bn hk hằng đẳng thức chưa ?

a) 4x²  - 2x + 3 - 4x.(x - 5) = 7x - 3

--> 4x² - 2x + 3 - 4x² + 20x = 7x - 3

--> 4x² - 2x - 4x² + 20x - 7x = -3 - 3

--> 11x = -6

--> x = \(\frac{-6}{11}\)

b) -3x.(x - 5) + 5.(x - 1) + 3x² = 4x

--> -3x² + 15x + 5x - 5 + 3x² = 4x

--> -3x²  + 15x + 5x + 3x² - 4x = 5 

--> 16x = 5

--> x = \(\frac{5}{16}\)

c) 7x.(x - 2) - 5.(x - 1) = 21x² - 14x² + 3

--> 7x² - 14x - 5x + 5 = 7x² + 3 

--> 7x²  - 14x - 5x - 7x²  = -5 + 3 

--> -19x = -2 

--> x = \(\frac{2}{19}\)

d) 3.(5x - 1) - x.(x - 2) + x² - 13x = 7

--> 15x - 3 - x² + 2x + x² - 13x = 7

--> 15x - x² + 2x + x² - 13x = 3 + 7

--> 4x = 10

--> x = \(\frac{5}{2}\)

e) \(\frac{1}{5}\)x.(10x - 15) - 2x.(x - 5) = 12

--> 2x² - 3x - 2x² + 10x = 12

--> 7x = 12

--> x = \(\frac{12}{7}\)

~ Học tốt ~

a) 4x² - 2x + 3 - 4x(x - 5) = 7x - 3

=> 4x² - 2x + 3 - 4x² + 20x = 7x - 3

=> 18x + 3 = 7x - 3

=> 18x - 7x = -3 - 3

=> 11x = -6

=>  x = -6/11

b) -3x(x - 5) + 5(x - 1) + 3x² = 4x

=> -3x² + 15x + 5x - 5 + 3x² = 4x

=> 20x - 5 = 4x

=> 20x - 4x = 5

=> 16x = 5

=> x = 5/16

\(c,7x\left(x-2\right)-5\left(x-1\right)=21x^2-14x^2+3\)

\(\Leftrightarrow7x^2-14x-5x+5=7x^2+3\)

\(\Leftrightarrow7x^2-7x^2-19x=3-5\)

\(\Leftrightarrow-19x=-2\)

\(\Leftrightarrow x=\frac{2}{19}\)

a) 4x² - 2x + 3 - 4x.(x - 5) = 7x - 3

<=> 18x + 3 = 7x - 3

<=> 18x = 7x - 3 - 3

<=> 18x = 7x - 6

<=> 18x - 7x = -6

<=> 11x = -6

<=> x = -6/11

=> x = -6/11

b) -3x.(x - 5) + 5.(x - 1) + 3x² = 4x

<=> 20x - 5 = 4x

<=> 20x = 4x + 5

<=> 20x - 4x = 5

<=> 16x = 5

<=> x = 5/16

=> x = 5/16

c) 7x.(x - 2) - 5.(x - 1) = 21x² - 14x² + 3

<=> 7x.(x - 2) - 5.(x - 1) = 7x² + 3

<=> 7x² - 19x + 5 = 7x² + 3

<=> 7x² - 19x = 7x² + 3 - 5

<=> 7x² - 19x = 7x² - 2 

<=> 7x² - 19x - 7x² = -2

<=> -19x = -2

<=> x = 2/19

=> x = 2/19

d) 3.(5x - 1) - x.(x - 2) + x² - 13x = 7

<=> 4x - 3 = 7

<=> 4x = 7 + 3

<=> 4x = 10

<=> x = 10/4

=> x = 5/2

e) 1/5x.(10x - 15) - 2x.(x - 5) = 12

<=> x(2x - 3) - 2x(x - 5) = 12

<=> 7x = 12

<=> x = 12/7

=> x = 12/7

a) Ta có: A = 0

=> x² + 2x - 3 = 0

=> x² + 3x - x - 3 = 0

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

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

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

=> \(\orbr{\begin{cases}x=1\\x=-3\end{cases}}\)

Vậy ...

