Giải các phương trình sau
1) ( x - 2 ) . ( x2 - 3x + 5 ) = x3 - 2x2
2) ( 2x - 7 )2 = 6 . ( 2x - 7 ) . ( x - 3 )
3) x3 + 1 = ( x + 1 ) . ( x - 5 )
4) x3 - 3x2 + 6x + 4 = 0
Bài 5: Giải các phương trình sau:
a. (3x - 1)2 - (x + 3)2 = 0
b. x3 = \(\dfrac{x}{49}\)
c. x2 - 7x + 12 = 0
d. 4x2 - 3x -1 = 0
e. x3 - 2x - 4 = 0
f. x3 + 8x2 + 17x +10 = 0
g. x3 + 3x2 + 6x + 4 = 0
h. x3 - 11x2 + 30x = 0
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. \(7x-1=0\Leftrightarrow7x=1\Leftrightarrow x=\dfrac{1}{7}\)
*Cách khác:
a) Ta có: \(\left(3x-1\right)^2-\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(3x-1\right)^2=\left(x+3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=-x-3\\3x-1=x+3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-2\\2x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=2\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{1}{2};2\right\}\)
1. (x3 – 3x2 + x – 3) : (x – 3) 2. (2x4 – 5x2 + x3 – 3 – 3x) : (x2 – 3) 3. (x – y – z)5 : (x – y – z)3 4. (x2 + 2x + x2 – 4) : (x + 2) 5. (2x3 + 5x2 – 2x + 3) : (2x2 – x + 1) 6. (2x3 – 5x2 + 6x – 15) : (2x – 5)
1: \(=x^2+1\)
3: \(=\left(x-y-z\right)^2\)
Cứu với ạ
Làm tính chia
1) (x3 – 3x2 + x – 3) : (x – 3) 2) (2x4 – 5x2 + x3 – 3 – 3x) : (x2 – 3)
3) (x – y – z)5 : (x – y – z)3 4) (x2 + 2x + x2 – 4) : (x + 2)
5) (2x3 + 5x2 – 2x + 3) : (2x2 – x + 1) | 6) (2x3 – 5x2 + 6x – 15):(2x – 5) |
Bài 1 Rút gọn biểu thức
a, [(3x - 2)(x + 1) - (2x + 5)(x2 - 1)] : (x + 1)
b, (2x + 1)2 - 2(2x + 1)(3 - x) + (3 - x)2
c, (x - 1)2 - (x + 1) (x2 - x + 1) - (3x + 1)(1 - 3x)
d, (x2 + 1)(x - 3) - (x - 3)(x2 + 3x + 9)
e, (3x +2)2 + (3x - 2)2 - 2(3x + 2)(3x - 2) + x
Bài 2 Phân tích các đa thức sau thành nhân tử
1, 3(x + 4) - x2 - 4x
2, x2 - xy + x - y
3, 4x2 -25 + (2x + 7)(5 - 2x)
4, x2 + 4x - y2 + 4
5, x3 - x2 - x + 1
6, x3 + x2y - 4x - 4y
7, x3 - 3x2 + 1 - 3x
8, 2x2 + 3x - 5
9, x2 - 7xy + 10y2
10, x3 - 2x2 + x - xy2
1. Tìm x,y:
a) (x+2)2 + (x-3)2 = 2x ( x+ 7)
b) x3- 3x2 + 3x - 126 = 0
c) x2 + y2 - 2x + 4y + 5 = 0
d) 2x2 - 2xy + y2 + 4x + 4 = 0
\(a.\left(x^2+4x+4\right)+\left(x^2-6x+9\right)=2x^2+14x\)
\(x^2+4x+4+x^2-6x+9-2x^2-14x=0\)
\(-18x+13=0\)
\(x=\dfrac{13}{18}\)
Vậy \(S=\left\{\dfrac{13}{18}\right\}\)
\(b.\left(x-1\right)^3-125=0\)
\(\left(x-1\right)^3=125\)
\(x-1=5\)
\(x=6\)
Vậy \(S=\left\{6\right\}\)
\(c.\left(x-1\right)^2+\left(y +2\right)^2=0\)
\(Do\left(x-1\right)^2\ge0\forall x;\left(y+2\right)^2\ge0\forall y\)
\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\forall x,y\)
Mà \(\left(x-1\right)^2+\left(y+2\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
Vậy \(S=\left\{1;-2\right\}\)
\(d.x^2-4x+4+x^2-2xy+y^2=0\)
\(\left(x-2\right)^2+\left(x-y\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-2\right)^2=0\\\left(x-y\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-y=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
Vậy \(S=\left\{2;2\right\}\)
Tìm x, biết:
a) 2(5x-8)-3(4x-5) = 4(3x-4) + 11;
b) 2 x ( 6 x - 2 x 2 ) + 3 x 2 ( x - 4 ) = 8;
c) 2 ( x 3 - 1 ) - 2 x 2 ( x + 2 x 4 ) + ( 4 x 5 + 4 ) x = 6;
d)(2x)2(4x-2)-(x3 -8x2) = 15.
a) x = 2 7 b) x = 2.
c) x = 2 d) x = 1.
