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A = ( 3x )3 + 23 - 27x3 + 6 = 27x3 + 8 - 27x3 + 6 = 14 ( đpcm )
B = x3 + 3x2 + 3x + 1 - ( x3 - 1 ) - 3x2 - 3x = x3 + 1 - x3 + 1 = 2 ( đpcm )
C = 6( x + 2 )( x2 - 2x )( x2 - 2x + 4 ) - 6x3 - 2 ( bạn xem lại đề bài nhé ._. )
D = 2[ ( 3x )3 + 13 ] - 54x3 = 2( 27x3 + 1 ) - 54x3 = 54x3 + 2 - 54x3 = 2 ( đpcm )
1) a) \(\left(3x-1\right)\left(9x^2+3x+1\right)-4x\left(x-5\right)\)
\(=27x^3+9x^2+3x-9x^2-3x-1-4x^2+20x\)
\(=27x^3+\left(9x^2-9x^2-4x^2\right)+\left(3x-3x+20x\right)+\left(-1\right)\)
\(=27x^3-4x^2+20x-1\)
b)\(\left(7x+2\right)\left(3-4x\right)-\left(x+3\right)\left(x^2-3x+9\right)\)
\(=21x-28x^2+6-8x-x^3+3x^2-9x-3x^2+9x-27\)
\(=\left(21x-8x-9x+9x\right)+\left(-28x^2+3x^2-3x^2\right)\)\(+\left(6-27\right)\)\(+\left(-x^3\right)\)
\(=13x-28x^2-21-x^3\)
c)\(\left(4x+3\right)\left(4x-3\right)-\left(2-x\right)\left(4+2x+x^2\right)\)
\(=16x^2-12x+12x-9-8-4x-2x^2+4x+2x^2+x^3\)
\(=\left(16x^2-2x^2+2x^2\right)+\left(-12x+12x-4x+4x\right)\)\(+\left(-9-8\right)\)\(+x^3\)
\(=16x^2-17+x^3\)
d)\(\left(3x-8\right)\left(-5x+6\right)-\left(4x+1\right)\left(3x-2\right)\)
\(=-15x^2+18x+40x-48-12x^2+8x-3x+2\)
\(=\left(-15x^2-12x^2\right)+\left(18x+40x+8x-3x\right)\)\(+\left(-48+2\right)\)
\(=-27x^2+63x-46\)
e)\(\left(3x-6\right)4x-2x\left(3x+5\right)-4x^2\)
\(=12x^2-24x-6x^2-10x-4x^2\)
\(=\left(12x^2-6x^2-4x^2\right)+\left(-24x-10x\right)\)
\(=2x^2-34x\)
f)\(\left(5x-6\right)\left(6x-5\right)-x\left(3x+10\right)\)
\(=30x^2-25x-36x+30-3x^2-10x\)
\(=\left(30x^2-3x^2\right)+\left(-25x-36x-10x\right)+30\)
\(=27x^2-71x+30\)
2) a)\(x\left(x+3\right)-x^2=6\)
\(\Rightarrow x^2+3x-x^2=6\)
\(\Rightarrow\left(x^2-x^2\right)+3x=6\)
\(\Rightarrow3x=6\)
\(\Rightarrow x=2\)
Vậy x=2
b) \(2x\left(x-5\right)+x\left(-2x-1\right)=6\)
\(\Rightarrow2x^2-10x-2x^2-x=6\)
\(\Rightarrow\left(2x^2-2x^2\right)+\left(-10x-x\right)=6\)
\(\Rightarrow-11x=6\)
\(\Rightarrow x=-\dfrac{6}{11}\)
\(\)Vậy \(x=-\dfrac{6}{11}\)
c) x(x+5)-(x+1)(x-2)=7
\(\Rightarrow x^2+5x-x^2+2x-x+2=7\)
\(\Rightarrow\left(x^2-x^2\right)+\left(5x+2x-x\right)=7-2\)
\(\Rightarrow6x=5\)
\(\Rightarrow x=\dfrac{5}{6}\)
Vậy x=\(\dfrac{5}{6}\)
