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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=5\end{cases}}}\)

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

\(\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\\ \Leftrightarrow\left(x-7\right)\left(x+6\right)=0\\ \Leftrightarrow\orbr{\begin{cases}x-7=0\\x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-6\end{cases}}}\)

3, x^3-5x^2+x-5=0

\(\Leftrightarrow x^2\left(x-5\right)+\left(x-5\right)=0\\ \Leftrightarrow\left(x^2+1\right)\left(x-5\right)=0\\ \Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x-5=0\end{cases}}\)

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

\(\left(x+1\right)^2=x+1\)

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

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

\(\left(x+1\right)x=0\)

\(\orbr{\begin{cases}x+1=0\\x=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=-1\\x=0\end{cases}}\)vậy.....

\(x\left(x-5\right)^2-4x+20=0\)

\(x\left(x-5\right)^2-4\left(x-5\right)=0\)

\(\left(x-5\right)\left[x\left(x-5\right)-4\right]=0\)

\(\left(x-5\right)\left(x^2-5x-4\right)=0\)

\(\orbr{\begin{cases}x-5=0\\x^2-5x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-0,7015621187\end{cases}}}\)vậy.........

\(x\left(x+6\right)-7x-42=0\)
\(x\left(x+6\right)-7\left(x+6\right)=0\)

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

\(\orbr{\begin{cases}x+6=0\\x-7=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-6\\x=7\end{cases}}}\) vậy....

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

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

\(\left(x-5\right)\left(x^2+1\right)=0\)

\(\orbr{\begin{cases}x-5=0\\x^2+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x^2=-1\Rightarrow x\in\Phi\end{cases}}}\)vậy........

\(x^4-2x^3+10x^2-20x=0\)

\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)

\(\left(x-2\right)\left(x^3+10x\right)=0\)

\(\orbr{\begin{cases}x-2=0\\x^3+10x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}}\)vậy..............

nhớ chọn mk nha

a) (2x - 3)² = (x + 5)²

=> 4x² - 12x + 9 = x² + 10x + 25

=> 4x² - 12x + 9 - (x² + 10x + 25) = 0

=> 3x² - 22x - 16 = 0

=> 3x² - 24x + 2x - 16 = 0

=> 3x(x - 8) + 2(x - 8) = 0

=> (3x + 2)(x - 8) = 0

=> \(\orbr{\begin{cases}3x+2=0\\x-8=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{2}{3}\\x=8\end{cases}}\)

b) x²(x - 1) - 4x² + 8x - 4 = 0

=> x²(x - 1) - (2x  - 2)² = 0

=> x²(x - 1) - [2(x- 1)]² = 0

=> x²(x - 1) - 4(x - 1)² = 0

=> (x - 1)(x² - 4(x - 1) = 0

=> (x - 1)(x² - 4x + 4) = 0

=> (x - 1)(x - 2)² = 0

=> \(\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)

c) x² + 7x + 12 = 0

=> x² + 3x + 4x + 12 = 0

=> x(x + 3) + 4(x + 3) = 0

=> (x + 4)(x + 3) = 0

=> \(\orbr{\begin{cases}x+4=0\\x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-4\\x=-3\end{cases}}\)

d) x² + 3x - 18 = 0

=> x² + 6x - 3x - 18 = 0

=> x(x + 6) - 3(x + 6) = 0

=> (x - 3)(x + 6) = 0

=> \(\orbr{\begin{cases}x-3=0\\x+6=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-6\end{cases}}\)

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

=> x(x + 6) - 7(x + 6) = 0

=> (x - 7)(x + 6) = 0

=> \(\orbr{\begin{cases}x-7=0\\x+6=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=7\\x=-6\end{cases}}\)

1. ( 2x - 3 )² = ( x + 5 )²

<=> ( 2x - 3 )² - ( x + 5 )² = 0

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

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

<=> ( x - 8 )( 3x + 2 ) = 0

<=> \(\orbr{\begin{cases}x-8=0\\3x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=-\frac{2}{3}\end{cases}}\)

2. x²( x - 1 ) - 4x² + 8x - 4 = 0

<=> x²( x - 1 ) - ( 4x² - 8x + 4 ) = 0

<=> x²( x - 1 ) - 4( x² - 2x + 1 ) = 0

<=> x²( x - 1 ) - 4( x - 1 )² = 0

<=> ( x - 1 )[ x² - 4( x - 1 ) ] = 0

<=> ( x - 1 )( x² - 4x + 4 ) = 0

<=> ( x - 1 )( x - 2 )² = 0

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

3. x² + 7x + 12 = 0

<=> x² + 3x + 4x + 12 = 0

<=> x( x + 3 ) + 4( x + 3 ) = 0

<=> ( x + 3 )( x + 4 ) = 0

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

4. x² + 3x - 18 = 0

<=> x² - 3x + 6x - 18 = 0

<=> x( x - 3 ) + 6( x - 3 ) = 0

<=> ( x - 3 )( x + 6 ) = 0

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

5. x( x + 6 ) - 7x - 42 = 0

<=> x( x + 6 ) - 7( x + 6 ) = 0

<=> ( x + 6 )( x - 7 ) = 0

<=> \(\orbr{\begin{cases}x+6=0\\x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-6\\x=7\end{cases}}\)

a,\(\left(2x-3\right)^2=\left(x+5\right)^2\)

