a) \(=x^3+27-54-x^3=-27\)

b) \(=8x^3+y^3\)

Đây là bài giải pt chứ có phải biểu thức đâu mà thu gọn hả bạn?

Lời giải:

a. ĐKXĐ: $x\neq 1$

PT $\Leftrightarrow \frac{x^2+x+1}{(x-1)(x^2+x+1)}+\frac{2x(x-1)}{(x-1)(x^2+x+1)}=\frac{3x^2}{(x-1)(x^2+x+1)}$

$\Leftrightarrow x^2+x+1+2x(x-1)=3x^2$

$\Leftrightarrow 3x^2-x+1=3x^2$

$\Leftrightarrow x=1$ (không thỏa đkxđ)

Vậy pt vô nghiệm.

b. ĐKXĐ: $x\neq \pm 3$

PT $\Leftrightarrow \frac{(x+2)(x+3)}{(x-3)(x+3)}=\frac{x^2+3x}{(x-3)(x+3)}$

$\Leftrightarrow (x+2)(x+3)=x^2+3x$

$\Leftrightarrow x^2+5x+6=x^2+3x$

$\Leftrightarrow 2x+6=0$

$\Leftrightarrow x=-3$ (không thỏa mãn đkxđ)

Do đó pt vô nghiệm.

c. ĐKXĐ: $x\neq \pm 2$

PT $\Leftrightarrow \frac{(x-2)^2-(x+2)^2}{(x+2)(x-2)}=\frac{-16}{(x-2)(x+2)}$

$\Leftrightarrow (x-2)^2-(x+2)^2=-16$

$\Leftrightarrow -8x=-16$

$\Leftrightarrow x=2$ (vi phạm đkxđ)

Do đó pt vô nghiệm.

a) \(\left(x+3\right)^2+\left(x-3\right)^2+2\left(x^2+9\right)\)

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

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

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

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

\(=4x^2\)

b) \(\left(4x-1\right)^3-\left(4x-3\right)\left(16x^2+3\right)\)

\(=\left(64x^3-48x^2+12x-1\right)-\left(64x^3+12x-48x^2-9\right)\)

\(=64x^3-48x^2+12x-1-64x^3-12x+48x^2+9\)

\(=\left(64x^3-64x^3\right)-\left(48x^2-48x^2\right)+\left(12x-12x\right)-\left(1-9\right)\)

\(=0-0+0+8\)

\(=8\)

a) (x + 3)² + (x - 3)² + 2(x² - 9)

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

= (x + 3 + x - 3)²

= (2x)²

= 4x²

b) (4x - 1)³ - (4x - 3)(16x² + 3)

= 64x³ - 48x² + 12x - 1 - 64x³ - 12x + 48x² + 9

= (64x³ - 64x³) + (-48x² + 48x²) + (12x - 12x) + (-1 + 9)

= 8

a) (x+9)(x-9)-x²=x²-81-x²=-81

b) (10x-1)(10x+1)-(10x-1)²=100x²-1-100x²+20x-1=20x-2

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

a: \(=\dfrac{x^2-x+x+1+2x}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{x-1}\)

b: \(=\dfrac{x^2+2x-4x-2x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2-4x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x-2}{x+2}\)

c: \(=\dfrac{2x^2-3x-9-x^2+3x+x^2+6x+9}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{2x^2+6x}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x}{x-3}\)

\(A=\left(\dfrac{2\sqrt{x}}{\sqrt{x}-3}+\dfrac{x}{3\sqrt{x}-x}\right).\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}\left(dkxd:x\ne0;\pm\sqrt{3}\right)\)

\(=\left(\dfrac{2}{\sqrt{x}-3}-\dfrac{x}{\sqrt{x}\left(\sqrt{x}-3\right)}\right).\left(\sqrt{x}-3\right)\)

\(=\left(\dfrac{2\sqrt{x}-x}{\sqrt{x}\left(\sqrt{x}-3\right)}\right).\left(\sqrt{x}-3\right)\)

\(=\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}}\)

\(=2-\sqrt{x}\)

Vậy \(A=2-\sqrt{x}\)

\(\dfrac{x}{2}-\dfrac{2}{9}=\dfrac{3}{2}-\left(\dfrac{-5}{4}\right)\\ \Rightarrow\dfrac{x}{2}-\dfrac{2}{9}=\dfrac{11}{4}\\ \Rightarrow\dfrac{x}{2}=\dfrac{11}{4}+\dfrac{2}{9}\\ \Rightarrow\dfrac{x}{2}=\dfrac{107}{36}\)

\(\Rightarrow x=\dfrac{107}{36}\times2\\ \Rightarrow x=\dfrac{107}{18}\)

a) M=
−
1
9
x4y3(2xy2)2=
−
1
9
x4y3(4x2y4)=
−
1
9
x6y7

b) y=
−
x
3
=> x=-3y

mà x+y=2

=>-3y+y=2 <=> -2y=2 => y=-1 => x=-3y=-3*-1=3

Thay x=3; y=-1 vào M...=>M=
−
1
9
(36)(-17)=81

nhớ nhé!

khó chưa học mới học l 3

Câu 1:

a: \(\dfrac{2}{5}\sqrt{75}-0,5\cdot\sqrt{48}+\sqrt{300}-\dfrac{2}{3}\cdot\sqrt{12}\)

\(=\dfrac{2}{5}\cdot5\sqrt{3}-0,5\cdot4\sqrt{3}+10\sqrt{3}-\dfrac{2}{3}\cdot2\sqrt{3}\)

\(=2\sqrt{3}-2\sqrt{3}+10\sqrt{3}-\dfrac{4}{3}\sqrt{3}\)

\(=10\sqrt{3}-\dfrac{4}{3}\sqrt{3}=\dfrac{26}{3}\sqrt{3}\)

b: \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\dfrac{3}{3+\sqrt{6}}\)

\(=\dfrac{\sqrt{3}\cdot3\sqrt{3}-2\sqrt{3}}{\sqrt{2}\left(3\sqrt{3}-2\right)}+\dfrac{3\left(3-\sqrt{6}\right)}{9-6}\)

\(=\dfrac{\sqrt{3}\left(3\sqrt{3}-2\right)}{\sqrt{2}\left(3\sqrt{3}-2\right)}+3-\sqrt{6}\)

\(=\dfrac{\sqrt{3}}{\sqrt{2}}+3-\sqrt{6}=3-\dfrac{\sqrt{6}}{2}\)

c: \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

=\(\sqrt{9-2\cdot3\cdot\sqrt{6}+6}+\sqrt{24-2\cdot2\sqrt{6}\cdot3+9}\)

\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)

\(=\left|3-\sqrt{6}\right|+\left|2\sqrt{6}-3\right|\)

\(=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)

Bài 2:

a: 

b: Phương trình hoành độ giao điểm là:

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

=>4x=-6

=>x=-3/2

Thay x=-3/2 vào y=-x-4, ta được:

\(y=-\left(-\dfrac{3}{2}\right)-4=\dfrac{3}{2}-4=-\dfrac{5}{2}\)

Vậy: \(A\left(-\dfrac{3}{2};-\dfrac{5}{2}\right)\)

c: Vì (d2)//(d) nên \(\left\{{}\begin{matrix}a=-1\\b\ne-4\end{matrix}\right.\)

Vậy: (d2): y=-x+b

Thay x=-2 và y=5 vào (d2), ta được:

\(b-\left(-2\right)=5\)

=>b+2=5

=>b=5-2=3

Vậy: (d2): y=-x+3