d)\(\left(3x+4\right)\left(6x-3\right)-\left(2x+1\right)\left(9x-2\right)=10\)
\(\Rightarrow18x^2-9x+24x-12-18x^2+4x-9x+2=10\)
\(\Rightarrow\left(18x^2-18x^2\right)+\left(-9x+24x+4x-9x\right)+\left(-12+2\right)=10\)
\(\Rightarrow10x-10=10\)
\(\Rightarrow10x=20\)
\(\Rightarrow x=2\)
Vậy x=2
a) \(\left(x+y\right)^2\) + \(\left(x-y\right)^2\) - \(2x^2\)
= \(x^2\) +2xy + \(y^2\) + \(x^2\) -2xy + \(y^2\) - \(2x^2\)
= 2\(y^2\)
Vậy giá trị của biểu thức ko phụ thuộc vào biến x
b) \(\left(x^2+4\right)\) (x+2) (x-2) - \(\left(x^2+3\right)\left(x^2-3\right)\)
= \(\left(x^2+4\right)\) \(\left(x^2-4\right)\) - \(\left(x^4-3x^2+3x^2-9\right)\)
= \(x^4-4x^2+4x^2-16-x^4+3x^2-3x^2+9\)
= -7
Vậy giá trị của biểu thức k phụ thuộc vào biến x
c: \(=27x^3-8-27x^3+6=-2\)
b: \(=\left(3x+5\right)^2-2\left(3x+5\right)\left(3x-2\right)+\left(3x-2\right)^2\)
\(=\left(3x+5-3x+2\right)^2=7^2=49\)
a, (x+y)2+(x-y)2-2x2=x2+2xy+y2+x2-2xy+y2-2x2=2y2
b,(x2+4)(x+2)(x-2)-(x2-3)(x2+3)=(x2+4)(x2-4)-(x4+3x2-3x2-9)=x4-4x2+4x2-16-x4-3x2+3x2+9=-7
____Tương tự nhân hết ra kq cuối cùng sẽ k còn biến x->K phụ thuộc----
a) \(4x^2-12x+9-4\left(x^2-4\right)-x=8\)
\(4x^2-12x+9-4x^2+16-x=8\)
\(-13x+25=8\)
\(-13x=-17\)
\(x=\dfrac{17}{13}\)
\(\left(2x+3\right)^2+2\left(2x+3\right)\left(x-2\right)+\left(2-x\right)^2=4\)
\(\left(2x+3\right)^2+2\left(2x+3\right)\left(x-2\right)+\left(x-2\right)^2=4\)
\(\left(2x+3+x-2\right)^2=\left(\pm2\right)^2\)
\(\left(3x+1\right)^2=\left(\pm2\right)^2\)
\(\left[\begin{array}{nghiempt}3x+1=2\\3x+1=-2\end{array}\right.\)
\(\left[\begin{array}{nghiempt}3x=2-1\\3x=-2-1\end{array}\right.\)
\(\left[\begin{array}{nghiempt}3x=1\\3x=-3\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=\frac{1}{3}\\x=-1\end{array}\right.\)
***
\(\left(x+3\right)\left(3-x\right)=5\)
\(3^2-x^2=5\)
\(x^2=9-5\)
\(x^2=4\)
\(x^2=\left(\pm2\right)^2\)
\(x=\pm2\)
***
\(\left(3x+1\right)\left(9x^2-3x+1\right)=2\)
\(27x^3+3=2\)
\(27x^3=2-3\)
\(\left(3x\right)^3=-1\)
\(3x=-1\)
\(x=-\frac{1}{3}\)
a: \(=\left(x^2+4\right)\left(x^2-4\right)-\left(x^4-9\right)\)
\(=x^4-16-x^4+9=-7\)
b: \(=27x^3-8-27x^3+6=-2\)
c: \(=\left(3x+5+2-3x\right)^2=7^2=49\)