hướng dẫn cách làm-tự làm tiếp nha :)

a) đặt \(k=x^2-4x\), ta có:\(k^2-2k=15\)\(\Rightarrow k^2-2x+1=16\Rightarrow\left(k-1\right)^2=4^2=\left(-4\right)^2\)

b) đặt \(A=x^2-3x\), ta có: \(A^2-2A-8=0\Rightarrow A^2-2A+1=9\Rightarrow\left(A-1\right)^2=3^2=\left(-3\right)^2\)

c)theo đề \(\Leftrightarrow\orbr{\begin{cases}x^2-4x+3=0\\x^2-8x+9=0\end{cases}}\)

\(x^2-4x+3=0\Leftrightarrow x^2-4x+4=1\Leftrightarrow\left(x-2\right)^2=1^2=\left(-1\right)^2\)

\(x^2-8x+9=0\Leftrightarrow x^2-8x+16=7\Leftrightarrow\left(x-4\right)^2=\pm\sqrt{7}^2\)

vt ko chi tiết bn ib là đc rùi, sai tớ làm gì T.T 

mà tớ làm mẫu 1 bài thui nha, bài còn lại có cách làm òi. bn tự dựa vô nha

\(\text{Đặt }k=x^2-4x,\text{ta có:}\)

\(\left(x^2-4x\right)^2-2.\left(x^2-4x\right)=15\)

\(\Leftrightarrow k^2-2k=0\)

\(\Leftrightarrow k^2-2k+1=16\)

\(\Leftrightarrow\left(k-1\right)^2=16\)

\(\Leftrightarrow\orbr{\begin{cases}k-1=4\\k-1=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}k=5\\k=-3\end{cases}}}\)

\(\text{Với }k=5,\text{Ta có: }x^2-4x=5\Rightarrow x^2-4x-5=0\Rightarrow x^2-5x+x-5=0\)

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

\(\text{Với }k=-3,\text{ta có: }x^2-4x=-3\Rightarrow x^2-4x+3=0\Rightarrow k^2-3x-x+3=0\)

\(\Rightarrow x.\left(x-3\right)-\left(x-3\right)=0\Rightarrow\left(x-1\right).\left(x-3\right)=0\Rightarrow\orbr{\begin{cases}x=1\\x=3\end{cases}}\)

Vậy...

\(x\left(3x-5\right)-9x+15=0\)

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

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

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

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

\(\Leftrightarrow3x\left(x-5\right)+2\left(x-5\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}3x+2=0\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-2}{3}\\x=5\end{cases}}\)

1)\(21x^2y-12xy^2=xy.\left(21x-12y\right)\)

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

3)\(3x.\left(x-1\right)+7x^2\left(x-1\right)=\left(x-1\right).\left(3x+7x^2\right)=x.\left(x-1\right)\left(3+7x\right)\)

15)\(\left(2a+3\right)^2-\left(2a+1\right)^2=\left(2a+3-2a-1\right)\left(2a+3+2a+1\right)=2.\left(4a+4\right)=8\left(a+1\right)\)

14) \(-4y^2+4y-1=-\left[\left(2y\right)^2-2.2y.1+1^2\right]=-\left(2y-1\right)^2\)

13) \(x^6+1=\left(x^2\right)^3+1=\left(x^2+1\right)\left(x^4-x^2+1\right)\)

12) \(\left(x+1\right)^2-\left(y+6\right)^2=\left(x+1-y-6\right)\left(x+1+y+6\right)=\left(x-y-5\right)\left(x+y+7\right)\)

4) \(3x\left(x-a\right)+4a\left(a-x\right)=3x.\left(x-a\right)-4a\left(x-a\right)=\left(x-a\right)\left(3x-4a\right)\)

Sao nhiều thế!

Đúng là nhiều thật , dù sao cx cảm ơn bn nhìn nha!!!

