a: =>7x=63

hay x=9

b: =>3x=-15

hay x=-5

d: =>-6x=-16

hay x=8/3

a) \(7x=63\Leftrightarrow x=9\)

b) \(3x=-15\Leftrightarrow x=-5\)

c) \(2x-5=0\Leftrightarrow2x=5\Leftrightarrow x=\dfrac{5}{2}\)

d) \(-6x=-16\Leftrightarrow x=\dfrac{8}{3}\)

Lời giải:

a) $0,2x^2+0,4x-7=0$

$\Leftrightarrow 2x^2+4x-70=0$

$\Leftrightarrow x^2+2x-35=0$

$\Leftrightarrow (x-5)(x+7)=0$

$\Rightarrow x=5$ hoặc $x=-7$

b) 

$\frac{1}{2}x^2+11x+60,5=0$

$\Leftrightarrow x^2+22x+121=0$

$\Leftrightarrow (x+11)^2=0\Leftrightarrow x=-11$

c) 

$5x^2+\sqrt{3}-1=0$

$\Leftrightarrow 5x^2=1-\sqrt{3}< 0$ (vô lý)

Vậy  PT vô nghiệm.

a)  x 4   –   5 x 2   +   4   =   0   ( 1 )

Đặt x 2   =   t, điều kiện t ≥ 0.

Khi đó (1) trở thành :  t 2   –   5 t   +   4   =   0   ( 2 )

Giải (2) : Có a = 1 ; b = -5 ; c = 4 ⇒ a + b + c = 0

⇒ Phương trình có hai nghiệm  t 1   =   1 ;   t 2   =   c / a   =   4

Cả hai giá trị đều thỏa mãn điều kiện.

+ Với t = 1 ⇒ x 2   =   1  ⇒ x = 1 hoặc x = -1;

+ Với t = 4 ⇒ x 2   =   4  ⇒ x = 2 hoặc x = -2.

Vậy phương trình (1) có tập nghiệm S = {-2 ; -1 ; 1 ; 2}.

b)  2 x 4   –   3 x 2   –   2   =   0 ;   ( 1 )

Đặt   x 2   =   t , điều kiện t ≥ 0.

Khi đó (1) trở thành :  2 t 2   –   3 t   –   2   =   0   ( 2 )

Giải (2) : Có a = 2 ; b = -3 ; c = -2

⇒   Δ   =   ( - 3 ) 2   -   4 . 2 . ( - 2 )   =   25   >   0

⇒ Phương trình có hai nghiệm

Chỉ có giá trị t 1   =   2  thỏa mãn điều kiện.

+ Với t = 2 ⇒ x 2   =   2  ⇒ x = √2 hoặc x = -√2;

Vậy phương trình (1) có tập nghiệm S = {-√2 ; √2}.

c)  3 x 4   +   10 x 2   +   3   =   0   ( 1 )

Đặt x 2   =   t , điều kiện t ≥ 0.

Khi đó (1) trở thành :  3 t 2   +   10 t   +   3   =   0   ( 2 )

Giải (2) : Có a = 3; b' = 5; c = 3

⇒  Δ ’   =   5 2   –   3 . 3   =   16   >   0

⇒ Phương trình có hai nghiệm phân biệt

Cả hai giá trị đều không thỏa mãn điều kiện.

Vậy phương trình (1) vô nghiệm.

c) Ta có: \(C=4x^2+y^2-4xy+8x-4y+4\)

\(=\left(2x-y\right)^2+2\cdot\left(2x-y\right)\cdot2+2^2\)

\(=\left(2x-y+2\right)^2\)

11) Ta có: \(a^6+a^4+a^2b^2+b^4-b^6\)

\(=a^6-b^6+a^4+a^2b^2+b^4\)

\(=\left(a^2-b^2\right)\left(a^4+a^2b^2+b^4\right)+\left(a^4+a^2b^2+b^4\right)\)

\(=\left(a^4+a^2b^2+b^4\right)\left(a^2-b^2+1\right)\)

12) Ta có: \(x^3+3xy+y^3-1\)

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

\(=\left[\left(x+y\right)^3-1\right]-3xy\left(x+y-1\right)\)

\(=\left(x+y-1\right)\left[x^2+2xy+y^2+x+y+1\right]-3xy\left(x+y-1\right)\)

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

14) Ta có: \(x^8+x+1\)

