b) \(x^2+2\sqrt{3}x-6=0\)

\(\Leftrightarrow\) \(x^2+2\sqrt{3}x+3-9=0\)

\(\Leftrightarrow\) \(\left(x+\sqrt{3}\right)^2-9=0\)

\(\Leftrightarrow\) \(\left(x+\sqrt{3}-3\right).\left(x+\sqrt{3}+3\right)=0\)

\(\Leftrightarrow\) \(\left[\begin{array}{} x+\sqrt{3}-3=0 \\ x+\sqrt{3}+3=0 \end{array} \right.\)\(\Leftrightarrow\) \(\left[\begin{array}{} x= 3-\sqrt{3} \\ x= -3-\sqrt{3} \end{array} \right.\)

Vậy phương trình có tập nghiệm là S={\(3-\sqrt{3};-3-\sqrt{3}\)}

hay ak m hjhj

rất cần có những bài như thế này để mn tham khảo, cám ơn bn

\(\left(x+y\right)^2-2xy=x^2+y^2=4^2-2.1=14\)

\(x^4+y^4=\left(x^2+y^2\right)^2-2x^2y^2=14^2-2=196-2=194\)

\(x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)=4\left(14-1\right)=52\)

\(\left(x^4+y^4\right)\left(x+y\right)=194.4=776\Leftrightarrow x^5+y^5+x^4y+y^4x=\left(x^5+y^5\right)+xy\left(x^3+y^3\right)=\left(x^5+y^5\right)+1.52=\left(x^5+y^5\right)+52=776\Rightarrow x^5+y^5=724\)

\(\left\{{}\begin{matrix}x+y=4\\xy=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2+2xy+y^2=16\\4xy=4\end{matrix}\right.\Rightarrow x^2+2xy-4xy+y^2=\left(x-y\right)^2=12mà:x>y\Leftrightarrow x-y>0\Rightarrow x-y=\sqrt{12}=2\sqrt{3};x+y=2.2\Rightarrow\left\{{}\begin{matrix}x=\sqrt{3}+2\\y=2-\sqrt{3}\end{matrix}\right.\)

\(x^2-y^2=\left(x-y\right)\left(x+y\right)=4.2\sqrt{3}=8\sqrt{3}\)

\(\left(x^2+y^2\right)\left(x^2-y^2\right)=8\sqrt{3}.14=112\sqrt{3}\Rightarrow x^4-y^4=112\sqrt{3}\)

\(\left(x^3-y^3\right)=\left(x-y\right)\left(x^2+xy+y^2\right);x^6-y^6=\left(x^3+y^3\right)\left(x^3-y^3\right)tựlm\)

Cái này mình biết chút... nhưng mà giải trên đây không tiện lắm bạn có chới zalo ko gửi ad qua cho mình để kp rồi mình gửi lời giải qua luôn...

ok pn. Số zalo của mk là: 037 678 1096. Cảm ơn bạn nhiều

2)Đầu tiên ta cm bđt:\(xy+yz+zx\le\dfrac{\left(x+y+z\right)^2}{3}\)

\(\Leftrightarrow3\left(xy+yz+zx\right)\le x^2+y^2+z^2+2\left(xy+yz+zx\right)\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx\ge0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)(luôn đúng)

\(\Rightarrow xy+yz+zx\le3\)

"="<=>x=y=z=1

Ta có:\(\left\{{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2=2009\end{matrix}\right.\)

\(\Rightarrow ab+bc+ca=\dfrac{-2009}{2}\)

\(\Rightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=1009020,25\)

\(\Rightarrow2\left(a^2b^2+b^2c^2+c^2a^2\right)=2018040,5\)

\(\Rightarrow a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2\left(a^2b^2+b^2c^2+c^2a^2\right)=\)2018040,5

a)Trừ theo vế của \(pt\left(2\right)\) cho \(pt\left(1\right)\):

\(\left(5x+3y\right)-\left(3x+2y\right)=-4-1\)

\(\Leftrightarrow2x+y=-5\). Khi đó

\(3x+2y=1\Leftrightarrow2\left(2x+y\right)-x=1\)

\(\Leftrightarrow2\cdot\left(-5\right)-x=1\)\(\Leftrightarrow x=-11\)

\(\Rightarrow3x+2y=1\Rightarrow y=\dfrac{1-3x}{2}=\dfrac{1-3\cdot\left(-11\right)}{2}=17\)

Vậy nghiệm hpt \(\left(x;y\right)=\left(-11;17\right)\)

b)\(2x^2+2\sqrt{3}x-3=0\)

\(\Delta=\left(2\sqrt{3}\right)^2-\left(4\cdot2\cdot\left(-3\right)\right)=36\)

\(\Rightarrow x_{1,2}=\dfrac{-2\sqrt{3}\pm\sqrt{36}}{4}\)

c)\(9x^4+8x^2-1=0\)

\(\Leftrightarrow9x^4-x^2+9x^2-1=0\)

\(\Leftrightarrow x^2\left(9x^2-1\right)+\left(9x^2-1\right)=0\)

\(\Leftrightarrow\left(9x^2-1\right)\left(x^2+1\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}3x-1=0\\3x+1=0\\x^2+1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\pm\dfrac{1}{3}\\x^2+1>0\left(loai\right)\end{matrix}\right.\)