cho x + y + z = 0 và x, y , z khác 0 hãy 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 = \(\dfrac{\left(x^2+y^2-z^2\right)\left(y^2+z^2-x^2\right)\left(z^2+x^2-y^2\right)}{16xyz}\)
Cho x+y+z=0 và x,y,z khác 0. Rút gọn:
a) A= \(\frac{x^2+y^2+z^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
b) B= \(\frac{\left(x^2+y^2-z^2\right)\left(y^2+z^2-x^2\right)\left(z^2+x^2-y^2\right)}{16xyz}\)
Cho x+y+z=0 và x,y,z khác 0. Rút gọn:
a) A= \(\frac{x^2+y^2+z^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
b) B= \(\frac{\left(x^2+y^2-z^2\right)\left(y^2+z^2-x^2\right)\left(z^2+x^2-y^2\right)}{16xyz}\)
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:
=>\(\dfrac{\left(x^2+y^2-z^2\right)\left(y^2+z^2-x^2\right)\left(z^2+x^2-y^2\right)}{16xyz}=\dfrac{\left(-2xy\right)\left(-2yz\right)\left(-2zx\right)}{16xyz}\\=\dfrac{\left(-2yz\right)\left(-2zx\right)}{-8z}\\ =\dfrac{y\left(-2zx\right)}{4}\\ =\dfrac{-2xyz}{4}\\ =-\dfrac{xyz}{2}\)
Rút gọn: \(\dfrac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}\), biết rằng: x+y+z=0
\(x+y+z=0\Leftrightarrow\left(x+y+z\right)^2\)
\(\Leftrightarrow x^2+y^2+z^2=-2\left(xy+yz+zx\right)\)
\(P=\dfrac{x^2+y^2+z^2}{2\left(x^2+y^2+z^2\right)-2\left(xy+yz+zx\right)}=\dfrac{x^2+y^2+z^2}{2\left(x^2+y^2+z^2\right)+x^2+y^2+z^2}=\dfrac{1}{3}\)
CẦN GẤP!!
Cho x+y+z=0 xà x,y,z khác 0 rút gọn
a)P=\(\frac{x^2+y^2+z^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
b)Q=\(\frac{\left(x^2+y^2-z^2\right)\left(y^2+z^2-x^2\right)\left(z^2+x^2-y^2\right)}{16xyz}\)
\(x+y+z=0\Rightarrow\hept{\begin{cases}x=-\left(y+z\right)\\y=-\left(z+x\right)\\z=-\left(x+y\right)\end{cases}}\)
\(\Rightarrow P=\frac{x^2+y^2+z^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}=\frac{\left[-\left(y+z\right)\right]^2+\left[-\left(z+x\right)\right]^2+\left[-\left(x+y\right)\right]^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}=\frac{\left(y+z\right)^2+\left(z+x\right)^2\left(x+y\right)^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}=\frac{-\left[\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2\right]}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}=-1\)
Mik mới biết làm câu a thôi còn câu b thì từ từ mik nghĩ đã nhé @-@
Chúc bn học giỏi nhoa!!!
Chứng minh đẳng thức:
a) \(\dfrac{y}{\left(x-y\right)\left(y-z\right)}+\dfrac{z}{\left(y-z\right)\left(z-x\right)}+\dfrac{x}{\left(z-x\right)\left(x-y\right)=0}\)
b) \(\dfrac{x^2}{\left(x-y\right)\left(y-z\right)}+\dfrac{y^2}{\left(y-z\right)\left(y-x\right)}+\dfrac{z^2}{\left(z-x\right)\left(z-y\right)=1}\)
c) \(\dfrac{1}{x\left(x-y\right)\left(x-z\right)}+\dfrac{1}{y\left(y-z\right)\left(y-x\right)}+\dfrac{1}{z\left(z-x\right)\left(z-y\right)}=\dfrac{1}{xyz}\)
a: \(\dfrac{y}{\left(x-y\right)\left(y-z\right)}-\dfrac{z}{\left(y-z\right)\left(x-z\right)}-\dfrac{x}{\left(x-y\right)\left(x-z\right)}\)
\(=\dfrac{xy-yz-xz+yz-xy+xz}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
=0
c: \(=\dfrac{1}{x\left(x-y\right)\left(x-z\right)}-\dfrac{1}{y\left(y-z\right)\left(x-y\right)}+\dfrac{1}{z\left(x-z\right)\left(y-z\right)}\)
\(=\dfrac{zy\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{zy^2-z^2y-x^2z+xz^2+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{1}{xyz}\)
Rút gọn:
A=\(\dfrac{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+2\left(\dfrac{1}{x-y}+\dfrac{1}{y-z}+\dfrac{1}{z-x}\right)\)
Quy đồng tính bình thường.
