Question

Answer 1

Bài 6:

\(1,\left(2x-1\right)^2-\left(4x^2-1\right)=0\\ \Leftrightarrow\left(4x^2-4x+1\right)-\left(4x^2-1\right)=0\\ \Leftrightarrow-4x=-2\\ \Leftrightarrow x=\dfrac{-2}{-4}=\dfrac{1}{2}\\ ---\\ 2,\left(x+2\right)^2-x\left(x-3\right)=2\\ \Leftrightarrow x^2+4x+4-x^2+3x-2=0\\ \Leftrightarrow7x=-2\\ \Leftrightarrow x=-\dfrac{2}{7}\\ ---\\3, \left(x-5\right)^2-x\left(x+2\right)=5\\ \Leftrightarrow\left(x^2-10x+25\right)-x^2-2x-5=0\\ \Leftrightarrow x^2-x^2-10x-2x=-25+5\\ \Leftrightarrow-12x=-20\\ \Leftrightarrow x=\dfrac{-20}{-12}=\dfrac{5}{3}\\ ---\\ 4,\left(x-1\right)^2+x\left(4-x\right)=11\\ \Leftrightarrow\left(x^2-2x+1\right)-x^2+4x-11=0\\ \Leftrightarrow x^2-x^2-2x+4x=11-1\\ \Leftrightarrow2x=10\\ \Leftrightarrow x=5\)

Answer 2

5:

1: \(\left(2x+1\right)^2+2\left(2x+1\right)+1\)

\(=\left(2x+1\right)^2+2\cdot\left(2x+1\right)\cdot1+1^2\)

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

2: \(\left(3x-2y\right)^2+4\left(3x-2y\right)+4\)

\(=\left(3x-2y\right)^2+2\cdot\left(3x-2y\right)\cdot2+2^2\)

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

3: \(\left(x+3\right)^2+\left(x-2\right)^2-2\left(x+3\right)\left(x-2\right)\)

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

\(=\left(x+3-x+2\right)^2=5^2=25\)

4: \(\left(3x-5\right)^2-2\left(3x-5\right)\left(3x+5\right)+\left(3x+5\right)^2\)

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

\(=\left(-10\right)^2=100\)

5: \(\left(x-y\right)^2+\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)\)

\(=\left(x-y\right)^2-2\left(x-y\right)\cdot\left(x+y\right)+\left(x+y\right)^2\)

\(=\left(x-y-x-y\right)^2=\left(-2y\right)^2=4y^2\)

6: \(\left(5-x\right)^2+\left(x+5\right)^2-\left(2x+10\right)\left(x-5\right)\)

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

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

\(=\left(-10\right)^2=100\)

7: \(\left(x-2\right)^2+\left(x+1\right)^2+2\left(x-2\right)\left(-1-x\right)\)

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

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

8: \(\left(2x+3y\right)^2+\left(2x-3y\right)^2-2\left(4x^2-9y^2\right)\)

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

\(=\left(2x+3y-2x+3y\right)^2=\left(6y\right)^2=36y^2\)

Answer 3

Bài 5:

\(1,\left(2x+1\right)^2+2\left(2x+1\right)+1\\ =\left(2x+1\right)^2+2.\left(2x+1\right).1+1^2\\ =\left(2x+1+1\right)^2=\left(2x+2\right)^2=\left[2.\left(x+1\right)\right]^2\\ =4.\left(x+1\right)^2\\ ----\\ 2,\left(3x-2y\right)^2+4.\left(3x-2y\right)+4\\ =\left(3x-2y\right)^2-2.\left(3x-2y\right).2+2^2\\ =\left(3x-2y-2\right)^2\\ ---\\ 3,\left(x+3\right)^2+\left(x-2\right)^2-2.\left(x+3\right)\left(x-2\right)\\ =\left[\left(x+3\right)-\left(x-2\right)\right]^2=5^2=25\\ ----\\ 4,\left(3x-5\right)^2-2\left(3x-5\right)\left(3x+5\right)+\left(3x+5\right)^2\\ =\left[\left(3x-5\right)-\left(3x+5\right)\right]^2\\ =\left[-10\right]^2=100\)

