x3 - 7x - 6 ( làm bằng nhiều cách )
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) 3 x 2 − 7 x − 10 ⋅ 2 x 2 + ( 1 − 5 ) x + 5 − 3 = 0
+ Giải (1):
3 x 2 – 7 x – 10 = 0
Có a = 3; b = -7; c = -10
⇒ a – b + c = 0
⇒ (1) có hai nghiệm x 1 = - 1 v à x 2 = - c / a = 10 / 3 .
+ Giải (2):
2 x 2 + ( 1 - √ 5 ) x + √ 5 - 3 = 0
Có a = 2; b = 1 - √5; c = √5 - 3
⇒ a + b + c = 0
⇒ (2) có hai nghiệm:
Vậy phương trình có tập nghiệm
b)
x 3 + 3 x 2 - 2 x - 6 = 0 ⇔ x 3 + 3 x 2 - ( 2 x + 6 ) = 0 ⇔ x 2 ( x + 3 ) - 2 ( x + 3 ) = 0 ⇔ x 2 - 2 ( x + 3 ) = 0
+ Giải (1): x 2 – 2 = 0 ⇔ x 2 = 2 ⇔ x = √2 hoặc x = -√2.
+ Giải (2): x + 3 = 0 ⇔ x = -3.
Vậy phương trình có tập nghiệm S = {-3; -√2; √2}
c)
x 2 − 1 ( 0 , 6 x + 1 ) = 0 , 6 x 2 + x ⇔ x 2 − 1 ( 0 , 6 x + 1 ) = x ⋅ ( 0 , 6 x + 1 ) ⇔ x 2 − 1 ( 0 , 6 x + 1 ) − x ( 0 , 6 x + 1 ) = 0 ⇔ ( 0 , 6 x + 1 ) x 2 − 1 − x = 0
+ Giải (1): 0,6x + 1 = 0 ⇔
+ Giải (2):
x 2 – x – 1 = 0
Có a = 1; b = -1; c = -1
⇒ Δ = ( - 1 ) 2 – 4 . 1 . ( - 1 ) = 5 > 0
⇒ (2) có hai nghiệm
Vậy phương trình có tập nghiệm
d)
x 2 + 2 x − 5 2 = x 2 − x + 5 2 ⇔ x 2 + 2 x − 5 2 − x 2 − x + 5 2 = 0 ⇔ x 2 + 2 x − 5 − x 2 − x + 5 ⋅ x 2 + 2 x − 5 + x 2 − x + 5 = 0 ⇔ ( 3 x − 10 ) 2 x 2 + x = 0
⇔ (3x-10).x.(2x+1)=0
+ Giải (1): 3x – 10 = 0 ⇔
+ Giải (2):
=x3-7x+6
=x3-2x2+2x2-4x-3x+6
=x2(x-2)+2x(x-2)-3(x-2)
=(x-2)(x2+2x-3)
=(x-2)(x2+2x+1-4)
=(x-2)[(x+1)2-4]
=(x-2)(x+1-2)(x+1+2)=(x-1)(x-2)(x+3)
x3 - 7x + 6
= x3 - 2x2 + 2x2 - 4x - 3x + 6
= x2 ( x - 2 ) + 2x ( x - 2 ) + 3 ( x - 2 )
= ( x2 + 2x + 3 ) ( x - 2 )
= ( x2 + 2x + 1 - 4 ) ( x - 2 )
= [ ( x + 1 )2 - 22 ] ( x - 2 )
= ( x + 1 - 2 ) ( x + 1 + 2 ) ( x - 2 )
= ( x - 1 ) ( x + 3 ) ( x - 2 )
Ta có:\(x^3-7x-6=\left(x^3-3x^2\right)+\left(3x^2-9x\right)+\left(2x-6\right)\)
\(=\left(x-3\right)\left(x^2+3x+2\right)=\left(x-3\right)\left(x^2+2x+x+2\right)\)
\(=\left(x-3\right)\left(x+2\right)\left(x+1\right)\)
=x3-x-6x-6
=(x3-x)-(6x-6)
=x(x2-1)-6(x-1)
=x(x-1)(x+1)-6(x-1)
=(x-1)(x2+1-6)
x^2 + 5x -6
= x^2 + 5x - (5+1)
= x^2 + 5x -5 -1
= 5(x-1) + (x^2 -1)
= 5(x-1) + (x-1) (x+1)
= (5+x+1) (x-1)
5x^2 + 5xy -x-y
= 5x(x+y) - (x+y)
= (5x -1) (x+y)
7x - 6x^2 - 1 (câu này tớ tự ý sửa đề chút ^^!)
