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A=5; B=3; C=24 không phụ thuộc x; câu D thì mong bạn xem lại đề
\(A=\left(x^3+x^2+x\right)-\left(x^3+x^2\right)-x+5\)5
\(A=x^3+x^2+x-x^3-x^2-x+5\)
=> A=5
=> A luôn = 5 với mọi x => A không phụ thuộc vào x
\(B=x\left(2x+1\right)-x^2\left(x+2\right)+x^3-x+3\)
\(B=\left(2x^2+x\right)-\left(x^3+2x^2\right)+x^3-x+3\)
\(B=2x^2+x-x^3-2x^2+x^3-x+3\)
=> B= 3
=> B luôn =3 với mọi x => B không phụ thuộc vào x
\(C=4\left(6-x\right)+x^2\left(2+3x\right)-x\left(5x-4\right)+3x^2\left(1-x\right)\)
\(C=24-4x+2x^2+3x^3-5x^2+4x+3x^2-3x^3\)
C=24
=> C=24 với mọi x => C không phụ thuộc vào x
Câu D kí tự cuối có vẻ bạn gõ sai nên mình không làm được, sorry nhiều
A = x(x2 + x + 1) - x2(x + 1) - x + 5
A = x.x2 + x.x + x.1 + (-x2).x + (-x2).1 - x + 5
A = x3 + x2 + x - x3 - x2 - x + 5
A = (x3 - x3) + (x2 - x2) + (x - x) + 5
A = 0 + 0 + 0 + 5
A = 5
Vậy: Biểu thức không phụ thuộc giá trị của biến.
B = x(2x + 1) - x2(x + 2) + x3 - x + 3
B = x.2x + x.1 + (-x2).x + (-x2).2 + x3 - x + 3
B = 2x2 + x - x3 - 2x2 + x3 - x + 3
B = (2x2 - 2x2) + (x - x) + (-x3 + x3) + 3
B = 0 + 0 + 0 + 3
B = 3
Vậy: Biểu thức không phụ thuộc giá trị của biến.
C = 4(6 - x) + x2(2 + 3x) - x(5x - 4) + 3x2(1 - x)
C = 4.6 + 4.(-x) + x2.2 + x2.3x + (-x).5x + (-x).(-4) + 3x2.1 + 3x2.(-x)
C = 24 - 4x + 2x2 + 3x3 - 5x2 + 4x + 3x2 - 3x3
C = 24 + (-4x + 4x) + (2x2 - 5x2 + 3x2) + (3x3 - 3x3)
C = 24 + 0 + 0 + 0
C = 24
Vậy: Biểu thức không phụ thuộc giá trị của biến.
D viết sai thì chịu
a: \(=12x^{n+2}+4x^2-8x^{n+2}\)
\(=4x^{n+2}+4x^2\)
b: \(=2x^{2n}+4x^ny^n+2y^{2n}-4x^ny^n-2y^{2n}\)
\(=2x^{2n}\)
c: \(=\left(x^{3n}-y^{3n}\right)\left(x^{3n}+y^{3n}\right)\)
\(=x^{6n}-y^{6n}\)
d: \(=4^n\cdot4-3\cdot4^n=4^n\)
a)6xn(x2-1)+2x(3xn-1+1)
=6xn(x2-1)+6xn+2x
=xn(6x2-6)+6xn+2x
=xn(6x2+6x-6)
b)3n+1-2*3n
=3n(3-1)
=3n
c)310*210-67(63-1)
=(3*2)10-67*63-67*(-1)
=610-610-(-67)
=67=279936
\(1,\left(x-3\right)\left(x-1\right)-3\left(x-3\right)\)
\(=\left(x-3\right)\left(x-1-3\right)\)
\(=\left(x-3\right)\left(x-4\right)\)
\(2,6x+3-\left(2x-5\right)\left(2x+1\right)\)
\(=3\left(2x+1\right)-\left(2x-5\right)\left(2x+1\right)\)
\(=\left(2x+1\right)\left(3-2x+5\right)\)
