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

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

\(=8x^2+2x-3\)

a)

$(2x+1)^2-(2x+1)(2x-1)=(2x+1)[(2x+1)-(2x-1)]$

$=2(2x+1)$

b)

$(4x+3)(x-1)-2x(2x+1)=4x^2-x-3-4x^2-2x=-3x-3=-3(x+1)$

c)

$(2x+3)^2-(4x+1)(x+5)=(4x^2+12x+9)-(4x^2+21x+5)$

$=-9x+4$

d)

$(x+2)^3-(x-1)(x^2+x+1)=(x^3+6x^2+12x+8)-(x^3-1)$

$=6x^2+12x+9$

e)

$(x+2)(x^2-2x+1)-(x+3)(x-3)=(x^3-3x+2)-(x^2-9)$

$=x^3-x^2-3x+11$

f)

$(x+3)(x^2-3x+9)-(x^2+2x+4)(x-2)$

$=x^3+3^3-(x^3-2^3)=3^3+2^3=35$

a: Ta có: \(2x\left(3x-1\right)-\left(x-3\right)\left(6x+2\right)\)

\(=6x^2-2x-6x^2-2x+18x+6\)

=14x+6

b: Ta có: \(\left(2x-3\right)^2-\left(2x+1\right)\left(2x-1\right)+3\left(2x-3\right)\)

\(=4x^2-12x+9-4x^2+1+6x-9\)

\(=-6x+1\)

c: Ta có: \(\left(x+y-1\right)^2-2\left(x+y-1\right)\left(x+y\right)+\left(x+y\right)^2\)

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

=1

a) \(2x\left(3x-1\right)-\left(x-3\right)\left(6x+2\right)=6x^2-2x-6x^2-2x+18x+6=14x+6\)

b) \(\left(2x-3\right)^2-\left(1+2x\right)\left(2x-1\right)+3\left(2x-3\right)=4x^2-12x+9-4x^2+1+6x-9=-6x+1\)

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

Bài 1

A= (x-2)(2x-1)-2x(x+3)=2x²-x-4x+2-2x²-6x=-11x+2

Bài 1:

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

\(A=2x^2-x-4x+2-2x^2-6x\)

\(A=-11x+2\)

b) \(B=\left(3x-2\right)\left(2x+1\right)-\left(6x-1\right)\left(x+2\right)\)

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

\(B=-12x\)

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

\(C=12x^2+18x-12x^2+8x+3x-2\)

\(C=29x-2\)

d) \(D=\left(2x+3\right)\left(5x-2\right)+\left(x+4\right)\left(2x-1\right)-6x\left(2x-3\right)\)

\(D=10x^2-4x+15x-6+2x^2-x+8x-4-12x^2+18x\)

\(D=36x-10\)

Bài 2: 

a: Ta có: \(2x\left(3x-5\right)\left(x+11\right)-3x\left(2x+3\right)\left(x+7\right)\)

\(=2x\left(3x^2+33x-5x-55\right)-3x\left(2x^2+14x+3x+21\right)\)

\(=6x^3+56x^2-110x-6x^2-51x^2-63x\)

\(=-117x\)

b: Ta có: \(\left(x^2+5x-6\right)\left(x-1\right)-\left(x+2\right)\left(x^2-x+1\right)-x\left(3x-10\right)\)

\(=x^3+4x^2-11x+6-\left(x^3-x^2+x+2x^2-2x+2\right)-3x^2+10x\)

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

=4

c: Ta có: \(\left(x^2+x+1\right)\left(x-1\right)-x^2\left(x+1\right)+x^2-5\)

\(=x^3-1-x^3-x^2+x^2-5\)

=-6

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

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

\(\Leftrightarrow2x\left(-2\right)=2\left(x+3\right)\)

\(\Leftrightarrow-4x=2x+6\)

\(\Leftrightarrow-6x=6\)

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

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

\(\Leftrightarrow4x\left(-2\right)-4x+12=0\)

\(\Leftrightarrow-12x=-12\)

\(\Leftrightarrow x=1\)

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

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

\(\Leftrightarrow7\left(2x+3\right)+x^2-4x+4=0\)

\(\Leftrightarrow x^2+10x+25=0\)

\(\Leftrightarrow\left(x+5\right)^2=0\)

\(\Leftrightarrow x=-5\)

4) \(8x^3-\left(x+1\right)^3=3x-3\)

\(\Leftrightarrow8x^3-\left(x^3+3x+3x^2+1\right)-3x+3=0\)

\(\Leftrightarrow7x^3-3x^2-6x+2=0\)

\(\Leftrightarrow\left(x-1\right)\left(7x^2+4x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{-2+3\sqrt{2}}{7}\\x=\frac{-2-3\sqrt{2}}{7}\end{matrix}\right.\)

5)\(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)

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

\(\Leftrightarrow27x^3-8-\left(27x^3-1\right)=x-4\)

