(14,78-a)/(2,87+a)=4/1

14,78+2,87=17,65

Tổng số phần bằng nhau là 4+1=5

Mỗi phần có giá trị bằng 17,65/5=3,53

=>2,87+a=3,53

=>a=0,66.

ta có : x^5+2x^4+3x^3+3x^2+2x+1=0

\(\Leftrightarrow\)x^5+x^4+x^4+x^3+2x^3+2x^2+x^2+x+x+1=0

\(\Leftrightarrow\)(x^5+x^4)+(x^4+x^3)+(2x^3+2x^2)+(x^2+x)+(x+1)=0

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

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

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

\(\Leftrightarrow\)(x+1)[x^2(x^2+x+1)+(x^2+x+1)]=0

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

VÌ x^2+x+1=(x+\(\dfrac{1}{2}\))^2+\(\dfrac{3}{4}\)\(\ne0\) và x^2+1\(\ne0\)

\(\Rightarrow\)x+1=0

\(\Rightarrow\)x=-1

CÒN CÂU B TỰ LÀM (02042006)

b: x^4+3x^3-2x^2+x-3=0

=>x^4-x^3+4x^3-4x^2+2x^2-2x+3x-3=0

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

=>x-1=0

=>x=1

a) \(2x^3-5x^2+3x=0\)

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

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

\(\Leftrightarrow x\left[2x\left(x-1\right)-3\left(x-1\right)\right]=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy .................

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

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

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

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

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

Vậy ...............

c) \(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2+2-7x+10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\)

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

P/s: tới đây bn tự giải tiếp nha

a) Ta có: \(|-5x|-16=3x\) 

Đk:  \(3x\ge0\Rightarrow x\ge0\)

\(\Rightarrow\orbr{\begin{cases}-5x-16=3x\\5x-16=3x\end{cases}}\Rightarrow\orbr{\begin{cases}-5x-3x=16\\5x-3x=16\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}-8x=16\\-2x=16\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\x=8\end{cases}}}\)

Mà x \(\ge0\)\(\Rightarrow x=8\)

b) \(|3x-2|=1-x\)

\(\Rightarrow\orbr{\begin{cases}3x-2=1-x\\3x-2=-1+x\end{cases}\Rightarrow}\orbr{\begin{cases}3x+x=1+2\\3x-x=-1+2\end{cases}}\)

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

Vậy: x = \(\frac{3}{4}\)hoặc x\(=\frac{1}{2}\)

c) Ta có: \(|-2x|=4x-10\)

Đk: \(4x-10\ge0\Rightarrow4x\ge10\Rightarrow x\ge\frac{5}{2}\)

\(\Rightarrow\orbr{\begin{cases}-2x=4x-10\\2x=4x-10\end{cases}}\Rightarrow\orbr{\begin{cases}-2x-4x=-10\\2x-4x=-10\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}-6x=-10\\-2x=-10\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=5\end{cases}}\)

mà x\(\ge\frac{5}{2}\)\(\Rightarrow x=5\)

a) 

Đặt x^2 + x - 5 = t.

Khi đó, pt đã cho trở thành :

t ( t + 9 ) = -18

<=> t^2 + 9t + 18 = 0

<=> ( t + 3 )( t + 6 ) = 0

Giải pt trên, ta được t = -3 và t = -6 là các nghiệm của pt.

+) t = -3 => x^2 + x - 5 = -3

           <=> x^2 + x - 2 = 0

          <=> ( x + 2 )( x - 1 ) = 0

Giải pt trên, ta được x = -2 ; x = 1 là các nghiệm của pt.

+) t = -6 => x^2 + x - 5 = -6

            <=> x^2 + x + 1 = 0

           <=> ( x + 1/2 )^2 + 3/4 = 0

=> Pt trên vô nghiệm.

Vậy..........

b)

x^3 - 7x + 6 = 0

<=> ( x^3 + 3x^2 ) - ( 3x^2 + 9x ) + ( 2x + 6 ) = 0

<=> x^2 . ( x + 3 ) - 3x . ( x + 3 ) + 2( x + 3 ) = 0

<=> ( x + 3 ) ( x^2 - 3x + 2 ) = 0

<=> ( x+ 3 )( x - 2 )( x - 1 ) = 0

Giải pt trên, ta được x = -3 ; x= 2 ; x= 1 là các nghiệm của pt.

