A = (3x - 5)(2x + 11) - (2x + 3)(3x + 7)

A = 3x(2x + 11) - 5(2x+  11) - 2x(3x + 7) - 3(3x + 7)

A=  6x² + 33x - 10x - 55 - 6x² - 14x - 9x - 21

A = (6x² - 6x²) + (33x - 10x - 14x - 9x) + (-55 - 21) = -76 => không phụ thuộc vào biến x (đpcm)

B = (2x + 3)(4x² - 6x + 9) - 2(4x³ - 1)

= 2x(4x² - 6x + 9) + 3(4x² - 6x + 9) - 8x³ + 2

= 8x³ - 12x² + 18x + 12x² - 18x - 27 - 8x³ + 2

= (8x³ - 8x³) + (-12x² + 12x²) + (18x - 18x) + (-27 + 2) = -25 => không phụ thuộc vào biến x (đpcm)

A= ( 3x - 5 ) ( 2x+11) - (2x+3)(3x+7) 

=\(6x^2+23x-55-\left(6x^2+23x+21\right)\) 

=\(6x^2+23x-55-6x^2-23x-21\)  

= -76 

Vậy A không phụ thuộc vào x

A = ( 3x - 5 )( 2x + 11 ) - ( 2x + 3 )( 3x + 7 )

A = 6x² + 23x - 55 - ( 6x² + 23x + 21 )

A = 6x² + 23x - 55 - 6x² - 23x - 21

A = -55 - 21 = -76

Vậy A không phụ thuộc vào biến ( đpcm )

B = ( 2x + 3 )( 4x² - 6x + 9 ) - 2( 4x³ - 1 )

B = ( 2x )³ + 3³ - 8x³ + 2

B = 8x³ + 27 - 8x³ + 2

B = 27 + 2 = 29

Vậy B không phụ thuộc vào biến ( đpcm )

\(E=\dfrac{\left(cosx-siny\right)\left(cosx+siny\right)}{sin^2x\cdot sin^2y}-\dfrac{cos^2x}{sin^2x}\cdot\dfrac{cos^2y}{sin^2y}\)

\(=\dfrac{cos^2x\left(1-cos^2y\right)-sin^2y}{sin^2x\cdot sin^2y}\)

\(=\dfrac{sin^2y\left(cos^2x-1\right)}{sin^2x\cdot sin^2y}=-1\)

\(\left(3x+2\right).\left(2x-1\right)-6x.\left(x-1\right)-7x+4\)

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

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

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

\(=2\)

Vậy giá trị biểu thức không phụ thuộc vào biến \(x\)

67996

\(A=27x^3-8-27x^3+3=-5\\ B=8x^3+y^3+8x^3-y^3-16x^3=0\)

Ta có:

\(M=3x\left(x-5y\right)+\left(y-5x\right)\left(-3y\right)-3\left(x^2-y^2\right)-1\)

\(M=3x^2-15xy-3y^2+15xy-3x^2+3y^2\)

\(M=0\left(đpcm\right)\)

M=3x²-15xy-3y²+15xy-3x²+3y²-1

M=-1

=2x^2+3x-10x-15-2x^2+6x+x+7

=-8

A = \(\left(\dfrac{\sqrt{x}+1}{2\sqrt{x}-2}+\dfrac{3}{x-1}-\dfrac{\sqrt{x}+3}{2\sqrt{x}+2}\right)\cdot\dfrac{4x-4}{5}\) (ĐK: x \(\ge\) 0; x \(\ne\) 1)

A = \(\left(\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}+\dfrac{3}{x-1}-\dfrac{\sqrt{x}+3}{2\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{4\left(x-1\right)}{5}\)

A = \(\left(\dfrac{\left(\sqrt{x}+1\right)^2}{2\left(x-1\right)}+\dfrac{6}{2\left(x-1\right)}-\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{2\left(x-1\right)}\right)\cdot\dfrac{4\left(x-1\right)}{5}\)

A = \(\left(\dfrac{x+2\sqrt{x}+1+6-x-3\sqrt{x}+\sqrt{x}+3}{2\left(x-1\right)}\right)\cdot\dfrac{4\left(x-1\right)}{5}\)

A = \(\dfrac{10}{2\left(x-1\right)}\cdot\dfrac{4\left(x-1\right)}{5}\)

A = 4

Vậy A không phụ thuộc vào x

Chúc bn học tốt!

Ta có: \(A=\left(\dfrac{\sqrt{x}+1}{2\sqrt{x}-2}+\dfrac{3}{x-1}-\dfrac{\sqrt{x}+3}{2\sqrt{x}+2}\right)\cdot\dfrac{4x-4}{5}\)

\(=\dfrac{x+2\sqrt{x}+1+6-\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{4\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{5}\)

\(=\dfrac{x+2\sqrt{x}+7-x-2\sqrt{x}+3}{1}\cdot\dfrac{2}{5}\)

\(=10\cdot\dfrac{2}{5}=4\)

\(M=\left(2x+5\right)^3-30x\left(2x+5\right)-8x^3\)

\(=\left(2x+5\right)\left(4x^2+20x+25-30x\right)-8x^3\)

\(=\left(2x+5\right)\left(4x^2-10x+25\right)-8x^3\)

\(=8x^3+125-8x^3\)

=125