câu a thiếu đề 

b) Đặt P = ... 

\(P=\sqrt{\left(\frac{1}{\sqrt{ab+a+2}}+\frac{1}{\sqrt{bc+b+2}}+\frac{1}{\sqrt{ca+c+2}}\right)^2}\)

\(\le\sqrt{3\left(\frac{1}{ab+1+a+1}+\frac{1}{bc+1+b+1}+\frac{1}{ca+1+c+1}\right)}\)

\(\le\sqrt{\frac{3}{4}\left(\frac{1}{ab+1}+\frac{1}{bc+1}+\frac{1}{ca+1}+\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)}\)

\(=\sqrt{\frac{3}{4}\left(\frac{a}{a+1}+\frac{1}{a+1}+\frac{b}{b+1}+\frac{1}{b+1}+\frac{c}{c+1}+\frac{1}{c+a}\right)}=\sqrt{\frac{9}{4}}=\frac{3}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=1\)

\(A=\frac{3}{x^4-x^3+x-1}-\frac{1}{x^4+x^3-x-1}-\frac{4}{x^5-x^4+x^3-x^2+x-1}\)

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

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

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

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

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

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

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

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

\(\Rightarrow A=\frac{2}{x^4+x^2+1}\left(x\ne\pm1\right)\)

Ta có: \(x^4+x^2+1=\left(x^2\right)^2+2.x^2.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(x^2+\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)

Vậy A > 0 với mọi \(x\ne\pm1\)

Đặt \(a=2x;\sqrt[3]{2-8x^3}=t\)

=> \(\hept{\begin{cases}\left(2a-1\right)t=a\\t^3+a^3=2\end{cases}}\)

<=>\(\hept{\begin{cases}a+t=2at\\\left(a+t\right)^3-3at\left(a+t\right)=2\end{cases}}\)=> \(\hept{\begin{cases}a+t=2at\\\left(a+t\right)^3-\frac{3}{2}\left(a+t\right)^2-2=0\end{cases}}\)

=> \(\hept{\begin{cases}a+t=2\\at=1\end{cases}}\)

=> \(a=t=1\)

=> \(x=\frac{1}{2}\)

Vậy \(x=\frac{1}{2}\)

a)  x=8 hoặc x=-1

Đặt ẩn phụ

g)  x=1 hoặc x=2 hoặc x=-3

Phân tích thành nhân tử rồi xét giá trị

e) 

\(\sqrt{2x+1}-\sqrt{3x}=x-1\) 1

<=>\(2x+1-3x=\left(x+1\right)^2\)

<=>\(2x+1-3x=x^2-2x+1\)

<=> \(2x-3x-x^2+2x=1-1\)

<=> \(x-x^2=0\)

<=> \(x\left(1-x\right)=0\)

<=> \(x=0\)Hoặc \(1-x=0\)

trg hợp 1 : \(x=0\)

th2: \(1-x=0\)<=>\(x=1\)

Từ D hạ DI vuông góc BC tại I. Có ngay I là trung điểm cạnh BC và AI = BI = CI

Áp dụng ĐL Pytagoras có DH² + AH² = DI² + IH² + AI² - IH² = DI² + BI² = DB² (đpcm).