Vậy..........

c)

( 3x^2 + 10x - 8 )^2 = ( 5x^2 - 2x + 10 )^2

<=> ( 3x^2 + 10x - 8 )^2 - ( 5x^2 - 2x + 10 )^2 = 0

<=> ( 3x^2 + 10x - 8 - 5x^2 + 2x - 10 )( 3x^2 + 10x - 8 + 5x^2 - 2x + 10 ) = 0

<=> ( -2x^2 + 12x - 18 )( 8x^2 + 8x + 2 ) = 0

<=> ( x^2 - 6x + 9 )( 4x^2 + 4x + 1 ) = 0

<=> ( x - 3 )^2 . ( 2x + 1 )^2 = 0.

Giải pt trên, ta được x = 3 và x = -1/2 là các nghiệm của pt.

Vậy..........

\(\Leftrightarrow x^3+x^2-2x+5x^2+5x-10=0\)

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

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

\(\Leftrightarrow\left(x+5\right)\left(x+2\right)\left(x-1\right)=0\)

b/ \(\Leftrightarrow x^3+5x^2+6x-x^2-5x-6=0\)

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

\(\Leftrightarrow\left(x-1\right)\left(x^2+5x+6\right)=0\)

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

\(x^3+6x^2+3x-10=0\)

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

\(\Leftrightarrow x^2\left(x-1\right)+7x\left(x-1\right)+10\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+7x+10\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+2x+5x+10\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[x\left(x+2\right)+5\left(x+2\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=-5\end{matrix}\right.\)

Vậy \(S=\left\{1;-2;-5\right\}\)

\(x^3+4x^2+x-6=0\)

\(\Leftrightarrow x^3-x^2+5x^2-5x+6x-6=0\)

\(\Leftrightarrow x^2\left(x-1\right)+5x\left(x-1\right)+6\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+5x+6\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+2x+3x+6\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[x\left(x+2\right)+3\left(x+2\right)\right]=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=-3\end{matrix}\right.\)

Vậy \(S=\left\{1;-2;-3\right\}\)

\(a.17+8x=10-6x\\\Leftrightarrow 8x+6x=-17+10\\\Leftrightarrow 2x=-7\\ \Leftrightarrow x=-\frac{7}{2}\)

Vậy nghiệm của phương trình trên là \(-\frac{7}{2}\)

\(b.3\left(x+5\right)+7=19-5\left(x-2\right)\\\Leftrightarrow 3x+15+7=19-5x+10\\ \Leftrightarrow3x+5x=-15-7+19+10\\ \Leftrightarrow8x=7\\\Leftrightarrow x=\frac{7}{8}\)

Vậy nghiệm của phương trình trên là \(\frac{7}{8}\)

\(c.3x-4\left(x+2\right)\left(x+3\right)=14-4\left(x^2-3x\right)\\ \Leftrightarrow3x-4\left(x^2+5x+6\right)=14-4x^2+12x\\ \Leftrightarrow4x^2-4x^2+3x-5x-12x=24+14\\ \Leftrightarrow-14x=38\\ \Leftrightarrow x=-\frac{19}{7}\)

Vậy nghiệm của phương trình trên là \(-\frac{19}{7}\)

\(d.x+\frac{3}{4}+3x+2=\frac{x}{3}-3x-\frac{2}{6}\\ \Leftrightarrow\frac{12x}{12}+\frac{9}{12}+\frac{36x}{12}+\frac{24}{12}=\frac{4x}{12}-\frac{36x}{12}-\frac{4}{12}\\ \Leftrightarrow12x+9+36x+24=4x-36x-4\\ \Leftrightarrow12x+36x+36x-4x=-24-9-4\\ \Leftrightarrow80x=-37\\ \Leftrightarrow x=-\frac{37}{80}\)

a, \(\frac{x+9}{x^2-3x-10}-\frac{x+15}{x^2-25}=\frac{1}{x+2}\left(ĐKXĐ:x\ne\pm2;\pm5\right)\)

\(\frac{x+9}{\left(x-5\right)\left(x+2\right)}-\frac{x+15}{\left(x+5\right)\left(x-5\right)}=\frac{1}{x+2}\)

\(\frac{\left(x+9\right)\left(x+5\right)}{\left(x-5\right)\left(x+2\right)\left(x+5\right)}-\frac{\left(x+15\right)\left(x+2\right)}{\left(x+5\right)\left(x-5\right)\left(x+2\right)}=\frac{\left(x+5\right)\left(x-5\right)}{\left(x+2\right)\left(x+5\right)\left(x-5\right)}\)