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

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

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

\(\Rightarrow\orbr{\begin{cases}x-8=0\\3x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=8\\x=-\frac{2}{3}\end{cases}}}\)

a)  -4x(x - 7) + 4x(x² - 5) = 28x² - 13

=> -4x² + 28x + 4x² - 20x = 28x² - 13

=> (-4x² + 4x²) + (28x - 20x) = 28x² - 13

=> 8x = 28x² - 13

=> 8x - 28x² + 13 = 0

=> phương trình vô nghiệm

b) (4x² - 5x)(3x + 2) - 7x(x + 5) = (-4 + x)(-2x - 3) + 12x² + 2x²

=> 4x²(3x + 2) - 5x(3x + 2) - 7x² - 35x = -4(-2x - 3) + x(-2x - 3) + 14x²

=> 12x³ + 8x² - 15x² - 10x - 7x² - 35x = 8x + 12 - 2x² - 3x + 14x²

=> 12x³ + (8x² - 15x² - 7x²) + (-10x - 35x) = (8x - 3x) + 12 + (-2x² + 14x²)

=> 12x³ - 14x² - 45x = 5x + 12 + 12x²

=> 12x³ - 14x² - 45x - 5x - 12 - 12x² = 0

=> 12x³ + (-14x² - 12x²) + (-45x - 5x) - 12 = 0

=> 12x³ - 26x² - 50x - 12 = 0

Làm nốt

Cái câu b sửa cái đề lại nhé dấu " = " ở chỗ (-2x = 3) là gì vậy?

a, \(-4x\left(x-7\right)+4x\left(x^2-5\right)=28x^2-13\)

\(\Leftrightarrow-4x^2+28x+4x^3-20x=28x^2-13\)

\(\Leftrightarrow-32x^2+8x+4x^3+13=0\)( vô nghiệm ) 

a: Ta có: \(\left(7x+4\right)^2-\left(7x-4\right)\left(7x+4\right)\)

\(=\left(7x+4\right)\left(7x+4-7x+4\right)\)

\(=8\left(7x+4\right)\)

=56x+32

b: Ta có: \(8\left(x-2\right)^2-3\left(x^2-4x-5\right)-5x^2\)

\(=8x^2-32x+32-3x^2+12x+15-5x^2\)

\(=-20x+47\)

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

\(=x^3+3x^2+3x+1-x^3+1-3x^2-3x\)

=2

\(a.\Leftrightarrow2|x-6|-|x-6|-2=0\)

\(\Leftrightarrow|x-6|-2=0\)

\(\Leftrightarrow|x-6|=2\)

\(+x-6=2\)

\(\Leftrightarrow x=8\)

\(+x-6=-2\)

\(\Leftrightarrow x=4\)

v...

\(b.\Leftrightarrow-4\left(5-x\right)-7\left(5-x\right)+10\left(5-x\right)=-3\)

\(\Leftrightarrow\left(5-x\right)\left(10-4-7\right)=-3\)

\(\Leftrightarrow-1.\left(5-x\right)=-3\)

\(\Leftrightarrow5-x=3\)

\(\Leftrightarrow x=2\)

v...

tách x ra 1 nhóm, tách số nguyên ra 1 nhóm là xong bạn

bạn làm đầy đủ nha

\(\frac{|x-2|}{12}\)\(+\)\(\frac{|x-2|}{20}+\)\(\frac{|x-2|}{30}+\)\(\frac{|x-2|}{42}\)\(=\frac{70^5}{2^3.21^6}\)

\(\Rightarrow|x-2|.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)=\frac{2^5.5^5.7^5}{2^3.7^6.3^6}\)

\(\Rightarrow|x-2|.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)=\frac{2^2.5^5}{7.3^6}\)

\(\Rightarrow|x-2|.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)=\frac{4.5^5}{21.3^5}\)

\(\Rightarrow|x-2|\left(\frac{1}{3}-\frac{1}{7}\right)=\frac{4.5^5}{21.3^5}\)\(\Rightarrow|x-2|=\frac{5^5}{3^5}\)

ĐẾN ĐÂY DỄ RÙI TỰ GIẢI TIẾP