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

=>\(6x^2+2x-6x-2-\left(6x^2-9x-4x+6\right)=5\)

=>\(6x^2-4x-2-6x^2+13x-6=5\)

=>9x-8=5

=>9x=13

=>\(x=\frac{13}{9}\)

2: \(\left(1-3x\right)\left(3x-5\right)-\left(2x-4\right)\left(2-3x\right)=x-6\)

=>\(3x-5-9x^2+15x+\left(2x-4\right)\left(3x-2\right)=x-6\)

=>\(-9x^2+18x-5+6x^2-4x-12x+8=x-6\)

=>\(-3x^2+2x+3-x+6=0\)

=>\(-3x^2+x+9=0\)

=>\(3x^2-x-9=0\)

=>\(x^2-\frac13x-3=0\)

=>\(x^2-2\cdot x\cdot\frac16+\frac{1}{36}-\frac{109}{36}=0\)

=>\(\left(x-\frac16\right)^2=\frac{109}{36}\)

=>\(x-\frac16=\pm\frac{\sqrt{109}}{6}\)

=>\(x=\frac16\pm\frac{\sqrt{109}}{6}\)

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

=>\(8x^3-1-8x^3-1=5x+6\)

=>5x+6=-2

=>5x=-8

=>\(x=-\frac85\)

ai giúp minh đi cần quá gấp

\(\left(3.x^2-2^4\right)2^3=2^8\\ \left(3.x^2-2^4\right)=2^8:2^3\\ \left(3.x^2-2^4\right)=2^5\\ 3.x^2-2^4=32\\ 3.x^2=32+2^4\\ 3.x^2=48\\ x^2=48:3\\ x^2=16\\ x=+-4\)

bn ghi lại đề câu b) nhé

Câu a:

(x + 2)(x -3) - (x -2)(x + 5) = 0

x^2 + 2x - 3x - 6 - x^2 - 5x + 2x + 10 = 0

(x^2 - x^2) + (2x - 3x - 5x + 2x) + (10 - 6) = 0

0 + (-x - 5x + 2x) + 4 = 0

-6x + 2x + 4 = 0

-4x + 4 = 0

4x = 4

x = 4 : 4

x = 1

Vậy x = 1

Câu b:

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

2x^2 - 8x + 3x - 12+ x^2 - 2x - 5x + 10 = 3x^2 - 12x - 5x + 20

(2x^2 + x^2) - (8x+5x-3x +2x) - (12 - 10) = 3x^2 - (12x + 5x) + 20

3x^2 - (13x - 3x + 2x) - 2 = 3x^2 - 17x + 20

3x^2 - (10x + 2x) - 2 = 3x^2 - 17x + 20

3x^2 - 12x - 2 = 3x^2 - 17x + 20

3x^2 - 12x - 2 - 3x^2 + 17x - 20 = 0

(3x^2 - 3x^2) + (-12x + 17x) - (2 + 20) = 0

0 + 5x - 22 = 0

5x = 22

x = 22/5

Vậy x = 22/5

a) ( x + 2 )( x + 3 ) - ( x - 2 )( x + 5 ) = 16

<=> x² + 5x + 6 - ( x² + 3x - 10 ) = 16

<=> x² + 5x + 6 - x² - 3x + 10 = 16

<=> 2x + 16 = 16

<=> 2x = 0

<=> x = 0

b) 3x( 2x - 4 ) - 2x( 3x + 5 ) = 44

<=> 6x² - 12x - 6x² - 10x = 44

<=> -22x = 44

<=> x = -2

c) 2( 5x - 8 - 3 )( 4x - 5 ) = 4( 3x - 4 )

<=> 2( 5x - 11 )( 4x - 5 ) = 4( 3x - 4 )

<=> 2( 20x² - 69x + 55 ) = 12x - 16

<=> 40x² - 138x + 110 = 12x - 16

<=> 40x² - 138x + 110 - 12x + 16 = 0

<=> 40x² - 150 + 126 = 0 ( chưa học nghiệm vô tỉ nên để vô nghiệm nha :) )

=> Vô nghiệm

a. (x+2) (x+3)-(x-2) (x+5)= 16

x²+5x+6-x²-3x+10=16

2x+16=16

2x=0

x=0

b,3x (2x-4)-2x (3x+5)= 44

6x²-12x-6x²-10x=44

-22x=44

x=-2

Ý c bạn tự lm,tương tự nhưa,b

a,\(\frac{1}{3}x=-2-\frac{2}{3}=\frac{-8}{3}\)

        \(x=\frac{-8}{3}:\frac{1}{3}=\frac{-8}{3}.\frac{3}{1}=-8\)

\(b,\left[x+\frac{1}{1}\right]^2+\frac{5}{6}=\frac{7}{8}\)

\(\Rightarrow\left[x+1\right]^2=\frac{7}{8}-\frac{5}{6}\)

\(\Rightarrow\left[x+1\right]^2=\frac{7\cdot3}{24}-\frac{5\cdot4}{24}\)

\(\Rightarrow\left[x+1\right]^2=\frac{21}{24}-\frac{20}{24}\)

\(\Rightarrow\left[x+1\right]^2=\frac{1}{24}\)

\(\Rightarrow x\in\left\{\varnothing\right\}\)

:)?

\(a.\left(x-2\right)^2-\left(x+3\right)^2-4\left(x+1\right)=5\)

\(\left(x^2-4x+4\right)-\left(x^2+6x+9\right)-4x-4=5\)

\(\left(-4x-6x\right)+\left(4-9\right)-4x-4=5\)

\(-10x-5-4x-4=5\)

\(-14x-9=5\)

\(-14x=14\Rightarrow x=-1\)

\(b.\left(2x-3\right)\left(2x+3\right)-\left(x-1\right)^2-3x\left(x-5\right)=-44\)

\(4x^2-9-\left(x^2-2x+1\right)-\left(3x^2-15x\right)=-44\)

\(4x^2-9-x^2+2x-1-3x^2+15x=-44\)

\(17x-10=-44\)

\(17x=-34\Rightarrow x=-2\)

\(c.\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)

\(25x^2+10x+1-\left(25x^2-9\right)=30\)

\(10x+10=30\)

\(10x=20\Rightarrow x=2\)

\(d.\left(x+3\right)^2+\left(x-2\right)\left(x+2\right)-2\left(x-1\right)^2=7\)

\(\left(x^2+6x+9\right)+\left(x^2-4\right)-2\left(x^2-2x+1\right)=7\)

\(2x^2+6x+5-2x^2+4x-2=7\)

\(10x+3=7\)

\(10x=4\Rightarrow x=\frac{4}{10}=\frac25\)

\(f.\left(3x-8\right)^2=0\)

\(3x-8=0\Rightarrow x=\frac83\)

\(e.6\left(x+1\right)^2-2\left(x+1\right)+2\left(x-1\right)\left(x^2+x+1\right)=0\)

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

\(6x^2+12x+6-2x-2+2x^3-2=0\)

\(2x^3+6x^2+10x+2=0\)

\(\Rightarrow x\approx-0,23\)