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

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

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

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

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

\(\Leftrightarrow\)\(x=\frac{1}{5}\)

Vậy \(x=\frac{1}{5}\)

\(b)\) \(x\left(x+1\right)\left(x+2\right)=0\)

Suy ra \(x=0\) hoặc \(x+1=0\) hoặc \(x+2=0\)

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

Vậy \(x=0\) hoặc \(x=-1\) hoặc \(x=-2\)

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

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

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}3x=-2\\x=3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-2}{3}\\x=3\end{cases}}}\)

Vậy \(x=\frac{-2}{3}\) hoặc \(x=3\)

Chúc bạn học tốt ~ 

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

<=> \(\left(5x-1\right)\left(5x-1-5x\right)=0\)

<=> \(-1\left(5x-1\right)=0\)

<=> \(5x-1=0\)

<=> \(5x=1\)

<=> \(x=\frac{1}{5}\)

b/ \(x\left(x+1\right)\left(x+2\right)=0\)

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

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

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

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

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

<=> \(\orbr{\begin{cases}3x=-2\\x=3\end{cases}}\)

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

Answer:

\(3x^2-4x=0\)

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

\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{4}{3}\end{cases}}\)

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

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

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)

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

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

\(\Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\)

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

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

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

\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}5x-1=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=3\end{cases}}\)

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

\(\Rightarrow\left(x^2-2x+1\right)+4=0\)

\(\Rightarrow\left(x-1\right)^2=-4\) (Vô lý)

Vậy không có giá trị \(x\) thoả mãn

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

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

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

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

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

Ta có x² + 5x - 3√(x² + 5x + 2) - 2 = 0

<=> x² + 5x - 2 = 3√(x² + 5x + 2)

<=> (x² + 5x - 2)² = [3√(x² + 5x + 2)]²

<=> x⁴ + 25x² + 4 + 10x³ - 20x - 4x³ = 9(x² + 5x + 2)

<=> x⁴ + 6x³ + 25x² - 20x + 4 = 9x² + 45x + 18

<=> x⁴ + 6x³ + 25x² - 20x + 4 - (9x² + 45x + 18) = 0

<=> x⁴ + 6x³ + 25x² - 9x² - 20x - 45x + 4 - 18 = 0

<=> x⁴ + 6x³ + 16x² - 65x - 14 = 0

Đến đây, ta phân tích đa thức thành nhân tử bằng phương pháp hệ số bất định

Ta có x⁴ + 6x³ + 16x² - 65x - 14 sau khia phân tích có dạng (x² + ax + b)(x² + cx + d) = x⁴ + (a+c)x³ + (ac+b+d)x² + (ad+bc)x + db

=> x⁴ + 6x³ + 16x² - 65x - 14 = x⁴ + (a+c)x³ + (ac+b+d)x² + (ad+bc)x + db

<=> a+c = 6 ; ac+b+d = 16 ; ad+dc = -65 ; db = -14

Sau đó bạn tìm ra a,b,c,d và giải ra phương trình.

Mình chỉ mới lớp 7 nên chưa tìm ra đươc a,b,c,d.Mong bạn thông cảm cho mình

a) 4x(x+1)=8(x+1)

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

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

<=>\(\left[\begin{array}{} 4x-8=0\\ x+1=0 \end{array} \right.\)

<=>\(\left[\begin{array}{} x=2\\ x=-1 \end{array} \right.\)

Vậy...

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

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

<=>\(\left[\begin{array}{} x+2=0\\ x-1=0 \end{array} \right.\)

<=>\(\left[\begin{array}{} x=-2\\ x=1 \end{array} \right.\)

Vậy...

c)5x(x-2)-(2-x)=0

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

<=>\(\left[\begin{array}{} 5x+1=0\\ x-2 \end{array} \right.\)

<=>\(\left[\begin{array}{} x=-1/5\\ x=2 \end{array} \right.\)

d)5x(x-200)-x+200=0

<=>(5x-1)(x-200)=0

<=>\(\left[\begin{array}{} 5x-1=0\\ x-200=0 \end{array} \right.\)

<=>\(\left[\begin{array}{} x=1/5\\ x=200 \end{array} \right.\)

e)\(x^3+4x=0 \)

\(\Leftrightarrow x(x^2+4)=0 \)

\(\Leftrightarrow \left[\begin{array}{} x=0\\ x^2+4=0 (loại vì x^2+4>=0 với mọi x) \end{array} \right.\)

Vậy x=0

f)\((x+1)=(x+1)^2\)

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

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

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

\(\Leftrightarrow \left[\begin{array}{} x=-1\\ x=0 \end{array} \right.\)

Vậy....

