\(a,\left(x^2-4x+11\right)\left(x^4-8x^2+21\right)=35\)

Phương trình trên tương đương với:

\(\left[\left(x-2\right)^2+7\right]\left[\left(x^2-4\right)^2+5\right]=35\left(1\right)\)

Do: \(\hept{\begin{cases}\left(x-2\right)^2+7\ge7\forall x\\\left(x^2-4\right)^2+5\ge5\forall x\end{cases}}\Rightarrow\left[\left(x+2\right)^2+7\right]\left[\left(x^2+4\right)^2+5\right]\ge35\forall x\)

Nên: \(\left(1\right)\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^2+7=7\\\left(x^2-4\right)^2+5=5\end{cases}\Leftrightarrow}x=2\)

Vậy ..................................

\(b,\sqrt{x}+\sqrt{1-x}+\sqrt{x\left(1-x\right)}=1\)

\(Đkxđ:0\le x\le1\) Đặt: \(0< a=\sqrt{x}+\sqrt{1-x}\Rightarrow\frac{a^2-1}{2}=\sqrt{x\left(1-x\right)}\)

\(+)\) Phương trình mới là: \(a+\frac{a^2-1}{2}=1\Leftrightarrow a^2+2a-3=0\Leftrightarrow\left(a-1\right)\left(a+3\right)=0\)

\(\Leftrightarrow a=\left\{-3;1\right\}\Rightarrow a=1>0\)

\(\sqrt{x}+\sqrt{1-x}=1\)

\(+)\) Nếu \(a=1\Leftrightarrow x+1-x+2\sqrt{x\left(1-x\right)}=1\Leftrightarrow\sqrt{x\left(1-x\right)}=0\)

\(\Rightarrow x=\left\{0;1\right\}\left(tm\right)\)

Vậy .............................

\(a,Đk:1\le x\le4\)

Đặt \(y=\sqrt{4-x}+\sqrt{2x-2}\)Ta có: \(y^2=4-x+2x-2+2\sqrt{\left(4-x\right)\left(2x-2\right)}\)

\(\Leftrightarrow x+2+2\sqrt{\left(4-x\right)\left(2x-2\right)}=y^2\Leftrightarrow x+2\sqrt{\left(4-x\right)\left(2x-2\right)}=y^2-2\)

Phương trình trở thành: \(5+y^2-2=4y\)

\(\Leftrightarrow y^2-4y+3=0\)

\(\Leftrightarrow\orbr{\begin{cases}y=1\\y=3\end{cases}}\) ( Vì \(a+b+c=0\))

• \(y=1.\) Ta có: \(\sqrt{4-x}+\sqrt{2x-2}=1\Leftrightarrow\sqrt{2x-2}=1-\sqrt{4-x}\)

\(\Leftrightarrow\hept{\begin{cases}1-\sqrt{4-x}\ge0\\2x-2=\left(1-\sqrt{4-x}\right)^2\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}\sqrt{4-x}\le1\\2x-2=1-2\sqrt{4-x}+4-x\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}0\le4-x\le1\\2\sqrt{4-x}=7-3x\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}3\le x\le4;7-3x\ge0\\4\left(4-x\right)=\left(7-3x\right)^2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\in\varnothing\\4\left(4-x\right)=\left(7-3x\right)^2\end{cases}}\) \(\Leftrightarrow x\in\varnothing\)

• \(y=3\)Ta có: \(\sqrt{4-x}+\sqrt{2x-2}=3\Leftrightarrow\sqrt{2x-2}=3-\sqrt{4-x}\)

\(\Leftrightarrow\hept{\begin{cases}3-\sqrt{4-x}\ge0\\2x-2=\left(3-\sqrt{4-x}\right)^2\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}\sqrt{4-x}\le3\\2x-2=9-6\sqrt{4-x}+4-x\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{4-x}\le3\\2\sqrt{4-x}=5-x\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}0\le4-x\le9;5-x\ge0\\4\left(4-x\right)=\left(5-x\right)^2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}-5\le x\le4\\x^2-6x+9=0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}-5\le x\le4\\\left(x-3\right)^2=0\end{cases}}\Leftrightarrow x=3\)

Vậy pt có nghiệm duy nhất là \(x=3\)

(Làm xong hoa mắt :((

\(\hept{\begin{cases}\left(x+\sqrt{x^2+2012}\right)\left(y+\sqrt{y^2+2012}\right)=2012\left(1\right)\\x^2+z^2-4\left(y+z\right)+8=0\left(2\right)\end{cases}}\)

Ta có:(1) \(\Leftrightarrow\left(x+\sqrt{x^2+2012}\right)\left(y+\sqrt{y^2+2012}\right)\left(\sqrt{y^2+2012}-y\right)\)\(=2012\left(\sqrt{y^2+2012}-y\right)\)(Do \(\sqrt{y^2+2012}-y\ne0\forall y\))