Khử mẫu : \(\left(x+9\right)\left(x+5\right)-\left(x+15\right)\left(x+2\right)=\left(x+5\right)\left(x-5\right)\)

\(x^2+14x+45-x^2-17x-30=x^2-25\)

\(-3x+15-x^2+25=0\)

\(-3x-x^2+40=0\)( giải delta ta đc )

\(x_1=-5;x_2=8\)

b, \(\frac{1}{3x-1}+\frac{2x+2}{x-1}-\frac{3x^2+1}{3x^2-4x+1}=1ĐKXĐ\left(x\ne1;\frac{1}{3}\right)\)

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

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

Khửi mẫu \(x-1+\left(2x+2\right)\left(3x-1\right)-3x^2-1=\left(3x-1\right)\left(x-1\right)\)( bn tự nốt nhé)

c, \(\left(x+3\right)^2-10\ge\left(x+3\right)\left(x+2\right)-4\)

\(x^2+6x+9-10\ge x^2+5x+6-4\)

\(x-3\ge0\Leftrightarrow x\ge3\)

a) \(\frac{x+9}{x^2-3x-10}-\frac{x+15}{x^2-25}=\frac{1}{x+2}\); ĐKXĐ: x # -2; x # +-5

<=> \(\frac{x+9}{\left(x+2\right)\left(x-5\right)}-\frac{x+15}{\left(x-5\right)\left(x+5\right)}=\frac{1}{x+2}\)

<=> \(\frac{\left(x+9\right)\left(x+5\right)-\left(x+15\right)\left(x+2\right)}{\left(x+2\right)\left(x-5\right)\left(x+5\right)}=\frac{\left(x-5\right)\left(x+5\right)}{\left(x+2\right)\left(x-5\right)\left(x+5\right)}\)

<=> (x + 9)(x + 5) - (x + 15)(x + 2) = (x - 5)(x + 5)

<=> -3x + 15 = x^2 - 25

<=> -3x + 15 - x^2 + 25 = 0

<=> -3x + 40 - x^2 = 0

<=> x^2 + 3x - 40 = 0

<=> (x - 5)(x + 8) = 0

<=> x - 5 = 0 hoặc x + 8 = 0

<=> x = 5 (ktm0 hoặc x = -8 (tm)

b) \(\frac{1}{3x-1}+\frac{2x+2}{x-1}-\frac{3x^2+1}{3x^2-4x+1}=1\); ĐKXĐ: x # 1/3; x # 1

<=> \(\frac{1}{3x-1}+\frac{2\left(x+1\right)}{x-1}-\frac{3x^2+1}{x\left(3x-1\right)-\left(3x-1\right)}=1\)

<=> \(\frac{1}{3x-1}+\frac{2\left(x+1\right)}{x-1}-\frac{3x^2+1}{\left(x-1\right)\left(3x-1\right)}=1\)

<=> \(\frac{x-1}{\left(x-1\right)\left(3x-1\right)}+\frac{2\left(x+1\right)\left(3x-1\right)}{\left(x-1\right)\left(3x-1\right)}-\frac{3x^2+1}{\left(x-1\right)\left(3x-1\right)}=\frac{\left(x-1\right)\left(3x-1\right)}{\left(x-1\right)\left(3x-1\right)}\)

<=> x - 1 + 2(x + 1)(3x - 1) - 3x^2 + 1 = (x - 1)(3x - 1)

<=> 5x - 4 + 3x^2 = 3x^2 - 4x + 1

<=> 5x - 4 = -4x + 1

<=> 5x + 4x = 1 + 4

<=> 9x = 5

<=> x = 5/9 (tm)

c) (x + 3)^2 - 10 >= (x + 3)(x + 2) - 4

<=> x^2 + 3x + 3x + 9 - 10 >=  x^2 + 2x + 3x + 6 - 4

<=> x^2 + 6x + 9 - 10 >= x^2 + 5x + 6 - 4

<=> x^2 + 6x - 1 >= x^2 + 5x + 2

<=> x^2 + 6x - 1 - x^2 - 5x - 2 >= 0

<=> x - 3 >= 0

<=> x >= 3

Viết lại đề bài hộ cái, đề sai rùi.

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

\(=x^2-4-4x^2-40x-100=2x-3x^2\)

\(\Leftrightarrow-42x-104=0\)

\(\Leftrightarrow x=-\dfrac{52}{21}\)