\(TH_1:x\ge0\Leftrightarrow x^3\ge0\Leftrightarrow VT>0\left(loại\right)\)

\(TH_2:x< 0\)

Với \(x=-1\Leftrightarrow VT=4\cdot9\cdot14\cdot29>0\left(loại\right)\)

Với \(x=-2\Leftrightarrow VT=-3\cdot2\cdot7\cdot23< 0\left(nhận\right)\)

Với \(x=-3\Leftrightarrow VT=-22\left(-17\right)\left(-12\right)\cdot3< 0\left(nhận\right)\)

Với \(x< -4\Leftrightarrow x^3< -64\Leftrightarrow x^3+5< x^3+10< x^3+15< x^3+30< 0\)

Do đó cả 4 thừa số trong tích đều âm nên tích này luôn dương

Vậy \(x\in\left\{-2;-3\right\}\)

a. 3​,6 : 1,2 x 3 : 0,3                          
​= 3 x 10 = 30
​b. 3,6 : 12 x 3 + 5 x 10 - 30 
= 0,3 x 3 + 50 - 30
​= 0,9 + 20 = 20,9
​c. 3,6 : (1,2 x 3) + 5 x (10:2x5) - 30
= 3,6 ; 3,6 + 5 x 25 -30 
​= 1 + 125 - 30 = 96 
d. 1500 + 4,5 : 25 - 1,5 
= 1500 + 0,18 - 1,5 
​= 1500,18 - 1,5 = 1498,68

Nhớ cho m nha.

a. 3,6:1,2x3:0,3=(3,6:1,2)x(3:0,3)=3x10=30

b. 3,6:12x3+5x10-30=(3,6:12)x3+50-30=0,3x3+20=0,9+20=19,1

c. 3,6:(1,2x3)+5x(10:2x5)-30=3,6:3,6+5x25-30=1+125-30=96

d. 1500+4,5:25-1,5=1500+4,5:5:5-1,5=1500+0,9:5-1,5=1500+0,18-1,5=1500,18-1,5=1499,68

phấn b.các bạn thử tính lại xem . Phần B được 4 hai bạn ạ

a. (3x - 1)² - (x + 3)² = 0

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

\(\Leftrightarrow\left(4x+2\right)\left(2x-4\right)=0\)

\(\Leftrightarrow4x+2=0\)  hoặc  \(2x-4=0\)

1. \(4x+2=0\Leftrightarrow4x=-2\Leftrightarrow x=-\dfrac{1}{2}\)

2. \(2x-4=0\Leftrightarrow2x=4\Leftrightarrow x=2\)

S=\(\left\{-\dfrac{1}{2};2\right\}\)

b. \(x^3=\dfrac{x}{49}\)

\(\Leftrightarrow49x^3=x\)

\(\Leftrightarrow49x^3-x=0\)

\(\Leftrightarrow x\left(49x^2-1\right)=0\)

\(\Leftrightarrow x\left(7x+1\right)\left(7x-1\right)=0\)

\(\Leftrightarrow x=0\) hoặc  \(7x+1=0\) hoặc \(7x-1=0\)

1. x=0

2. \(7x+1=0\Leftrightarrow7x=-1\Leftrightarrow x=-\dfrac{1}{7}\)

3. \(7x-1=0\Leftrightarrow7x=1\Leftrightarrow x=\dfrac{1}{7}\)

*Cách khác:

a) Ta có: \(\left(3x-1\right)^2-\left(x+3\right)^2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=-x-3\\3x-1=x+3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-2\\2x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=2\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{1}{2};2\right\}\)

 a) Ta thấy \(xy=\dfrac{\left(x+y\right)^2-\left(x^2+y^2\right)}{2}=\dfrac{3^2-5}{2}=2\)

\(\Rightarrow x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)\) \(=3\left(5-2\right)=9\)

 b) Ta thấy \(xy=\dfrac{-\left(x-y\right)^2+\left(x^2+y^2\right)}{2}=\dfrac{15-5^2}{2}=-5\)

\(\Rightarrow x^3-y^3=\left(x-y\right)\left(x^2+y^2+xy\right)\) \(=5\left(15-5\right)=50\)

\(C=\left(5^3-1\right)\cdot\left(5^3-2\right)\cdot...\cdot\left(5^3-125\right)\cdot...\cdot\left(5^3-2014\right)\cdot\left(5^3-2015\right)=0\)

i) x³- 11x² + 30x

=\(x\left(x^2-11x+30\right)\)

=\(x\left(x-6\right)\left(x-5\right)\)

j) 4x⁴- 21x²y² + y⁴

=4x^4+4x^2y^2+y^4-25x^2y^2

=(2x^2+y^2)^2-(5xy)^2

=(2x^2+y^2-5xy)(2x^2+y^2+5xy)

              k,x³ + 4x²- 7x - 10

\(a.x^2-11x+15=-15.\Leftrightarrow x^2-11x+30=0.\)

\(\Leftrightarrow\left(x-6\right)\left(x-5\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=6.\\x=5.\end{matrix}\right.\)

\(b.2x-3x+10=x.\Leftrightarrow-2x+10=0.\Leftrightarrow x=5.\)

\(c.x^3-4=4.\Leftrightarrow x^3=8.\Leftrightarrow x^3=2^3.\Rightarrow x=2.\)

\(d.x^4+x^3-x^2-x=0.\Leftrightarrow x^2\left(x^2+x\right)-\left(x^2+x\right)=0.\Leftrightarrow\left(x^2-1\right)\left(x^2+x\right)=0.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0.\\x+1=0.\\x=0.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=-1.\\x=0.\end{matrix}\right.\)

a: \(=\dfrac{x^3-3x^2-7x+x^2-3x-7}{x^2-3x-7}=x+1\)

b:\(=\dfrac{x^3+x^2+3x^2+3x+5x+5}{x+1}=x^2+3x+5\)

c:\(=\dfrac{x^3-3x^2-7x+2x^2-6x-14}{x^2-3x-7}=x+2\)

d: \(=\dfrac{x^2\left(x+5\right)+5x+25-25}{x+5}=x^2+5-\dfrac{25}{x+5}\)