Đặt \(x^2+3x+1=t\)

\(\left(x^2+3x+1\right)\left(x^2+3x-3\right)-5\)

\(=t\left(t-4\right)-5\)

\(=t^2-4t-5\)

tự làm nốt ý này nhé.

những ý kia lát nx mình làm.

d) \(x^4+5x^2+9\).Đặt \(x^2=t\) thì:

\(x^4+5x^2+9=t^2+5t+9\)

Làm nốt ý này nhé bạn! Ý kia chút nữa rảnh làm!

VietJack môn nào cũng có nha 

\(M=x^2-4x+y^2-3y+2018\)

\(M=x^2-4x+4+y^2-3y+\frac{9}{4}+2015,75\)

\(M=\left(x^2-2\cdot x\cdot2+2^2\right)+\left[y^2-2\cdot y\cdot\frac{3}{2}+\left(\frac{3}{2}\right)^2\right]+2015,75\)

\(M=\left(x-2\right)^2+\left(y-\frac{3}{2}\right)^2+2015,75\)

Vì \(\left(x-2\right)^2\ge0\forall x;\left(y-\frac{3}{2}\right)^2\ge0\forall y\)

\(\Rightarrow M\ge0+0+2015,75=2015,75\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-2=0\\y-\frac{3}{2}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=\frac{3}{2}\end{cases}}}\)

Vậy \(M_{min}=2015,75\Leftrightarrow\hept{\begin{cases}x=2\\y-\frac{3}{2}\end{cases}}\)

\(M=x^2-4x+y^2-3y+2018\)

\(M=\left(x^2-2.x.2+2^2\right)+\left(y^2-2.y.1,5+1,5^2\right)+2011,75\)

\(M=\left(x-2\right)^2+\left(y-1,5\right)^2+2011,75\)

Ta có: \(\hept{\begin{cases}\left(x-2\right)^2\ge0\forall x\\\left(y-1,5\right)^2\ge0\forall y\end{cases}}\Rightarrow\left(x-2\right)^2+\left(y-1,5\right)^2+2011,75\ge2011,75\)

\(M=2011,75\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^2=0\\\left(y-1,5\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x-2=0\\y-1,5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=1,5\end{cases}}\)

Vậy \(M_{min}=2011,75\Leftrightarrow\hept{\begin{cases}x=2\\y=1,5\end{cases}}\)

ĐKXĐ bạn tự xét nhé

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

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

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

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

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

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

\(M=\frac{a^2+a+1}{a-1}\)

Để M thuộc Z thì \(a^2+a+1⋮a-1\)

\(\Leftrightarrow a^2-a+2a-2+3⋮a-1\)

\(\Leftrightarrow a\left(a-1\right)+2\left(a-1\right)+3⋮a-1\)

\(\Leftrightarrow\left(a-1\right)\left(a+2\right)+3⋮a-1\)

Mà \(\left(a-1\right)\left(a+2\right)⋮a-1\)

\(\Rightarrow3⋮a-1\)

\(\Rightarrow a-1\inƯ\left(3\right)=\left\{1;3;-1;-3\right\}\)

\(\Rightarrow a\in\left\{2;4;0;-2\right\}\)

Để M = 7 thì :

\(\frac{a^2+a+1}{a-1}=7\)

\(\Leftrightarrow a^2+a+1=7\left(a-1\right)\)

\(\Leftrightarrow a^2+a+1=7a-7\)

\(\Leftrightarrow a^2-6a+8=0\)

\(\Leftrightarrow a^2-2a-4a+8=0\)

\(\Leftrightarrow a\left(a-2\right)-4\left(a-2\right)=0\)

\(\Leftrightarrow\left(a-2\right)\left(a-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}a-2=0\\a-4=0\end{cases}\Rightarrow\orbr{\begin{cases}a=2\\a=4\end{cases}}}\)

Để M > 0 thì :

\(\frac{a^2+a+1}{a-1}>0\)

Vì \(a^2+a+1>0\forall a\), do đó để M > 0 thì : \(a-1>0\Leftrightarrow a>1\)

Chứng minh \(a^2+a+1>0\):

Đặt \(B=a^2+a+1\)

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

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

Vì \(\left(a+\frac{1}{2}\right)^2\ge0\forall a\)

\(\Rightarrow B\ge0+\frac{3}{4}=\frac{3}{4}>0\)

\(\Rightarrow B>0\left(đpcm\right)\)

Dấu "=" xảy ra \(\Leftrightarrow a+\frac{1}{2}=0\Leftrightarrow a=\frac{-1}{2}\)

\(A=\left(1+\frac{x^2}{y^2}\right)\left(1+\frac{y^2}{x^2}\right)\ge2\sqrt{\frac{x^2}{y^2}}.2\sqrt{\frac{y^2}{x^2}}=2.\frac{x}{y}.2.\frac{y}{x}=4\) ( Cosi ) 

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=1\)

... 

xin chào bạn khỏe không mình đang tập nói tiếng việt

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

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

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

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

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

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

\(M=\frac{a^2+1+a}{a^2+1}.\frac{\left(a-1\right)\left(a^2+1\right)}{\left(a-1\right)^2}\)

\(M=\frac{a^2+1+a}{a-1}\)

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

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

\(M=\frac{a^2+a+1}{a^2+1}\cdot\frac{a^2+1}{a-1}=\frac{a^2+a+1}{a-1}\)

p/s: đề là tính M pk ko??

a, ĐỂ A có nghĩa :

\(\Rightarrow x-2\ne0\)

\(\Rightarrow x\ne2\)

\(a,\text{để a xác định thì }\hept{\begin{cases}x-2\ne0\\2-x\ne0\end{cases}\Rightarrow x\ne2}\)

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

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

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

Vậy với \(x=\frac{1}{2}\text{ }\Rightarrow A=\frac{-\frac{4.1}{2}}{\frac{1}{2}-2}=\frac{4}{3}\)