tìm x biết :
a) x^3 = -125
b) x^2 = 81
c) (x-1) . (x-3) = 0
d) x^2 -4x = 0
Bài 1: Tìm x biết a) x^3 - 4x^2 - x + 4= 0 b) x^3 - 3x^2 + 3x + 1=0 c) x^3 + 3x^2 - 4x - 12=0 d) (x-2)^2 - 4x +8 =0
a: \(x^3-4x^2-x+4=0\)
=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)
=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)
=>\(\left(x-4\right)\left(x^2-1\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)
b: Sửa đề: \(x^3+3x^2+3x+1=0\)
=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)
=>\(\left(x+1\right)^3=0\)
=>x+1=0
=>x=-1
c: \(x^3+3x^2-4x-12=0\)
=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)
=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)
=>\(\left(x+3\right)\left(x^2-4\right)=0\)
=>\(\left(x+3\right)\left(x-2\right)\left(x+2\right)=0\)
=>\(\left[{}\begin{matrix}x+3=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)
d: \(\left(x-2\right)^2-4x+8=0\)
=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)
=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)
=>\(\left(x-2\right)\left(x-2-4\right)=0\)
=>(x-2)(x-6)=0
=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)
1) phân tích thành phân tử:
a) x^3y^3 + 1/125
b) (x+5)^3-(x-5)^3
c) (6-x)^3+(6+x)^3
d) 27x^3 -135x^2+225x-125
2) tìm x:
a)4x^2-25x^4=0
b) x^2-8x+16=0
c)x^3-3x^2+3x-1=0
1.
a) = (xy + \(\frac{1}{5}\)) (x2y2 - \(\frac{xy}{5}\)+ \(\frac{1}{25}\))
b) = (x + 5 - x + 5) [(x+5)2 + (x+5)(x-5) + (x-5)2] = 10 (x2 + 10x + 25 + x2 - 25 + x2 - 10x + 25) = 10 (3x2 +25)
c) = (6 - x + 6 + x) [(6-x)2 - (6-x)(6+x) + (6+x)2] = 12 (36 - 12x + x2 - 26 + x2 + 36 + 12x + x2) = 12 (3x2 + 36) = 12. 3(x2 + 12) = 36(x2 +12)
d) = (3x - 5)3
2.
a) => (2x - 5x2)(2x + 5x2) = 0 ............. giải ra
b) => (x-4)2 = 0 => x - 4 = 0 => x= 4
c) => (x - 1)3 = 0 => x - 1 = 0 => x = 1
Tìm x, biết:
a) 2-x = 2 ( x - 2 ) 3 ; b) 8 x 3 - 72x = 0;
c) ( x - 1 , 5 ) 6 + 2 ( 1 , 5 - x ) 2 = 0; d) 2 x 3 +3 x 2 +3 + 2x = 0;
e) x 3 - 4x- 14x(x - 2) = 0; g) x 2 (x + 1)- x(x + 1) + x(x - 1) = 0.
1,phân tích thành nhân tử
a. x3-x2-5x+125
b.5x2-5xy-3x+3y
c. x2-2x-4y2+1
d. x3-x+y3-y
2,tìm x
a.x3-1/4x=0
b.4x4+4x3-x2-x=0
c.x4-4x3+8x2-16x+16=0
a) \(x^3-x^2-5x+125\)
\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
b) \(5x^2-5xy-3x+3y\)
\(=5x\left(x-y\right)-3\left(x-y\right)\)
\(=\left(x-y\right)\left(5x-3\right)\)
c) \(x^2-2x-4y^2+1\)
\(=\left(x-1\right)^2-4y^2\)
\(=\left(x-2y-1\right)\left(x+2y-1\right)\)
tìm x: part 1 : a,(x^3)^2-(x+1)(x-1)=1 b,(x-2)^2-3(x-2)=0 c,(x+2)(x^2-2x+4)-x(x^2+2)=15 d,(x+1)^2-(x+1)(x-2)=0 e,4x(x-2017)-x+2017=0 f,(x+4)^2-16=0 part 2: a,x^3+27+(x+3)(x-9)=0 b,(2x-1)^2-4x^2+1=0 c,2(x-3)+x^2-3x=0 d,x^2-2x+1=6x-6 e,x^3-9x=0
tìm x biết
a/ x^3-x^2-x+1=0
b/(2x^3-3)^2-(4x^2-9)=0
c/x^4+2x^3-6x-9=0
d/2(x+5)-x^2-5x=0
\(a)\)\(x^3-x^2-x+1=0\)
\(\Leftrightarrow\)\(x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\)\(\left(x-1\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\)\(\left(x-1\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\)\(\left(x-1\right)^2\left(x+1\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}\left(x-1\right)^2=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}}\)
Vậy \(x=1\) hoặc \(x=-1\)
Chúc bạn học tốt ~
a) x3-x2-x+1 = 0 \(\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\)\(\Leftrightarrow x^2-1=0\)hoặc x-1=0
\(\Leftrightarrow x=1\)
\(c)\)\(x^4+2x^3-6x-9=0\)
\(\Leftrightarrow\)\(\left(x^4-9\right)+\left(2x^3-6x\right)=0\)
\(\Leftrightarrow\)\(\left(x^2-3\right)\left(x^2+3\right)+2x\left(x^2-3\right)=0\)
\(\Leftrightarrow\)\(\left(x^2-3\right)\left(x^2+3+2x\right)=0\)
\(\Leftrightarrow\)\(x^2-3=0\)
Hoặc \(x^2+3+2x=0\)
\(\Leftrightarrow\)\(x^2=3\)
Hoặc \(x\left(x+2\right)=-3\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
Hoặc \(x;\left(x-2\right)\inƯ\left(-3\right)\)
Ta có bảng :
\(x\) | \(1\) | \(-3\) | \(-1\) | \(3\) |
\(x-2\) | \(-3\) | \(1\) | \(3\) | \(-1\) |
\(x\) | \(1\) | \(-3\) | \(-1\) | \(3\) |
\(x\) | \(-1\) | \(3\) | \(5\) | \(1\) |
Vậy \(x\in\left\{1;-1;3;-3;5\right\}\)
Chúc bạn học tốt ~
Tìm x biết.
