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b)\(\frac{x^2-25}{5x-x^2}\)=\(\frac{\left(x-5\right)\left(x+5\right)}{x\left(5-x\right)}\)=\(\frac{\left(x-5\right)\left(x+5\right)}{-x\left(x-5\right)}\)=\(\frac{-x-5}{x}\)
c)\(\frac{y^2-xy}{4xy-4y^2}\)=\(\frac{y\left(y-x\right)}{4y\left(x-y\right)}=\frac{-y\left(x-y\right)}{4y\left(x-y\right)}=\frac{-1}{4}\)
d)\(\frac{x^2+xz-xy-yz}{x^2+xz+xy+yz}=\frac{x\left(x+z\right)-y\left(x+z\right)}{x\left(x+z\right)+y\left(x+z\right)}=\frac{\left(x+z\right)\left(x-y\right)}{\left(x+z\right)\left(x+y\right)}=\frac{x-y}{x+y}\)
a) \(\frac{3m-6n}{10n-5m}\)
\(=\frac{-3\left(2n-m\right)}{5\left(2n-m\right)}=\frac{-3}{5}\)
b) \(\frac{y^3+y^2+4y+4}{y^2+2y-8}\)
\(=\frac{y^2\left(y+1\right)+4\left(y+1\right)}{y^2+2y+1-9}\)
\(=\frac{\left(y^2+4\right)\left(y+1\right)}{\left(y+1\right)^2-9}\)
\(=\frac{\left(y^2+4\right)\left(y+1\right)}{\left(y-2\right)\left(y+4\right)}\)
c) \(\frac{x^2-xy-xz+yz}{x^2+xy-xz-yz}\)
\(=\frac{x\left(x-y\right)-z\left(x-y\right)}{x\left(x+y\right)-z\left(x+y\right)}\)
\(=\frac{\left(x-z\right)\left(x-y\right)}{\left(x-z\right)\left(x+y\right)}\)
\(=\frac{x-y}{x+y}\)
#)Góp ý :
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tk cho mk nhé
Chứng minh một số bất đẳng thức phụ:
1. \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\Rightarrow x^2+y^2+z^2\ge xy+yz+zx\ge3\)
2. \(2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+zx\right)\text{ (vừa chứng minh ở trên)}\)
\(\Rightarrow3\left(x^2+y^2+z^2\right)\ge x^2+y^2+z^2+2\left(xy+yz+zx\right)=\left(x+y+z\right)^2\)
3. \(x^2+y^2+z^2\ge xy+yz+zx\Rightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)\ge3\left(xy+y+zx\right)\)
\(\Rightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\)
\(\Rightarrow x+y+z\ge\sqrt{3\left(xy+yz+zx\right)}\ge\sqrt{3.3}=3\)
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{x^4}{y+3z}+\frac{y^4}{z+3x}+\frac{z^4}{x+3y}\ge\frac{\left(x^2+y^2+z^2\right)^2}{y+3z+z+3x+x+3y}=\frac{\left(x^2+y^2+z^2\right)\left(x^2+y^2+z^2\right)}{4\left(x+y+z\right)}\)
\(\ge\frac{3.\frac{1}{3}\left(x+y+z\right)^2}{4\left(x+y+z\right)}=\frac{x+y+z}{4}\ge\frac{3}{4}\)
Dấu "=" xảy ra khi và chỉ khi x = y = z = 1.
C2: Áp dụng Co6si:
\(\frac{x^4}{y+3z}+\frac{y+3z}{16}+\frac{1}{4}+\frac{1}{4}\ge4\sqrt[4]{\frac{x^4}{y+3z}.\frac{y+3z}{16}.\frac{1}{4}.\frac{1}{4}}=x\)
\(\Rightarrow\frac{x^4}{y+3z}\ge x-\frac{y+3z}{16}-\frac{1}{2}\)
Tương tự \(\frac{y^4}{z+3x}\ge y-\frac{z+3x}{16}-\frac{1}{2};\frac{z^4}{x+3y}\ge z-\frac{x+3y}{16}-\frac{1}{2}\)
\(\Rightarrow\frac{x^4}{y+3z}+\frac{y^4}{z+3x}+\frac{z^4}{x+3y}\ge\frac{3}{4}\left(x+y+z\right)-\frac{3}{2}\ge\frac{3}{4}.3-\frac{3}{2}=\frac{3}{4}\)
(\(\left(x+y+z\right)^2=x^2+y^2+z^2+2\left(xy+yz+zx\right)\ge xy+yz+zx+2\left(xy+yz+zx\right)\)
\(=3\left(xy+yz+zy\right)\ge9\)
\(\Rightarrow x+y+z\ge3\))
Dấu "=" xảy ra khi x = y = z = 1.