Bài 6: Giải các phương trình sau:
2) |
3) |
4) |
5) |
6) |
7) |
8)
9)
10)
11)
12)
13)
14) x2 – 2x + 1 = 0
15) 1 + 3x + 3x2 + x3 = 0
4) Ta có: \(\dfrac{2x-5}{5}-\dfrac{x+3}{3}=\dfrac{2-3x}{2}-x-2\)
\(\Leftrightarrow\dfrac{6\left(2x-5\right)}{30}-\dfrac{10\left(x+3\right)}{30}=\dfrac{15\left(2-3x\right)}{30}-\dfrac{30\left(x+2\right)}{30}\)
\(\Leftrightarrow12x-30-10x-30=30-45x-30x-60\)
\(\Leftrightarrow-22x-60=-75x-30\)
\(\Leftrightarrow-22x+75x=-30+60\)
\(\Leftrightarrow53x=30\)
\(\Leftrightarrow x=\dfrac{30}{53}\)
Vậy: \(S=\left\{\dfrac{30}{53}\right\}\)
5) Ta có: \(\dfrac{5x-3}{6}-\dfrac{7x-1}{4}=5\)
\(\Leftrightarrow\dfrac{2\left(5x-3\right)}{12}-\dfrac{3\left(7x-1\right)}{12}=\dfrac{60}{12}\)
\(\Leftrightarrow10x-6-21x+3=60\)
\(\Leftrightarrow-11x-3=60\)
\(\Leftrightarrow-11x=63\)
\(\Leftrightarrow x=-\dfrac{63}{11}\)
Vậy: \(S=\left\{-\dfrac{63}{11}\right\}\)
`9,x^3+x^2-2=0`
`x^3-x^2+2x^2-2=0`
`<=>x^2(x-1)+2(x-1)(x+1)=0`
`<=>(x-1)(x^2+2x+2)=0`
`<=>x=1`
`14,x^2-2x+1=0`
`<=>(x-1)^2=0`
`<=>x-1=0`
`<=>x=1`
`15,x^3+3x^2+3x+1=0`
`<=>(x+1)^3=0`
`<=>x+1=0`
`<=>x=-1`
Bài 6: Giải các phương trình sau:
1) |
2) |
3) |
4) |
5) |
6) |
7) |
8)
9)
10)
11)
12)
13)
14) x2 – 2x + 1 = 0
15) 1 + 3x + 3x2 + x3 = 0
Bài 6:
1) Ta có: \(2x\left(x-5\right)-\left(x+3\right)^2=3x-x\left(5-x\right)\)
\(\Leftrightarrow2x^2-10x-\left(x^2+6x+9\right)=3x-5x+x^2\)
\(\Leftrightarrow2x^2-10x-x^2-6x-9-3x+5x-x^2=0\)
\(\Leftrightarrow-14x-9=0\)
\(\Leftrightarrow-14x=9\)
\(\Leftrightarrow x=-\dfrac{9}{14}\)
Vậy: \(S=\left\{-\dfrac{9}{14}\right\}\)
`1)2x(x-5)-(x+3)^2=3x-x(5-x)`
`<=>2x^2-10x-x^2-6x-9=3x-5x+x^2`
`<=>x^2-16x-9=x^2-2x`
`<=>14x=-9`
`<=>x=-9/14`
Giải các phương trình sau:
a, x2 - 9x +20 = 0
b, x2 - 3x - 18 = 0
c, 2x2 - 9 x + 9 = 0
d, 3x2 - 8x + 4 = 0
e, 3x3 - 6x2 - 9x = 0
f, x(x - 5) - 2 + x = 0
g, x3 + 32 + 6x +8 = 0
h, 2x(x - 2) - 2 + x = 0
i, 5x(1 - x) + x - 1 = 0
k, 4 - 9(x - 1)2 = 0
l, (x - 2)2 - 36(x + 3)2 = 0
\(a)x^2-9x+20=0 \\<=>(x-4)(x-5)=0 \\<=>x=4\ hoặc\ x=5 \\b)x^2-3x-18=0 \\<=>(x+3)(x-6)=0 \\<=>x=-3\ hoặc\ x=6 \\c)2x^2-9x+9=0 \\<=>(x-3)(2x-3)=0 \\<=>x=3\ hoặc\ x=\dfrac{3}{2}\)
d: \(\Leftrightarrow3x^2-6x-2x+4=0\)
=>(x-2)(3x-2)=0
=>x=2 hoặc x=2/3
e: \(\Leftrightarrow3x\left(x^2-2x-3\right)=0\)
=>x(x-3)(x+1)=0
hay \(x\in\left\{0;3;-1\right\}\)
f: \(\Leftrightarrow x^2-5x-2+x=0\)
\(\Leftrightarrow x^2-4x-2=0\)
\(\Leftrightarrow\left(x-2\right)^2=6\)
hay \(x\in\left\{\sqrt{6}+2;-\sqrt{6}+2\right\}\)
Giải phương trình :
1) √x2+x+2 + 1/x= 13-7x/2
2) x2 + 3x = √1-x + 1/4
3) ( x+3)√48-x2-8x= 28-x/ x+3
4) √-x2-2x +48= 28-x/x+3
5) 3x2 + 2(x-1)√2x2-3x +1= 5x + 2
6) 4x2 +(8x - 4)√x -1 = 3x+2√2x2 +5x-3
7) x3/ √16-x2 + x2 -16 = 0