\(A=\dfrac{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+2\left(\dfrac{1}{x-y}+\dfrac{1}{y-z}+\dfrac{1}{z-x}\right)\)\(=\dfrac{2x^2+2y^2+2z^2-2xy-2yz-2xz}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+\dfrac{2yz+2xz+2xy-2x^2-2y^2-2z^2}{ }\)
=0
1. Cho các số x, y, z thỏa mãn : (x + y)(y + z)(z + x) = 4. CMR: \(\left(x^2-y^2\right)^3\)+ \(\left(y^2-z^2\right)^3\)+ \(\left(z^2-x^2\right)^3\)= 12 (x - y)(y - z)(z - x)
2. Rút gọn: \(\dfrac{\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3}{\left(x^2-y^2\right)^3+\left(y^2-z^2\right)^3+\left(z^2-x^2\right)^3}\) biết (x + y)(y + z)(z + x) = 1
3. Cho a, b, c ≠ 0 thỏa mãn: a + b + c = \(a^2+b^2+c^2\) = 2. CMR: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{abc}\)
MONG MN GIẢI GIÚP EM Ạ!!! EM ĐANG CẦN GẤP ! CẢM ƠN MN NHIỀU
Hầy mình không nghĩ lớp 7 đã phải làm những bài biến đổi như thế này. Cái này phù hợp với lớp 8-9 hơn.
1.
Đặt $x^2-y^2=a; y^2-z^2=b; z^2-x^2=c$.
Khi đó: $a+b+c=0\Rightarrow a+b=-c$
$\text{VT}=a^3+b^3+c^3=(a+b)^3-3ab(a+b)+c^3$
$=(-c)^3-3ab(-c)+c^3=3abc$
$=3(x^2-y^2)(y^2-z^2)(z^2-x^2)$
$=3(x-y)(x+y)(y-z)(y+z)(z-x)(z+x)$
$=3(x-y)(y-z)(z-x)(x+y)(y+z)(x+z)$
$=3.4(x-y)(y-z)(z-x)=12(x-y)(y-z)(z-x)$
Ta có đpcm.
Bài 2:
Áp dụng kết quả của bài 1:
Mẫu:
$(x^2-y^2)^3+(y^2-z^2)^3+(z^2-x^2)^3=3(x-y)(y-z)(z-x)(x+y)(y+z)(z+x)=3(x-y)(y-z)(z-x)(1)$
Tử:
Đặt $x-y=a; y-z=b; z-x=c$ thì $a+b+c=0$
$(x-y)^3+(y-z)^3+(z-x)^3=a^3+b^3+c^3$
$=(a+b)^3-3ab(a+b)+c^3=(-c)^3-3ab(-c)+c^3=3abc$
$=3(x-y)(y-z)(z-x)(2)$
Từ $(1);(2)$ suy ra \(\frac{(x-y)^3+(y-z)^3+(z-x)^3}{(x^2-y^2)^3+(y^2-z^2)^3+(z^2-x^2)^3}=1\)
Bài 3:
\(ab+bc+ac=\frac{(a+b+c)^2-(a^2+b^2+c^2)}{2}=\frac{2^2-2}{2}=1\)
Do đó:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{ab+bc+ac}{abc}=\frac{1}{abc}\)
Ta có đpcm.
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:
\(\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}=\dfrac{1}{x-y}+\dfrac{1}{y-z}\)
\(\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}=\dfrac{1}{y-z}+\dfrac{1}{z-x}\)
\(\Rightarrow\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}+\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}+\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}\) \(=\dfrac{2}{x-y}+\dfrac{2}{y-z}+\dfrac{2}{z-x}\) \(\left(đpcm\right)\)