Đặt \(A=\sqrt{x^2+2002}\)thì \(a^2=x^2+2002\Leftrightarrow a^2-x^2=2002\)

pt: \(\Leftrightarrow x^4+a=a^2-x^2\Leftrightarrow x^4-a^2+x^2+a=0\Leftrightarrow\left(x^2-a\right)\left(x^2+a\right)+\left(x^2+a\right)=0\)

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

\(x^2>0;a\ge\sqrt{2002}\)nên: \(x^2-a+1=0\Leftrightarrow x^2+1=\sqrt{x^2+2002}\)

Do 2 vế đều không âm nên ta bình phương 2 vế:\(x^4+2x^2+1=x^2+2002\Leftrightarrow x^4+x^2-2001=0\)

Tới đây pt trùng phương giải tiếp đi bn.

\(\frac{2002x^4+x^4\sqrt{x^2+2002}+x^2}{2001}=2002\)

\(\frac{x^2\left(x^2+2002\right)+x^4\sqrt{x^2+2002}}{2001}=2002\)

\(x^2\sqrt{x^2+2002}\left(\sqrt{x^2+2002}+x^2\right)=2002.2001\)

đặt x^2+2002=a

a-2002=x^2

pt \(\left(a-2002\right)\sqrt{a}\left(\sqrt{a}+a-2002\right)=2002.2001\)

\(x^4+\sqrt{x^2+2002}=2002\) (DKXĐ: xác định vs mọi x)

\(\Leftrightarrow x^4+x^2+\frac{1}{4}+\sqrt{x^2+2002}=x^2+2002+\frac{1}{4}\)

\(\Leftrightarrow x^4+x^2+\frac{1}{4}=x^2+2002-\sqrt{x^2+2002}+\frac{1}{4}\)

\(\Leftrightarrow\left(x^2+\frac{1}{2}\right)^2=\left(\sqrt{x^2+2002}-\frac{1}{2}\right)^2\)

xét \(x^2+\frac{1}{2}=\sqrt{x^2+2002}-\frac{1}{2}\Leftrightarrow x^2+1=\sqrt{x^2+2002}\)

\(\Leftrightarrow x^4+2x^2+1=x^2+2002\Leftrightarrow x^4+x^2-2001=0\)

đặt x²=a(a>0) => a²+a-2001=0

\(\Delta=1+4.2001=8005\rightarrow\left[\begin{matrix}a=\frac{\sqrt{8005}-1}{2}\\a=\frac{-\sqrt{8005}-1}{2}\end{matrix}\right.\)

mà a>0 \(\rightarrow a=\frac{\sqrt{8005}-1}{2}\Leftrightarrow x=\pm\sqrt{\frac{\sqrt{8005}-1}{2}}\)

xét\(x^2+\frac{1}{2}=\frac{1}{2}-\sqrt{x^2+2002}\Leftrightarrow x^2=-\sqrt{x^2+2002}\)(vô nghiệm)

vậy pt có 2 nghiệm là...

bạn ly ơi alibaa làm liệu có đúng k o toán math y

\(x^4+\sqrt{x^2+2002}=2002\)

Đặt \(\sqrt{x^2+2002}=a^2>0\)

\(\Rightarrow\hept{\begin{cases}x^4+a^2=2002\left(1\right)\\a^4-x^2=2002\left(2\right)\end{cases}}\)

Lấy (1) - (2) ta được

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

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

\(\Leftrightarrow x^2+1=a^2=\sqrt{x^2+2002}\)

\(\Leftrightarrow x^4+2x^2+1=x^2+2002\)

\(\Leftrightarrow x^4+x^2-2001=0\)

Tới đây thì đơn giản rồi

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

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

\(\Leftrightarrow x^2=8\)

\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{8}\\x=-\sqrt{8}\end{cases}}\)

\(\sqrt{2002+x^2}=2002-x^4\)

\(\Leftrightarrow x^8-4004x^4-x^2+4006002=0\)

\(\Leftrightarrow\left(x^4-x^2-2002\right)\left(x^4+x^2-2001\right)=0\)

Làm tiếp nhé

\(Pt\Leftrightarrow2002x^4+x^4\sqrt{x^2+2002}+x^2-2002.2001=0\)

\(\Leftrightarrow x^4\left(\sqrt{x^2+2002}+2002\right)+x^2-2002.2001=0\)

\(\Leftrightarrow\dfrac{x^4}{\sqrt{x^2+2002}-2002}\left(x^2+2002-2002^2\right)+\left(x^2-2001.2002\right)=0\)

\(\Leftrightarrow\left(x^2-2001.2002\right)\left(\dfrac{x^4}{\sqrt{x^2+2002}-2002}+1\right)=0\)

Done !

Đặt \(\left\{{}\begin{matrix}\sqrt[3]{3x^2-x+2001}=a\\\sqrt[3]{3x^2-7x+2002}=b\\\sqrt[3]{6x-2003}=c\end{matrix}\right.\)

\(\Rightarrow a^3-b^3-c^3=2002\) từ đây ta có:

\(a-b-c=\sqrt[3]{a^3-b^3-c^3}\)

\(\Leftrightarrow\left(a-b-c\right)^3=\sqrt[3]{a^3-b^3-c^3}\)

\(\Leftrightarrow\left(a-c\right)\left(a-b\right)\left(b+c\right)=0\)

Tự làm nốt nhé

Xem lại đề nhé bạn: \(\sqrt[3]{6x-2003}\) mới đúng chứ nhỉ?

a picece of cake

\(\frac{\sqrt{x-2002}}{x-2002}-\frac{1}{x-2002}+\frac{\sqrt{y-2003}}{y-2003}-\frac{1}{y-2003}+\frac{\sqrt{z-2004}}{z-2004}-\frac{1}{z-2004}=\frac{3}{4}\)

\(1-\frac{1}{x-2002}+1-\frac{1}{y-2003}+1-\frac{1}{z-2004}=\frac{3}{4}\)

\(3-\frac{1}{x-2002}-\frac{1}{y-2003}-\frac{1}{z-2004}=\frac{3}{4}\)

\(\frac{1}{x-2002}+\frac{1}{y-2003}+\frac{1}{z-2004}=3-\frac{3}{4}=\frac{9}{4}\)

=> không có giá trị x,y,z thỏa mãn đề