a) 5x( x - 1 ) = x - 1

<=> 5x² - 5x = x - 1

<=> 5x² - 5x - x + 1 = 0

<=> 5x² - 6x + 1 = 0

<=> 5x² - 5x - x + 1 = 0

<=> 5x( x - 1 ) - 1( x - 1 ) = 0

<=> ( x - 1 )( 5x - 1 ) = 0

<=> \(\orbr{\begin{cases}x-1=0\\5x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)

b) 2( x + 5 ) - x² - 5x = 0

<=> 2x + 10 - x² - 5x = 0

<=> -x² - 3x + 10 = 0

<=> -x² - 5x + 2x + 10 = 0

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

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

<=> \(\orbr{\begin{cases}x+5=0\\2-x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)

c) x² - 2x - 3 = 0

<=> x² + x - 3x - 3 = 0

<=> x( x + 1 ) - 3( x + 1 ) = 0

<=> ( x + 1 )( x - 3 ) = 0

<=> \(\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)

d) 2x² + 5x - 3 = 0

<=> 2x² - x + 6x - 3 = 0

,<=> x( 2x - 1 ) + 3( 2x - 1 ) = 0

<=> ( 2x - 1 )( x + 3 ) = 0

<=> \(\orbr{\begin{cases}2x-1=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-3\end{cases}}\)

a) 5x ( x - 1 ) = x - 1 <=> 5x² - 5x - x + 1 = 0

<=> 5x² - 6x + 1 = 0 <=> 5x² - x - ( 5x - 1 ) = 0 

<=> x ( 5x - 1 ) - ( 5x - 1 ) = 0 <=> ( x - 1 )( 5x - 1 ) = 0

<=> x = 1 hoặc x = 1/5

b) 2 ( x + 5 ) - x² - 5x = 0 <=> 2 ( x + 5 ) - x ( x + 5 ) = 0

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

c) x² - 2x - 3 = 0 <=> x² + x - 3x - 3 = 0 

<=> x ( x + 1 ) - 3 ( x + 1 ) = 0 <=> ( x - 3 ) ( x + 1 ) = 0 

<=> x = 3 hoặc x = -1

d) 2x²  + 5x - 3 = 0

Ta có : delta = 5² - 4.2.3 = 25 - 24 = 1

Khi đó : x = -1 hoặc x = 3/2  

a) 5x(x-1)=x-1

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

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-1=0\\5x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)

vậy x=1 hoặc x=1/5

b) 2(x+5)-x²-5x=0

\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)

Vậy x=2 hoặc x=-5

c) x²-2x-3=0

\(\Leftrightarrow x^2-3x+x-3=0\)\(\Leftrightarrow x\left(x-3\right)+\left(x-3\right)=0\)\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)

vậy...

`a)5x(x-1)-(x+2)(5x-7)=6`

`<=>5x^2-5x-(5x^2-7x+10x-14)=6`

`<=>5x^2-5x-(5x^2+3x-14)=6`

`<=>-8x+14=6`

`<=>8x=8<=>x=1`

Vậy `x=1`

`b)(x+2)^2-(x^2-4)=0`

`<=>x^2+4x+4-x^2+4=0`

`<=>4x+8=0`

`<=>4x=-8`

`<=>x=-2`

Vậy `x=-2`

a)x=5/2

b)x=-2

a)5x.(x-1)-(x+2).(5x-7)=6

<=> 5x²-5x-(5x²-7x+10x-14)=6

<=>  5x²-5x-5x²+7x-10x+14=6

<=> -8x+14=6

<=> -8x=-8 => x=1

Vậy x=1

b) (x+2)²-(x²-4)=0

<=> (x+2)²-(x²-2²)=0 <=> (x+2)²-(x-2)(x+2)=0

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

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

<=> (x+2).4=0

=> x+2=0

=> x=-2

Vậy x=-2

b) 5x(x-2000)-x+2000=0

\(\Rightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\\ \Rightarrow\left(x-2000\right)\left(5x-1\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x-2000=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0+2000\\5x=0+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2000\\5x=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2000\\x=\dfrac{1}{5}\end{matrix}\right.\)

Ai giúp minh làm bài 5 phía trên với

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

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

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

d) Ta có: \(5x^2+x=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-1}{5}\end{matrix}\right.\)

a)

\(\Rightarrow x\left(x-5\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x-5=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=5\end{array}\right.\)

b)

\(\Rightarrow3x\left(x-2\right)-2\left(x-2\right)=0\)

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

\(\Rightarrow\left[\begin{array}{nghiempt}x-2=0\\3x-2=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=2\\x=\frac{2}{3}\end{array}\right.\)

c)

\(\Rightarrow\left(3x-1\right)\left(5x+x-2\right)=0\)

\(\Rightarrow\left(3x-2\right)^2.2=0\)

\(\Rightarrow3x-2=0\)

\(\Rightarrow x=\frac{2}{3}\)

a)x^2-5=0

x^2=5

x=2.236

b) 3x^3-27x=0

=)x=3

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

=)x=1

D)2(x+5)-x^2-5×=0

=)x=2