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<=> \(\dfrac{x+2}{x-2}\)-\(\dfrac{1}{x}\)=\(\dfrac{2}{x\left(x-2\right)}\)
<=> \(\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{x-2}{x\left(x-2\right)}=\dfrac{2}{x\left(x-2\right)}\)
ok, ở đây đã có mẫu chung rồi, em cứ vậy làm tiếp thôi :D
\(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x^2-2x}\) (ĐKXĐ: \(x\ne0;x\ne2\))
\(\Leftrightarrow x\left(x+2\right)-\left(x-2\right)=2\)
\(\Leftrightarrow x^2+2x-x+2=2\)
\(\Leftrightarrow x^2+x+2-2=0\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
\(\Rightarrow S=\left\{-1\right\}\)
Bài 2 : Phân tích đa thức thành nhân tử
a) \(8x^2-2\)
\(=2\left(4x^2-1\right)\)
\(=2.\left(2x-1\right)\left(2x+1\right)\)
b) \(x^2-6x-y^2+9\)
\(=\left(x^2-6x+9\right)-y^2\)
\(=\left(x-3\right)^2-y^2\)
\(=\left(x-3+y\right)\left(x-3-y\right)\)
1. Tính giá trị biểu thức :
\(Q=x^2-10x+1025\)
\(Q=\left(x^2-2.x.5+25\right)+1000\)
\(Q=\left(x-5\right)^2+1000\)
Thay x=1005 vào biểu thức trên ta có :
\(Q=\left(1005-5\right)^2+1000\)
\(Q=1000000+1000\)
\(Q=1001000\)
a: \(B=\left(\dfrac{4x}{x+2}-\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\cdot\dfrac{4\left(x^2-2x+4\right)}{\left(x-2\right)\left(x+2\right)}\right)\cdot\dfrac{x+2}{16}\cdot\dfrac{\left(x+2\right)\left(x+1\right)}{x^2+x+1}\)
\(=\left(\dfrac{4x}{x+2}-\dfrac{4\left(x^2+2x+4\right)}{\left(x+2\right)^2}\right)\cdot\dfrac{x+2}{16}\cdot\dfrac{\left(x+2\right)\left(x+1\right)}{x^2+x+1}\)
\(=\dfrac{4x^2+8x-4x^2-8x-16}{\left(x+2\right)^2}\cdot\dfrac{\left(x+2\right)^2\cdot\left(x+1\right)}{16\left(x^2+x+1\right)}\)
\(=\dfrac{-16}{16\left(x^2+x+1\right)}\cdot\left(x+1\right)=-\dfrac{x+1}{x^2+x+1}\)
b: \(B=\dfrac{\left(x+2\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x+2}{x^2+x+1}\)
\(P=A+B=\dfrac{-x-1+x+2}{x^2+x+1}=\dfrac{1}{x^2+x+1}=\dfrac{1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}< =1:\dfrac{3}{4}=\dfrac{4}{3}\)
Dấu = xảy ra khi x=-1/2
b,\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
=>\(\dfrac{bc}{abc}+\dfrac{ac}{bac}+\dfrac{ab}{abc}=0\)
=>\(\dfrac{ab+ac+bc}{abc}=0\)
=>ab+ac+bc=0
=>ab=-ac-bc
ac=-ab-bc
bc=-ab-ac
N=\(\dfrac{1}{a^2+2bc}+\dfrac{1}{b^2+2ca}+\dfrac{1}{c^2+2ab}\)
N=\(\dfrac{1}{a^2+bc+bc}+\dfrac{1}{b^2+ca+ca}+\dfrac{1}{c^2+ab+ab}\)
N=\(\dfrac{1}{a^2-ab-ac+bc}+\dfrac{1}{b^2-ab-bc+ca}+\dfrac{1}{c^2-ac-bc+ab}\)
N=\(\dfrac{1}{a\left(a-b\right)-c\left(a-b\right)}+\dfrac{1}{b\left(b-a\right)-c\left(b-a\right)}+\dfrac{1}{c\left(c-a\right)-b\left(c-a\right)}\)
N=\(\dfrac{1}{\left(a-c\right)\left(a-b\right)}+\dfrac{1}{\left(b-c\right)\left(b-a\right)}+\dfrac{1}{\left(c-b\right)\left(c-a\right)}\)
N=\(\dfrac{b-c}{\left(a-c\right)\left(b-c\right)\left(a-b\right)}-\dfrac{a-c}{\left(b-c\right)\left(a-b\right)\left(a-c\right)}+\dfrac{a-b}{\left(b-c\right)\left(a-c\right)\left(a-b\right)}\)
N=\(\dfrac{b-c-a+c+a-b}{\left(a-c\right)\left(b-c\right)\left(a-b\right)}\)=0
2.a)\(\dfrac{3\text{x}-2}{2}\)=\(\dfrac{1-2\text{x}}{3}\)
<=>\(\dfrac{9\text{x}-6}{6}\)=\(\dfrac{2-4\text{x}}{6}\)
<=>9x-6=2-4x
<=>9x+4x=2+6
<=>13x=8
<=>x=\(\dfrac{8}{13}\)
1.a)2(x-0,5)+3=0,25(4x-1)
<=>2x-1+3=x-1phần4
