\(\left(x+1\right)^2+\left(\frac{x+1}{x+2}\right)^2=8\)

\(\left(x+1\right)^2+\frac{\left(x+1\right)^2}{\left(x+2\right)^2}=8\)

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

\(x^4+6x^3+6x^2-18x-27=0\)

\(\left(x+3\right)\left(x^2-3\right)\left(x+3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=-3\\x^2=3\end{cases};x=-3\Rightarrow\orbr{\begin{cases}x=-3\\x=\pm\sqrt{3}\end{cases}}}\)

e thử cách khác :

\(\left(x+1\right)^2+\left(\frac{x+1}{x+2}\right)^2=8\)

\(x^2+2x+1+\frac{\left(x+1\right)^2}{\left(x+2\right)^2}=8\)

\(x^2+2x+1+\frac{x+1}{x+2}=8\)

\(x^2+2x+1\left(x+1\right):\left(x+2\right)=8\)

\(x^2+2x+x+1.\frac{1}{x+2}=8\)

\(x^2+3x+\frac{1}{x+2}=8\)

\(x^3+5x^2-2x-15=0\)

sai rồi :((

\(\left(2x+7\right)^2=9\left(x+2\right)^2\Leftrightarrow\left(2x+7\right)^2=\left(3x+3.2\right)^2=\left(3x+6\right)^2\)

Có \(\left(2x+7\right)^2=\left(3x+6\right)^2\Leftrightarrow2x+7=3x+6\). Áp dụng quy tắc chuyển vế

Ta có : \(2x+7=3x+6\Leftrightarrow7-6=3x-2x\Leftrightarrow1=x\) 

Hay \(x=1\).                  Vậy \(x=1\)

\(\left(2x+7\right)^2=9\left(x+2\right)^2\)

\(4x^2+28x+49=9x^2+36x+36\)

\(4x^2+28x+49-9x^2-36x-36=0\)

\(-5x^2-8x+13=0\)

\(x\left(5x+13\right)-\left(5x+13\right)=0\)

\(\left(x-1\right)\left(5x+13\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=1\\5x=13\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-\frac{13}{5}\end{cases}}}\)

4a^2 : (a+3)=1

=>a+3=4a² : 1

a+3=4a.a

<=> 4a.a - a-3=0

a.(4a-1)-3=0

=>a.(4a-1)=3

=>a và 4a-1 thuộc Ư₍₃₎={1;-1;3;-3}

với a =1=>4a-1=3<=>a=1(tm)

với a=-1=>4a-1=-3<=>a=-1/2(ktm)

với a=3=>4a-1=1<=>a=1/2(ktm)

với a= -3=>4a-1=-1<=>a=0(ktm)

vậy ....

đáp án: -(x-1)

bạn đổi dấu vế phải, và phân tích hằng đẳng thức vế trái

a) \(ĐKXĐ:\hept{\begin{cases}a\ne-3\\a\ne\pm2\end{cases}}\)

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

\(\Leftrightarrow M=\frac{a\left(2-a\right)}{a+3}\cdot\frac{\left(a-2\right)^2-\left(a+2\right)^2-4a^2}{\left(a-2\right)\left(a+2\right)}\)

\(\Leftrightarrow M=\frac{a\left(2-a\right)}{a+3}\cdot\frac{a^2-4a+4-a^2-4a-4-4a^2}{\left(a-2\right)\left(a+2\right)}\)

\(\Leftrightarrow M=\frac{a\left(2-a\right)}{a+3}\cdot\frac{-4a^2-8a}{\left(a-2\right)\left(a+2\right)}\)

\(\Leftrightarrow M=\frac{a\left(2-a\right)}{a+3}\cdot\frac{-4a\left(a+2\right)}{\left(a-2\right)\left(a+2\right)}\)

\(\Leftrightarrow M=\frac{a\left(2-a\right)}{a+3}\cdot\frac{-4a}{a-2}\)

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

\(\Leftrightarrow M=\frac{4a^2}{a+3}\)

b) Để M = 1

\(\Leftrightarrow\frac{4a^2}{a+3}=1\)

\(\Leftrightarrow4a^2=a+3\)

\(\Leftrightarrow4a^2-a-3=0\)

\(\Leftrightarrow\left(4a+3\right)\left(a-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}4a+3=0\\a-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}a=-\frac{3}{4}\left(tm\right)\\a=1\left(tm\right)\end{cases}}\)

Vậy để \(M=1\Leftrightarrow a\in\left\{-\frac{3}{4};1\right\}\)

c) Để M > 0

\(\Leftrightarrow\frac{4a^2}{a+3}>0\)

\(\Leftrightarrow a+3>0\)(Vì 4a² > 0, loại trường hợp = 0)

\(\Leftrightarrow a>-3\)

Vậy để \(M>0\Leftrightarrow a>-3\)

Để M < 0

\(\Leftrightarrow\frac{4a^2}{a+3}< 0\)

\(\Leftrightarrow a+3< 0\)(Vì 4a² > 0, loại trường hợp = 0)

\(\Leftrightarrow a< -3\)

Vậy để \(M< 0\Leftrightarrow a< -3\)

Ta có: \(\overline{abc}⋮37\Leftrightarrow100a+10b+c⋮37\)(1)

+) (1) => \(10\left(100a+10b+c\right)⋮37\)

<=> \(100b+10c+a+999a⋮37\) mà \(999a=37.27a⋮37\)

=> \(100b+10c+a⋮37\Leftrightarrow\overline{bca}⋮37\)

+) (1) => \(100\left(100a+10b+c\right)⋮37\)

<=> \(\left(100c+10a+b\right)+999\left(10a+b\right)⋮37\)mà \(999\left(10a+b\right)=37.27\left(10a+b\right)⋮37\)

=> \(\overline{cab}=100c+10a+b⋮37\)