\(\left(\sqrt[3]{3x+1}\right)^2+\left(\sqrt[3]{3x-1}\right)^2+\sqrt[3]{9x^2-1}=1\)

\(\Leftrightarrow\left(\sqrt[3]{3x+1}-\sqrt[3]{3x-1}\right)\left[\left(\sqrt[3]{3x+1}\right)^2+\left(\sqrt[3]{3x-1}\right)^2+\sqrt[3]{9x^2-1}\right]=\sqrt[3]{3x+1}-\sqrt[3]{3x-1}\)

Hay \(3x+1-\left(3x-1\right)=\sqrt[3]{3x+1}-\sqrt[3]{3x-1}\)

Vậy \(\sqrt[3]{3x+1}-\sqrt[3]{3x-1}=2\)

Đặt \(\hept{\begin{cases}\sqrt[3]{3x+1}=a\\\sqrt[3]{3x-1}=b\end{cases}\Rightarrow\hept{\begin{cases}a^2+ab+b^2=1\\a-b=2\end{cases}}}\)\(\Rightarrow a=b+2\text{ thế vào phương trình đầu ta có :}\)

\(\left(b+2\right)^2+b\left(b+2\right)+b^2=1\Leftrightarrow3b^2+6b+3=0\Leftrightarrow b=-1\Leftrightarrow\sqrt[3]{3x-1}==-1\)

\(\Leftrightarrow3x-1=-1\Leftrightarrow x=0\)

Minh Quang ơi tại sao \(3x+1-\left(3x-1\right)=\sqrt[3]{3x+1}-\sqrt[3]{3x-1}\)

\(\Leftrightarrow\sqrt[3]{3x+1}-\sqrt[3]{3x-1}=1\)

\(\sqrt{x+2-4\sqrt{x-2}}+\sqrt{x+7-6\sqrt{x-2}}=1\)ĐK : x >= 2 

\(\Leftrightarrow\sqrt{x-2-4\sqrt{x-2}+4}+\sqrt{x-2-6\sqrt{x-2}+9}=1\)

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

\(\Leftrightarrow\sqrt{x-2}-2+\sqrt{x-2}-3=1\Leftrightarrow2\sqrt{x-2}=6\)

\(\Leftrightarrow\sqrt{x-2}=3\Leftrightarrow x-2=9\Leftrightarrow x=11\)(tmđk)

ĐK : x >= 2

\(\Leftrightarrow\sqrt{\left(x-2\right)-4\sqrt{x-2}+4}+\sqrt{\left(x-2\right)-6\sqrt{x-2}+9}=1\)

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

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

Với 2 ≤ x < 4 (1) trở thành \(2-\sqrt{x-2}+3-\sqrt{x-2}=1\Leftrightarrow-2\sqrt{x-2}=-4\Leftrightarrow\sqrt{x-2}=2\Leftrightarrow x=6\left(ktm\right)\)

Với 4 ≤ x < 11 (1) trở thành \(\sqrt{x-2}-2+3-\sqrt{x-2}=1\Leftrightarrow1=1\left(luondung\right)\)(2)

Với x ≥ 11 (1) trở thành \(\sqrt{x-2}-2+\sqrt{x-2}-3=1\Leftrightarrow2\sqrt{x-2}=6\Leftrightarrow\sqrt{x-2}=3\Leftrightarrow x=11\)( tm ) (3)

Từ (2) và (3) => S = { x | 4 ≤ x ≤ 11 }

a, ĐK : \(4-x^2\ge0\Leftrightarrow\left(2-x\right)\left(x+2\right)\ge0\Leftrightarrow-2\le x\le2\)

b, ĐK : \(16-x^2\ge0\Leftrightarrow\left(4-x\right)\left(x+4\right)\ge0\Leftrightarrow-4\le x\le4\)

c, ĐK : \(x^2-3\ge0\Leftrightarrow\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\ge0\Leftrightarrow x\le-\sqrt{3};x\ge\sqrt{3}\)

\(A=\left|3-0,7.9\right|+\frac{2}{5}=\left|3-2,7\right|+0,4=\left|0,3\right|+0,4=0,3+0,4=0,7\)

\(B=\left|x\right|+x-5=-x+x-5=-5\)

\(C=\left|2-x\right|+\left|x-3\right|=x-2+3-x=1\)

\(\hept{\begin{cases}3x^2-5xy-4y^2=-3\\-8x^2+11xy+9y^2=6\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}6x^2-10xy-8y^2=-6\\-8x^2+11xy+9y^2=6\end{cases}}\)

\(\Rightarrow-2x^2+xy+y^2=0\)

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

\(\Leftrightarrow-2x\left(x-y\right)-y\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(-2x-y\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=y\\y=2x\end{cases}}\)

th1 : x = y  ta có :

3x^2 - 5x^2 - 4x^2 = -3

=>-6x^2 = -3

=> x^2 = 1/2

=> x = +- căn 1/2 = y

th2 : y =2x ta có 

3x^2 - 10x^2 - 4(2x)^2 = -3

=> 3x^2 - 10x^2 - 16x^2 = -3

=> -23x^2 = -3

=> x^2 = 3/23

=> x = +- căn 3/23 

tính nốt y đi

viết này thì sao biết được bạn

a, \(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)

\(=\sqrt{4+\sqrt{15}}\sqrt{4-\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4+\sqrt{15}}\)

\(=\sqrt{16-15}\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4+\sqrt{15}}\)

\(=\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8+2\sqrt{5.3}}=\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\)

\(=\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)=5-3=2\)