\(=\left(2x+1\right)\left(-2-2x\right)\)
\(3,\left(x-1\right)\left(2x+1\right)+3\left(x-1\right)\left(x+2\right)\left(2x+1\right)\)\(=\left(x-1\right)\left(2x+1\right)\left(1+3x+6\right)\)
\(=\left(x-1\right)\left(2x+1\right)\left(3x+7\right)\)
\(4,\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)\(=\left(3x-2\right)\left(4x-3\right)+\left(3x-2\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)\(=\left(3x-2\right)\left(4x-3+x-1-2x-2\right)\)
\(=\left(3x-2\right)\left(3x-6\right)\)
\(5,\left(x-5\right)^2+\left(x+5\right)\left(x-5\right)-\left(5-x\right)\left(2x+1\right)\)\(=\left(x-5\right)^2+\left(x+5\right)\left(x-5\right)+\left(x-5\right)\left(2x+1\right)\)\(=\left(x-5\right)\left(x-5+x+5+2x+1\right)\)
\(=\left(x-5\right)\left(4x+1\right)\)
6, Tương tự
a) \(\left(3x^{n+1}-y^{n-1}\right)-3\left(x^{n+1}+5y^{n-1}\right)-4\left(x^{n+1}+2y^{n-1}\right)\)
\(=3x^{n+1}-y^{n-1}-3x^{n+1}-15y^{n-1}+4x^{n+1}+8y^{n-1}\)
\(=-8y^{n-1}+4x^{n+1}\)
b) \(\left(\dfrac{3}{4}x^{n+1}-\dfrac{1}{2}y^n\right)\cdot2xy-\left(\dfrac{2}{3}x^{n+1}-\dfrac{5}{6}y^n\right)\cdot7xy\)
\(=\dfrac{3}{2}x^{n+2}y-xy^{n+1}+\left(-\dfrac{2}{3}x^{n+1}-\dfrac{5}{6}y^n\right)\cdot7xy\)
\(=\dfrac{3}{2}x^{n+2}y-xy^{n+1}-\dfrac{14}{3}x^{n+2}y+\dfrac{35}{6}xy^{n+1}\)
\(=-\dfrac{19}{6}x^{n+2}y+\dfrac{29}{6}xy^{n+1}\)
a)\(\left(3x^{n+1}-y^{n-1}\right)-3\left(x^{n+1}+5y^{n-1}\right)+4\left(x^{n+1}+2y^{n-1}\right)\)
\(=3x^{n+1}-y^{n-1}-3x^{n+1}-15y^{n-1}+4x^{n+1}+8y^{n-1}\)
\(=4x^{n+1}-8y^{n-1}\) \(\left(=4\left(x^{n+1}-2y^{n-1}\right)\right)\)
a: =2x^5-15x^3-x^2-2x^5-x^3=-16x^3-x^2
b: =x^3+3x^2-2x-3x^2-9x+6
=x^3-11x+6
c: \(=\dfrac{4x^3+2x^2-6x^2-3x-2x-1+5}{2x+1}\)
\(=2x^2-3x-1+\dfrac{5}{2x+1}\)
a) \(6x^3\left(\dfrac{1}{3}x^2-\dfrac{5}{2}-\dfrac{1}{6}\right)-2x^5-x^3\)
\(=6x^3\left(\dfrac{1}{3}x^2-\dfrac{16}{6}\right)-2x^5-x^3\)
\(=2x^5-16x^3-2x^5-x^3\)
\(=-17x^3\)
b) \(\left(x+3\right)\left(x^2+3x-2\right)\)
\(=x^3+3x^2-2x+3x^2+9x-6\)
\(=x^3+6x^2+7x-6\)
c) \(\left(4x^3-4x^2-5x+4\right):\left(2x+1\right)\)
\(=2x^2+4x^3-2x-4x^2-\dfrac{5}{2}-5x+\dfrac{2}{x}+4\)
\(=4x^3-2x^2-7x+\dfrac{2}{x}+\dfrac{3}{2}\)