\(\Leftrightarrow2012\left(x+\sqrt{x^2+2012}\right)=2012\left(\sqrt{y^2+2012}-y\right)\)

\(\Leftrightarrow x+\sqrt{x^2+2012}=\sqrt{y^2+2012}-y\)\(\Leftrightarrow x+y=\sqrt{y^2+2012}-\sqrt{x^2+2012}\)

\(\Leftrightarrow x+y=\)\(\frac{\left(\sqrt{y^2+2012}+\sqrt{x^2+2012}\right)\left(\sqrt{y^2+2012}-\sqrt{x^2+2012}\right)}{\sqrt{y^2+2012}+\sqrt{x^2+2012}}\)

\(\Leftrightarrow x+y=\frac{y^2-x^2}{\sqrt{y^2+2012}+\sqrt{x^2+2012}}\)\(\Leftrightarrow\left(x+y\right)\frac{\sqrt{y^2+2012}-y+\sqrt{x^2+2012}+x}{\sqrt{y^2+2012}+\sqrt{x^2+2012}}=0\)

Do \(\hept{\begin{cases}\sqrt{y^2+2012}>\sqrt{y^2}=\left|y\right|\ge y\forall y\\\sqrt{x^2+2012}>\sqrt{x^2}=\left|x\right|\ge-x\forall x\end{cases}}\)\(\Rightarrow\sqrt{y^2+2012}-y+\sqrt{x^2+2012}+x>0\forall x,y\Rightarrow x+y=0\)

\(\Rightarrow y=-x\)

Thay y = -x vào (2), ta được: \(x^2+z^2+4x-4z+8=0\)

\(\Leftrightarrow\left(x+2\right)^2+\left(z-2\right)^2=0\Leftrightarrow\hept{\begin{cases}x=-2\\z=2\end{cases}}\Rightarrow y=-x=2\)

Vậy hệ có nghiệm \(\left(x;y;z\right)=\left(-2;2;2\right)\)

b, \(x^3+3x^2y-4y^3+x-y=0\)

\(\Leftrightarrow x^3-x^2y+4x^2y-4xy^2+4xy^2-4y^3+x-y=0\)

\(\Leftrightarrow x^2\left(x-y\right)+4xy\left(x-y\right)+4y^2\left(x-y\right)+\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x^2+4xy+4y^2+1\right)=0\)

\(\Leftrightarrow x-y=0\Leftrightarrow x=y\)

Khi đó pt (2) của hệ trở thành: 

\(\left(x^2+3x+2\right)\left(x^2+7x+12\right)=24\)

\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=24\)

\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)=24\)

\(\Leftrightarrow\left(x^2+5x+5\right)^2-1=24\)

\(\Leftrightarrow\left(x^2+5x+5\right)^2-5^2=0\)

\(\Leftrightarrow\left(x^2+5x\right)\left(x^2+5x+10\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)

Vậy hệ có nghiệm \(\left(x;y\right)\in\left\{\left(0;0\right),\left(-5;-5\right)\right\}\)

\(a,\hept{\begin{cases}x^2-3y=2\\9y^2-8x=8\end{cases}}\)

\(x^2-3y=2\)

\(y=\frac{1^2-2}{3}\)

\(9-\left(\frac{x^2-2}{3}\right)^2-8x=8\)

\(\Rightarrow x^4-4x^2+4-8x-8=0\)

\(\Rightarrow x^4-4x^2-8x-4=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=1+\sqrt{3}\\x=1-\sqrt{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}y=\frac{2+2\sqrt{3}}{3}\\y=\frac{2-2\sqrt{3}}{3}\end{cases}}\)

Vậy ................................

Ta có: \(4x^2+y^2< 2xy+2x+y+1\)

\(\Leftrightarrow8x^2+2y^2-4xy-4x-2y-2< 0\)

\(\Leftrightarrow\left(4x^2-4xy+y^2\right)+\left(4x^2-4x+1\right)+\left(y^2-2y+1\right)< 4\)

\(\Leftrightarrow\left(2x-y\right)^2+\left(2x-1\right)^2+\left(y-1\right)^2< 4\)

Lại có: \(x,y\in Z^+\Rightarrow2x-1\ne0\)

\(\Rightarrow0< \left(2x-1\right)^2< 4\Rightarrow\left(2x-1\right)^2=1\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=-1\\2x-1=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\left(l\right)\\x=1\end{cases}}\Rightarrow x=1\)

\(0\le\left(y-1\right)^2< 4\Rightarrow\hept{\begin{cases}y-1=0\\y-1=-1\\y-1=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}y=1\Rightarrow\left(2x-y\right)^2=1\left(tm\right)\\y=2\Rightarrow\left(2x-y\right)^2=0\left(tm\right)\end{cases}}\) (Ngoặc nhọn bạn chuyển thành ngoặc vuông nha tại olm không có như h ý.

\(\Rightarrow\left(x,y\right)=\left\{\left(1;1\right);\left(1;0\right)\right\}\)

Vậy ....................