\(\left(3x-2\right)\left(x+1\right)^2\left(3x+8\right)=-16\)

<=>   \(\left(3x-2\right)\left(x+1\right)^2.3^2.\left(3x+8\right)+144=0\)

<=>  \(\left(3x-2\right)\left(3x+3\right)^2\left(3x+8\right)+144=0\)   (*)

Đặt  \(3x+3=t\) Khi đó pt (*) trở thành: 

   \(\left(t-5\right)t^2\left(t+5\right)+144=0\)

<=>  \(t^4-25t^2+144=0\)

<=>  \(\left(t-4\right)\left(t-3\right)\left(t+3\right)\left(t+4\right)=0\)

đến đây bn tự giải tiếp nhé

\(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)

<=>  \(\frac{x-5}{100}-1+\frac{x-4}{101}-1+\frac{x-3}{102}-1=\frac{x-100}{5}-1+\frac{x-101}{4}-1+\frac{x-102}{3}-1\)

<=>  \(\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}=\frac{x-105}{5}+\frac{x-105}{4}+\frac{x-105}{3}\)

<=>  \(\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)

Nhận thấy:   \(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}+\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\ne0\)

=>  \(x-105=0\)

<=>  \(x=105\)

\(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)

\(\Leftrightarrow\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}-\frac{x-100}{5}-\frac{x-101}{4}-\frac{x-102}{3}=0\)

\(\Leftrightarrow\left(\frac{x-5}{100}-1\right)+\left(\frac{x-4}{101}-1\right)+\left(\frac{x-3}{102}-1\right)-\left(\frac{x-100}{5}-1\right)-\left(\frac{x-101}{4}-1\right)-\left(\frac{x-102}{3}-1\right)=0\)

\(\Leftrightarrow\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}-\frac{x-105}{5}-\frac{x-105}{4}-\frac{x-105}{3}=0\)

\(\Leftrightarrow\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)

\(\Leftrightarrow x-105=0\left(Vì\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\ne0\right)\)

\(\Leftrightarrow x=105\)

\(\frac{1-2x}{1-x}=1\)

\(\Leftrightarrow1-x=1-2x\)

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

\(\Leftrightarrow x=0\)

Tương tự ta cũng có \(y=0\)

Khi đó : \(x^2+y^2-xy=0^2+0^2-0\cdot0=0=0^2\left(đpcm\right)\)

Sai đề ạ:

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

à mình ghi nhầm ở câu a , chứng minh BIO bằng CMO chứ ko phải BEO nha

\(x^2+y^2=325\)

<=>  \(\left(x+y\right)^2-2xy=325\)

Đặt:  \(x+y=a;\)\(xy=b\)Khi đó ta có:

\(a-b=155\)   (1)

và  \(a^2-2b=325\)

Từ (1) ta có:   \(b=a-155\) thay vào (2) ta được:

\(a^2-2\left(a-155\right)=325\)

giải ra tìm được:  \(\orbr{\begin{cases}a=5\\a=-3\end{cases}}\)  =>  \(\orbr{\begin{cases}a=5;b=-150\\a=-3;b=-158\end{cases}}\)

TH1:  \(\hept{\begin{cases}a=5\\b=-150\end{cases}}\) ,=>  \(\hept{\begin{cases}x+y=5\\xy=-150\end{cases}}\)

\(x^2+y^2=325\) 

<=>   \(\left(x-y\right)^2+2xy=325\)

<=>  \(\left(x-y\right)^2=325-2xy=625\)

<=>  \(\left|x-y\right|=25\)

=>  \(\left|x^3-y^3\right|=\left|\left(x-y\right)\left(x^2+y^2+xy\right)\right|=\left|x-y\right|\left(x^2+y^2+xy\right)=4375\)

TH2: bn tự lm tiếp nhé

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

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

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

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

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

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

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