c ) (3x)3 = (2x + 1 )3
Bài 1: giải các pt sau:
1,(x-1)^2-(x+1)^2=2(x+3)
2,(2x-1)^2-(2x+1)^2=4(x-3)
3,(2x+3)^2-(2x+3).(2x-4)=-(x-2)^2
4,8x^3-(x+1)^3=3x-3
5,(3x-2).(9x^2+6x+4)-(3x+1).(9x^2-3x+1)=(2x+1).(2x-1)-4x(x-3)
\(\left(x-1\right)^2-\left(x+1\right)^2=2\left(x+3\right)\)
\(\Leftrightarrow\left(x-1+x+1\right)\left(x-1-x-1\right)=2\left(x+3\right)\)
\(\Leftrightarrow2x\left(-2\right)=2\left(x+3\right)\)
\(\Leftrightarrow-4x=2x+6\)
\(\Leftrightarrow-6x=6\)
\(\Leftrightarrow x=-1\)
2) \(\left(2x-1\right)^2-\left(2x+1\right)^2=4\left(x-3\right)\)
\(\Leftrightarrow\left(2x-1+2x+1\right)\left(2x-1-2x-1\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow4x\left(-2\right)-4x+12=0\)
\(\Leftrightarrow-12x=-12\)
\(\Leftrightarrow x=1\)
3)\(\left(2x+3\right)^2-\left(2x+3\right)\left(2x-4\right)+\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(2x+3\right)\left(2x+3-2x+4\right)+\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow7\left(2x+3\right)+x^2-4x+4=0\)
\(\Leftrightarrow x^2+10x+25=0\)
\(\Leftrightarrow\left(x+5\right)^2=0\)
\(\Leftrightarrow x=-5\)
4) \(8x^3-\left(x+1\right)^3=3x-3\)
\(\Leftrightarrow8x^3-\left(x^3+3x+3x^2+1\right)-3x+3=0\)
\(\Leftrightarrow7x^3-3x^2-6x+2=0\)
\(\Leftrightarrow\left(x-1\right)\left(7x^2+4x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{-2+3\sqrt{2}}{7}\\x=\frac{-2-3\sqrt{2}}{7}\end{matrix}\right.\)
5)\(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)
\(\Leftrightarrow\left(3x\right)^3-2^3-\left(\left(3x\right)^3-1^3\right)=x-4\)
\(\Leftrightarrow27x^3-8-\left(27x^3-1\right)=x-4\)
\(\Leftrightarrow-7=x-4\)
\(\Leftrightarrow x=-3\)
Bài 1: Thực hiện phép tính:
a) 2x.(3x + 3) b) 5x.(3x2-2x + 1) c) 3x2(2x +4)
d) 5x2.(3x2 + 4x – 1) e) (x-1).(2x +3) f) (x+2).(3x-5)
Bài 2: Tìm x, biết:
a) 3x(x+1) – 3x2 = 6
b) 3x(2x+1) – (3x +1).(2x-3) = 10
Bài 1:
\(a,=6x^2+6x\\ b,=15x^3-10x^2+5x\\ c,=6x^3+12x^2\\ d,=15x^4+20x^3-5x^2\\ e,=2x^2+3x-2x-3=2x^2+x-3\\ f,=3x^2-5x+6x-10=3x^2+x-10\)
Bài 2:
\(a,\Leftrightarrow3x^2+3x-3x^2=6\\ \Leftrightarrow3x=6\Leftrightarrow x=2\\ b,\Leftrightarrow6x^2+3x-6x^2+9x-2x-3=10\\ \Leftrightarrow10x=13\Leftrightarrow x=\dfrac{13}{10}\)
Tìm nghiệm : a) (2x-3).(2x+3) B)(x-4).(x-1).(x-2) C)2x(3x-1)-3x(5+2x) D)(3x-2).(3x+2)-4.(x-1)
a) \(\left(2x-3\right)\left(2x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=3\\2x=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
b) \(\left(x-4\right)\left(x-1\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=1\\x=2\end{matrix}\right.\)
c) \(2x\left(3x-1\right)-3x\left(5+2x\right)=0\)
\(\Rightarrow x\left[2\left(3x-1\right)-3\left(5+2x\right)\right]=0\)
\(\Rightarrow x\left(6x-2-15-6x\right)\)
\(\Rightarrow-16x=0\)
\(\Rightarrow x=0\)
d) \(\left(3x-2\right)\left(3x+2\right)-4\left(x-1\right)=0\)
\(\Rightarrow9x^2-4-4x+4=0\)
