4x^2+2z^2-4zx-2x+1
4x^2+2z^2-4zx-2z+z
BÀI 8: THU GỌN VÀ TÌM BẬC CỦA MỖI ĐA THỨC:
A= -2xy + 3/2xy^2 + 1/2xy^2 + xy
B= xy^2z + 2xy^2z - xyz - 3xy^2z + xy^2z
C= 4x^2y^3 + x^4 - 2x^2 + 6x^4 - x^2y^3
D= 3/4xy^2 - 2xy - 1/2xy^2 + 3xy
E= 2x^2 - 3y^3 - z^4 - 4x^2 + 2y^3 + 3z^4
F= 3xy^2z + xy^2z - xyz + 2xy^2z -3xyz
0,2:x=1,03+3,97
a: A=-2xy+xy+xy^2=-xy+xy^2
Bậc là 3
b: \(B=xy^2z+2xy^2z-3xy^2z+xy^2z-xyz=-xyz+xy^2z\)
Bậc là 4
c: \(C=4x^2y^3-x^2y^3+x^4+6x^4-2x^2=3x^2y^3+7x^4-2x^2\)
Bậc là 5
d: \(D=\dfrac{3}{4}xy^2-\dfrac{1}{2}xy^2+xy=\dfrac{1}{4}xy^2+xy\)
bậc là 3
e: \(E=2x^2-4x^2+3z^4-z^4-3y^3+2y^3\)
=-2x^2+2z^4-y^3
Bậc là 4
f: \(=3xy^2z+xy^2z+2xy^2z-4xyz=6xy^2z-4xyz\)
Bậc là 4
Cho các số thực x,y,z đôi 1 khác nhau và x+y+z=0 tính giá trị
P=\(\frac{\left(4yz-x^2\right)\left(4zx-y^2\right)\left(4xy-z^2\right)}{\left(yz+2x^2\right)\left(zx+2y^2\right)\left(xy+2z^2\right)}\)
Do \(x+y+z=0\)
\(\Rightarrow x=-\left(y+z\right)\Rightarrow x^2=\left(y+z\right)^2\Rightarrow4yz-x^2=4yz-\left(y+z^2\right)=-\left(y-z\right)^2\)
Tương tự \(4zx-y^2=-\left(z-x\right)^2\)
\(4xy-z^2=-\left(x-y\right)^2\)
Ta lại có: \(yz+2x^2=yz+x^2-x\left(y+z\right)=yz+x^2-xy-xz=\left(x-y\right)\left(x-z\right)\)
Tương tự: \(zx+2y^2=\left(y-x\right)\left(y-z\right)\)
\(xy+2z^2=\left(y-z\right)\left(y-y\right)\)
\(P=\frac{\left(4yz-x^2\right)\left(4zx-y^2\right)\left(4xy-z^2\right)}{\left(yz+2x^2\right)\left(zx+2y^2\right)\left(xy+2z^2\right)}=\frac{-\left(y-z\right)^2\left(z-x\right)^2\left(x-y^2\right)}{\left(x-y\right)\left(x-z\right)\left(y-x\right)\left(y-z\right)\left(z-x\right)\left(z-y\right)}\)
\(=\frac{-\left(y-z\right)^2\left(z-x\right)^2\left(x-y\right)^2}{-\left(y-z\right)^2\left(z-x\right)^2\left(x-y\right)^2}=1\)
Cho x, y, z là 3 số thỏa mãn điều kiện:
\(4x^2+2y^2+2z^2-4xy-4zx+2yz-6y-10z+34=0\)Tính
\(S=\left(x-4\right)^{2017}+\left(y-4\right)^{2017}+\left(z-4\right)^{2017}\)
Ta có : \(4x^2+2y^2+2z^2-4xy-4zx+2yz-6y-10z+34=0\)
\(\Rightarrow\left(4x^2+y^2+z^2-4xy-4zx+2yz\right)+\left(y^2-6y+9\right)+\left(z^2-10z+25\right)=0\)
\(\Rightarrow\left(2x-y-z\right)^2+\left(y-3\right)^2+\left(z-5\right)^2=0\)
