Tính:
\(D=\dfrac{4x^2-1}{\left(x-y\right)\cdot\left(x+y\right)}+\dfrac{4y^2-1}{\left(y-z\right)\cdot\left(y-x\right)}+\dfrac{4z^2-1}{\left(z-x\right)\cdot\left(z-y\right)}\)
cho 3 số x,y,z đôi 1 khác nhau và chứng minh rằng :
\(\dfrac{y-z}{\left(x-y\right)\cdot\left(x-z\right)}+\dfrac{z-x}{\left(y-z\right)\cdot\left(y-x\right)}+\dfrac{y-x}{\left(z-x\right)\cdot\left(z-y\right)}=\dfrac{2}{x-y}+\dfrac{2}{y-z}+\dfrac{2}{z-x}\)
Ta có: \(\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}=\dfrac{y-x+x-z}{\left(x-y\right)\left(x-z\right)}\)\(=\dfrac{y-x}{\left(x-y\right)\left(x-z\right)}+\dfrac{x-z}{\left(x-y\right)\left(x-z\right)}\) \(=\dfrac{1}{z-x}+\dfrac{1}{x-y}\)
Tương tự:
\(\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}=\dfrac{1}{x-y}+\dfrac{1}{y-z}\)
\(\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}=\dfrac{1}{y-z}+\dfrac{1}{z-x}\)
\(\Rightarrow\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}+\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}+\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}\) \(=\dfrac{2}{x-y}+\dfrac{2}{y-z}+\dfrac{2}{z-x}\) \(\left(đpcm\right)\)
Cho x + y + z + 0 và x, y, z \(\ne\) 0. Rút gọn :
a/ \(P=\dfrac{x^2+y^2+z^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
b/ \(Q=\dfrac{\left(x^2+y^2-z^2\right)\cdot\left(y^2+z^2-x^2\right)\cdot\left(z^2+x^2-y^2\right)}{16\cdot x\cdot y\cdot z}\)
Sửa lại đề nha: x+y+z=0
a)
Xét x+y+z=0
(x+y+z)2=02
x2+y2+z2+2xy+2yz+2zx=0
=> x2+y2+z2=-2xy-2yz-2zx
Xét \(\dfrac{x^2+y^2+z^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
= \(\dfrac{x^2+y^2+z^2}{\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)}\)
=\(\dfrac{x^2+y^2+z^2}{x^2-2xy+y^2+y^2-2yz+z^2+z^2-2zx+x^2}\)
=\(\dfrac{x^2+y^2+z^2}{2x^2+2y^2+2z^2-2xy-2yz-2zx}\)(1)
Thay x2+y2+z2=-2xy-2yz-2zx vào (1)
=>\(\dfrac{x^2+y^2+z^2}{2x^2+2y^2+2z^2+x^2+y^2+z^2}\\=\dfrac{x^2+y^2+z^2}{3x^2+3y^2+3z^2}\\ =\dfrac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)}\\ =\dfrac{1}{3}\)
b)
Xét x+y+z=0 ba lần:
- Lần 1:x+y+z=0
<=> x+y=0-z
<=>(x+y)2=(0-z)2
<=>x2+2xy+y2=z2
<=>x2+y2-z2=-2xy(1)
-Lần 2: x+y+z=0
<=> y+z=0-x
<=>(y+z)2=(0-x)2
<=>y2+2yz+z2=x2
<=>y2+z2-x2=-2yz(2)
-Lần 3: x+y+z=0
<=>z+x=0-y
<=>(z+x)2=(0-y)2
<=>z2+2zx+x2=y2
<=> z2+x2-y2=-2zx(3)
Thay (1),(2),(3) vào Q, ta có:
=>\(\dfrac{\left(x^2+y^2-z^2\right)\left(y^2+z^2-x^2\right)\left(z^2+x^2-y^2\right)}{16xyz}=\dfrac{\left(-2xy\right)\left(-2yz\right)\left(-2zx\right)}{16xyz}\\=\dfrac{\left(-2yz\right)\left(-2zx\right)}{-8z}\\ =\dfrac{y\left(-2zx\right)}{4}\\ =\dfrac{-2xyz}{4}\\ =-\dfrac{xyz}{2}\)
1)
a, Cho x,y với xy lớn hơn hoặc bằng 0. Cm \(\left(x^2-y^2\right)^2\) lớn hơn hoặc bằng \(\left(x-y\right)^2\)
b, Cho \(x\cdot y\cdot z=1\) và \(x+y+z>\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\). Cm \(\left(x-1\right)\cdot\left(y-1\right)\cdot\left(z-1\right)>0\)
\(\left(x^2-y^2\right)^2=\left(x-y\right)^2\left(x+y\right)^2\) \(\Rightarrow\left\{{}\begin{matrix}x;y>0\\x+y< 1\end{matrix}\right.\)=> dccm sai = > người ra đề sai họăc người chép đề sai ;
Cho 3 số x,y,z thỏa mãn điều kiện
\(\dfrac{y+z-x}{x}=\dfrac{z+x-y}{y}=\dfrac{5x+7y-7}{4x}\)
Hãy tính giá trị biểu thức : B=\(\left(1+\dfrac{x}{y}\right)\cdot\left(1+\dfrac{y}{z}\right)\cdot\left(1+\dfrac{z}{x}\right)\)
Thu gọn các đơn thức sau, xác định hệ số, phần biến và bậc của đơn thức
A=\(\left(\dfrac{-3}{7}\cdot x^3\cdot y^2\right)\cdot\left(\dfrac{-7}{9}\cdot y\cdot z^2\right)\cdot\left(6\cdot x\cdot y\right)\)