\(=x^8+x^7-x^7-x^6+x^6+x^5-x^5-x^4+x^4+x^3-x^3+x^2-x^2+x+1\)

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

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

15) Ta có: \(x^8+3x^4+4\)

\(=x^8+4x^4+4-x^4\)

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

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

1)  \(x^2-7x+6=x^3+1-7x-7=\left(x^3+1\right)-7\left(x+1\right)=\left(x+1\right)\left(x^2-x-6\right)\)

2)  \(x^3-9x^2+6x+16\)

\(\left(x^3+1\right)-\left[\left(9x^2-6x+1\right)-16\right]\)

\(=\left(x^3+1\right)-\left[\left(3x-1\right)^2-16\right]=\left(x^3+1\right)-\left(3x-1+4\right)\left(3x-1-4\right)\)\(=\left(x^3+1\right)-3\left(3x-5\right)\left(x+1\right)\)\(=\left(x+1\right)\left[x^2-x+1-9x+15\right]=\left(x+1\right)\left(x^2-10x+16\right)\)

\(=\left(x+1\right)\left[x\left(x-2\right)-8\left(x-2\right)\right]\)\(\left(x+1\right)\left(x-2\right)\left(x-8\right)\)

3)   \(x^3-6x^2-x+30\)

\(=x^3-5x^2-x^2+5x-6x+30\)

\(=x^2\left(x-5\right)-x\left(x-5\right)-6\left(x-5\right)\)

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

4)  \(2x^3-x^2+5x+3=\left(2x^3+x^2\right)-\left(2x^2+x\right)+\left(6x+3\right)\)

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

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

5) \(27x^3-27x^2+18x-4=\left(27x^3-1\right)-\left(27x^2-18x+3\right)\)

\(=\left(3x-1\right)\left(9x^2+3x+1\right)-3\left(9x^2-6x+1\right)\)

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

\(=\left(3x-1\right)\left(9x^2+3x+1-9x+3\right)=\left(3x-1\right)\left(9x^2-6x+4\right)\)

gửi phần này trước còn lại làm sau !!! tk mk nka !!!

nhiều thế

6) \(\left(x+y\right)^2-\left(x+y\right)-12\)\(=\left(x+y\right)^2-2\cdot\frac{1}{2}\left(x+y\right)+\frac{1}{4}-\frac{49}{4}\)

\(=\left(x+y-\frac{1}{2}\right)^2-\left(\frac{7}{2}\right)^2\)\(=\left(x+y-\frac{1}{2}-\frac{7}{2}\right)\left(x+y-\frac{1}{2}+\frac{7}{2}\right)\)

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

7)   \(\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)          (NHÂN x + 2 vs x +  5  và  x + 3 vs x + 4 )

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

ĐẶT   \(x^2+7x+11=y\)   ta được :  

\(\left(y+1\right)\left(y-1\right)-24=y^2-1-24\)

\(=y^2-25=\left(y-5\right)\left(y+5\right)\)

8)  \(4x^4-32x^2+1=4x^4+4x^2+1-36x^2\)

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

9) sai đề rùi bạn ơi ! đề đúng nè 

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

Ta thấy :  

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

Thay vào biểu thức bài cho ta được : 

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

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

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

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

bài ở trên câu 3 : kết luận là  \(\left(x-3\right)\left(x^2-x-6\right)\)bạn sửa lại giúp mk nka !!! Th@nk !!! Tk Mk vs  

1) \(\left\{{}\begin{matrix}2x+y=10\\5x-3y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}10x+5y=50\\10x-6y=6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}11y=44\\2x+y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=4\\x=3\end{matrix}\right.\)

Vậy hpt có nghiệm (x;y) = (3;4)

2)

a) 3x² - 2x - 1 = 0

\(\Leftrightarrow3x^2-3x+x-1=0\)

\(\Leftrightarrow3x\left(x-1\right)+\left(x-1\right)=0\)

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

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

Vậy pt có nghiệm x = 1 hoặc x = 3

b) Đặt x² = t (t \(\ge\) 0)

Pt trở thành: t² - 20t + 4 = 0

\(\Delta\) = (-20)² - 4.1.4 = 400 - 16 = 384

=> pt có 2 nghiệm phân biệt t₁ = \(\dfrac{20+8\sqrt{6}}{2}=10+4\sqrt{6}\)

t₂ = \(\dfrac{20-8\sqrt{6}}{2}=10-4\sqrt{6}\)

