Gi ải các phương trình sau
e) x3-7x+6=0
f) x4-4x3+12x-9=0
g)x5-5x3+4x=0
h) x4-4x3+3x2+4x-4=0
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, \(\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)-\left(3x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(\left(9x^2-4\right)-\left(\left(3x+2\right)\left(x-1\right)\right)\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-\left(3x^2-x-2\right)\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-3x^2+x+2\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x+1\right)=0;3x^2+x-2=0\)
=> x=-1
với \(3x^2+x-2=0\)
ta sử dụng công thức bậc 2 suy ra : \(x=\dfrac{2}{3};x=-1\)
Vậy ghiệm của pt trên \(S\in\left\{-1;\dfrac{2}{3}\right\}\)
b: \(\Leftrightarrow x^2-2x+1-1+x^2=x+3-x^2-3x\)
\(\Leftrightarrow2x^2-2x=-x^2-2x+3\)
\(\Leftrightarrow3x^2=3\)
hay \(x\in\left\{1;-1\right\}\)
c: \(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x-3\right)-\left(x-1\right)\left(x-2\right)\left(x+2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+1\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-2x-3-x^2-3x+10\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(-5x+7\right)=0\)
hay \(x\in\left\{1;-2;\dfrac{7}{5}\right\}\)
a) (5x3 – 2x2 + 4x – 4) . ( x3 + 3x2 – 5)
= 5x3 . ( x3 + 3x2 – 5) - 2x2 . ( x3 + 3x2 – 5) + 4x . ( x3 + 3x2 – 5) – 4 . ( x3 + 3x2 – 5)
= 5x3 . x3 + 5x3 . 3x2 + 5x3 . (-5) – [ 2x2 . x3 + 2x2 . 3x2 +2x2 . (-5)] + [4x . x3 + 4x. 3x2 + 4x . (-5)] – [ 4x3 + 4.3x2 + 4.(-5)]
= 5x6 + 15x5 – 25x3 – (2x5 + 6x4 – 10x2) + 4x4 + 12x3 – 20x – (4x3 + 12x2 – 20)
= 5x6 + 15x5 – 25x3 – 2x5 - 6x4 + 10x2 + 4x4 + 12x3 – 20x – 4x3 - 12x2 + 20
= 5x6 + (15x5 – 2x5 ) + (- 6x4 + 4x4 ) + (-25x3 + 12x3 – 4x3 ) + (10x2 - 12x2 ) – 20x + 20
= 5x6 + 13x5 – 2x4 – 17x3 -2x2 – 20x + 20
b) (-2,5.x4 + 0,5x2 + 1) . (4x3 – 2x + 6)
= -2,5.x4 . (4x3 – 2x + 6) + 0,5x2 . (4x3 – 2x + 6) + 1. (4x3 – 2x + 6)
= (-2,5.x4) . 4x3 + (-2,5.x4 ) . (-2x) + (-2,5.x4 ) . 6 + 0,5x2 . 4x3 + 0,5x2 . (-2x) + 0,5x2 . 6 + 4x3 – 2x + 6
= -10x7 + 5x5 – 15x4 + 2x5 – x3 + 3x2 + 4x3 – 2x + 6
= -10x7 + ( 5x5 + 2x5 ) - 15x4 + (– x3 + 4x3 ) + 3x2 – 2x + 6
= -10x7 +7x5 - 15x4 + 3x3 + 3x2 – 2x + 6
Giải phương trình
e) x4 -4x3-8x2+8x=0
f) 2x2+3xy+y2=0
g) 2x4-x3-9x2+13x-5=0
h) (x+1)(x+3)(x+5)(x+7)+15=0
e: =>x(x^3-4x^2-8x+8)=0
=>x[(x^3+8)-4x(x+2)]=0
=>x(x+2)(x^2-2x+4-4x)=0
=>x(x+2)(x^2-6x+4)=0
=>\(x\in\left\{0;-2;3+\sqrt{5};3-\sqrt{5}\right\}\)
g: =>2x^4+5x^3-6x^3-15x^2+6x^2+15x-2x-5=0
=>(2x+5)(x^3-3x^2+3x-1)=0
=>(2x+5)(x-1)^3=0
=>x=1 hoặc x=-5/2
h: =>(x^2+8x+7)(x^2+8x+15)+15=0
=>(x^2+8x)^2+22(x^2+8x)+120=0
=>(x^2+8x+10)(x^2+8x+12)=0
=>(x^2+8x+10)(x+2)(x+6)=0
=>\(x\in\left\{-2;-6;-4+\sqrt{6};-4-\sqrt{6}\right\}\)
f(x) = x5 + 3x2 − 5x3 − x7 + x3 + 2x2 + x5 − 4x2 + x7
= (x5 + x5) + (3x2 + 2x2 – 4x2) + (-5x3 + x3) + (-x7 + x7)
= 2x5 + x2 – 4x3.
