\(\left(1-3x^2\right)^2=\left(5x+2\right)^2\\ =>\left[{}\begin{matrix}1-3x^2=5x+2\\1-3x^2=-\left(5x+2\right)\end{matrix}\right.< =>\left[{}\begin{matrix}3x^2+5x+1=0\\3x^2-5x-3=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=\dfrac{-5\pm\sqrt{13}}{6}\\x=\dfrac{5\pm\sqrt{61}}{6}\end{matrix}\right.\\ =>S=\left\{\dfrac{-5+\sqrt{13}}{6};\dfrac{-5-\sqrt{13}}{6};\dfrac{5+\sqrt{61}}{6};\dfrac{5-\sqrt{61}}{6}\right\}\)

= (3x)⁴ + 4(3x)³(4y) + 6(3x)²(4y)² + 4(3x)(4y)³ + (4y)³

= 81x⁴ + 432x³y + 864x²y² + 768xy³ + 64y³

a + b = 8 ⇔ (a+b)³ = 8³ = 512

nếu a + b = 8 thì ( a + b ) mũ 3 + 8 mũ 3

8 mũ 3 = 8 x 8 x 8 =512

mình nghĩ ab = - 3 đó là gợi ý đánh lừa người làm bài

\(x^2+4y^2=115-2x\\ < =>\left(x^2+2x+1\right)+4y^2=116\\ < =>\left(x+1\right)^2+\left(2y\right)^2=116=\left(\pm10\right)^2+\left(\pm4\right)^2\\ \)

TH1 : `x+1=10;2y=4<=>x=9;y=2`

TH2 : `x+1=10;2y=-4<=>x=9;y=-2`

TH3 : `x+1=-10;2y=4<=>x=-11;y=2`

TH4 : `x+1=-10;2y=-4<=>x=-11;y=-2`

TH5 : `x+1=4;2y=10<=>x=3;y=5`

TH6 : `x+1=4;2y=-10<=>x=3;y=-5`

TH7 : `x+1=-4;2y=10<=>x=-5;y=5`

TH8 : `x+1=-4;2y=-10<=>x=-5;y=-5`

a) \(\left|x+2,7\right|\ge0\forall x=>4,5+\left|x+2,7\right|\ge4,5\\ =>A\ge4,5=>GTNN=4,5\)

Dấu ''='' xảy ra khi :

\(\left|x+2,7\right|=0< =>x=-2,7\)

b) \(\left|3x+3,4\right|\ge0\forall x=>\left|3x+3,4\right|-14\ge-14\\ =>B\ge-14=>GTNN=-14\)

Dấu ''='' xảy ra khi :

\(\left|3x+3,4\right|=0< =>3x=-3,4< =>x=-\dfrac{17}{15}\)

\(8x^2\left(x-5\right)-4x\left(5-x\right)\\ =8x^2\left(x-5\right)+4x\left(x-5\right)\\ =\left(x-5\right)\left(8x^2+4x\right)\\ =4x\left(x-5\right)\left(2x+1\right)\)

= (8x² - 4x) (x-5)