<=>2x-x=-1/4+1-3
<=>x=-3/4
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a: ĐK của A là x<>-3; x<>2
ĐKXĐ của B là x<>3
DKXĐ của C là x<>0; x<>4/3
ĐKXĐ của D là x<>-2
ĐKXĐ của E là x<>2; x<>-2
ĐKXĐ của F là x<>2
b,c:
\(A=\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\dfrac{2}{x-2}\)
Để A=0 thì 2=0(loại)
\(B=\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x-3\right)}=\dfrac{x+3}{x-3}\)
Để B=0 thì x+3=0
=>x=-3
\(C=\dfrac{\left(3x-4\right)\left(3x+4\right)}{x\left(3x-4\right)}=\dfrac{3x+4}{x}\)
Để C=0 thì 3x+4=0
=>x=-4/3
\(D=\dfrac{\left(x+2\right)^2}{2\left(x+2\right)}=\dfrac{x+2}{2}\)
Để D=0 thì x+2=0
=>x=-2(loại)
\(E=\dfrac{x\left(2-x\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{-x}{x+2}\)
Để E=0 thì x=0
\(F=\dfrac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{3}{x-2}\)
Để F=0 thì 3=0(loại)
Bải 3a
\(\dfrac{-a+b+c}{2a}+\dfrac{-b+c+a}{2b}+\dfrac{-c+a+b}{2c}\ge\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{-a}{2a}+\dfrac{b+c}{2a}+\dfrac{-b}{2b}+\dfrac{c+a}{2b}+\dfrac{-c}{2c}+\dfrac{a+b}{2c}\ge\dfrac{3}{2}\)
\(\Leftrightarrow-\dfrac{3}{2}+\dfrac{b+c}{2a}+\dfrac{c+a}{2b}+\dfrac{a+b}{2c}\ge\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{b+c}{2a}+\dfrac{c+a}{2b}+\dfrac{a+b}{2c}\ge3\)
\(\Leftrightarrow\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}\ge6\)
\(\Leftrightarrow\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}\right)\ge6\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{ab}{ba}}=2\\\dfrac{b}{c}+\dfrac{c}{b}\ge2\sqrt{\dfrac{bc}{cb}}=2\\\dfrac{c}{a}+\dfrac{a}{c}\ge2\sqrt{\dfrac{ca}{ac}}=2\end{matrix}\right.\)
\(\Rightarrow\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}\right)\ge2+2+2=6\)
\(\Leftrightarrow\dfrac{-a+b+c}{2a}+\dfrac{-b+c+a}{2b}+\dfrac{-c+a+b}{2c}\ge\dfrac{3}{2}\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c\)
Bài 3b)
\(P=\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}\)
\(P=\dfrac{x^2}{xy+xz}+\dfrac{y^2}{xy+yz}+\dfrac{z^2}{xz+yz}\)
Áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức
\(\Rightarrow\dfrac{x^2}{xy+xz}+\dfrac{y^2}{xy+yz}+\dfrac{z^2}{xz+yz}\ge\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)}\)( 1 )
Theo hệ quả của bất đẳng thức Cauchy
\(\Rightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\)
\(\Rightarrow\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)}\ge\dfrac{3\left(xy+yz+xz\right)}{2\left(xy+yz+xz\right)}=\dfrac{3}{2}\) ( 2 )
Từ ( 1 ) và ( 2 )
\(\Rightarrow\)\(\dfrac{x^2}{xy+xz}+\dfrac{y^2}{xy+yz}+\dfrac{z^2}{xz+yz}\ge\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}\ge\dfrac{3}{2}\)
\(\Leftrightarrow P\ge\dfrac{3}{2}\)
Vậy \(P_{min}=\dfrac{3}{2}\)
Dấu " = " xảy ra khi \(a=b=c\)
Bài 2:
a) \(\dfrac{x-17}{33}+\dfrac{x-21}{29}+\dfrac{x}{25}=4\)
\(\Rightarrow\left(\dfrac{x-17}{33}-1\right)+\left(\dfrac{x-21}{29}-1\right)+\left(\dfrac{x}{25}-2\right)=0\)
\(\Rightarrow\dfrac{x-50}{33}+\dfrac{x-50}{29}+\dfrac{x-50}{25}=0\)
\(\Rightarrow\left(x-50\right)\left(\dfrac{1}{33}+\dfrac{1}{29}+\dfrac{1}{25}\right)=0\)
Mà \(\dfrac{1}{33}+\dfrac{1}{29}+\dfrac{1}{25}\ne0\)
\(\Rightarrow x-50=0\)
\(\Rightarrow x=50\)
Vậy x = 50
