Lời giải:

$x^3+9x=0$

$\Leftrightarrow x(x^2+9)=0$

$\Leftrightarrow x=0$ hoặc $x^2+9=0$

Xét TH $x^2+9=0\Leftrightarrow x^2=-9<0$ (vô lý)

Vậy $x=0$

S(x) = x^9(x - 12) -x^8(x - 12) + x^7(x - 12) + . . . +x(x-12) - (x - 12) - 2

Suy ra:    S(x) = -2 

Nhầm đề

x¹⁰ - 13x⁹ + 13x⁸ - ... - 13x + 13

= (x¹⁰ - 12x⁹) + (- x⁹ + 12x⁸) + ... + (- x + 12) + 1

= x⁹(x - 12) + x⁸(- x + 12) +...+ (- x + 12) + 1 = 1

Bạn tự sửa nhé mình nhầm  - 14 thành 13

x = 79 => 80 = x + 1 thay vòa Px ta có 

P(x) = x^7 - ( x + 1) x^6 + .... + (x + 1) x + 15

         = x^7 - x^7- x^6 + ... + x^2 + x  +15

          = x + 15

          = 79 + 15 

           = 94

Ý B tương tự

Ta có:

C = 13x² + 4y² - 12xy - 2x - 4y + 10

C = (9x² - 12xy + 4y²) + 2(3x - 2y) + 1 + (4x² - 8x + 4) + 5

C = (3x - 2y)² + 2(3x - 2y) + 1 + 4(x² - 2x + 1) + 5

C = (3x - 2y + 1)² + 4(x - 1)² + 5 \(\ge\)5 \(\forall\)x; y

Dấu "=" xảy ra <=> \(\hept{\begin{cases}3x-2y+1=0\\x-1=0\end{cases}}\) <=> \(\hept{\begin{cases}2y=3x+1\\x=1\end{cases}}\) <=> \(\hept{\begin{cases}2y=3.1+1=4\\x=1\end{cases}}\)<=> \(\hept{\begin{cases}y=2\\x=1\end{cases}}\)

Vậy MinC = 5 <=> x = 1 và y = 2

SOS dao lam có thể sử dụng trong bài này!

Chú ý:

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

+) \(C=8\left(x-1\right)^2+5+\left(x-2y+3\right)\left(5x-2y-1\right)\)

Vậy ta tìm được: \(C=\frac{C+C}{2}=\frac{2\left(3x-2y+1\right)^2+8\left(x-1\right)^2+10}{2}\)

\(=\left(3x-2y+1\right)^2+4\left(x-1\right)^2+5\ge5\)

bạn vừa đăng câu này r mà

a: Ta có: \(\left(2x-3\right)^2+6\left(2x-1\right)=7\)

\(\Leftrightarrow\left(2x-3\right)^2+6\left(2x-1\right)-7=0\)

\(\Leftrightarrow4x^2-12x+9+12x-6-7=0\)

\(\Leftrightarrow4x^2=4\)

\(\Leftrightarrow x^2=1\)

hay \(x\in\left\{1;-1\right\}\)

b: Ta có: \(x^2-7x+10=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)

bạn học sd máy tính tìm nghiệm chưa?

a) \(\left(2x-3\right)^2+6\left(2x-1\right)=7\\ \Rightarrow4x^2-12x+9+12x-6-7=0\\ \Rightarrow4x^2-4=0\\ \Rightarrow x^2-1=0\\ \Rightarrow x^2=1\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)

b) \(x^2-7x+10=0\\ \Rightarrow\left(x^2-2x\right)-\left(5x-10\right)=0\\ \Rightarrow\left(x-2\right)\left(x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)

c) \(-6x^2+13x-5=0\\ \Rightarrow-\left(6x^2-13x+5\right)=0\\ \Rightarrow-\left[\left(6x^2-10x\right)-\left(3x-5\right)\right]=0\\ \Rightarrow-\left[2x\left(3x-5\right)-\left(3x-5\right)\right]=0\\ \Rightarrow-\left(2x-1\right)\left(3x-5\right)=0\)

    \(\Rightarrow\left[{}\begin{matrix}-\left(2x-1\right)=0\\3x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\3x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\)

d) \(x^4+7x^2-18=0\\ \Rightarrow\left(x^4-4\right)+\left(7x^2-14\right)=0\\ \Rightarrow\left(x^2-2\right)\left(x^2+2\right)+7\left(x^2-2\right)=0\\ \Rightarrow\left(x^2-2\right)\left(x^2+9\right)=0\\ \Rightarrow\left[{}\begin{matrix}x^2-2=0\\x^2+9=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\pm\sqrt{2}\\x^2=-9\left(loại\right)\end{matrix}\right.\)

\(x^3+2x^2-13x+10=0\)

\(\Rightarrow x^3+2x^2-13x=-10\)

\(\Rightarrow x\times\left(x^2+2x-13\right)=-10\)

\(\Rightarrow x;x^2+2x-13\inƯ\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)

Mà \(x^2+2x-13\)lẻ

\(\Rightarrow x^2+2x-13\in\left\{\pm5\right\}\)

Lập bảng làm tiếp nhé, em ms lớp 7 nên có gì sai sót mong chị bỏ qua.

~Std well~

#Dư Khả

\(x^3+2x^2-13x+10=0\)

\(\Rightarrow x^3-x^2+3x^2-3x-10x+10=0\)

\(\Rightarrow x^2\left(x-1\right)+3x\left(x-1\right)-10\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(x^2+3x-10\right)=0\)

bn lớp 8 rồi nên lm theo kiểu lớp 8 

  x^3 +2x^2+4x^2-8x-5x+10 = 0

x^2 (x-2)+4x(x-2) - 5(x-2) = 0

(x-2)( x^2+4x-5) = 0

(x-2) (x^2-5x+x-5) =0

(x-2) [ x(x-5)+ (x-5)] = 0

(x-2) ( x+1) (x-5) = 0 

=> x có 3 TH :  x-2 =0        => x = 2

                        x+1 = 0       => x = -1

                        x - 5=0        +> x= 5

        vậy c thuộc ( 2 ; -1 ; 5)

a) \(-45:5.\left(-3-2x\right)=3\)

\(-9.\left(-3-2x\right)=3\)

\(-3-2x=\left(3:-9\right)\)

\(-3-2x=\dfrac{-1}{3}\)

\(-2x=-3-\dfrac{1}{3}\)

-2x=\(\dfrac{-10}{3}\)

\(x=\dfrac{-10}{3}:-2\)

\(x=\dfrac{5}{3}\)

b) 

3x - 28 = x + 36

<=> 3x - x = 36 + 28

<=> 2x = 64

<=> x = 32

Vậy x = 32

c) 

(-12)2.x = 56 + 10.13.x

144.x = 56 + 130.x

144x – 130 x = 56

14x = 56

x = 56: 14

x = 4

Vậy x = 4

\(a,-45:5\cdot\left(-3-2x\right)=3\\ \Leftrightarrow-9\cdot\left(-3-2x\right)=3\\ \Leftrightarrow-2x-3=-\dfrac{1}{3}\\ \Leftrightarrow2x+3=\dfrac{1}{3}\\ \Leftrightarrow2x=-\dfrac{8}{3}\\ \Leftrightarrow x=-\dfrac{4}{3}\\ b,3x-28=x+36\\ \Leftrightarrow2x=64\\ \Leftrightarrow x=32\\ c,\left(-12\right)^2x=56+10\cdot13x\\ \Leftrightarrow144x=56+130x\\ \Leftrightarrow14x=56\\ \Leftrightarrow x=4\)