a) (x - 1/2)² = 4

<=> (x - 1/2)² = 2²

<=> x - 1/2 = -2; 2

<=> x - 1/2 = 2 hoặc x - 1/2 = -2

       x = 2 + 1/2         x = -2 + 1/2

       x = 5/2               x = -3/2

=> x = 5/2 hoặc x = -3/2

b) 10/1/2 - (x + 1/3)² = 1/1/2

<=> -(x + 1/3)² = 1/1/2 - 10/1/2

<=> -(x + 1/3)² = 1/2 - 5

<=> -(x + 1/3)² = -5.2 + 1/2

<=> -(x + 1/3)² = -9/2

<=> (x + 1/3)² = 9/2

<=> x + 1/3 = \(\sqrt{\frac{9}{2}}\)  hoặc x + 1/3 = \(-\sqrt{\frac{9}{2}}\)

       x = \(\frac{3\sqrt{2}}{2}\) - 1/3         x = \(-\frac{3\sqrt{2}}{2}\) -1/3

=> x = \(\frac{3\sqrt{2}}{2}\) - 1/3 hoặc x = \(-\frac{3\sqrt{2}}{2}\) -1/3

c) (x - 1/5)² + 17/25 = 26/25

<=> (x - 1/5)² = 26/25 - 17/25

<=> (x - 1/5)² = (3/5)²

<=> x - 1/5 = -3/5; 3/5

<=> x - 1/5 = 3/5 hoặc x - 1/5 = -3/5

       x = 3/5 + 1/5         x = -3/5 + 1/5

       x = 4/5                  x = -2/5

=> x = 4/5 hoặc x = -2/5

a, \(2\left(x+3\right)\left(x-4\right)=\left(2x-1\right)\left(x+2\right)-27\)

\(\Leftrightarrow2\left(x^2-4x+3x-12\right)=2x^2+4x-x-2-27\)

\(\Leftrightarrow2x^2-2x-24=2x^2+3x-29\Leftrightarrow-5x+5=0\Leftrightarrow x=1\)

b, \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x-3\right)\left(x+3\right)=26\)

\(\Leftrightarrow x^3-8-x\left(x^2-9\right)=26\Leftrightarrow-8+9x=26\)

\(\Leftrightarrow9x=18\Leftrightarrow x=2\)

1: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5x+2\right)=0\)

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

hay \(x\in\left\{3;\dfrac{1}{4}\right\}\)

2: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)-\left(x-1\right)\left(x^2-2x+16\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1-x^2+2x-16\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(3x-15\right)=0\)

hay \(x\in\left\{1;5\right\}\)

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

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

hay \(x\in\left\{1;\dfrac{1}{2};-\dfrac{1}{2}\right\}\)

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

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

hay \(x\in\left\{-4;3;-3\right\}\)

5: \(\Leftrightarrow\left[{}\begin{matrix}3x+5=x-1\\3x+5=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-6\\4x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)

6: \(\Leftrightarrow\left(6x+3\right)^2-\left(2x-10\right)^2=0\)

\(\Leftrightarrow\left(6x+3-2x+10\right)\left(6x+3+2x-10\right)=0\)

\(\Leftrightarrow\left(4x+13\right)\left(8x-7\right)=0\)

hay \(x\in\left\{-\dfrac{13}{4};\dfrac{7}{8}\right\}\)

1.

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

\(\Leftrightarrow x+3=5x-2\)

\(\Leftrightarrow4x=5\Leftrightarrow x=\dfrac{5}{4}\)

2.

\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=\left(x-1\right)\left(x^2-2x+16\right)\)

\(\Leftrightarrow x^2+x+1=x^2-2x+16\)

\(\Leftrightarrow3x=15\Leftrightarrow x=5\)

3.

\(\Leftrightarrow4x^2\left(x-1\right)-\left(x-1\right)=0\)

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

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

7.

\(\Leftrightarrow x^2+2x-15=0\)

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

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

8.\(\Leftrightarrow x^4+x^3+4x^3+4x^2=0\)

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

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

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

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

\(\Leftrightarrow x+2=3-2x\)

\(\Leftrightarrow3x=1\Leftrightarrow x=\dfrac{1}{3}\)

Bài 1:

\(A=26^2-24^2=\left(26-24\right)\left(26+24\right)=2\cdot50=100\)

\(B=27^2-25^2=\left(27-25\right)\left(27+25\right)=2\cdot52=104\)

=>A<B

Bài 2:

\(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)=11\)

=>\(4\left(x^2+2x+1\right)+4x^2-4x+1-8\left(x^2-1\right)=11\)

=>\(4x^2+8x+4+4x^2-4x+1-8x^2+8=11\)

=>4x+13=11

=>4x=-2

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

1) Ta có: \(3\left(x-1\right)-5\left(x-2\right)=4\left(x+1\right)\)

\(\Leftrightarrow3x-5-5x+10-4x-4=0\)

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

\(\Leftrightarrow-6x=-1\)

hay \(x=\dfrac{1}{6}\)

2) Ta có: \(-2\left(x-2\right)-4\left(x+1\right)=-3\left(x+3\right)\)

\(\Leftrightarrow-2x+4-4x-4+3x+9=0\)

\(\Leftrightarrow-3x=-9\)

hay x=3

3) Ta có: \(3x^2+2x=0\)

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

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

4) Ta có: \(x^2-5x=0\)

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

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

5) Ta có: \(\left(2x-3\right)^2=36\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

6) Ta có: \(\left(5x-1\right)^3=125\)

\(\Leftrightarrow5x-1=5\)

\(\Leftrightarrow5x=6\)

hay \(x=\dfrac{6}{5}\)

7) Ta có: \(3^{x+1}=27\)

\(\Leftrightarrow x+1=3\)

hay x=2

1/

\(3\left(-1-4x^2+5x\right)+4\left(3x^2+7x-6\right)=-27\)

\(\Leftrightarrow-3-12x^2+15x+12x^2+28x-24=-27\)

\(\Leftrightarrow43x=0\Rightarrow x=0\)

2/

\(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2-1\right)=27\)

\(\Leftrightarrow x^3+27-x^3+x=27\)

\(\Leftrightarrow x=0\)