1. x\(^4\)-x\(^3\)+2x\(^2\)-x+1=0

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

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

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

\(\Leftrightarrow\)\([\)(x^2-x+1/4)+3/4\(]\)(x^2+1)=0

\(\Leftrightarrow\)\([\)(x-1/2)\(^2\)+3/4\(]\)(x^2+1)=0  

VÌ (x-1/2)\(^2\)+3/4>0\(\forall\)x

x^2+1>0\(\forall\)x

\(\Rightarrow\)Phương trình đã cho vô nghiệm

1)x^4 - x^3 + 2x^2 - x + 1 = 0

  (x^4 + 2x^2 +1) - (x^3+x)= 0

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

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

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

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

               x^2+1                =  x (vô lí)

==> PT vô nghiệm

Ví dụ cho bạn một bài, còn lại tương tự.

a)Ta có: \(3x^4-5x^3+8x^2-5x+3\)

\(=3x^2\left(x-\frac{5}{6}\right)^2+\frac{71}{12}\left(x-\frac{30}{71}\right)^2+\frac{138}{71}>0\)

Vậy phương trình vô nghiệm.

tth_new bạn làm hết ra đc ko. mình đọc không hiểu đc

Bài 2: 

a: \(a^3+b^3+c^3-3abc\)

\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\)

\(=0\cdot\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

\(\Leftrightarrow a^3+b^3+c^3=3abc\)

b: Đặt 4x+3=a; 3x-8=b

Theo đề, ta có có phương trình:

\(a^3+b^3-\left(a+b\right)^3=0\)

\(\Leftrightarrow3ab\left(a+b\right)=0\)

\(\Leftrightarrow\left(4x+3\right)\left(7x-5\right)\left(3x-8\right)=0\)

hay \(x\in\left\{-\dfrac{3}{4};\dfrac{5}{7};\dfrac{8}{3}\right\}\)

1) tìm x : 

5x. (x - 3 ) + 7.(x - 3 ) = 0

<=> ( x -3 ) . ( 5x +7 ) = 0

<=> x - 3 = 0 hoặc 5x + 7 = 0 

<=> x = 3 hoặc x = -7/5

Vậy x € { 3 ; -7/5 }

3 ) chứng mình rằng : 

7 ¹⁹⁹⁶ + 7¹⁹⁹⁵ + 7¹⁹⁹⁴ chia hết cho 57 

7¹⁹⁹⁶ + 7¹⁹⁹⁵ + 7¹⁹⁹⁴ 

<=> 7¹⁹⁹⁴  . 7² + 7¹⁹⁹⁴ .7 + 7¹⁹⁹⁴

<=> 7¹⁹⁹⁴ . ( 7²  + 7 + 1 ) 

<=> 7¹⁹⁹⁴ .  57 chia  hết cho 57 ( vì 57 chia hết cho 57 )  ( đ..p.c.m ) 

Bài 1 : \(5x\left(x-3\right)+7\left(x-3\right)=0.\)

\(\Rightarrow5x^2-15x+7x-21=0\)

\(\Rightarrow5x^2-8x-21=0\)

\(\Rightarrow5x^2-15x+7x-21=0\)

\(\Rightarrow5x\left(x-3\right)+7\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right)\left(5x-7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\5x-7=0\end{cases}\Rightarrow\hept{\begin{cases}x=3\\x=\frac{7}{5}\end{cases}}}\)

Bài 2 : \(a,A=0\Rightarrow x^2-3x=0\Rightarrow x\left(x-3\right)=0\Rightarrow x\in\left\{0;3\right\}\)

\(b,A>0\Rightarrow x^2-3x>0\Rightarrow x\left(x-3\right)>0\)

TH1 : \(\hept{\begin{cases}x>0\\x-3>0\end{cases}\Rightarrow\hept{\begin{cases}x>0\\x>3\end{cases}\Rightarrow}x>3}\)

TH2 : \(\hept{\begin{cases}x< 0\\x-3< 0\end{cases}\Rightarrow\hept{\begin{cases}x< 0\\x< 3\end{cases}\Rightarrow}x< 3}\)

C, tương tự 

Bài 3 : \(7^{1996}+7^{1995}+7^{1994}=7^{1994}\left(7^2+7+1\right)\)

\(=7^{1994}.57\)\(⋮\)\(7\)

\(\Rightarrow7^{1996}+7^{1995}+7^{1994}⋮\)\(7\)

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

\(VP=x\left(x^2+3x+2\right)\)

\(VP=x\left[\left(x^2+x\right)+\left(2x+2\right)\right]\)

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

\(VP=x\left(x+1\right)\left(x+2\right)\) ( đpcm ) 

Chúc bạn học tốt ~ 

a) x(x+1)(x+2)=(x²+x)(x+2)=x³+2x²+x²+2x=x³+3x²+3x

b)

(3x - 2)(4x - 5) - (2x - 1)(6x + 1) = 0

12x2 - 15x - 8x + 10 - 12x2 - 2x + 6x + 1 = 0

- 19x = - 11

x = 11/19

\(\text{CM vô nghiệm}\)
\(\text{a) }\left(x-2\right)^3=\left(x-2\right).\left(x^2+2x+4\right)-6\left(x-1\right)^2\)
\(\Leftrightarrow x^3-6x^2+12x-8=x^3-8-6\left(x^2-2x+1\right)\)
\(\Leftrightarrow x^3-6x^2+12x-8=x^3-8-6x^2+12x-6\)
\(\Leftrightarrow x^3-6x^2+12x-x^3+6x-12x=-8+8-6\)
\(\Leftrightarrow0x=-6\text{ (vô lí)}\)
\(\text{Vậy }S=\varnothing\)

\(\text{b) }4x^2-12x+10=0\)
\(\Leftrightarrow\left(4x^2-12x+9\right)+1=0\)
\(\Leftrightarrow\left(2x-3\right)^2+1=0\)
\(\Leftrightarrow\left(2x-3\right)^2=-1\text{ (vô lí)}\)
\(\text{Vậy }S=\varnothing\)

\(\text{CM vô số nghiệm}\)
       \(\left(x+1\right)\left(x^2-x+1\right)=\left(x+1\right)^3-3x\left(x+1\right)\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)=\left(x+1\right)\left[\left(x+1\right)^2-3x\right]\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)=\left(x+1\right)\left(x^2+2x+1-3x\right)\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)=\left(x+1\right)\left(x^2-x+1\right)\text{ (luôn luôn đúng)}\)
\(\text{Vậy }S\inℝ\)