Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=2y-x-\left\{2x-y-\left[y+3x-\left(5y-x\right)\right]\right\}\)
\(=2y-x-\left\{2x-y-\left[y+3x-5y+x\right]\right\}\)
\(=2y-x-\left\{2x-y-y-3x+5y-x\right\}\)
\(=2y-x-2x+y+y+3x-5y+x\)
\(=\left(2y+y+y-5y\right)+\left(-x-2x+3x+x\right)\)
= \(-y+x\)
Thay \(x=a^2+2ab+b^2,y=a^2-2ab+b^2\) vào đa thức -y + x :
\(-\left(a^2-2ab+b^2\right)+\left(a^2+2ab+b^2\right)\)
\(=-a^2+2ab-b^2+a^2+2ab+b^2\)
\(=\left(-a^2+a^2\right)+\left(2ab+2ab\right)+\left(-b^2+b^2\right)\)
= 4ab
\(A=2y-x-\left\{2x-y-\left[y+3x-\left(5y-x\right)\right]\right\}\\ =2y-x-\left\{2x-y-y-3x+5y-x\right\}\\ =2y-x-2x+y+y+3x-5y+x\\ =-y+x=-\left(a^2-2ab+b^2\right)+\left(a^2+2ab+b^2\right)\\ =-a^2+2ab-b^2+a^2+2ab+b^2=4ab\)
a: \(M+N-P=2a^2-3a+1+5a^2+a-a^2+4=6a^2-2a+5\)
b: \(=2y-x-\left\{2x-y-\left[3x+y-5y+x\right]\right\}\)
\(=2y-x-\left\{2x-y-\left[4x-4y\right]\right\}\)
\(=2y-x-\left\{2x-y-4x+4y\right\}\)
\(=2y-x-\left[-2x+3y\right]\)
\(=-x+2y+2x-3y=x-y=\left(a-b\right)^2-\left(a-b\right)^2\)
=4ab
c: TH1: x>=1/2
A=5x-3-2x+1=3x-2
TH2: x<1/2
A=5x-3+2x-1=7x-4
143. a) \(-6x^n.y^n.\left(-\dfrac{1}{18}x^{2-n}+\dfrac{1}{72}y^{5-n}\right)\)
\(=-6.\left(-\dfrac{1}{18}\right)x^n.x^{2-n}.y^n+\left(-6\right).\dfrac{1}{27}x^n.y^n.y^{5-n}\)
\(=\dfrac{1}{3}x^{n+2-n}y^n-\dfrac{2}{9}x^n.y^{n+5-n}\)
\(=\dfrac{1}{3}x^2y^n-\dfrac{2}{9}x^ny^5\)
b) Ta có: \(\left(5x^2-2y^2-2xy\right)\left(-xy-x^2+7y^2\right)\)
\(=5x^2\left(-xy\right)+5x^2.\left(-x^2\right)+5x^2.7y^2-2y^2.\left(-xy\right)-2y^2.\left(-x^2\right)-2y^2.7y^2-2xy.\left(-xy\right)-2xy\left(-x^2\right)-2xy.7y^2\)
\(=-5x^3y-5x^4+35x^2y^2+2xy^3+2x^2y^2-14y^4+2x^2y^2+2x^3y-14xy^3\)
Rút gọn các đa thức đồng dạng, ta có kết quả:
\(-5x^4-3x^3y+39x^2y^2-12xy^3-14y^4\)
Kết quả đã được xếp theo lũy thừa giảm dần của x
\(a,\left(3x+5\right)^2+\left(3x-5\right)^2-\left(3x+2\right)\left(3x-2\right)=9x^2+30x+25+9x^2-30x+25-9x^2+4=9x^2+54\)
\(b,BT=2x\left(4x^2-4x+1\right)-3x\left(x^2-9\right)-4x\left(x^2+2x+1\right)=8x^3-8x^2+2x-3x^3+27x-4x^3-8x^2-4x=x^3-16x^2+25x\)
\(c,BT=\left(x+y-z\right)^2-2\left(x+y-z\right)\left(x+y\right)+\left(x+y\right)^2=\left(x+y-z-x-y\right)^2=z^2\)
a)
\(A=\left(x+3\right)\left(x^2-3x+9\right)-\left(54+x^3\right)\)
\(=x^3-3x^2+9x+3x^2-9x+27-54-x^3\)
\(=-27\)
or
\(A=x^3+27-54-x^3=-27\)