\(1.\text{ }\text{ }\text{ }\dfrac{\left(x^2+2\right)^2-4x^2}{y\left(x^2+2\right)-2xy-\left(x-1\right)^2-1}\\ =\dfrac{\left(x^2+2-2x\right)\left(x^2+2+2x\right)}{x^2y+2y-2xy-x^2+2x-1-1}\\ =\dfrac{\left(x^2+2-2x\right)\left(x^2+2+2x\right)}{\left(x^2y-x^2\right)-\left(2xy-2x\right)+\left(2y-2\right)}\\ =\dfrac{\left(x^2+2-2x\right)\left(x^2+2+2x\right)}{x^2\left(y-1\right)-2x\left(y-1\right)+2\left(y-1\right)}\\ =\dfrac{\left(x^2+2-2x\right)\left(x^2+2+2x\right)}{\left(x^2-2x+2\right)\left(y-1\right)}\\ =\dfrac{x^2+2x+2}{y-1}\)
\(2.\text{ }\text{ }\text{ }\text{ }\dfrac{x^2+5x+6}{x^2+3x+2}\\ =\dfrac{x^2+3x+2x+6}{x^2+2x+x+2}\\ =\dfrac{\left(x^2+3x\right)+\left(2x+6\right)}{\left(x^2+2x\right)+\left(x+2\right)}\\ =\dfrac{x\left(x+3\right)+2\left(x+3\right)}{x\left(x+2\right)+\left(x+2\right)}\\ =\dfrac{\left(x+2\right)\left(x+3\right)}{\left(x+2\right)\left(x+1\right)}\\ =\dfrac{x+3}{x+1}\)
\(3.\text{ }\text{ }\text{ }\dfrac{x^2+y^2-z^2-2zt+2xy-t^2}{x^2-y^2+z^2-2yt+2xz-t^2}\text{ ( Chữa đề ) }\\ =\dfrac{\left(x^2+2xy+y^2\right)-\left(z^2+2zt+t^2\right)}{\left(x^2+2xz+z^2\right)-\left(y^2+2yt+t^2\right)}\\ =\dfrac{\left(x+y\right)^2-\left(z+t\right)^2}{\left(x+z\right)^2-\left(y+t\right)^2}\\ =\dfrac{\left(x+y+z+t\right)\left(x+y-z-t\right)}{\left(x+z+y+t\right)\left(x+z-y-t\right)}\\ =\dfrac{x+y-z-t}{x+z-y-t}\)
\(4.\text{ }\text{ }\text{ }\dfrac{\left(n+1\right)!}{\left(n+1\right)!+\left(n+2\right)!}=\dfrac{\left(n+1\right)!}{\left(n+1\right)!\left(1+n+2\right)}=\dfrac{1}{n+3}\)
\(5.\text{ }\text{ }\text{ }\dfrac{x^2+5x+4}{x^2-1}\\ =\dfrac{x^2+x+4x+4}{\left(x+1\right)\left(x-1\right)}\\ =\dfrac{\left(x^2+x\right)+\left(4x+4\right)}{\left(x+1\right)\left(x-1\right)}\\ =\dfrac{x\left(x+1\right)+4\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\\ =\dfrac{\left(x+1\right)\left(x+4\right)}{\left(x+1\right)\left(x-1\right)}\\ =\dfrac{x+4}{x-1}\)
\(6.\text{ }\text{ }\text{ }\dfrac{x^2-3x}{2x^2-7x+3}\\ =\dfrac{x\left(x-3\right)}{2x^2-6x-x+3}\\ =\dfrac{x\left(x-3\right)}{\left(2x^2-6x\right)-\left(x-3\right)}\\ =\dfrac{x\left(x-3\right)}{2x\left(x-3\right)-\left(x-3\right)}\\ =\dfrac{x\left(x-3\right)}{\left(2x-1\right)\left(x-3\right)}\\ =\dfrac{x}{2x-1}\)