B= \(-4\cdot x\cdot y^3\cdot\left(-x^2\cdot y\right)^3\cdot\left(-2\cdot x\cdot y\cdot z^3\right)^2\)
HELP ME
\(A=\left(\dfrac{-3}{7}.x^3.y^2\right).\left(\dfrac{-7}{9}.y.z^2\right).\left(6.x.y\right)\)
\(A=\left(\dfrac{-3}{7}x^3y^2\right).\left(\dfrac{-7}{9}yz^2\right).6xy\)
\(A=\left(\dfrac{-3}{7}.\dfrac{-7}{9}.6\right).\left(x^3.x\right)\left(y^2.y.y\right).z^2\)
\(A=2x^4y^4z^2\)
\(B=-4.x.y^3\left(-x^2.y\right)^3.\left(-2.x.y.z^3\right)^2\)
\(B=\left[\left(-4\right).\left(-2\right)\right].\left(x.x^6.x^2\right)\left(y^3.y^3.y^2\right)\left(z^6\right)\)
\(B=8x^7y^{y^8}z^6\)
Cho x, y, z thỏa mãn \(\dfrac{x}{2013}=\dfrac{y}{2014}=\dfrac{z}{2015}\). Chứng minh rằng: \(\left(x-z\right)^3=8\cdot\left(x-y\right)^2\left(y-z\right)\)
Áp dụng tc dtsbn:
\(\dfrac{x}{2013}=\dfrac{y}{2014}=\dfrac{z}{2015}=\dfrac{x-z}{-2}=\dfrac{y-z}{-1}=\dfrac{x-y}{-1}\\ \Leftrightarrow\dfrac{x-z}{2}=\dfrac{y-z}{1}=\dfrac{x-y}{1}\\ \Leftrightarrow x-z=2\left(y-z\right)=2\left(x-y\right)\\ \Leftrightarrow\left(x-z\right)^3=8\left(x-y\right)^3=8\left(x-y\right)^2\left(x-y\right)=8\left(x-y\right)^2\left(y-z\right)\)
Thu gọn biểu thức :
1, \(\left(2x-y\right)^2+2\cdot\left(2x-y\right)\cdot\left(y-x\right)+\left(x-y\right)^2\)
2, \(\left(x-y+z\right)^2+2\cdot\left(x-y+z\right)\cdot\left(y-z\right)+\left(y-z\right)^2\)
1, đa thức đã cho \(\Leftrightarrow\left(2x-y\right)^2-2\left(2x-y\right)\left(x-y\right)+\left(x-y\right)^2=\left[\left(2x-y\right)-\left(x-y\right)\right]^2=\left(2x-y-x+y\right)^2=x^2\)
2, đa thức đã cho \(\Leftrightarrow\left(x-y+z\right)^2+2\left(x-y+z\right)\left(y-z\right)+\left(y-z\right)^2=\left[\left(x-y+z\right)+\left(y-z\right)\right]^2=\left(x-y+z+y-z\right)^2=x^2\)
--- giải chi tiết lắm rồi đó---
a, \(\left(2x-y\right)^2+2\left(2x-y\right)\left(y-x\right)+\left(x-y\right)^2\)
\(=4x^2-4xy+y^2+2\left(2xy-2x^2-y^2+xy\right)+x^2-2xy+y^2\)
\(=4x^2-4xy+y^2+4xy-4x^2-2y^2+2xy+x^2-2xy+y^2\)
\(=x^2\)
b, \(\left(x-y+z\right)^2+2\left(x-y+z\right)\left(y-z\right)+\left(y-z\right)^2\)
\(=\left(x-y+z\right)\left[1+2\left(y-z\right)\right]+y^2-2yz+z^2\)
\(=\left(x-y+z\right)\left(1+2y-2z\right)+y^2-2yz+z^2\)
\(=x+2xy-2xz-y-2y^2+2yz+z+2yz-2z^2+y^2-2yz+z^2\)
\(=x-y+z+2xy-2xz+2yz-y^2-z^2\)
Chúc bạn học tốt!!!
Tính tổng:
\(S=\frac{x+1}{x\cdot\left(x-y\right)\cdot\left(x-z\right)}+\frac{y+1}{y\cdot\left(y-z\right)\cdot\left(y-x\right)}+\frac{z+1}{z\cdot\left(z-x\right)\left(z-y\right)}\)
\(S=\frac{yz\left(x+1\right)\left(y-z\right)-zx\left(y+1\right)\left(x-z\right)+xy\left(z+1\right)\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
+ \(yz\left(x+1\right)\left(y-z\right)-zx\left(y+1\right)\left(x-z\right)+xy\left(z+1\right)\left(x-y\right)\)
\(=yz\left(x+1\right)\left(y-z\right)-zx\left(y+1\right)\left[\left(y-z\right)+\left(x-y\right)\right]\)
\(+xy\left(z+1\right)\left(x-y\right)\)
\(=\left(y-z\right)\left[yz\left(x+1\right)-zx\left(y+1\right)\right]+\left(x-y\right)\left[xy\left(z+1\right)-zx\left(y+1\right)\right]\)
\(=\left(y-z\right)\left[z\left(y-x\right)\right]+\left(x-y\right)\cdot x\cdot\left(y-z\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)
\(\Rightarrow S=\frac{1}{xyz}\)
1, Giải hệ phương trình:
\(\hept{\begin{cases}x\cdot\left|x\right|-\left(x+10\right)\cdot\left|x+10\right|=y\cdot\left|y\right|\\y\cdot\left|y\right|-\left(y+10\right)\cdot\left|y+10\right|=z\cdot\left|z\right|\\z\cdot\left|z\right|-\left(z+10\right)\cdot\left|z+10\right|=x\cdot\left|x\right|\end{cases}}\)
Giải hộ mk nhoa mk tick cho !!!!!!!!!