\(\Rightarrow9x^2-4x=0\)
\(\Rightarrow x\left(9x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\9x-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{9}\end{matrix}\right.\)
\(a,\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ b,\left(x-4\right)\left(x-1\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\\x=2\end{matrix}\right.\)
\(c,2x\left(3x-1\right)-3x\left(5+2x\right)=0\\ \Leftrightarrow6x^2-2x-15x-6x^2=0\\ \Leftrightarrow-17x=0\\ \Leftrightarrow x=0\\ d,\left(3x-2\right)\left(3x+2\right)-4\left(x-1\right)=0\\ \Leftrightarrow9x^2-4-4x+4=0\\ \Leftrightarrow9x^2-4x=0\\ \Leftrightarrow x\left(9x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\9x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{9}\end{matrix}\right.\)
Tìm x biết:
a; 3x.(2x+3)-(2x+5x).(3x-2)=8
b;4.(x-1)-3.(x^2-5)_x^2=(x-3)-(x+4)
c; 2.(3x-1).(2x+5)-6.(2x-1).(x+2)=-6
d; 3.(2x-1).(3x-1)-(2x-3.(9x-1)-3=3
a, 3-3/2.(2x-1)-1/2.(3x+2)=2x-3
b, 7/2+1/2.(4x-3/2)-1/3.(3-6x)=1/2.(x-1)
c, 3/2.(2x-1/2)-7/2.(3x+1/3)-1/2.(2x-1)=3x+5
Bài I. Rút gọn các biểu thức sau:
a) 3x(2x+1)+ (2x - 3)(x+1),
b) x(3x - 2)2 + 3(x-2)(x+2)
c) (2x+1)(4x² - 2x+1)-2x(2x+3)(2x - 3)-(x-3)²
a: Ta có: \(3x\left(2x+1\right)+\left(2x-3\right)\left(x+1\right)\)
\(=6x^2+3x+2x^2+2x-3x-3\)
\(=8x^2+2x-3\)
Giải các phương trình
A) 3x - 2 = 2x - 3 B) -3x + 1/3 = 0 c ) 1/2x + 2 = 0 d) -1/3x +3=0
a) 3x - 2 = 2x - 3
⇔ 3x - 2x = -3 + 2
⇔ x = -1
Vậy x = -1 là nghiệm của PT
b ) -3x + \(\dfrac{1}{3}=0\)
⇔\(\dfrac{-3x\cdot3}{3}+\dfrac{1}{3}=0\)
⇔-9x +1 = 0
⇔-9x = -1
⇔ x = \(\dfrac{1}{9}\)
Vậy x = \(\dfrac{1}{9}\)là nghiệm của PT
c) \(\dfrac{1}{2x}+2=0\)
⇔\(\dfrac{1}{2x}+\dfrac{2\cdot2x}{2x}=0\)
⇔1 + 4x = 0
⇔ 4x = -1
⇔ x = \(\dfrac{-1}{4}\)
Vậy x =\(\dfrac{-1}{4}\) là nghiệm của PT
d)\(\dfrac{-1}{3x}+3=0\)
⇔\(\dfrac{-1}{3x}+\dfrac{3\cdot3x}{3x}=0\)
⇔ -1 + 9x = 0
⇔ 9x = 1
⇔ x = \(\dfrac{1}{9}\)
Vậy x = \(\dfrac{1}{9}\)là nghiệm của PT
1) Thực hiện phép tính :
a) -(5x - 4)(2x+3)
b) ( x - y)( x + xy+ y)
c) 7x( x - 4) - ( 7x +3)(2x - x+4)
2) Chứng minh rằng giá trị của biểu thức không phụ thuộc vào giá trị của biến x:
a) x(3x +12) - ( 7x - 20) + x(2x - 3) - x( 2x +5)
b) 3( 2x-1) - 5( x-3) + 6( 3x - 4) - 19x
3) tìm x:
a) 3x( x - 2) - x( 1+3x) = 14
b) (2x - 1)( x + 5) - (2x +1)( x + 4,5)=3,5
c) 3x - 3x( x - 3) = 36
d) (3x + 1)(x - 1) + x( 4 - 3x )= 5
Bài 3:
a: =>3x^2-6x-x-3x^2=14
=>-7x=14
=>x=-2
b: \(\Leftrightarrow2x^2+10x-x-5-2x^2-9x-x-4.5=3.5\)
=>-x-9,5=3,5
=>-x=12
=>x=-12
c: =>\(3x-3x^2+9x=36\)
=>-3x^2+12x-36=0
=>x^2-6x+12=0(loại)
d: \(\Leftrightarrow3x^2-3x+x-1+4x-3x^2=5\)
=>2x=6
=>x=3
Mọi người giúp tới gấp nhé:
1. Tìm x, biết:
a/ 3(2x - 3) + 2(2 - x) = -3
b/ 2x(x2 - 2) + x2(1 - 2x) - x2 = -12
2. Tìm x, biết:
a/ 3x(2x + 3) - (2x + 5)(3x - 2) = 8
b/ 4x(x - 1) - 3(x2 - 5) - x2 = (x - 3) - (x + 4)