\(a,\left(x-3\right)^2-4=0\)

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

\(\Rightarrow x-3=\pm2\)

\(\hept{\begin{cases}x-3=2\Rightarrow x=5\\x-3=-2\Rightarrow x=1\end{cases}}\)

Vậy \(x=5\)hoặc \(x=1\)

\(b,x^2-2x=24\)

\(\Leftrightarrow x^2-2x+1-1=24\)

\(\Leftrightarrow\left(x-1\right)^2=24+1=25\)

\(\Leftrightarrow x-1=\pm5\)

\(\hept{\begin{cases}x-1=5\Rightarrow x=6\\x-1=-5\Rightarrow x=-4\end{cases}}\)

Vậy \(x=6\) hoặc \(x=-4\)

\(c,\left(2x+1\right)^2+\left(x+3\right)^2-5\left(x-7\right)\left(x+7\right)=0\)

\(\Leftrightarrow4x^2+4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)

\(\Leftrightarrow4x^2+4x+1+x^2+6x+9-5x^2+245=0\)

\(\Leftrightarrow10x+255=0\)

\(\Leftrightarrow10x=-255\)

\(\Leftrightarrow x=\frac{-51}{2}\)

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

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

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

\(\Leftrightarrow4x-27=1\)

\(\Leftrightarrow4x=28\)

\(\Leftrightarrow x=7\)

Ít thôi -..-

a) ( 3x + 2 )( 2x + 9 )  - ( x + 3 )( 6x + 1 ) = ( x + 1 )² - ( x + 2 )( x - 2 )

<=> 6x² + 31x + 18 - ( 6x² + 19x + 3 ) = x² + 2x + 1 - ( x² - 4 )

<=> 6x² + 31x + 18 - 6x² - 19x - 3 = x² + 2x + 1 - x² + 4

<=> 12x + 15 = 2x + 5

<=> 12x - 2x = 5 - 15

<=> 10x = -10

<=> x = -1

b) ( 2x + 3 )( x - 4 ) + ( x - 5 )( x - 2 ) = ( 3x - 5 )( x - 4 )

<=> 2x² - 5x - 12 + x² - 7x + 10 = 3x² - 17x + 20

<=> 3x² - 12x - 2 = 3x² - 17x + 20

<=> 3x² - 12x - 3x² + 17x = 20 + 2

<=> 5x = 22

<=> x = 22/5

c) ( x + 2 )³ - ( x - 2 )³ - 12x( x - 1 ) = -8

<=> x³ + 6x² + 12x + 8 - ( x³ - 6x² + 12x - 8 ) - 12x² + 12x = -8

<=>  x³ + 6x² + 12x + 8 - x³ + 6x² - 12x + 8 - 12x² + 12x = -8

<=> 12x + 16 = -8

<=> 12x = -24

<=> x = -2

d) ( 3x - 1 )² - 5( x + 1 ) + 6x - 3.2x + 1 - ( x - 1 )² = 16

<=> 9x² - 6x + 1 - 5x - 5 + 6x - 6x + 1 - ( x² - 2x + 1 ) = 16

<=> 9x² - 11x - 3 - x² + 2x - 1 = 16

<=> 8x² - 9x - 4 = 16

<=> 8x² - 9x - 4 - 16 = 0

<=> 8x² - 9x - 20 = 0

( Đến đây bạn có hai sự lựa chọn : 1 là vô nghiệm

                                                         2 là nghiệm vô tỉ =) )

a) (3x + 2)(2x + 9) - (x + 3)(6x + 1) = (x + 1)² - (x + 2)(x - 2)

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

=> 6x² + 27x + 4x + 18 - 6x² - x - 18x - 3 = x² + 2x + 1 - x² + 2x - 2x + 4

=> (6x² - 6x²) + (27x + 4x - x - 18x) + (18 - 3) = (x² - x²) + (2x + 2x - 2x) + (1 + 4)

=> 12x + 15 = 2x + 5

=> 12x + 15  - 2x - 5 = 0

=> 10x + 10 = 0

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

b) (2x + 3)(x - 4) + (x - 5)(x - 2) = (3x - 5)(x - 4)

=> 2x(x - 4) + 3(x - 4) + x(x - 2) - 5(x - 2) = 3x(x - 4) - 5(x - 4)