= 2x5 - 4x3 + x2
Đa thức có bậc là 5
g(x) = x4 + 4x3 – 5x8 – x7 + x3 + x2 – 2x7 + x4 – 4x2 – x8
= (x4 + x4) + (4x3 + x3) – (5x8 + x8) – (x7 + 2x7) + (x2 – 4x2)
= 2x4 + 5x3 – 6x8 – 3x7 – 3x2
= -6x8 - 3x7 + 2x4 + 5x3 - 3x2.
Đa thức có bậc là 8.
a) Đa thức thương 4x – 11 và đa thức dư 26x – 10.
b) Đa thức thương 2 x 2 – 3x + 5 và đa thức dư 3x + 4.
a. (3x - 1)2 - (x + 3)2 = 0
\(\Leftrightarrow\left(3x-1+x+3\right)\left(3x-1-x-3\right)=0\)
\(\Leftrightarrow\left(4x+2\right)\left(2x-4\right)=0\)
\(\Leftrightarrow4x+2=0\) hoặc \(2x-4=0\)
1. \(4x+2=0\Leftrightarrow4x=-2\Leftrightarrow x=-\dfrac{1}{2}\)
2. \(2x-4=0\Leftrightarrow2x=4\Leftrightarrow x=2\)
S=\(\left\{-\dfrac{1}{2};2\right\}\)
b. \(x^3=\dfrac{x}{49}\)
\(\Leftrightarrow49x^3=x\)
\(\Leftrightarrow49x^3-x=0\)
\(\Leftrightarrow x\left(49x^2-1\right)=0\)
\(\Leftrightarrow x\left(7x+1\right)\left(7x-1\right)=0\)
\(\Leftrightarrow x=0\) hoặc \(7x+1=0\) hoặc \(7x-1=0\)
1. x=0
2. \(7x+1=0\Leftrightarrow7x=-1\Leftrightarrow x=-\dfrac{1}{7}\)
3. \(7x-1=0\Leftrightarrow7x=1\Leftrightarrow x=\dfrac{1}{7}\)
a.
\(x^3-7x+6=0\)
\(\Leftrightarrow x^3-3x^2+2x+3x^2-9x+6=0\)
\(\Leftrightarrow x\left(x^2-3x+2\right)+3\left(x^2-3x+2\right)=0\)
\(\Leftrightarrow\left(x^2-3x+2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x^2-x-2x+2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[x\left(x-1\right)-2\left(x-1\right)\right]\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-3\end{matrix}\right.\)
f.
\(x^4-4x^3+12x-9=0\)
\(\Leftrightarrow x^4-4x^3+3x^2-3x^2+12x-9=0\)
\(\Leftrightarrow x^2\left(x^2-4x+3\right)-3\left(x^2-4x+3\right)=0\)
\(\Leftrightarrow\left(x^2-4x+3\right)\left(x^2-3\right)=0\)
\(\Leftrightarrow\left(x^2-x-3x+3\right)\left(x^2-3\right)=0\)
\(\Leftrightarrow\left[x\left(x-1\right)-3\left(x-1\right)\right]\left(x^2-3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(x^2-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=3\\x=\pm\sqrt{3}\end{matrix}\right.\)