2:

=>x^3-1-2x^3-4x^6+4x^6+4x=6

=>-x^3+4x-7=0

=>x=-2,59

4: =>8x-24x^2+2-6x+24x^2-60x-4x+10=-50

=>-62x+12=-50

=>x=1

Trả lời:

a, ( x² - 4x + 16 )( x + 4 ) - x ( x + 1 )( x + 2 ) + 3x² = 0

<=> x³ + 4x² - 4x² - 16x + 16x + 64 - x ( x² + 3x + 2 ) + 3x² = 0 

<=> x³ + 64 - x³ - 3x² - 2x + 3x² = 0

<=> 64 - 2x = 0

<=> 2x = 64

<=> x = 32

Vậy x = 32 là nghiệm của pt.

b, ( 8x + 2 )( 1 - 3x ) + ( 6x - 1 )( 4x - 10 ) = - 50

<=> 8x - 24x² + 2 - 6x + 24x² - 60x - 4x + 10 = - 50

<=> - 62x + 12 = - 50

<=> - 62x = - 62

<=> x = 1

Vậy x = 1 là nghiệm của pt.

a) Thực hiện rút gọn VT = -2x – 64

Giải phương trình -2x – 64 = 0 thu được x = -32.

b) Thực hiện rút gọn VT = -62 x +12

Giải phương trình -62x + 12 = -50 thu được x = 1.

1)(x²-4x+16)(x+4)-x(x+1)(x+2)+3x²=0

\(\Rightarrow\)(x³+64)-x(x²+2x+x+2)+3x²=0

\(\Rightarrow\)x³+64-x³-2x²-x²-2x+3x²=0

\(\Rightarrow\)-2x+64=0

\(\Rightarrow\)-2x=-64

\(\Rightarrow\)x=\(\dfrac{-64}{-2}\)

\(\Rightarrow x=32\)

2)(8x+2)(1-3x)+(6x-1)(4x-10)=-50

\(\Rightarrow\)8x-24x²+2-6x+24x²-60x-4x+10=50

\(\Rightarrow\)-62x+12=50

\(\Rightarrow\)-62x=50-12

\(\Rightarrow\)-62x=38

\(\Rightarrow\)x=\(-\dfrac{38}{62}=-\dfrac{19}{31}\)

\(\left(4-3x\right)\left(10x-5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)

\(\left(7-2x\right)\left(4+8x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)

rồi thực hiện đến hết ... 

Brainchild bé ngây thơ qus e , ko thực hiện đến hết như thế đc đâu :>

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

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

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

\(-2x^2-11x+6=0\)

\(2x^2+11x-6=0\)

\(2x^2+12x-x-6=0\)

\(2x\left(x+6\right)-\left(x+6\right)=0\)

\(\left(x+6\right)\left(2x-1\right)=0\)

\(x+6=0\Leftrightarrow x=-6\)

\(2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)

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

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

\(x=0\)

\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)

Tự lm tiếp nha 

a) \(\sqrt[]{x^2-2x+4}=2x-2\)

\(\Leftrightarrow\sqrt[]{x^2-2x+4}=2\left(x-1\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x-1\right)\ge0\\x^2-2x+4=4\left(x-1\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\x^2-2x+4=4x^2-8x+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\3x^2-6x=0\end{matrix}\right.\) \(\left(1\right)\)

Giải pt \(3x^2-6x=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=2\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow x=2\)

c) \(\sqrt{x^2-3x+2}=\sqrt[]{x-1}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\x^2-3x+2=x-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x^2-4x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x=1\cup x=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

1: \(x^2+3x+2\)

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

=x(x+1)+2(x+1)

=(x+1)(x+2)

2: \(x^2+4x+3\)

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

=x(x+1)+3(x+1)

=(x+1)(x+3)

3: \(x^2+5x+4\)

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

=x(x+1)+4(x+1)

=(x+1)(x+4)

4: \(x^2-4x+3\)

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

=x(x-1)-3(x-1)

=(x-1)(x-3)

5: \(x^2-4x+4=x^2-2\cdot x\cdot2+2^2=\left(x-2\right)^2\)

6: \(x^2-5x+4\)

\(=x^2-x-4x+4\)

=x(x-1)-4(x-1)

=(x-1)(x-4)

7: \(x^2-5x+6\)

