Em xin phép làm bài EZ nhất :)

4,ĐK :\(\forall x\in R\)

Đặt \(x^2+x+2=t\) (\(t\ge\dfrac{7}{4}\))

\(PT\Leftrightarrow\sqrt{t+5}+\sqrt{t}=\sqrt{3t+13}\)

\(\Leftrightarrow2t+5+2\sqrt{t\left(t+5\right)}=3t+13\)

\(\Leftrightarrow t+8=2\sqrt{t^2+5t}\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge-8\\\left(t+8\right)^2=4t^2+20t\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\3t^2+4t-64=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left(t-4\right)\left(3t+16\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left[{}\begin{matrix}t=4\left(tm\right)\\t=-\dfrac{16}{3}\left(l\right)\end{matrix}\right.\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

Vậy ....

Ta có:

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

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

\(\Leftrightarrow\left(xy-1\right)^2+3=4x\)

Mà \(\left(xy-1\right)^2+3>0\)

Nên 4x>0

x>0

Ta có:

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

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

Mà \(x^2y^2+4>0\forall x,y\)

Nên \(2x\left(y+2\right)>0\)

Mặt khác x>0

nên y+2>0

=> y>-2 (1)

Áp dụng bđt Cosi ta có:

\(x^2y^2+4\ge4xy\)

Mà \(\Leftrightarrow x^2y^2+4=2x\left(y+2\right)\)

Nên \(2x\left(y+2\right)\ge4xy\)

\(\Rightarrow y+2\ge2y\)

\(\Leftrightarrow y\le2\) (2)

Do y \(\in Z\) và ta đã có (1), (2)

Nên \(y\in\left\{-1;0;1;2\right\}\)

Th1: y = -1

\(\Rightarrow x^2-2x\left(-1+2\right)+4=0\)

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

\(\Leftrightarrow\left(x-1\right)^2+3=0\left(vl\right)\)

Th2: y = 0

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

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

\(\Rightarrow x=2\) (nhận)

Th3: y = 1

\(\Rightarrow x^2-2x\left(1+2\right)+4=0\)

\(\Leftrightarrow x^2-6x+4=0\)

\(\Leftrightarrow\left(x-3\right)^2=5\)

\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{5}+3\\x=-\sqrt{5}+3\end{matrix}\right.\)

Loại do x \(\in Z\)

Th4: y = 2

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

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

\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{12}+3\\x=-\sqrt{12}+3\end{matrix}\right.\)

Loại do x \(\in Z\)

Vậy \(\left(x;y\right)\in\left\{2;0\right\}\)

4 Th sai cả rồi

do mình thế ngu

ra y \(\in\left\{-1;0;1;2\right\}\) thì bạn thế vô tính x nhé

Th1 và Th3 thì mình làm đúng rồi

Th2 : y=0

\(\Rightarrow-2x\left(0+2\right)+4=0\)

\(\Leftrightarrow4x=4\Leftrightarrow x=1\) (nhận)

Th4: y=2

\(\Rightarrow4x^2-2x\left(2+2\right)+4=0\)

\(\Leftrightarrow4x^2-8x+4=0\)

\(\Rightarrow x=1\) (nhận)

Vậy \(\left(x;y\right)\in\left\{\left(1;0\right),\left(1;2\right)\right\}\)

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

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

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

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

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

\(\Leftrightarrow2x+1=\left(x^2+1\right)\left(2x+1\right)\)

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

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

\(\Leftrightarrow x^2\left(2x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=0\end{cases}}\) (nhận)

Vậy .....

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

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

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

Vì VT > 0 nên VP >0

\(\Leftrightarrow\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\ge0\)

\(\Leftrightarrow x\ge-\frac{1}{2}\)

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

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

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

                   \(\Leftrightarrow\left(2x-3\right)\left(2x+1\right)-\frac{1}{4}\left(x^2+1\right)^2\left(2x+1\right)^2=0\)

                 \(\Leftrightarrow\left(2x+1\right)\left(2x-3-\frac{1}{4}\left(x^2+1\right)^2\left(2x+1\right)\right)=0\)

                \(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\2x-3=\frac{1}{4}\left(x^2+1\right)^2\left(2x+1\right)\end{cases}}\)

 Cần cù bù thông minh , phá tung pt dưới ra được cái phương trình bậc 5, sau đó dùng Wolfram|Alpha: Computational Intelligence để tính nghiệm rồi phân tích nhân tử =))

X = 0

100% đúng kq

x=0

nhanh nhất có thể

X=0

100% đúng

0% sai