\(a,(x-2)^2-25=0\\\Leftrightarrow (x-2)^2=25\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)

\(---\)

\(b,4x(x-2)+x-2=0\\\Leftrightarrow4x(x-2)+(x-2)=0\\\Leftrightarrow(x-2)(4x+1)=0\)

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

\(---\)

\(c,4x(x-2)-x(3+4x)(?)\)

\(d,(2x-5)^2-3x(5-2x)=0\\\Leftrightarrow(2x-5)^2+3x(2x-5)=0\\\Leftrightarrow(2x-5)(2x-5+3x)=0\\\Leftrightarrow(2x-5)(5x-5)=0\)

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

\(---\)

\(e,x^2-25-(x+5)=0(sửa.đề)\\\Leftrightarrow(x^2-5^2)-(x+5)=0\\\Leftrightarrow (x-5)(x+5)-(x+5)=0\\\Leftrightarrow(x+5)(x-5-1)=0\\\Leftrightarrow(x+5)(x-6)=0\)

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

\(---\)

\(f,5x(x-3)-x+3=0\\\Leftrightarrow5x(x-3)-(x-3)=0\\\Leftrightarrow(x-3)(5x-1)=0\)

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

\(Toru\)

a/

\(\left(x-1\right)^2-\left(x+1\right)^2=2x-6\\ x^2-2x+1-\left(x^2+2x+1\right)=2x-6\\ \)

\(\Leftrightarrow x^2-2x+1-x^2-2x-1-2x+6=0\)

\(\Leftrightarrow6-6x=0\)

=> x=1

b, \(4x^2+12x+9-3\left(x^2-16\right)=x^2-4x+4\)

\(\Leftrightarrow12x+9+48=-4x+4\Leftrightarrow16x=-53\Leftrightarrow x=-\dfrac{53}{16}\)

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

\(\Leftrightarrow\left(x-3\right)\left(x+3-5x-2\right)=0\Leftrightarrow\left(x-3\right)\left(-4x+1\right)=0\Leftrightarrow x=3;x=\dfrac{1}{4}\)

d, \(x^2\left(x+4\right)-9\left(x+4\right)=0\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x+4\right)=0\Leftrightarrow x=-3;3;-4\)

rút gọn à banj

đúng rồi á

Bài 1: Rút gọn

a) Ta có: \(2x\left(x-5\right)-\left(x-2\right)^2-\left(x+3\right)\left(x-3\right)\)

\(=2x^2-10x-\left(x^2-4x+4\right)-\left(x^2-9\right)\)

\(=2x^2-10x-x^2+4x-4-x^2+9\)

\(=-6x+5\)

b) Ta có: \(\left(2x-3\right)^2+3-x^2+\left(4x-6\right)\left(x-3\right)\)

\(=4x^2-12x+9+3-x^2+4x^2-12x-6x+18\)

\(=7x^2-30x+30\)

Bài 2: Tìm x

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

\(\Leftrightarrow x^2-4x+4-\left(x^2-9\right)=0\)

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

\(\Leftrightarrow-4x+13=0\)

\(\Leftrightarrow-4x=-13\)

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

Vậy: \(x=\frac{13}{4}\)

b) Ta có: \(\left(2x+1\right)^2+2\left(4x^2-1\right)+\left(2x-1\right)^2=0\)

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

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

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

\(\Leftrightarrow16x^2=0\)

mà 16≠0

nên \(x^2=0\)

hay x=0

Vậy: x=0

Bài 3:

Ta có: \(A=\left(3x-y\right)^2-\left(3x+y\right)^2\)

\(=\left[3x-y-\left(3x+y\right)\right]\cdot\left(3x-y+3x+y\right)\)

\(=\left(3x-y-3x-y\right)\cdot6x\)

\(=6x\cdot\left(-2y\right)=-12xy\)

Thay \(x=\frac{1}{2}\) và \(y=\frac{1}{3}\) vào biểu thức A=-12xy, ta được:

\(A=-12\cdot\frac{1}{2}\cdot\frac{1}{3}=-2\)

Vậy: -2 là giá trị của biểu thức \(A=\left(3x-y\right)^2-\left(3x+y\right)^2\) tại \(x=\frac{1}{2}\) và \(y=\frac{1}{3}\)

