Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) 2x(3x+1) – (2x+3)(3x-2) = 12
\(\Leftrightarrow6x^2+2x-\left(6x^2-4x+9x-6\right)=12\)
\(\Leftrightarrow6x^2+2x-6x^2+4x-9x+6=12\)
\(\Leftrightarrow-3x+6=12\)
\(\Leftrightarrow-3x=6\)
\(\Leftrightarrow x=-2\)
vậy x = -2
b) (x+2)2 – (x-3)(x+3) = 5
\(\Leftrightarrow\left(x+2\right)^2-\left(x^2-9\right)=5\)
\(\Leftrightarrow x^2+4x+4-x^2+9-5=0\)
\(\Leftrightarrow4x+8=0\)
\(\Leftrightarrow4x=-8\)
\(\Leftrightarrow x=-2\)
Vậy x = -2
a) (x-2)3+6(x+1)2-x3+12=0
\(\Rightarrow\)x3-6x2+12x-8+6(x2+2x+1)-x3+12=0
\(\Rightarrow\)x3-6x2+12x-8+6x2+12x+6-x3+12=0
\(\Rightarrow\)24x+10=0
\(\Rightarrow\)24x=-10
\(\Rightarrow\)x=\(\dfrac{-10}{24}=\dfrac{-5}{12}\)
b)(x-5)(x+5)-(x+3)2+3(x-2)2=(x+1)2-(x-4)(x+4)+3x2
\(\Rightarrow\)x2-25-(x2+6x+9)+3(x2-4x+4)=x2+2x+1-(x2-16)+3x2
\(\Rightarrow\)x2-25-x2-6x-9+3x2-12x+12=x2+2x+1-x2+16+3x2
\(\Rightarrow\)3x2-18x-22=3x2+2x+17
\(\Rightarrow\)3x2-18x-22-3x2-2x-17=0
\(\Rightarrow\)-20x-39=0
\(\Rightarrow\)-20x=39
\(\Rightarrow\)x=\(-\dfrac{39}{20}\)
c: Ta có: \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)
\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\)
\(\Leftrightarrow3x^2+26x=0\)
\(\Leftrightarrow x\left(3x+26\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\)
\(a,\Leftrightarrow x^2+8x+16-x^3-12x^2=16\\ \Leftrightarrow x^3+11x^2-8x=0\\ \Leftrightarrow x\left(x^2+11x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+11x-8=0\left(1\right)\end{matrix}\right.\\ \Delta\left(1\right)=121+32=153\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11-3\sqrt{17}}{2}\\x=\dfrac{-11+3\sqrt{17}}{2}\end{matrix}\right.\\ S=\left\{0;\dfrac{-11-3\sqrt{17}}{2};\dfrac{-11+3\sqrt{17}}{2}\right\}\)
\(c,\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\\ \Leftrightarrow3x^2+26x=0\\ \Leftrightarrow x\left(3x+26\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\\ d,\Leftrightarrow x^3-6x^2+12x-8-x^3-125-6x^2=11\\ \Leftrightarrow-12x^2+12x-144=0\\ \Leftrightarrow x^2-x+12=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\end{matrix}\right.\)
a)(x+2).(x+3)-(x-2).(x+5)=10
( x^2 +3x+2x+6)-(x^2 +5x-2x-10)=10
x^2 +3x+2x+6-x^2 -5x+2x+10-10=0
2x+6=0
2x=-6
x=-3
a/ \(x=\dfrac{-5}{12}\)
b/ \(x\approx-1,9526\)
c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)
d/ \(x=\dfrac{-20}{13}\)
Bài 2:
a: Ta có: \(A=\left(x+1\right)^3+\left(x-1\right)^3\)
\(=x^3+3x^2+3x+1+x^3-3x^2+3x-1\)
\(=2x^3+6x\)
b: Ta có: \(B=\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)
\(=x^3-9x^2+27x-27-x^3-27+9x^2-1\)
\(=27x-55\)
\(a,5\left(3x+5\right)-4\left(2x-3\right)=5x+8\left(2x+12\right)+1\)
\(\Rightarrow5\left(3x+5\right)-4\left(2x-3\right)-5x-8\left(2x+12\right)-1=0\)
\(\Rightarrow15x+25-8x+12-5x-16x-96-1=0\)
\(\Rightarrow-14x-60=0\)
\(\Rightarrow-14x=60\) \(\Rightarrow x=-\frac{60}{14}=\frac{-30}{7}\)
\(b,\left(2x+3\right)\left(x-4\right)-\left(3x-5\right)\left(x-4\right)=\left(5-x\right)\left(x-2\right)\)
\(\Rightarrow2x^2+3x-8x-12-3x^2+5x+12x-20=5x-x^2-10+2x\)
\(\Rightarrow-x^2+12x-32=7x-x^2-10\)
\(\Rightarrow-x^2+12x-32-7x+x^2+10=0\)
\(\Rightarrow5x-22=0\)
\(\Rightarrow5x=22\Rightarrow x=\frac{22}{5}\)
a) 5(3x+5)-4(2x-3) = 5x+8(2x+12)+1
15x + 25 - 8x + 12 = 5x + 16x + 96 + 1
15x - 8x - 5x - 16x = 96 + 1 - 25 - 12
-14x = 60
x = \(\frac{60}{-14}\)
x = \(-\frac{30}{7}\)
b) (2x+3)(x-4)-(3x-5)(x-4) = (5-x).(x-2)
(x - 4)(2x + 3 - 3x +5) = 5x - 10 - x2 + 2x
(x - 4)[(2x - 3x) + (3 + 5)] = 5x - 10 - x2 + 2x
(x - 4)(-x + 8) = 5x - 10 - x2 + 2x
-x2 + 8x + 4x - 32 = 5x - 10 - x2 + 2x
(-x2 + x2) + (8x + 4x - 5x - 2x) = -10 + 32
5x = 22
x = \(\frac{22}{5}\)
a/3x(12x - 4 ) -9x (4x -3 ) = 30
<=> 36x^2 - 12x - 36x²+27x = 30
<=> 15x = 30
<=> x=2
b/ => 5x -2. x^2 + 2.x^2 -2x = 15
=> 5x -2x = 15
=> 3x = 15 => x= 5
a/3x(12x - 4 ) -9x (4x -3 ) = 30
36x^2 - 12x - 36x²+27x = 30
15x = 30
x=2
b/ 5x -2. x^2 + 2.x^2 -2x = 15
5x -2x = 15
3x = 15 => x= 5
\(\left(x-3\right)\left(x+x^2\right)+\left(x-5\right)\left(x+1\right)-x^3=12\)
\(\Leftrightarrow x^2+x^3-3x-3x^2+x^2+x-5x-5-x^3=12\)
\(\Leftrightarrow-x^2-7x-5=12\Leftrightarrow-x^2-7x-17=0\)
Ta có : \(\left(-7\right)^2-4\left(-17\right)\left(-1\right)< 0\)Vậy phương trình vô nghiệm