Vì \(\hept{\begin{cases}\left(2x-y-z\right)^2\ge0\forall x,y,z\\\left(y-3\right)^2\ge0\forall y\\\left(z-5\right)^2\ge0\forall z\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}\left(2x-y-z\right)^2=0\\\left(y-3\right)^2=0\\\left(z-5\right)^2=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}2x-y-z=0\\y-3=0\\z-5=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}2x-3-5=0\\y=3\\z=5\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}2x=8\\y=3\\z=5\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=4\\y=3\\z=5\end{cases}}\left(1\right)\)
Lại có : \(S=\left(x-4\right)^{2017}+\left(y-4\right)^{2017}+\left(z-4\right)^{2017}\)
Thay \(\left(1\right)\)vào \(S\),ta được :
\(S=0^{2017}+\left(-1\right)^{2017}+1^{2017}\)
\(=0-1+1=0\)
Vậy \(S=0\)
Ai đó giúp mình với
Cho 1/x + 2/y + 3/z = 0
Tính S= 9xy/2z2 +yz/6x2 +4zx/3y2
Cho các phân thức: A=\(\frac{4xy-z}{xy+2z^2}\);B=\(\frac{4yz-x^2}{yz+2x^2}\);C=\(\frac{4zx-y^2}{zx+2y}\)
C/m với x khác y;y khác z; z khác x và x+y+z=0 thì A.B.C=1, A+B+C=3
PP nhóm hạng tử chung
1)2x+2y-x(x+y)
2)5x^2-5xy-10x+10y
3)4x^2+8xy-3x-6y
4)2x^2+2y^2-x^2z+z-y^2z-2
5)x^2+xy-5x-5y
6)x(2x-7)-4x+14
7)x^2-3x+xy-3y
1) 2x + 2y - x(x+y)
= 2(x + y) - x(x + y)
= (2 - x)(x + y)
2/ 5x2 - 5xy -10x + 10y
= 5x(x - y) - 10(x - y)
= (5x - 10(x - y)
3/ 4x2 + 8xy - 3x - 6y
= 4x(x + 2y) - 3(x + 2y)
= (4x - 3)(x + 2y)
1) 2x + 2y - x(x + y)
= 2(x + y) - x(x + y)
= (2 - x)(x + y)
2) 5x2 - 5xy - 10x + 10y
= 5x(x - y) - 10(x - y)
= (5x - 10)(x - y)
= 5(x - 2)(x - y)
3) 4x2 + 8xy - 3x - 6y
= 4x(x + 2y) - 3(x + 2y)
= (4x - 3)(x + 2y)
4) 2x2 + 2y2 - x2z + z - y2z - 2
= 2(x2 + y2 - z(x2 + y2) - (2 - z)
= (2 - z)(x2 + y2) - (2 - z)
= (2 - z)(x2 + y2)
5) x2 + xy - 5x - 5y
= x(x + y) - 5(x + y)
= (x - 5)(x + y)
6) x(2x - 7) - 4x + 14
= x(2x - 7) - 2(2x - 7)
= (x - 2)(2x - 7)
7)x2 - 3x + xy - 3y
= x(x + y) - 3(x + y)
= (x - 3)(x + y)
5/ x2 + xy - 5x - 5y
= x(x + y) - 5(x + y)
= (x - 5)(x + y)
6/ x(2x - 7) - 4x + 14
= 2x2 - 7x - 4x + 14
= (2x2 - 4x) - (7x - 14)
= 2x(x - 2) -7(x - 2)
= (2x - 7)(x - 2)
7/ x2 - 3x + xy - 3y
= x(x - 3) + y(x - 3)
= (x + y)(x - 3)
rút gọn biểu thức sau 1/2x^2yz-4x^2z-2x^2z
\(\dfrac{1}{2}x^2yz-4x^2z-2x^2z\)
= \(\dfrac{1}{2}x^2yz-(4x^2z-2x^2z)\)
= \(\dfrac{1}{2}x^2yz-2x^2z\)
Tính và thu gon:
\(\left(2x+1\right)^2-\left(4x-3\right)\left(x+7\right)-22\)
\(69x\left(3x^2-5x\right)-\left(3x+1\right)\left(9x^2-18x-1\right)\)
\(\left(1-2x\right)^3-4x^2\left(3-2x\right)+24x^2\)
\(\left(24x^2y^2z-36x^2y^2z^2-12x^2y^3z\right):12x^2yz\)
\(\left(2x+1\right)^2-\left(4x-3\right).\left(x+7\right)-22\)
\(=4x^2+4x+1-4x^2-28x+3x+21-22\)
\(=-21x\)
mấy câu khác tương tự