3+x\right)}{\left(4x+15-x\right)\left(4x+15+x\right)}\)

\(=\dfrac{4\left(x+3\right)^2}{5\left(x+1\right)\left(x+5\right)}-\dfrac{x^2-25}{\left(x-5\right)\cdot5\left(x+1\right)}-\dfrac{\left(2x-3\right)\left(4x-3\right)}{3\left(x+5\right)\cdot5\left(x+3\right)}\)

\(=\dfrac{4\left(x+3\right)^2}{5\left(x+1\right)\left(x+5\right)}-\dfrac{\left(x+5\right)}{5\left(x+1\right)}-\dfrac{\left(2x-3\right)\left(4x-3\right)}{15\left(x+5\right)\left(x+3\right)}\)

\(=\dfrac{4\left(x+3\right)^2-\left(x+5\right)\left(x+1\right)}{5\left(x+1\right)\left(x+5\right)}-\dfrac{\left(2x-3\right)\left(4x-3\right)}{15\left(x+3\right)\left(x+5\right)}\)

\(=\dfrac{4x^2+24x+36-x^2-6x-5}{5\left(x+1\right)\left(x+5\right)}-\dfrac{\left(2x-3\right)\left(4x-3\right)}{15\left(x+3\right)\left(x+5\right)}\)

\(=\dfrac{3x^2+18x+31}{5\left(x+1\right)\left(x+5\right)}-\dfrac{\left(2x-3\right)\left(4x-3\right)}{15\left(x+3\right)\left(x+5\right)}\)

\(=\dfrac{3\left(x+3\right)\left(3x^2+18x+31\right)-\left(2x-3\right)\left(4x-3\right)\left(x+1\right)}{15\left(x+3\right)\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{\left(3x+9\right)\left(3x^2+18x+31\right)-\left(8x^2-18x+9\right)\left(x+1\right)}{15\left(x+3\right)\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{9x^3+81x^2+255x+279-\left(8x^3+8x^2-18x^2-18x+9x+9\right)}{15\left(x+3\right)\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{9x^3+81x^2+255x+279-\left(8x^3-10x^2-9x+9\right)}{15\left(x+3\right)\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{x^3+91x^2+264x+270}{15\left(x+3\right)\left(x+5\right)\left(x+1\right)}\)

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)(5x+3-3x+8)=0

=(x-1)(2x+11)=0

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

\(\Leftrightarrow\) x=1 hoặc x=\(\dfrac{-11}{2}\)

Vậy S={1;\(\dfrac{-11}{2}\)}

b) 3x(25x+15)-35(5x+3)=0

=3x.5(5x+3)-35(5x+3)=0

=15x(5x+3)-35(5x+3)=0

=(5x+3)(15x-35)=0

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

\(\Leftrightarrow\) x=\(\dfrac{-3}{5}\) hoặc x=\(\dfrac{7}{3}\)

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

c) (2-3x)(x+11)=(3x-2)(2-5x)

=(2-3x)(x+11)-(3x-2)(2-5x)=0

=(3x-2)[(x+11)-(2-5x)]=0

=(3x-2)(x+11-2+5x)=0

=(3x-2)(6x+9)=0

\(\Leftrightarrow\) 3x-2=0 hoặc 6x+9=0

\(\Leftrightarrow\) x=\(\dfrac{2}{3}\) hoặc x=\(\dfrac{-3}{2}\)

Vậy S={\(\dfrac{2}{3};\dfrac{-3}{2}\)}

d) (2x²+1)(4x-3)=(2x²+1)(x-12)

=(2x²+1)(4x-3)-(2x²+1)(x-12)=0

=(2x²+1)[(4x-3)-(x-12)=0

=(2x²+1)(4x-3-x+12)=0

=(2x²+1)(3x+9)=0

\(\Leftrightarrow\)2x²+1=0 hoặc 3x+9=0

\(\Leftrightarrow\)x=\(\dfrac{1}{2}\)hoặc x=\(\dfrac{-1}{2}\) hoặc x=-3

Vậy S={\(\dfrac{1}{2};\dfrac{-1}{2};-3\)}

e) (2x-1)²+(2-x)(2x-1)=0

=(2x-1)[(2x-1)+(2-x)=0

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

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

\(\Leftrightarrow\) 2x-1=0 hoặc x+1=0

\(\Leftrightarrow\) x=\(\dfrac{-1}{2}\) hoặc x=-1

Vậy S={\(\dfrac{-1}{2}\);-1}

f)(x+2)(3-4x)=x²+4x+4

=(x+2)(3-4x)=(x+2)²

=(x+2)(3-4x)-(x+2)²=0

=(x+2)[(3-4x)-(x+2)]=0

=(x+2)(3-4x-x-2)=0

=(x+2)(-5x+1)=0

\(\Leftrightarrow\) x+2=0 hoặc -5x+1=0

\(\Leftrightarrow\) x=-2 hoặc x=\(\dfrac{1}{5}\)

Vậy S={-2;\(\dfrac{1}{5}\)}

<=>(x³-9x²+27x-27)-(x³-3³+6(x²+2x+1)=15

<=>x³-9x²+27x-27-x³+27+6x²+12x+6=15

<=>-3x²+39x+9=0

<=>x²-13x+3=0

<=>(x²-2.x.13/2+169/4)-157/4=0

<=>(x-13/2)²=157/4

<=>x-13/2=\(\sqrt{\frac{157}{2}}\)hoặc x=13/2= - \(\sqrt{\frac{157}{2}}\)

<=>x=(13+\(\sqrt{\frac{157}{2}}\))hoặc x=\(\frac{13-\sqrt{\frac{157}{2}}}{2}\)

(x-3)³ - (x-3)(x²+3x+9) + 6(x+1)² = 15

 x³ -9x² + 27x - 27 - (x³-27) + 6( x²+ 2x + 1)  =15

 x³ -9x² + 27x - 27 - x³+ 27 + 6x² + 12x + 6 = 15

-3x² + 39x -9 = 0

-3(x² - 13x + 3) = 0

x² - 13x + 3 = 0

=> x=0,2350179569 ( chỗ này bấm máy tính)

còn giải thì làm theo mấy cách trong đây 

BÀI 3 – 4

Phương trình bậc hai một ẩn – Công thức nghiệm

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