= 6x + x - 6x^2 -1
= 6x (1-x) - (1-x)
= (6x -1) (1-x)
x2 + 5x - 6= x2 - x + 6x -6= x(x-1) + 6(x-1)= (x+6)(x-1)
5x2 + 5xy - x - y= 5x2 - x + 5xy - y= x(5x-1) + y(5x - 1)= ( x +y)( 5x -1)
7x - 6x2 - 2= -6x2 + 3x + 4x -2= -3x(2x - 1) + 2(2x -1)= (2 - 3x)(2x-1)
\(c,20=2^2\cdot5\\ 45=3^2\cdot5\\ ƯCLN\left(20,45\right)=5\\ \RightarrowƯC\left(20,45\right)=Ư\left(5\right)=\left\{-5;-1;1;5\right\}\\ C=Ư\left(5\right)=\left\{-5;-1;1;5\right\}\)
\(d,\left(6x^2-7x+1\right)\left(x^3-x\right)=0\\ \Leftrightarrow\left(x-1\right)\left(6x-1\right)x\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{1}{6}\\x=1\\x=-1\end{matrix}\right.\)
\(\Leftrightarrow D=\left\{-1;0;\dfrac{1}{6};1\right\}\)
Sửa: \(\left[{}\begin{matrix}x=0\\x=\dfrac{1}{6}\\x=1\\x=-1\end{matrix}\right.\Leftrightarrow D=\left\{-1;0;\dfrac{1}{6};1\right\}\)
Ta có: P – Q = x4 + 3x3 – 5x2 + 7x – (-x3 + 4x2 – 2x +1)
= x4 + 3x3 – 5x2 + 7x + x3 - 4x2 - 4x2 + 2x – 1
= x4 + (3x3+ x3 ) + (– 5x2 - 4x2 ) + (7x + 2x ) – 1
= x4 + 4x3 – 9x2 + 9x – 1
Ta có: \(x^3-7x-6=0\)
\(\Leftrightarrow x^3-x-6x-6=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x-6\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\\x=-2\end{matrix}\right.\)
\(x^3-7x-6=0\)
\(\Leftrightarrow x^3-x-6x-6=0\)
\(\Leftrightarrow x\left(x^2-1\right)-6\left(x+1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left[x\left(x-1\right)-6\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+2x-3x-6\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left[x\left(x+2\right)-3\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=-2\\x=3\end{matrix}\right.\)
Vậy...
a: \(\Leftrightarrow8x^2+16x+14x+7=0\)
=>(2x+1)(8x+7)=0
=>x=-1/2 hoặc x=-7/8
b: \(=x^3-x-6x-6\)
\(=x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x-6\right)=\left(x+1\right)\left(x-3\right)\left(x+2\right)\)
\(a,\Rightarrow8x^2+2x+28x+7=0\\ \Rightarrow2x\left(4x+1\right)+7\left(4x+1\right)=0\\ \Rightarrow\left(2x+7\right)\left(4x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=-\dfrac{1}{4}\end{matrix}\right.\\ b,Sửa:x^3-7x-6=0\\ \Rightarrow x^3-x-6x-6=0\\ \Rightarrow x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)=0\\ \Rightarrow\left(x+1\right)\left(x^2-x-6\right)=0\\ \Rightarrow\left(x+1\right)\left(x^2-3x+2x-6\right)=0\\ \Rightarrow\left(x+1\right)\left(x-3\right)\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=3\\x=-2\end{matrix}\right.\)
Cách 1: x3-7x-6=x3+x2-x2-x-6x-6=x2(x+1)-x(x+1)-6(x-1)=(x-1)(x2-x-6)
Cách 2: x3-7x-6=x3-x-6x-6=x(x2-1)-6(x+1)=x(x-1)(x+1)-6(x+1)=(x+1)[x(x-1)-6]=(x+1)(x2-x-6)
Cách 3: x3-7x-6=x3+1-7x-7=(x+1)(x2-x+1)-7(x+1)=(x+1)(x2-x+1-7)=(x-1)(x2-x-6)
Chúc bạn học tốt !