1: Ta có: \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\)

\(\Leftrightarrow x^2+6x+9-x^2+4-4x=17\)

\(\Leftrightarrow x=2\)

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

\(\Leftrightarrow2x^2-2x+3x-3+2x-2x^2-3+3x=0\)

\(\Leftrightarrow6x=6\)

hay x=1

Thế mày làm đi

cho ít thôi thì làm

\(1,\\ a,A=4x^2\left(-3x^2+1\right)+6x^2\left(2x^2-1\right)+x^2\\ A=-12x^4+4x^2+12x^2-6x^2+x^2=-x^2=-\left(-1\right)^2=-1\\ b,B=x^2\left(-2y^3-2y^2+1\right)-2y^2\left(x^2y+x^2\right)\\ B=-2x^2y^3-2x^2y^2+x^2-2x^2y^3-2x^2y^2\\ B=-4x^2y^3-4x^2y^2+x^2\\ B=-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^3-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^2+\left(0,5\right)^2\\ B=\dfrac{1}{8}-\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{8}\)

\(2,\\ a,\Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ b,\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3=8=-2^3\\ \Leftrightarrow x=2\\ c,\Leftrightarrow4x^2\left(4x-2\right)-x^3+8x^2=15\\ \Leftrightarrow16x^3-8x^2-x^3+8x^2=15\\ \Leftrightarrow15x^3=15\\ \Leftrightarrow x^3=1\Leftrightarrow x=1\)

\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)

\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)

Bài 4:

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

\(\Leftrightarrow6x-9-2x+4=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

hay \(x=\dfrac{13}{3}\)

c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

hay x=1

b: =x-2

d: \(=-x^3+\dfrac{3}{2}-2x\)

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

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

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

Tìm x, biết:

1) 2x ( x - 5)  - x ( 2x - 4 ) = 15

<=> 2x² - 10x - 2x² + 4x - 15 = 0

<=> -6x - 15 = 0

<=> -6x = 15

<=> x = -15/6

2)  ( x +1)( x + 2 ) - ( x + 4 ) ( x + 3 ) = 6

<=> x² + 2x + x + 2 - x² - 3x - 4x - 12 - 6 = 0

<=> -4x = -16

<=> x = 4

3)  4x² - 4x + 5 - x ( 4x - 3) = 1 - 2x

<=> 4x² - 4x + 5 - 4x² + 3x - 1 + 2x = 0

<=> x + 4 = 0

<=> x = -4

4) ( x + 3 ) ( 2x + 1 ) - 2x² = 4x - 5

<=> 2x² + x + 6x + 3 - 2x² - 4x + 5 = 0

<=> 3x + 8 = 0

<=> 3x = -8

<=> x = -8/3

5) -4 ( 2x - 8 ) + ( 2x - 1 )( 4x + 3 ) = 0

<=> - 8x + 32 + 8x² + 6x - 4x - 3 = 0

.......

6) -3 . (x-2) + 4 . (2x-6) - 7 . (x-9)= 5 . (3-2)

<=> -3x + 6 + 8x - 24 - 7x + 63 - 5 = 0

<=> -2x + 40 = 0

<=> -2x = -40

<=> x = 20

Còn lại tương tự ....

1)2x^2-10x-2x^2+14x=15

4x=15

x=15/4

1: \(\frac{3x-2}{3}-2=\frac{4x+1}{4}\)

=>\(\frac{3x-2-6}{3}=\frac{4x+1}{4}\)

=>\(\frac{3x-8}{3}=\frac{4x+1}{4}\)

=>3(4x+1)=4(3x-8)

=>12x+3=12x-32

=>3=-32(vô lý)

=>Phương trình vô nghiệm

2: \(\frac{x-3}{4}+\frac{2x-1}{3}=\frac{2-x}{6}\)

=>\(\frac{3\left(x-3\right)+4\left(2x-1\right)}{12}=\frac{2\left(2-x\right)}{12}\)

=>3(x-3)+4(2x-1)=2(2-x)

=>3x-9+8x-4=4-2x

=>11x-13=4-2x

=>13x=17

=>\(x=\frac{17}{13}\)

3: \(\frac12\left(x+1\right)+\frac14\left(x+3\right)=3-\frac13\left(x+2\right)\)

=>\(\frac12x+\frac12+\frac14x+\frac34+\frac13x+\frac23=3\)

=>\(x\left(\frac12+\frac14+\frac13\right)+\frac{6}{12}+\frac{9}{12}+\frac{8}{12}=3\)

=>\(x\left(\frac{6}{12}+\frac{3}{12}+\frac{4}{12}\right)=3-\frac{23}{12}=\frac{36}{12}-\frac{23}{12}=\frac{13}{12}\)

=>\(x\cdot\frac{13}{12}=\frac{13}{12}\)

=>x=1

4: \(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\)

=>\(\frac{x+4}{5}+\frac{5\left(-x+4\right)}{5}=\frac{2x-3\left(x-2\right)}{6}\)

=>\(\frac{x+4-5x+20}{5}=\frac{2x-3x+6}{6}\)

=>\(\frac{-4x+24}{5}=\frac{-x+6}{6}\)

=>6(-4x+24)=5(-x+6)

=>-24x+144=-5x+30

=>-19x=-114

=>x=6

5: \(\frac{4-5x}{6}=\frac{2\left(-x+1\right)}{2}\)

=>\(\frac{4-5x}{6}=-x+1\)

=>6(-x+1)=-5x+4

=>-6x+6=-5x+4

=>-6x+5x=4-6

=>-x=-2

=>x=2

6: \(-\left(\frac{x-3}{2}-2\right)=\frac{5\left(x+2\right)}{4}\)

=>\(-\frac{x-3-4}{2}=\frac{5\left(x+2\right)}{4}\)

=>\(\frac{-2\left(x-7\right)}{4}=\frac{5\left(x+2\right)}{4}\)

=>5(x+2)=-2(x-7)

=>5x+10=-2x+14

=>7x=4

=>x=4/7

7: \(\frac{2\left(2x+1\right)}{5}-\frac{6+x}{3}=\frac{5-4x}{15}\)

=>\(\frac{6\left(2x+1\right)-5\left(x+6\right)}{15}=\frac{5-4x}{15}\)

=>6(2x+1)-5(x+6)=-4x+5

=>12x+6-5x-30=-4x+5

=>7x-24=-4x+5

=>7x+4x=5+24

=>11x=29

=>\(x=\frac{29}{11}\)

8: \(\frac{7-3x}{2}-\frac{5+x}{5}=1\)

=>\(\frac{5\left(7-3x\right)-2\left(x+5\right)}{10}=1\)

=>5(7-3x)-2(x+5)=10

=>35-15x-2x-10=10

=>-17x+25=10

=>-17x=-15

=>x=15/17