Mình nghĩ bạn viết hơi sai đề bài.
\(x^2+xz-y^2-yz=\left(x^2-y^2\right)+xz-yz=\left(x-y\right)\left(x+y\right)+z\left(x-y\right)=\left(x-y\right)\left(x+y+z\right)\)
Tương tự: \(y^2+xy-z^2-xz=\left(y-z\right)\left(x+y+z\right)\)
\(z^2+yz-x^2-xy=\left(x+y+z\right)\left(z-x\right)\)
Khi đó:
\(P=\frac{1}{\left(y-z\right)\left(x-y\right)\left(x+y+z\right)}+\frac{1}{\left(z-x\right)\left(y-z\right)\left(x+y+z\right)}+\frac{1}{\left(x-y\right)\left(x+y+z\right)\left(z-x\right)}\)
\(=\frac{z-x+x-y+y-z}{\left(x-y\right)\left(y-z\right)\left(z-x\right)\left(x+y+z\right)}=0\)
B3) a) x(x-5)-4(x-5)=0
<=> (x-4)(x-5)=0
TH1 :x-4=0
<=.x=4
TH2 : x-5=0
<=>x=5
b) x(x-6)-7x-42=0
<=>x(x+6)-7(x+6)=0
<=>(x-7)(x+6)=0
th1;x-7=0
<=>x=7
th2; x+6=0
<=>x=-6
c)x^3-5x^2+x-5=0
<=> x(x^2+1)-5(x^2+1)=0
<=> (x-5)(x^2+1)=0
th1:x-5=0
<=>x=5
TH2 : x^2+1=0
<=> x^2=-1 ( vo li )
=> th2 ko tồn tại
nho thick nha
Bài 3
a, x(x-5)-4(x-5)=0
(x-4)(x-5)=0
=>\(\orbr{\begin{cases}x-4=0\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=5\end{cases}}\)
b,x(x+6)-7(x+6)=0
(x-7)(x+6)=0\(\Rightarrow\orbr{\begin{cases}x-7=0\\x+6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-6\end{cases}}\)
c,x^2(x-5)+(x-5)=0
(x^2+1)(x-5)=0
\(\Rightarrow\orbr{\begin{cases}x^2+1=0\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\in\Phi\\x=5\end{cases}}\)
a) Áp dụng bài toán sau : a + b + c = 0 \(\Rightarrow\)a3 + b3 + c3 = 3abc
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)\(\Rightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=3.\frac{1}{x}.\frac{1}{y}.\frac{1}{z}\)
Ta có : \(A=\frac{yz}{x^2}+\frac{xz}{y^2}+\frac{xy}{z^2}=\frac{xyz}{x^3}+\frac{xyz}{y^3}+\frac{xyz}{z^3}\)
\(A=xyz.\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=xyz.3.\frac{1}{xyz}=3\)
b) x2 + y2 + z2 - xy - 3y - 2z + 4 = 0
4x2 + 4y2 + 4z2 - 4xy - 12y - 8z + 16 = 0
( 4x2 - 4xy + y2 ) + ( 3y2 - 12y + 12 ) + ( 4z2 - 8z + 4 ) = 0
( 2x - y )2 + 3 ( y - 2 )2 + 4 ( z - 1 )2 = 0
Ta có : ( 2x - y )2 \(\ge\)0 ; 3 ( y - 2 )2 \(\ge\)0 ; 4 ( z - 1 )2 \(\ge\)0
Mà ( 2x - y )2 + 3 ( y - 2 )2 + 4 ( z - 1 )2 = 0
\(\Rightarrow\)\(\hept{\begin{cases}2x-y=0\\y-2=0\\z-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=2\\z=1\end{cases}}}\)
Vậy ....