*vn:vô nghiệm.

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2=0\\x^2+x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vn\right)\end{matrix}\right.\)

\(\Leftrightarrow x=\pm\sqrt{2}\)

-Vậy \(S=\left\{\pm\sqrt{2}\right\}\).

b. \(16x^2-8x+5=0\)

\(\Leftrightarrow16x^2-8x+1+4=0\)

\(\Leftrightarrow\left(4x-1\right)^2+4=0\) (vô lí)

-Vậy S=∅.

c. \(2x^3-x^2-8x+4=0\)

\(\Leftrightarrow x^2\left(2x-1\right)-4\left(2x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x^2-4\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\pm2\end{matrix}\right.\)

-Vậy \(S=\left\{\dfrac{1}{2};\pm2\right\}\).

d. \(3x^3+6x^2-75x-150=0\)

\(\Leftrightarrow3x^2\left(x+2\right)-75\left(x+2\right)=0\)

\(\Leftrightarrow3\left(x+2\right)\left(x^2-25\right)=0\)

\(\Leftrightarrow3\left(x+2\right)\left(x+5\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\pm5\end{matrix}\right.\)

-Vậy \(S=\left\{-2;\pm5\right\}\)

*\(\left(2x-3\right)^2=\left(x+5\right)^2\)

\(\Rightarrow\left(2x-3\right)^2-\left(x+5\right)^2=0\)

\(\Rightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\)

\(\Rightarrow\left(x-8\right)\left(3x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)

* \(x^3-16x=0\)

\(\Rightarrow x\left(x^2-16\right)=0\)

\(\Rightarrow x\left(x^2-4^2\right)=0\)

\(\Rightarrow x\left(x-4\right)\left(x+4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

*\(x^4-2x^3+10x^2-20x=0\)

\(\Rightarrow\left(x^4+10x^2\right)-\left(2x^3+20x\right)=0\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x=\varnothing\\x=2\end{matrix}\right.\)

a) x³ - 16x = 0

x(x² - 16) = 0

=> x = 0 hoặc x² - 16 = 0

x = 4

Vậy x = 0 hoặc x = 4

b) x⁴ -2x³ + 10x² - 20x = 0

x³ (x - 2) + 10x(x - 2) = 0

(x - 2)(x³ + 10x) = 0

=> x - 2 = 0 hoặc x³ + 10x = 0

x = 2 x(x² + 10) = 0

+ TH1: x = 0

+ TH2: x² + 10 = 0

x² = -10 (vô lí)

Vậy x = 2 hoặc x = 0

c) (2x - 3)² = (x + 5)²

(2x)² + 2 . 2x . 3 + 3² = x² + 2.x.5 + 5²

4x² + 12x + 9 = x² + 10x + 25

4x² + 12x - x² - 10x = 25 - 9

3x² + 2x = 16

x(3x + 2) = 16

Đến đây bạn làm nốt câu c nhé!

a. \(\left(2x-1\right)\left(3x+2\right)\left(5-x\right)=0\)

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

\(\Rightarrow S=\left\{\dfrac{1}{2};\dfrac{-2}{3};5\right\}\)

b. \(\left(2x+5\right)\left(x-4\right)=\left(x-5\right)\left(4-x\right)\)

\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-5\right)\left(x-4\right)\)

\(\Leftrightarrow3x\left(x-4\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

\(\Rightarrow S=\left\{0;4\right\}\)

c. \(16x^2-8x+1=4\left(x+3\right)\left(4x-1\right)\)

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

\(\Leftrightarrow\left(4x-1\right)\left(4x-1-4x-3\right)=0\)

\(\Leftrightarrow-4\left(4x-1\right)=0\Leftrightarrow4x-1=0\Leftrightarrow x=\dfrac{1}{4}\)

d. \(27x^2\left(x+3\right)-12\left(x^2+3x\right)=0\)

\(\Leftrightarrow27x^2\left(x+3\right)-12x\left(x+3\right)=0\)

\(\Leftrightarrow x\left(27x-12\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\27x-12=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{9}\\x=-3\end{matrix}\right.\)

\(\Rightarrow S=\left\{0;\dfrac{4}{9};-3\right\}\)

e. \(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)

\(\Leftrightarrow2\left(3x+1\right)^2-\left(3x+1\right)\left(x-2\right)=0\)

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

\(\Leftrightarrow\left(3x+1\right)\left(7x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\7x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=\dfrac{-3}{7}\end{matrix}\right.\)

\(\Rightarrow S=\left\{\dfrac{-1}{3};\dfrac{-3}{7}\right\}\)

g. \(\left(2x-1\right)^2=49\)

\(\Leftrightarrow2x-1=7\Leftrightarrow x=4\)

Tìm GTNN của biểu thức :

\(x^2+2x+4\)

Đặt A = \(x^2+2x+4\)

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

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

Ta luôn có : \(\left(x+1\right)^2\ge0\forall x\)

Suy ra : \(\left(x+1\right)^2+3\ge3\forall x\)

Hay A\(\ge3\) với mọi x

Dấu "=" xảy ra khi \(x+1=0\Rightarrow x=-1\)

Nên : \(A_{min}=3khix=-1\)

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

\(4+2x.2x+8x=-x\)

\(4x+8x+x=-4\)

\(13x=-4\)

\(x=-\frac{4}{13}\)

 Vậy pt có nghiệm là { -4/13 }

2) mình nghĩ thế này

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

Đẻ 2 cái trên = nhau thfi 

2x-3=1

=> x=2

mấy cái kia cũng làm giống vậy

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

\(\orbr{\begin{cases}x=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)

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

\(\Leftrightarrow\hept{\begin{cases}x=0\\x=1\\x=-1\end{cases}}\)