b)
\(B=\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-8x^3+y^3=2y^3\)
c)
\(C=\left(2x+1\right)^2+\left(1-3x\right)^2+2\left(2x+1\right)\left(3x-1\right)\)
\(=\left(2x+1+3x-1\right)^2=\left(5x\right)^2=25x^2\)
d)
\(D=\left(x-2\right)\left(x^2+2x+4\right)-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)
\(=x^3-8-\left(x-1\right)^3+3\left(x-1\right)\left(x+1\right)\)
\(=6x^2-3x-10\)
2:
a: A(x)=0
=>5x-10-2x-6=0
=>3x-16=0
=>x=16/3
b: B(x)=0
=>5x^2-125=0
=>x^2-25=0
=>x=5 hoặc x=-5
c: C(x)=0
=>2x^2-x-3=0
=>2x^2-3x+2x-3=0
=>(2x-3)(x+1)=0
=>x=3/2 hoặc x=-1
a) Ta có: \(5x^2-3x\left(x+2\right)\)
\(=5x^2-3x^2-6x\)
\(=2x^2-6x\)
b) Ta có: \(3x\left(x-5\right)-5x\left(x+7\right)\)
\(=3x^2-15x-5x^2-35x\)
\(=-2x^2-50x\)
c) Ta có: \(3x^2y\left(2x^2-y\right)-2x^2\left(2x^2y-y^2\right)\)
\(=3x^2y\left(2x^2-y\right)-2x^2y\left(2x^2-y\right)\)
\(=x^2y\left(2x^2-y\right)=2x^4y-x^2y^2\)
d) Ta có: \(3x^2\left(2y-1\right)-\left[2x^2\cdot\left(5y-3\right)-2x\left(x-1\right)\right]\)
\(=6x^2y-3x^2-\left[10x^2y-6x^2-2x^2+2x\right]\)
\(=6x^2y-3x^2-10x^2y+6x^2+2x^2-2x\)
\(=-4x^2y+5x^2-2x\)
e) Ta có: \(4x\left(x^3-4x^2\right)+2x\left(2x^3-x^2+7x\right)\)
\(=4x^4-16x^3+4x^4-2x^3+14x^2\)
\(=8x^4-18x^3+14x^2\)
f) Ta có: \(25x-4\left(3x-1\right)+7x\left(5-2x^2\right)\)
\(=25x-12x+4+35x-14x^3\)
\(=-14x^3+48x+4\)
a ) \(A=\frac{ax^2\left(a-x\right)-a^2x\left(x-a\right)}{3a^2-3x^2}=\frac{ax\left(a-x\right)\left(a+x\right)}{3\left(a-x\right)\left(a+x\right)}=\frac{ax}{3}\)
Thay \(a=\frac{1}{2};x=-3\), ta có :
\(A=\frac{\frac{1}{2}.-3}{3}=-\frac{1}{2}\)
b ) \(B=\frac{\left(ab+bc+cd+da\right)abcd}{\left(c+d\right)\left(a+b\right)+\left(b-c\right)\left(a-d\right)}=\frac{\left[\left(ab+ad\right)+\left(bc+cd\right)\right]abcd}{ca+cb+da+db+ba-bd-ca+cd}\)
\(=\frac{\left[a\left(b+d\right)+c\left(b+d\right)\right]abcd}{ba+da+cb+cd}=\frac{\left(b+d\right)\left(a+c\right)abcd}{\left(b+d\right)\left(a+c\right)}=abcd\)
Thay \(a=-3;b=-4;c=2;d=3\), ta có :
\(B=\left(-3\right).\left(-4\right).2.3=72\)
a: M+N-P
\(=7a^2-2a+1-a^2+4\)
\(=6a^2-2a+5\)
b: \(=2y-x-2x+y+y+3x-5y+x\)
\(=-3x+3y-4y+4x=x-y\)
\(=a^2+2ab+b^2-a^2+2ab-b^2=4ab\)
c: \(=\left[{}\begin{matrix}5x-3-2x+1=3x-2\left(x>=\dfrac{1}{2}\right)\\5x-3+2x-1=7x-4\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\)