b) Ta có: B = 0

=> -3x² + 12x - 9 = 0

=> -3x² + 3x + 9x - 9 = 0

=> -3x(x - 1) + 9(x - 1) = 0

=> (-3x + 9)(x - 1) = 0

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

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

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

=> \(\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

Vậy ...

c) C = 0

=>  10x² - 7x - 3 = 0

=> 10x² - 10x + 3x - 3 = 0

=> 10x(x - 1) + 3(x - 1) = 0

=> (10x  + 3)(x - 1) = 0

=> \(\orbr{\begin{cases}10x+3=0\\x-1=0\end{cases}}\)

=> \(\orbr{\begin{cases}10x=-3\\x=1\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{3}{10}\\x=1\end{cases}}\)

d) D = 0

=> -7x⁴ + 10x³ - 3x² = 0

=> x²(-7x² + 10x - 3) = 0

=> x²(-7x² + 7x + 3x - 3) = 0

=> x².[-7x(x - 1) + 3(x - 1)] = 0

=> x².(-7x + 3)(x - 1) = 0

=> x^2 = 0

-7x + 3 = 0

hoặc  x - 1 = 0

=> x=  0

-7x = -3

hoặc x = 1

=> x = 0

hoặc x = 3/7

hoặc x = 1

Vậy ...

a) \(A=x^2+2x-3\)

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

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

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

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

b) \(B=-3x^2+12x-9\)

\(-3x^2+12x-9=0\)

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

c) \(C=10x^2-7x-3\)

\(10x^2-7x-3=0\)

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

\(\Leftrightarrow\left(10x+3\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}10x+3=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{10}\\x=1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-\frac{3}{10}\\x=1\end{cases}}\)

d) \(D=-7x^4+10x^3-3x^2\)

\(-7x^4+10x^3-3x^2=0\)

\(\Leftrightarrow-x^2\left(7x^2-10x+3\right)=0\)

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

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

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

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

\(\Leftrightarrow\hept{\begin{cases}x^2=0\\7x-3=0\\x-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\x=\frac{3}{7}\\x=1\end{cases}}\)(thay ngoặc nhọn bằng ngoặc vuông nhé, phần kl cũng thay luôn như thế nhé)

\(\Rightarrow\hept{\begin{cases}x=0\\x=\frac{3}{7}\\x=1\end{cases}}\)

mấy cái này thường thì phân tích thành bình phương thừa là xong.

bài này mà lớp 7 hả??

6x³ - 7x² + 5x - 2
= 6x³ - 4x² - 3x² + 2x + 3x - 2
= 6x²(x - 2/3) - 3x(x - 2/3) + 3(x - 2/3)
= (x - 2/3)(6x² - 3x + 3)
= 3(x - 2/3)(2x² - x + 1)

4x³ + 5x² + 10x - 12
= 4x³ - 3x² + 8x² - 6x + 16x - 12
= 4x²(x - 3/4) + 8x(x - 3/4) + 16(x - 3/4)
= (x - 3/4)(4x² + 8x + 16)
= 4(x - 3/4)(x² + 2x + 4)

4x³ - 7x² - x + 3
= 4x³ - 3x² - 4x² + 3x - 4x + 3
= 4x²(x - 3/4) - 4x(x - 3/4) - 4(x - 3/4)
= (x - 3/4)(4x² - 4x - 4)
= 4(x - 3/4)(x² - x - 1)

4x³ - 5x² + 6x + 9
= 4x³ + 3x² - 8x² - 6x + 12x + 9
= 4x²(x + 3/4) - 8x(x + 3/4) + 12(x + 3/4)
= (x + 3/4)(4x² - 8x + 12)
= 4(x + 3/4)(x² - 2x + 3)

3x³ - 5x² + 5x - 2
= 3x³ - 2x² - 3x² + 2x + 3x - 2
= 3x²(x - 2/3) - 3x(x - 2/3) + 3(x - 2/3)
= (x - 2/3)(3x² - 3x + 3)
= 3(x - 2/3)(x² - x + 1)

a) x=6.

b) x=-3/7.

c) x=1/15.

d) x=\(\pm\)2.

e) x=1.

f) Vô nghiệm.