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\(a.\left(2x-3\right)\left(4x^2+6x+9\right)-\left(2x+3\right)\left(4x^2-6x+9\right)\\ =\left(2x\right)^3-3^3-\left[\left(2x\right)^3+3^3\right]\\ =8x^3-9-\left(8x^3+9\right)\\ =8x^3-9-8x^3-9=-18\)
\(b.\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\\ =x^3+1-\left(x^3-1\right)\\ =x^3+1-x^3+1=2\)
\(c.\left(3x-1\right)\left(3x+1\right)-\left(3x-2\right)^2\\ =9x^2-1-\left(9x^2-12x+4\right)\\ =9x^2-1-9x^2+12x-4\\ =12x-5\)
\(d.\left(2x-3\right)^2-\left(2x+3\right)\left(2x-3\right)\\ =\left(2x-3\right)\cdot\left[\left(2x-3\right)-\left(2x+3\right)\right]\\ =\left(2x-3\right)\cdot\left(2x-3-2x-3\right)\\ =\left(2x-3\right)\cdot\left(-6\right)\\ =-12x\cdot18\)
\(e.\left(3x-4\right)^2-\left(2x+4\right)^2\\ =9x^2-24x+16-\left(4x^2+16x+16\right)\\ =9x^2-24x+16-4x^2-16x-16\\ =5x^2-40x\)
\(f.\left(3x-5\right)^3-\left(3x+5\right)^3\\ =27x^3-135x^2+225x-125-\left(27x^3+135x^2+225x+125\right)\\ =27x^3-135x^2+225x-125-27x^3-135x^2-225x-125\\ =-270x^2-250\)
\(g.\left(2x-1\right)^2-\left(3x-1\right)^2\\ =4x^2-4x+1-\left(9x^2-6x+1\right)\\ =4x^2-4x+1-9x^2+6x-1\\ =-5x^2+2x\)
\(h.\left(x-2y\right)\left(x^2+2xy+4y^2\right)+\left(x^3-6y^3\right)\\ =x^3-8y^3+x^3-6y^3\\ =2x^3-14y^3\)
\(a.x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=3\)
\(\Leftrightarrow x\left(x^2-5^2\right)-\left(x^3+2^3\right)=3\)
\(\Leftrightarrow x^3-25x-x^3-8=3\)
\(\Leftrightarrow x^3-x^3-25x=8+3\)
\(\Leftrightarrow x=\frac{11}{-25}\)
Vậy x có nghiệm là \(\frac{-11}{25}.\)
\(\)
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
\(\left(x^2+3\right)\left(3-x^2\right)\)
\(\left(x^2+3\right)\left(-x^2+3\right)\)
\(\left(-x^2+3\right).x^2+3\left(-x^2+3\right)\)
\(-x^2.x^2+3x^2+3\left(-x^2+3\right)\)
\(-x^2.x^2+3x^2-3x^2+9\)
\(-x^2.x^2+9\)
1/ \(A=3\left(x+1\right)^2-\left(x+3\right)^2\)
\(=3\left(x^2+2x+1\right)-\left(x^2+6x+9\right)\)
\(=3x^2+6x+3-x^2-6x-9\)
\(=2x^2-6\)
Vậy biểu thức A vẫn phụ thuộc vào biến -_-
2/ \(B=\left(x-2\right)^2-\left(x-4\right)x\)
\(=x^2-4x+4-x^2-4x\)
\(=4\)
Vậy biểu thức B không phụ thuộc vào biến (đpcm)
3/ \(C=3\left(x+2\right)^2-3\left(x^2-4x\right)\)
\(=3\left(x^2+4x+4\right)-3x^2+12x\)
\(=3x^2+12x+12-3x^2+12x\)
\(=24x+12\)
Vậy biểu thức C vẫn phụ thuộc vào biến -_-
4/ \(D=3x\left(x-2\right)\left(x+2\right)-x\left(3x+3\right)\)
\(=3x\left(x^2-4\right)-3x^2-3x\)
\(=3x^3-12x-3x^2-3x\)
\(=3x^3-3x^2-15x\)
Vậy biểu thức D vẫn phụ thuộc vào biến -_-
5/ \(E=x^2-\left(x+1\right)\left(x-1\right)+5\)
\(=x^2-\left(x^2-1\right)+5\)
\(=x^2-x^2+1+5\)
\(=6\)
Vậy biểu thức E không phụ thuộc vào biến.
Bài 1
a) (x5 + 4x3 - 6x2) : 4x2
= 4x2(\(\dfrac{1}{4}\)x3 + x - \(\dfrac{3}{2}\)) : 4x2
= \(\dfrac{1}{4}\)x3 + x - \(\dfrac{3}{2}\)
b) (x3 - 8) : (x2 + 2x + 4)
= (x - 2)(x2 + 2x + 4) : (x2 + 2x + 4)
= x - 2
c) (3x2 - 6x) : (2 - x)
= -(6x - 3x2) : (2 - x)
= -3x(2 - x) : (2 - x)
= -3x
d) (x3 + 2x2 - 2x - 1) : (x2 + 3x + 1)
= [(x3 - 1) + (2x2 - 2x)] : (x2 + 3x + 1)
= [(x - 1)(x2 + x + 1) + 2x(x - 1)] : (x2 + 3x + 1)
= (x - 1)(x2 + x + 1 + 2x) : (x2 + 3x + 1)
= (x - 1)(x2 + 3x + 1) : (x2 + 3x + 1)
= x - 1
Bài 2
a) (x - 4)2 - (x - 2)(x + 2) = 6
x2 - 8x + 16 - (x2 - 4) = 6
x2 - 8x + 16 - x2 + 4 = 6
-8x + 20 = 6
\(\Rightarrow\) -8x = - 14
\(\Rightarrow\) x = \(\dfrac{7}{4}\)
b) 9(x + 1)2 - (3x - 2)(3x + 2) = 10
9(x2 + 2x + 1) - (9x2 - 4) = 10
9x2 + 18x + 9 - 9x2 + 4 = 10
18x + 13 = 10
\(\Rightarrow\) 18x = -3
\(\Rightarrow\) x = \(\dfrac{-1}{6}\)
Nhớ tik mik nha
không lần sau mik ko giúp đâu
AK... có j ko hiểu thì bn cứ bình luận bên dưới
Tick Là Nhấn Vào Nút Phải ko bạn!
nhanh nhanh tik cho mik đi
PLEASE
Rồi bn nhé
uk. Thanks
