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

<=> \(9x^2-9x+2=9x^2+6x+1\)

<=>  \(15x=1\) <=> \(x=\frac{1}{15}\)

b) \(\left(4x-1\right)\left(x+1\right)=\left(2x-3\right)^2\)

<=> \(4x^2+3x-1=4x^2-12x+9\)

<=> \(15x^2=10\) <=> \(x=\frac{2}{3}\)

c) \(\left(5x+1\right)^2=\left(7x-3\right)\left(7x+2\right)\) <=> \(25x^2+10x+1=49x^2-7x-6\)

<=> \(24x^2-17x-7=0\) <=> \(24x^2-24x+7x-7=0\)

<=> \(\left(24x+7\right)\left(x-1\right)=0\) <=> \(\orbr{\begin{cases}x=-\frac{7}{24}\\x=1\end{cases}}\)

d) (4 - 3x)(4 + 3x) = (9x - 3)(1 - x)

<=> 16 - 9x² = 12x - 9x2 - 3

<=> 12x = 19

<=> x = 19/12

e) x(x + 1)(x + 2)(x + 3) = 24

<=> (x² + 3x)(x² + 3x + 2) = 24

<=> (x² + 3x)²  + 2(x² + 3x) - 24 = 0

<=> (x² + 3x)² + 6(x² + 3x) - 4(x² + 3x) - 24 = 0

<=> (x² + 3x + 6)(x² + 3x - 4) = 0

<=> \(\orbr{\begin{cases}x^2+3x+6=0\\x^2+3x-4=0\end{cases}}\)

<=> \(\orbr{\begin{cases}\left(x+\frac{3}{2}\right)^2+\frac{15}{4}=0\left(vn\right)\\\left(x+4\right)\left(x-1\right)=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-4\\x=1\end{cases}}\)

g) (7x - 2)² = (7x - 3)(7x + 2)

<=> 49x² - 28x + 4 = 49x² - 7x - 6

<=> 21x = 10 <=> x = 10/21

a,  \((\frac{3}{7}-\frac{2}{3})\)  .x  =\(\frac{10}{21}\)                                                                   

  \(\frac{-5}{21}\).x=\(\frac{10}{21}\)

x= -2

 Mk chỉ làm 1 phần  các phằn còn lại tương tự

Bài rút gọn biểu thức

a) M=|2x-3|+|x-1| với x > 1,5

b) N=|2-x|-3|x+1| với x < -1

c) P=|3x-5|+|x-2|

d) Q=|x-3|-2.|-5x|

a, \(\frac{3}{7}x-\frac{2}{3}x=\frac{10}{21}\Rightarrow\left(\frac{3}{7}-\frac{2}{3}\right)x=\frac{10}{21}\Rightarrow x=\frac{10}{21}.\left(\frac{-21}{5}\right)\Rightarrow x=-2\)

b, \(\frac{7}{35}:\left(x-\frac{1}{3}\right)=\frac{2}{25}\Rightarrow x-\frac{1}{3}=\frac{1}{5}:\frac{2}{25}\Rightarrow x=\frac{1}{10}-\frac{1}{3}=\frac{13}{30}\)

c, \(|2x-4|+1=5\)

\(\Rightarrow\orbr{\begin{cases}2x-4=4\\2x-4=-4\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=0\end{cases}}\)

d, \(3.|3-2x|-1=\frac{2}{5}\)

\(\Rightarrow3.|3-2x|=\frac{7}{5}\)

\(\Rightarrow\orbr{\begin{cases}3-2x=\frac{7}{15}\\3-2x=-\frac{7}{15}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{19}{15}\\x=\frac{26}{15}\end{cases}}\)

f, \(\left(2x-1\right).\left(x+\frac{2}{3}\right)=0\)

\(\Rightarrow\hept{\begin{cases}2x-1=0\\x+\frac{2}{3}=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\x=-\frac{2}{3}\end{cases}}\)

đăng ít một thôi bạn

Bỏ câu c,d đi ạ 

\(pt\Leftrightarrow x^3+3x^2-7x-1+\left(3x+1\right)\sqrt{x^3-7x+6}=0\)

\(\Leftrightarrow\left(x^3+3x^2-7x-1+\left(3x+1\right)\sqrt{x^3-7x+6}\right)\left(x^3+3x^2-7x-1-\left(3x+1\right)\sqrt{x^3-7x+6}\right)=0\)

\(\Leftrightarrow\left(\left(x^3+3x^2-7x-1\right)^2-\left(3x+1\right)^2\left(x^3-7x+6\right)=0\right)\)

Sau đó em giải tiếp đc r ^^ Phá bình phương rồi đặt nhân tử chung.

\(\Rightarrow7x^3+7x^2-4x^2-4x+x+1=0\\ \Rightarrow\left(x+1\right)\left(7x^2-4x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-1\\7x^2-4x+1=0\left(1\right)\end{matrix}\right.\\ \left(1\right)\Rightarrow7\left(x^2-2\cdot\dfrac{2}{7}x+\dfrac{4}{49}\right)+\dfrac{3}{7}=0\\ \Rightarrow7\left(x-\dfrac{2}{7}\right)^2+\dfrac{3}{7}=0\left(\text{vô lí}\right)\)

Vậy x=-1

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

a: \(\Leftrightarrow9x^2-9x+2=9x^2+6x+1\)

=>-3x=-1

hay x=1/3

b: \(\Leftrightarrow4x^2+4x-x-1=4x^2-12x+9\)

=>3x-1=-12x+9

=>15x=10

hay x=2/3

c: \(\Leftrightarrow25x^2+10x+1=25x^2+25x-x-1=24x-1\)

=>10x-24x=-1-1

=>-14x=-2

hay x=1/7

d: \(\Leftrightarrow49x^2-28x+4=49x^2+14x-21x-6\)

=>-28x+4=-7x-6

=>-21x=-10

hay x=10/21