\(\Leftrightarrow-7=x-4\)

\(\Leftrightarrow x=-3\)

\(a,=\dfrac{4x+8}{x^2+2x}=\dfrac{4\left(x+2\right)}{x\left(x+2\right)}=\dfrac{4}{x}\\ b,=\dfrac{\left(2x-3\right)-\left(2x-4\right)}{x-2}=\dfrac{2x-3-2x+4}{x-2}=\dfrac{1}{x-2}\\ c,=\dfrac{2x-1-3x-2}{x+3}=\dfrac{-x-3}{x+3}=\dfrac{-\left(x+3\right)}{x+3}=-1\\ d,=\dfrac{11x-18+x}{2x-3}=\dfrac{12x-18}{2x-3}=\dfrac{6\left(2x-3\right)}{2x-3}=6\)

\(e,=\dfrac{3x-6-9x+3}{2x+1}=\dfrac{-6x-3}{2x+1}=\dfrac{-3\left(2x+1\right)}{2x+1}=-3\)

a \(2x+2>4\\ \Leftrightarrow2\left(x+1\right)>4\\ \Leftrightarrow x+1>2\\ \Leftrightarrow x>1\)

b \(3x+2>-5\\ \Leftrightarrow3x>-7\\ \Leftrightarrow x>\dfrac{-7}{3}\)

c \(10-2x>2\\ \Leftrightarrow2\left(5-x\right)>2\\ \Leftrightarrow5-x>1\\ \Leftrightarrow-x>-4\\ \Leftrightarrow x< 4\)

d \(1-2x< 3\\ \Leftrightarrow-2x< 2\\ \Leftrightarrow2x>2\\ \Leftrightarrow x>1\)

a)2x+2>4

<=> 2x>4-2

<=>2x>2

<=>x>1

Vậy...

b)3x+2>-5

<=>3x>-5-2

<=>3x>-7

<=>x>\(\dfrac{-7}{3}\)

Vậy...

c)10-2x>2

<=>-2x>-10+2

<=>-2x>-8

<=>x<4

Vậy...

d)1-2x<3

<=>-2x<3-1

<=>-2x<2

<=>x>-1

Vậy...

e)10x+3-5\(\le\)14x+12

<=>10x-2\(\le\)14x+12

<=>10x-14x\(\le\)2+12

<=>-4x\(\le\)14

<=>x\(\ge\)\(\dfrac{-7}{2}\)

Vậy...

f)(3x-1)<2x+4

<=> 3x-2x<1+4

<=>x<5

Vậy...

1) \(5^{x+1}-5^x=20\Leftrightarrow5^x\left(5-1\right)=20\Leftrightarrow5^x=5\Leftrightarrow x=1\)

2) \(2^x+2^{x+4}=544\Leftrightarrow2^x\left(1+2^4\right)=544\Leftrightarrow2^x=32\Leftrightarrow x=5\)

3) \(4^{2x+1}+4^{2x}=80\Leftrightarrow4^{2x}\left(4+1\right)=80\Leftrightarrow16^x=16\Leftrightarrow x=1\)

4) \(3^{2x+2}+3^{2x+1}=108\Leftrightarrow3^{2x}\left(3^2+3\right)=108\Leftrightarrow9^x=9\Leftrightarrow x=1\)

5) \(7^{x+3}-7^{x+1}=16464\Leftrightarrow7^x\left(7^3-7\right)=16464\Leftrightarrow7^x=49\Leftrightarrow x=2\)

c:Ta có: \(5^{x+1}-5^x=20\)

\(\Leftrightarrow5^x\cdot5-5^x=20\)

\(\Leftrightarrow5^x\cdot4=20\)

\(\Leftrightarrow5^x=5\)

hay x=1

c: Ta có: \(2^x+2^{x+4}=544\)

\(\Leftrightarrow2^x+2^x\cdot16=544\)

\(\Leftrightarrow2^x\cdot17=544\)

\(\Leftrightarrow2^x=32\)

hay x=5

c: Ta có: \(4^{2x+1}+4^{2x}=80\)

\(\Leftrightarrow16^x\cdot4+16^x=80\)

\(\Leftrightarrow16^x\cdot5=80\)

\(\Leftrightarrow16^x=16\)

hay x=1

c: Ta có: \(3^{2x+2}+3^{2x+1}=108\)

\(\Leftrightarrow9^x\cdot9+9^x\cdot3=108\)

\(\Leftrightarrow9^x\cdot12=108\)

\(\Leftrightarrow9^x=9\)

hay x=1

c: Ta có: \(7^{x+3}-7^{x+1}=16464\)

\(\Leftrightarrow7^x\cdot343-7^x\cdot7=16464\)

\(\Leftrightarrow7^x\cdot336=16464\)

\(\Leftrightarrow7^x=49\)

hay x=2

b: =x-2

d: \(=-x^3+\dfrac{3}{2}-2x\)