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21 tháng 12 2020

Mk sai từ dòng 3 nhá -- 

\(=\left(x^2-1\right)\left(\frac{2-\left(x^2-1\right)}{\left(x-1\right)\left(x+1\right)}\right)\)

\(=\frac{\left(x^2-1\right)\left(2-\left(x^2-1\right)\right)}{\left(x-1\right)\left(x+1\right)}=2-x^2+1=3-x^2\)

21 tháng 12 2020

\(\left(x^2-1\right)\left(\frac{1}{x-1}-\frac{1}{x+1}-1\right)\)

\(=\left(x^2-1\right)\left(\frac{x+1}{\left(x-1\right)\left(x+1\right)}-\frac{x-1}{\left(x+1\right)\left(x-1\right)}-\frac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\right)\)

\(=\left(x^2-1\right)\left(\frac{-\left(x^2-1\right)}{\left(x-1\right)\left(x+1\right)}\right)\)

\(=\frac{-\left(x-1\right)^2\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}=-\left(x-1\right)\left(x+1\right)=-x^2+1\)

13 tháng 2 2020

Mình thử nha :33

ĐKXĐ : \(x\ne-3,x\ne-26,x\ne-6,x\ne1\)

Ta có :

\(A=\left[\frac{3}{2}-\left(\frac{x^4\left(x^2+1\right)-x^4-1}{x^2+1}\right)\cdot\frac{x^3-4x^2+\left(x-4\right)}{x^6\left(x+6\right)-\left(x+6\right)}\right]:\frac{\left(x+3\right)\left(x+26\right)}{3\left(x-2\right)\left(x+6\right)}\)

\(=\left[\frac{3}{2}-\left(\frac{x^6-1}{x^2+1}\right)\cdot\frac{\left(x-4\right)\left(x^2+1\right)}{\left(x+6\right)\left(x^6-1\right)}\right]\cdot\frac{3\left(x-2\right)\left(x+6\right)}{\left(x+3\right)\left(x+26\right)}\)

\(=\left[\frac{3}{2}-\frac{x-4}{x+6}\right]\cdot\frac{3\left(x-2\right)\left(x+6\right)}{\left(x+3\right)\left(x+26\right)}\)

\(=\frac{x+26}{2\left(x+6\right)}\cdot\frac{3\left(x-2\right)\left(x+6\right)}{\left(x+3\right)\left(x+26\right)}\)

\(=\frac{3\left(x-2\right)}{2\left(x+3\right)}\)

Vậy : \(A=\frac{3\left(x-2\right)}{2\left(x+3\right)}\left(x\ne-3,x\ne-26,x\ne-6,x\ne1\right)\)

28 tháng 11 2016

\(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x^2-4x-1}{x^2-1}\right):\left(\frac{x+2006}{x}\right)\)

\(=\left(\frac{x^2+2x+1-x^2+2x-1+x^2-4x-1}{x^2-1}\right):\left(\frac{x+2006}{x}\right)\)

\(=\frac{x^2-1}{x^2-1}:\frac{x+2006}{x}=\frac{x}{x+2006}\)

23 tháng 4 2018

Ta có : 

\(A=\frac{x^2+x+1}{\left(x+1\right)^2}\)

\(A=\frac{x^2+2x+1-x-1+1}{x^2+2x+1}\)

\(A=\frac{x^2+2x+1}{\left(x+1\right)^2}+\frac{-x-1}{\left(x+1\right)^2}+\frac{1}{\left(x+1\right)^2}\)

\(A=\frac{\left(x+1\right)^2}{\left(x+1\right)^2}-\frac{x+1}{\left(x+1\right)^2}+\frac{1^2}{\left(x+1\right)^2}\)

\(A=1-\frac{1}{x+1}+\left(\frac{1}{x+1}\right)^2\)

Đặt \(a=\frac{1}{x+1}\) ta có : 

\(A=1-a+a^2\)

\(A=a^2-a+1\)

\(A=\left(a^2-a+\frac{1}{4}\right)+\frac{3}{4}\)

\(A=\left(a-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Dấu "=" xảy ra khi và chỉ khi \(\left(a-\frac{1}{2}\right)^2=0\)

\(\Leftrightarrow\)\(a-\frac{1}{2}=0\)

\(\Leftrightarrow\)\(a=\frac{1}{2}\)

Do đó : 

\(a=\frac{1}{x+1}\)

\(\Leftrightarrow\)\(\frac{1}{2}=\frac{1}{x+1}\)

\(\Leftrightarrow\)\(x+1=2\)

\(\Leftrightarrow\)\(x=1\)

Vậy GTNN  của \(A\) là \(\frac{3}{4}\) khi \(x=1\)

Chúc bạn học tốt ~ 

16 tháng 10 2015

\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1+\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+1+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}+1\)

\(=3+\left(\frac{x}{y}+\frac{y}{x}\right)+\left(\frac{x}{z}+\frac{z}{x}\right)+\left(\frac{y}{z}+\frac{z}{y}\right)\)

Áp dụng BĐT cô-si cho hai số không âm ta có:

\(\frac{x}{y}+\frac{y}{x}\ge2\sqrt{\frac{x}{y}.\frac{y}{x}}=2\sqrt{1}=2\)

\(\frac{x}{z}+\frac{z}{x}\ge2\sqrt{\frac{x}{z}.\frac{z}{x}}=2\sqrt{1}=2\)

\(\frac{y}{z}+\frac{z}{y}\ge2\sqrt{\frac{y}{z}.\frac{z}{y}}=2\sqrt{1}=2\)

Suy ra: \(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge3+2+2+2=9\)

=>Điều phải chứng minh

16 tháng 10 2015

đặt A= vế trái

nhân phá ngoặc A ta đc:

A=1+x/y+x/z+y/x+1+y/z+z/x+z/y+1

=3+(x/y+y/x)+(x/z+z/x)+(y/z+z/y)

áp dụng BĐT:a/b+b/a>=2

=>A>=3+2+2+2=9

vậy...

đơn giản 

nhưng trả lời câu hỏi của tớ đã

24 tháng 3 2020

\(\frac{1}{x^2+3}+\frac{1}{x^2+9x+18}+\frac{1}{x^2+15x+54}=\frac{1}{2}\left(27-\frac{1}{x+9}\right)\)

\(\Leftrightarrow\frac{3}{x\left(x+3\right)}+\frac{3}{\left(x+3\right)\left(x+6\right)}+\frac{3}{\left(x+6\right)\left(x+9\right)}=27-\frac{1}{x+9}\)

Mà 

\(\frac{3}{x\left(x+3\right)}+\frac{3}{\left(x+3\right)\left(x+6\right)}+\frac{3}{\left(x+6\right)\left(x+9\right)}\)

\(=\frac{1}{x}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+9}\)

\(=\frac{1}{x}-\frac{1}{x+9}\)

\(\Rightarrow\frac{1}{x}=27\Rightarrow x=\frac{1}{27}\)