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

=x(x-2)-3(x-2)

=(x-2)(x-3)

8: \(x^2+6x+5\)

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

=x(x+1)+5(x+1)

=(x+1)(x+5)

9: \(x^2-7x+10\)

\(=x^2-2x-5x+10\)

=x(x-2)-5(x-2)

=(x-2)(x-5)

10: \(x^2+8x+12\)

\(=x^2+2x+6x+12\)

=x(x+2)+6(x+2)

=(x+2)(x+6)

11: \(x^2-8x+16=x^2-2\cdot x\cdot4+4^2=\left(x-4\right)^2\)

12: \(x^2+8x+15\)

\(=x^2+3x+5x+15\)

=x(x+3)+5(x+3)

=(x+3)(x+5)

13: \(x^2-8x+7\)

\(=x^2-x-7x+7\)

=x(x-1)-7(x-1)

=(x-1)(x-7)

14: \(x^2+9x+8\)

\(=x^2+x+8x+8\)

=x(x+1)+8(x+1)

=(x+1)(x+8)

15: \(x^2-9x+14\)

\(=x^2-2x-7x+14\)

=x(x-2)-7(x-2)

=(x-2)(x-7)

16: \(x^2+9x+18\)

\(=x^2+3x+6x+18\)

=x(x+3)+6(x+3)

=(x+3)(x+6)

17: \(x^2-9x+20\)

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

=x(x-4)-5(x-4)

=(x-4)(x-5)

18: \(2x^2-3x+1\)

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

=2x(x-1)-(x-1)

=(x-1)(2x-1)

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

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

3. \(x^2+5x+4=\left(x+1\right)\left(x+4\right)\)

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

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

6. \(x^2-5x+4=\left(x-1\right)\left(x-4\right)\)

7. \(x^2-5x+6=\left(x-2\right)\left(x-3\right)\)

8. \(x^2+6x+5=\left(x+1\right)\left(x+5\right)\)

9. \(x^2-7x+10=\left(x-2\right)\left(x-5\right)\)

10. \(x^2+8x+12=\left(x+2\right)\left(x+6\right)\)

11. \(x^2-8x+16=\left(x-4\right)^2\)

12. \(x^2+8x+15=\left(x+3\right)\left(x+5\right)\)

13. \(x^2-8x+7=\left(x-1\right)\left(x-7\right)\)

14. \(x^2+9x+8=\left(x+1\right)\left(x+8\right)\)

15. \(x^2-9x+14=\left(x-2\right)\left(x-7\right)\)

16. \(x^2+9x+18=\left(x+3\right)\left(x+6\right)\)

17. \(x^2-9x+20=\left(x-4\right)\left(x-5\right)\)

\(18.2x^2-3x+1=2x^2-x-2x+1\)

\(=x\cdot\left(2x-1\right)-\left(2x-1\right)=\left(2x-1\right)\left(x-1\right)\)

1) Ta có: \(5\left(x-3\right)\left(x-7\right)-\left(5x+1\right)\left(x-2\right)=-8\)

\(\Leftrightarrow5\left(x^2-10x+21\right)-\left(5x^2-10x+x-2\right)=-8\)

\(\Leftrightarrow5x^2-50x+105-5x^2+9x+2+8=0\)

\(\Leftrightarrow-41x=-115\)

hay \(x=\dfrac{115}{41}\)

2) Ta có: \(x\left(x+1\right)\left(x+2\right)-\left(x+4\right)\left(3x-5\right)=84-5x\)

\(\Leftrightarrow x\left(x^2+3x+2\right)-\left(3x^2+7x-20\right)=84-5x\)

\(\Leftrightarrow x^3+3x^2+2x-3x^2-7x+20-84+5x=0\)

\(\Leftrightarrow x^3=64\)

hay x=4

3) Ta có: \(\left(9x^2-5\right)\left(x+3\right)-3x^2\left(3x+9\right)=\left(x-5\right)\left(x+4\right)-x\left(x-11\right)\)

\(\Leftrightarrow9x^3+27x^2-5x-15-9x^3-27x^2=x^2-x-20-x^2+11x\)

\(\Leftrightarrow-5x-15=10x-20\)

\(\Leftrightarrow-5x-10x=-20+15\)

\(\Leftrightarrow x=\dfrac{-5}{-15}=\dfrac{1}{3}\)