Answer 4

\(B6\\ 5,\left(x-3\right)\left(x+3\right)=\left(x-5\right)^2\\ \Leftrightarrow\left(x^2-9\right)-\left(x^2-10x+25\right)=0\\ \Leftrightarrow x^2-x^2+10x=25+9\\ \Leftrightarrow10x=34\Leftrightarrow x=\dfrac{34}{10}=\dfrac{17}{5}\)

\(---\\ 6,\left(2x+1\right)^2-4x.\left(x-1\right)=11\\ \Leftrightarrow\left(4x^2+4x+1\right)-4x^2+4x-11=0\\ \Leftrightarrow4x^2-4x^2+4x+4x=11-1\\ \Leftrightarrow8x=10\Leftrightarrow x=\dfrac{10}{8}=\dfrac{5}{4}\)

\(---\\ 7,\left(3x+1\right)^2-9x\left(x-2\right)=25\\ \Leftrightarrow\left(9x^2-6x+1\right)-9x^2+18x=25\\ \Leftrightarrow9x^2-9x^2-6x+18x=25-1\\ \Leftrightarrow12x=24\Leftrightarrow x=2\)

\(----\\ 8,\left(3x-2\right)\left(3x+2\right)-9\left(x-1\right)x=0\\ \Leftrightarrow\left(9x^2-4\right)-9x^2+9x=0\\ \Leftrightarrow9x^2-9x^2+9x=4\\ \Leftrightarrow9x=4\Leftrightarrow x=\dfrac{4}{9}\)

Answer 5

\(9,\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\\ \Leftrightarrow\left(x+2\right)\left[\left(x+2\right)-\left(x-2\right)\right]=0\\ \Leftrightarrow\left(x+2\right).4=0\\ \Leftrightarrow x+2=0\Leftrightarrow x=-2\\ ---\\ 10,\left(x+2\right)^2-\left(x-3\right)\left(x+3\right)=-3\\ \Leftrightarrow\left(x^2+4x+4\right)-\left(x^2-9\right)+3=0\\ \Leftrightarrow x^2-x^2+4x=-4-9-3\\ \Leftrightarrow4x=-16\\ \Leftrightarrow x=\dfrac{-16}{4}=-4\\ ---\\ 11,\left(3x+2\right)^2-\left(3x-5\right)\left(3x+2\right)=0\\ \Leftrightarrow\left(3x+2\right)\left[\left(3x+2\right)-\left(3x-5\right)\right]=0\\ \Leftrightarrow\left(3x+2\right).7=0\\ \Leftrightarrow3x+2=0\\ \Leftrightarrow x=-\dfrac{2}{3}\)

Answer 6

\(12,\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\\ \Leftrightarrow\left(x^2+6x+9\right)-\left(x^2-4\right)-4x-17=0\\ \Leftrightarrow x^2-x^2+6x-4x=17-4-9\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=\dfrac{4}{2}=2\\ ----\\ 13,3\left(x-1\right)^2+\left(x+5\right)\left(2-3x\right)=-25\\ \Leftrightarrow3\left(x^2-2x+1\right)+\left(-3x^2-13x+10\right)+25=0\\ \Leftrightarrow3x^2-3x^2-6x-13x=-25-3-10\\ \Leftrightarrow-19x=-38\\ \Leftrightarrow x=\dfrac{-38}{-19}=2\)

\(---\\ 14,\left(x+3\right)^2+\left(x-2\right)^2=2x^2\\ \left(x^2+6x+9\right)+\left(x^2-4x+4\right)-2x^2=0\\ \Leftrightarrow x^2+x^2-2x^2+6x-4x=-9-4\\ \Leftrightarrow2x=-13\\ \Leftrightarrow x=-\dfrac{13}{2}\)