Bạn hỏi sớm hơn nữa nhé hỏi mụn lúc này ít ai tloi lắm

a) \(A=\frac{1}{4}x^2+x-2\)

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

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

Vì \(\left(\frac{1}{2}x+1\right)^2\ge0;\forall x\)

\(\Rightarrow\left(\frac{1}{2}x+1\right)^2-3\ge0-3;\forall x\)

Hay \(A\ge-3;\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\left(\frac{1}{2}x+1\right)^2=0\)

                        \(\Leftrightarrow x=-2\)

Vậy MIN A=-3 \(\Leftrightarrow x=-2\)

Các câu khác cứ việc khai triển ra hằng đẳng thức mũ chẵn mà làm nhé

Để P=-1 \(\Rightarrow\frac{4x-2}{4x^2-9}=-1\)

\(\Leftrightarrow4x-2=-\left(4x^2-9\right)\)

\(\Leftrightarrow4x-2=9-4x^2\)

\(\Leftrightarrow4x^2+4x-2-9=0\)

\(\Leftrightarrow4x^2+4x-11=0\)

????

\(P=-1\Rightarrow P=\frac{4x-2}{4x^2-9}=-1\)

\(P=\frac{4x-2}{4x^2-9}=-1\)

<=> \(\frac{2\left(2x-1\right)}{\left(2x\right)^2-3^2}=-1\)

<=> \(\frac{2\left(2x-1\right)}{\left(2x+3\right)\left(2x-3\right)}=-1\)

<=> \(2\left(2x-1\right)=-\left(2x+3\right)\left(2x-3\right)\)

<=> \(4x-2=-4x^2-6x+6x+9\)

<=> \(4x-2=-4x^2+9\)

<=> \(4x-2+4x^2-9=0\)

<=> \(4x-11+4x^2=0\)

<=> \(\orbr{\begin{cases}x=\frac{-4+8\sqrt{3}}{8}\\x=\frac{-4-8\sqrt{3}}{8}\end{cases}}\)<=> \(\orbr{\begin{cases}x=\frac{-1-2\sqrt{3}}{2}\\x=\frac{-1+2\sqrt{3}}{2}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{-1-2\sqrt{3}}{2}\\x=\frac{-1+2\sqrt{3}}{2}\end{cases}}\)

Ta có: \(P=\frac{1}{x+5}=-3\) 

               \(\Rightarrow x+5=\frac{-1}{3}\Rightarrow x=\frac{-16}{3}\) 

Ta lại có:\(Q=9x^2-42x+49=\left(3x-7\right)^2\)

                     \(=\left(3.\frac{-16}{3}-7\right)^2=529\)

Vậy......

Ta có: P = -3

=> \(\frac{1}{x+5}=-3\)

=> \(-3\left(x+5\right)=1\)

=> -3x - 15 = 1

=> -3x = 1 + 15

=> -3x = 16

=> x = 16 : (-3) = -16/3

Với x = -16/3 thay vào Q, ta được:

Q = 9.(-16/3)² - 42.(-16/3) + 49

Q = 9. 256/9 + 224 + 49

Q = 256  + 224 + 49

Q = 529

Vậy ...

a) \(x^3+2x^2-4x+1\)

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

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

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

c) cho da thuc P(x) =2x^4-7x^3 -2x^2 +13x +6? | Yahoo Hỏi & Đáp

Tham khảo

a) (x - 1)³ + (2 - x)(4 + 2x + x²) + 3x(x + 2) = 12

<=> x³ - 2x² + x - x² + 2x - 1 + 8 + 4x + 2x² - 4x - 2x² + 3x² + 6x = 17

<=> 9x + 7 = 17

<=> 9x = 17 - 7

<=> 9x = 10

<=> x = \(\frac{10}{9}\)

b) (x + 2)(x² - 2x + 4) - x(x² - 2) = 15

<=> x³ - 2x² + 4x + 2x² - 4x + 8 - x³ + 2x = 15

<=> 2x + 8 = 15

<=> 2x = 15 - 8

<=> 2x = 7

<=> x = \(\frac{7}{2}\)

c) (x - 3)³ - (x - 3)(x² + 3x + 9) + 9(x² + 1)² = 15

<=> x³ + 45x - 18 - x³ - 3x² - 9x + 3x² + 9x + 27 = 15

<=> 45x + 9 = 15

<=> 45x = 15 - 9

<=> 45x = 6

<=> x = \(\frac{6}{45}\)

d) x(x - 5)(x + 5) - (x + 2)(x² - 2x + 4) = 3

<=> x³ - 25x - x³ + 2x² - 4x - 8 = 3

<=> -25x - 8 = 3

<=> -25x = 3 + 8

<=> -25x = 11

<=> x = \(-\frac{11}{25}\)

a)\(\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=17\)

\(=>x^3-3x^2+3x-1+8-x^3+3x^2+6x=17\)

\(=>9x+7=17=>9x=10=>x=\frac{10}{9}\)

Ưm 

đề rồi sai bạn

Vui lòng sửa đề , hok tốt!

Ta có: \(x^4+6^4=x^4+72x^2+6^4-72x^2\)  

\(=\left(x^2+36\right)-72x^2\) 

\(=\left(x^2+36-\sqrt{72}x\right)\left(x^2+36+\sqrt{72}x\right)\)