=> x₁ = \(\sqrt{10+4\sqrt{6}}=\sqrt{\left(2+\sqrt{6}\right)^2}=2+\sqrt{6}\)

x₂ = \(2-\sqrt{6}\)

a,\(x^3-7x+6\)

\(=x^3-2x^2+2x^2-4x-3x+6\)

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

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

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

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

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

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

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

b,\(x^3-9x^2+6x+16\)

\(=x^3-8x^2-x^2+8x-2x+16\)

\(=\left(x^3-8x^2\right)-\left(x^2-8x\right)-\left(2x-16\right)\)

\(=x^2.\left(x-8\right)-x.\left(x-8\right)-2.\left(x-8\right)\)

\(=\left(x-8\right).\left(x^2-x-2\right)\)

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

\(=\left(x-8\right).\left[\left(x^2+x\right)-\left(2x+2\right)\right]\)

\(=\left(x-8\right).\left[x.\left(x+1\right)-2.\left(x+1\right)\right]\)

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

c,\(x^3-6x^2-x+30\)

\(=x^3-5x^2-x^2+5x-6x+30\)

\(=\left(x^3-5x^2\right)-\left(x^2-5x\right)-\left(6x-30\right)\)

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

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

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

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

\(=\left(x-5\right).\left[x.\left(x+2\right)-3.\left(x+2\right)\right]\)

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

Chúc bạn học tốt!!!

d,\(2x^3-x^2+5x+3\)

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

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

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

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

e, \(27x^3-27x^2+18x-4\)

\(=27x^3-9x^2-18x^2+6x+12x-4\)

\(=\left(27x^2-9x^2\right)-\left(18x^2-6x\right)+\left(12x-4\right)\)

\(=9x^2.\left(3x-1\right)-6x.\left(3x-1\right)+4.\left(3x-1\right)\)

\(=\left(3x-1\right).\left(9x^2-6x+4\right)\)

Chúc bạn học tốt!!!

7, \(\left(x+2\right).\left(x+3\right).\left(x+4\right).\left(x+5\right)-24\)

\(=\left[\left(x+2\right).\left(x+5\right)\right].\left[\left(x+3\right).\left(x+4\right)\right]-24\)

\(=\left(x^2+5x+2x+10\right).\left(x^2+4x+3x+12\right)-24\)

\(=\left(x^2+7x+10\right).\left(x^2+7x+12\right)-24\)(1)

Đặt \(t=x^2+7x+10\Rightarrow t+2=x^2+7x+12\)

\(\Rightarrow\left(1\right)=t.\left(t+2\right)-24\)

\(=t^2+2t-24=t^2-4t+6t-24\)

\(=\left(t^2-4t\right)+\left(6t-24\right)=t.\left(t-4\right)+6.\left(t-4\right)\)

\(=\left(t-4\right).\left(t+6\right)\) (2)

Vì \(t=x^2+7x+10\) nên:

(2) \(=\left(x^2+7x+10-4\right).\left(x^2+7x+10+6\right)\)

\(=\left(x^2+7x+6\right).\left(x^2+7x+16\right)\)

\(=\left(x^2+x+6x+6\right).\left(x^2+7x+16\right)\)

\(=\left[\left(x^2+x\right)+\left(6x+6\right)\right].\left(x^2+7x+16\right)\)

\(=\left[x.\left(x+1\right)+6.\left(x+1\right)\right].\left(x^2+7x+16\right)\)

\(=\left(x+1\right).\left(x+6\right).\left(x^2+7x+16\right)\)

Chúc bạn học tốt!!!

c)   \(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)=40\)

\(\Leftrightarrow\)\(\left(x^2+6x+5\right)\left(x^2+6x+8\right)-40=0\)

Đặt      \(x^2+6x+5=t\)   ta có:

                       \(t\left(t+3\right)-40=0\)

          \(\Leftrightarrow\)\(t^2+3t-40=0\)

          \(\Leftrightarrow\)\(\left(t-5\right)\left(t+8\right)=0\)

        \(\Leftrightarrow\)\(\orbr{\begin{cases}t-5=0\\t+8=0\end{cases}}\)

Thay trở lại ta có:      \(\orbr{\begin{cases}x^2+6x=0\\x^2+6x+13=0\end{cases}}\)

(*)     \(x^2+6x=0\)

 \(\Leftrightarrow\)\(x\left(x+6\right)=0\)

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

(*)   \(x^2+6x+13=0\)

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

Vậy......