Bài 4: Chứng minh

a) Ta có: \(x^2-4x+5\)

\(=x^2-4x+4+1\)

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

Ta có: \(\left(x-2\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-2\right)^2+1\ge1>0\forall x\)

hay \(x^2-4x+5>0\forall x\)

- Đặt lẻ câu hỏi bạn nhớ không nên đặt quá nhiều như vậy nha

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hãy giải mã ròi bạn sẽ có kết quả

a, (3x-2)(4x+5)=0

↔ TH1: 3x-2 = 0 ↔ x = 2/3

  TH2 : 4x+5 = 0 ↔ x = -5/4

Vậy PT có tập no S = ( 2/3; -5/4)

b,(2,3x-6,9)(0,1x+2)=0

↔ TH1: 2,3x - 6,9 = 0 ↔ x = 3

    TH2 : 0,1x + 2 = 0 ↔ x = -20

Vậy PT có tập no S = ( 3; -20)

c, (4x+2)(x^2 +1)=0

TH1: 4x+2=0 ↔ x = -1/2

Th2 : x^2 +1≠0 ( vô lí)

Vậy PT có tập no S = (-1/2)

d, (2x+7)(x-5)(5x+1)=0

↔ TH1: 2x+7 = 0 ↔ x = -7/2

 TH2: x-5 = 0 ↔ x = 5

TH3 :  5x+1 = 0 ↔ x = -1/5

Vậy PT có tập no S = ( -7/2 ; 5 ; -1/5

a, \(\left(3x-2\right)\left(4x+5\right)=0\Leftrightarrow x=\frac{2}{3};x=-\frac{5}{4}\)

b, \(\left(2,3-6,9\right)\left(0,1x+2\right)=0\Leftrightarrow\frac{x}{10}+2=0\Rightarrow x=-20\)

c, \(\left(4x+2\right)\left(x^2+1>0\right)=0\Leftrightarrow x=-\frac{1}{2}\)

a) Đặt  \(A=4x-x^2-5\)

\(-A=x^2-4x+5\)

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

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

Mà  \(\left(x-2\right)^2\ge0\forall x\)

\(\Rightarrow-A\ge1\)

\(\Leftrightarrow A\le-1< 0\left(đpcm\right)\)

b) Đặt  \(B=x^2-2x+5\)

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

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

Mà  \(\left(x-1\right)^2\ge0\forall x\)

\(\Rightarrow B\ge4>0\left(đpcm\right)\)

a)4x-x²-5 = -(x²-4x+4)-1= -(x-2)^2 -1 < 0 với mọi x (đpcm)

b) x² -2x+5= (x²-2x+1)+4=(x-1)^2 +4 >0  với mọi x (đpcm)

\(1,4x\left(1-x\right)-8=1-\left(4x^2+3\right)\\ \Leftrightarrow4x-4x^2-8=1-4x^2-3\\ \Leftrightarrow4x-4x^2-8-1+4x^2+3=0\\ \Leftrightarrow4x-6=0\\ \Leftrightarrow x=\dfrac{3}{2}\)

\(2,\left(2-3x\right)\left(x+11\right)=\left(3x-2\right)\left(2-5x\right)\\ \Leftrightarrow\left(2-3x\right)\left(x+11\right)-\left(2-3x\right)\left(5x-2\right)=0\\ \Leftrightarrow\left(2-3x\right)\left(x+11-5x+2\right)=0\\ \Leftrightarrow\left(2-3x\right)\left(-4x+13\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{13}{4}\end{matrix}\right.\)

\(x^8+x^7+1\)

\(=x^8+x^7+x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x-xx+1\)

\(=\left(x^8-x^6+x^5-x^3+x^2\right)\)

\(+\left(x^7-x^5+x^4-x^2+x\right)\)

\(+\left(x^6-x^4+x^3-x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)

\(x^5+x+1\)

\(=x^5-x^2+x^2+x+1\)

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

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

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

\(x^4+2x^2-24\)

\(=x^4+2x^2+1-25\)

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

\(=\left(x^2+6\right)\left(x^2-4\right)\)

\(=\left(x^2+6\right)\left(x+2\right)\left(x-2\right)\)