c/ 2(3x - 1)(2x + 5) - 6(2x - 1)(x + 2) = -6
d/ 3(2x - 1)(3x - 1) - (2x - 3)(9x -1) - 3 = -3
e/ (3x - 1)(2x + 7) - (x + 1)(6x - 5) = (x + 2) - (x - 5)
f/ 3xy(x + y) - (x + y)(x2 + y2 + 2xy) + y3 = 27
3. Chứng minh rằng giá trị của các biểu thức sau không phụ thuộc vào x:
a/ A = 2x(x - 1) - x(2x + 1) - (3 - 3x)
b/ B = 2x(x - 3) - (2x - 2)(x - 2)
c/ C = (3x - 5)(2x + 11) - (2x + 3)(3x + 7)
d/ D = (2x + 11)(3x - 5) - (2x + 3)(3x + 7)
f/ \(3xy\left(x+y\right)-\left(x+y\right)\left(x^2+y^2+2xy\right)+y^3=27\)
\(3x^2y+3xy^2-\left(x+y\right)\left(x+y\right)^2+y^3=27\)
\(3x^2y+3xy^3-\left(x+y\right)^3+y^3=27\)
\(3x^2y+3xy^3-\left(x^3+3x^2y+3xy^2+b^3\right)+y^3=27\)
\(-x^3=27\)
\(x=-3\)
Bài 1:
a/ \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(6x-9+4-2x=-3\)
\(4x=-2\)
\(x=-\frac{1}{2}\)
b/ \(2x\left(x^2-2\right)+x^2\left(1-2x\right)-x^2=-12\)
\(2x^3-4x+x^2-2x^3-x^2=-12\)
\(-4x=-12\)
\(x=\frac{1}{3}\)
Bài 2:
a/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)
\(6x^2+9x-6x^2-15x+4x+10=8\)
\(-2x=8\)
\(x=-4\)
b/ \(4x\left(x-1\right)-3\left(x^2-5\right)-x^2=\left(x-3\right)-\left(x+4\right)\)
\(4x^2-4x-3x^2+15-x^2=-7\)
\(-4x=-22\)
\(x=\frac{11}{2}\)
c/ \(2\left(3x-1\right)\left(2x+5\right)-6\left(2x-1\right)\left(x+2\right)=-6\)
\(6x-2\left(2x+5\right)-12x+6\left(x+2\right)=-6\)
\(6x-4x-10-12x+6x+12=-6\)
\(-4x=-8\)
\(x=2\)
Bài 1: Rút gọn các biểu thức sau:
a, A = (x-2).(2x-1) - 2x (x+3)
b, B = (3x-2).(2x+1) - (6x-1).(x+2)
c, C = 6x.(2x+3) - (4x-1).(3x-2)
d, D = (2x+3).(5x-2)+(x+4).(2x-1) - 6x.(2x-3)
Bài 2: Chứng tỏ rằng các đa thức không phụ thuộc vào biến.
a, 2x(3x-5).(x+11) - 3x.(2x+3).(x+7)
b, (x2+5x-6).(x-1) - (x+2).(x2-x+1) - x(3x-10)
c, (x2+x+1).(x-1) - x2(x+1) + x2 - 5
Bài 1
A= (x-2)(2x-1)-2x(x+3)=2x2-x-4x+2-2x2-6x=-11x+2
Bài 1:
a) \(A=\left(x-2\right)\left(2x-1\right)-2x\left(x+3\right)\)
\(A=2x^2-x-4x+2-2x^2-6x\)
\(A=-11x+2\)
b) \(B=\left(3x-2\right)\left(2x+1\right)-\left(6x-1\right)\left(x+2\right)\)
\(B=6x^2+3x-4x-2-6x^2-12x+x+2\)
\(B=-12x\)
c) \(C=6x\left(2x+3\right)-\left(4x-1\right)\left(3x-2\right)\)
\(C=12x^2+18x-12x^2+8x+3x-2\)
\(C=29x-2\)
d) \(D=\left(2x+3\right)\left(5x-2\right)+\left(x+4\right)\left(2x-1\right)-6x\left(2x-3\right)\)
\(D=10x^2-4x+15x-6+2x^2-x+8x-4-12x^2+18x\)
\(D=36x-10\)
Bài 2:
a: Ta có: \(2x\left(3x-5\right)\left(x+11\right)-3x\left(2x+3\right)\left(x+7\right)\)
\(=2x\left(3x^2+33x-5x-55\right)-3x\left(2x^2+14x+3x+21\right)\)
\(=6x^3+56x^2-110x-6x^2-51x^2-63x\)
\(=-117x\)
b: Ta có: \(\left(x^2+5x-6\right)\left(x-1\right)-\left(x+2\right)\left(x^2-x+1\right)-x\left(3x-10\right)\)
\(=x^3+4x^2-11x+6-\left(x^3-x^2+x+2x^2-2x+2\right)-3x^2+10x\)
\(=x^3+x^2-x+6-x^3-x^2+x-2\)
=4
c: Ta có: \(\left(x^2+x+1\right)\left(x-1\right)-x^2\left(x+1\right)+x^2-5\)
\(=x^3-1-x^3-x^2+x^2-5\)
=-6