a) 4x^2 - 49 = 0 b) x^2 + 36 = 12x
c) 1/16x^2 - x + 4 = 0 d) x^3 -3√3x2 + 9x - 3√3 = 0
e) (x - 2)^2 - 16 = 0 f) x^2 - 5x - 14 = 0
g) 8x(x - 3) + x - 3 = 0
a, 4x2 - 49 = 0
⇔⇔ (2x)2 - 72 = 0
⇔⇔ (2x - 7)(2x + 7) = 0
⇔{2x−7=02x+7=0⇔⎧⎪ ⎪⎨⎪ ⎪⎩x=72x=−72⇔{2x−7=02x+7=0⇔{x=72x=−72
b, x2 + 36 = 12x
⇔⇔ x2 + 36 - 12x = 0
⇔⇔ x2 - 2.x.6 + 62 = 0
⇔⇔ (x - 6)2 = 0
⇔⇔ x = 6
e, (x - 2)2 - 16 = 0
⇔⇔ (x - 2)2 - 42 = 0
⇔⇔ (x - 2 - 4)(x - 2 + 4) = 0
⇔⇔ (x - 6)(x + 2) = 0
⇔{x−6=0x+2=0⇔{x=6x=−2⇔{x−6=0x+2=0⇔{x=6x=−2
f, x2 - 5x -14 = 0
⇔⇔ x2 + 2x - 7x -14 = 0
⇔⇔ x(x + 2) - 7(x + 2) = 0
⇔⇔ (x + 2)(x - 7) = 0
⇔{x+2=0x−7=0⇔{x=−2x=7
a,\(4x^2-49=0\)
\(\Leftrightarrow\left(2x\right)^2-7^2=0\)
\(\Leftrightarrow\left(2x-7\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-7=0\\2x+7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=7\\2x=-7\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{7}{2}\end{cases}}}\)
b.\(x^2+36=12x\)
\(\Leftrightarrow x^2-12x+36=0\)
\(\Leftrightarrow\left(x-6\right)^2=0\)
\(\Leftrightarrow x-6=0\Leftrightarrow x=6\)
c.\(\frac{1}{16x^2}-x+4=0\)
\(\Leftrightarrow\left(\frac{1}{4x}\right)^2-2.\frac{1}{4x}.2+2^2=0\)
\(\Leftrightarrow\left(\frac{1}{4x}-2\right)^2=0\)
........
Bài 1 : Tìm số nguyên x , biết :
a, x . ( x + 7 ) = 0
b, ( 2 + 2x ) . ( 7 - x ) = 0
c, x^3 - 4x = 0
d, ( x^2 - 9 ) . ( 3x + 15 ) = 0
e, ( 4x - 8 ) . ( x^2 + 1 ) = 0
a,x.(x+7)=0
suy ra x=o hoặc x+7=0
vs x+7=0
x=0+7
x=7
vậy x=0 hoặc x=7
b(2+2x)(7-x)=0
suy ra 2+2x=0 hoặc 7-x=0
vs2+2x=0 vs7-x=0
2x =0-2 x=0+7
2x =(-2) x=7
x=(-2);2
x=-1
vậy x=-1 hoặc x=7
d(x^2-9)(3x+15)=0
suy ra x^2-9=0 hoặc 3x+15=0
vsx^2-9=0 vs 3x+15=0
x^2 =0+9 3x =0-15
x^2 =9 3x =-15
x^2 =3^2 x=(-15):3
suy ra x=3 hoặc x=-3 x=-5
vậy x=3 x=-3 hoặc x=-5
e,(4x-8)(x^2+1)=0
suy ra4x-8=0 hoặc x^2+1=0
vs 4x-8=0 vs x^2+1=0
4x =0+8 x^2 =0-1
4x =8 x^2 =-1
x =8:4 x^2 =-1^2 hoặc 1^2
x =2 suy ra x=-1 hoặc x=1
vậy x=2, x=-1 hoặc x=1
tìm x biết
a/ x^3-x^2-x+1=0
b/(2x^3-3)^2-(4x^2-9)=0
c/x^4+2x^3-6x-9=0
d/2(x+5)-x^2-5x=0