=> 2x² - 8x + 3x - 12 + x² - 2x - 5x + 10 = 3x² - 12x - 5x + 20

=> (2x² + x²) + (-8x + 3x - 2x - 5x) + (-12 + 10) = 3x² - 17x + 20

=> 3x² - 12x - 2 = 3x² - 17x + 20

=> 3x² - 12x - 2 - 3x² + 17x - 20 = 0

=> (3x² - 3x²) + (-12x + 17x) + (-2 - 20) = 0

=> 5x - 22 = 0

=> 5x = 22 => x = 22/5

c) (x + 2)³ - (x - 2)³ - 12x(x - 1) = -8

=> x³ + 6x² + 12x + 8 - (x³  - 6x² + 12x - 8) - 12x² + 12x = -8

=> x³ + 6x² + 12x + 8 -x³ + 6x² - 12x + 8 - 12x² + 12x = -8

=> (x³ - x³) + (6x² + 6x² - 12x²) + (12x - 12x + 12x) + (8 + 8) = -8

=> 12x + 16 = -8

=> 12x = -24

=> x = -2

Còn bài cuối làm nốt

Mình ko ghi lại đề , bạn ghi ra xong rồi suy ra như mình nha .

1) \(=>A=\left(6x^2+3x-10x-5\right)-\left(6x^2+14x-9x-21\right)\)

\(=>A=-12x+16\)

2) \(=>B=8x^3+27-8x^3+2=29\)

3)\(=>C=[\left(x-1\right)-\left(x+1\right)]^3=\left(-2\right)^3=-8\)

4)\(=>D=[\left(2x+5\right)-\left(2x\right)]^3=5^3=125\)

5)\(=>E=\left(3x+1\right)^2-\left(3x+5\right)^2+12x+2\left(6x+3\right)\)

\(=>E=\left(3x+1+3x+5\right)\left(3x+1-3x-5\right)+12x+12x+6\)

\(=>E=\left(6x+6\right)\left(-4\right)+24x+6=-24x-24+24x+6=-18\)

6)\(=>F=\left(2x^2+3x-10x-15\right)-\left(2x^2-6x\right)+x+7=-8\)

k cho mik nha , 

a) (x-2)³+6(x+1)²-x³+12=0

\(\Rightarrow\)x³-6x²+12x-8+6(x²+2x+1)-x³+12=0

\(\Rightarrow\)x³-6x²+12x-8+6x²+12x+6-x³+12=0

\(\Rightarrow\)24x+10=0

\(\Rightarrow\)24x=-10

\(\Rightarrow\)x=\(\dfrac{-10}{24}=\dfrac{-5}{12}\)

b)(x-5)(x+5)-(x+3)²+3(x-2)²=(x+1)²-(x-4)(x+4)+3x²

\(\Rightarrow\)x²-25-(x²+6x+9)+3(x²-4x+4)=x²+2x+1-(x²-16)+3x²

\(\Rightarrow\)x²​-25-x²-6x-9+3x²-12x+12=x²+2x+1-x²+16+3x²

\(\Rightarrow\)3x²-18x-22=3x²+2x+17

\(\Rightarrow\)3x²-18x-22-3x²-2x-17=0

\(\Rightarrow\)-20x-39=0

\(\Rightarrow\)-20x=39

\(\Rightarrow\)x=\(-\dfrac{39}{20}\)

\(a.\frac{4x-8}{2x^2+1}=0\\ \Leftrightarrow4x-8=0\\ \Leftrightarrow4\left(x-2\right)=0\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\)

Vậy nghiệm của phương trình trên là \(2\)

\(b.\frac{x^2-x-6}{x-3}=0\left(x\ne3\right)\\\Leftrightarrow x^2-x-6=0\\ \Leftrightarrow x^2+2x-3x-6=0\\\Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\\\Leftrightarrow \left(x-3\right)\left(x+2\right)=0\\\Leftrightarrow \left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\left(ktm\right)\\x=-2\left(tm\right)\end{matrix}\right.\)

Vậy nghiệm của phương trình trên là \(-2\)

b)(2x - 1)^2 - (2x + 5) (2x - 5 ) = 18

4x 2 -4x+1-4x 2+25=18

26-4x=18

4x=8

x=2

a,27x-18=2x-3x^2

<=> 3x^2-2x+27-18x=0

<=> 3x^2-20x+27=0

\(\Delta\)= 20^2-4-12.27

tính \(\Delta\)rồi tìm x1 ,x2