Bạn nên gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề và hỗ trợ bạn tốt hơn nhé.

a) A = (x - 5)(x² + 5x + 25) - (x - 2)(x + 2) + x(x² + x + 4)

= x³ - 125 - x² + 4 + x³ + x² + 4x

= (x³ + x³) + (-x² + x²) + 4x + (-125 + 4)

= 2x³ + 4x - 121

b) Tại x = -2 ta có:

A = 2.(-2)³ + 4.(-2) - 121

= 2.(-8) - 8 - 121

= -16 - 129

= -145

c) x² - 1 = 0

x² = 1

x = -1; x = 1

*) Tại x = -1 ta có:

A = 2.(-1)³ + 4.(-1) - 121

= 2.(-1) - 4 - 121

= -2 - 125

= -127

*) Tại x = 1 ta có:

A = 2.1³ + 4.1 - 121

= 2.1 + 4 - 121

= 2 - 117

= -115

a) Ta có: \(B=\dfrac{x^2}{5x+25}+\dfrac{2\left(x+5\right)}{x}+\dfrac{50+5x}{x\left(x+5\right)}\)

\(=\dfrac{x^2}{5\left(x+5\right)}+\dfrac{2\left(x+5\right)}{x}+\dfrac{50+5x}{x\left(x+5\right)}\)

\(=\dfrac{x^3}{5x\left(x+5\right)}+\dfrac{10\left(x+5\right)^2}{5x\left(x+5\right)}+\dfrac{250+25x}{5x\left(x+5\right)}\)

\(=\dfrac{x^3+10x^2+100x+250+250+25x}{5x\left(x+5\right)}\)

\(=\dfrac{x^3+10x^2+125x+500}{5x\left(x+5\right)}\)

\(=\dfrac{x^3+5x^2+5x^2+25x+100x+500}{5x\left(x+5\right)}\)

\(=\dfrac{x^2\left(x+5\right)+5x\left(x+5\right)+100\left(x+5\right)}{5x\left(x+5\right)}\)

\(=\dfrac{\left(x+5\right)\left(x^2+5x+100\right)}{5x\left(x+5\right)}\)

\(=\dfrac{x^2+5x+100}{5x}\)

b) Thay x=-2 vào biểu thức \(B=\dfrac{x^2+5x+100}{5x}\), ta được:

\(B=\dfrac{\left(-2\right)^2+5\cdot\left(-2\right)+100}{-5\cdot2}=\dfrac{4+100-10}{-10}=\dfrac{94}{-10}=-\dfrac{94}{10}=\dfrac{-47}{5}\)

Vậy: Khi x=-2 thì \(B=-\dfrac{47}{5}\)

b: \(x-2\sqrt{xy}+y=\left(\sqrt{x}-\sqrt{y}\right)^2\)

a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne25\end{cases}}\)

\(A=\frac{x+3\sqrt{x}}{x-25}+\frac{1}{\sqrt{x}+5}\)

\(=\frac{x+3\sqrt{x}+\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)

\(=\frac{x+4\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)

\(=\frac{\sqrt{x}-1}{\sqrt{x}-5}\)

\(\Rightarrow P=\frac{\sqrt{x}-1}{\sqrt{x}-5}:\frac{\sqrt{x}+2}{\sqrt{x}-5}=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)

b) Để P nguyên

\(\Leftrightarrow\sqrt{x}-1⋮\sqrt{x}+2\)

\(\Leftrightarrow3⋮\sqrt{x}+2\)

\(\Leftrightarrow\sqrt{x}+2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{-3;-1;-5;1\right\}\)

Mà \(\sqrt{x}\ge0,\forall x\)

\(\Leftrightarrow\sqrt{x}=1\)

\(\Leftrightarrow x=1\)

Vậy để P nguyên \(\Leftrightarrow x=1\)

a: \(=3\sqrt{3}-2\sqrt{3}+4\sqrt{3}-5\sqrt{3}=2\sqrt{3}\)

:) bóc lột !

Câu 1: 

a) 2x(3x+2) - 3x(2x+3) = 6x^2+4x - 6x^2-9x = -5x

b) \(\left(x+2\right)^3+\left(x-3\right)^2-x^2\left(x+5\right)\)

\(=x^3+6x^2+12x+8+x^2-6x+9-x^3-5x^2\)

\(=2x^2+6x+17\)

c) \(\left(3x^3-4x^2+6x\right)\div\left(3x\right)=x^2-\dfrac{4}{3}x+2\)

Câu 2: 

\(2x^3-12x^2+18x=2x\left(x^2-6x+9\right)=2x\left(x^2-2.x.3+3^2\right)=2x\left(x-3\right)^2\)

/x-25 và /x-2 đấy ạ,máy em bị đánh lỗi. :((

\(5\sqrt{x}-\frac{\left(x+10\sqrt{x}+25\right)\left(\sqrt{x}-5\right)}{x-25}=5\sqrt{x}-\frac{\left(\sqrt{x}+5\right)^2\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)

\(=5\sqrt{x}-\left(\sqrt{x}+5\right)=4\sqrt{x}-5\)

\(\frac{\sqrt{x^2-4x+4}}{x-2}=\frac{\sqrt{\left(x-2\right)^2}}{x-2}=\frac{\left|x-2\right|}{x-2}=\orbr{\begin{cases}\frac{x-2}{x-2}\left(x>2\right)\\\frac{2-x}{x-2}\left(x< 2\right)\end{cases}=\orbr{\begin{cases}1\left(x>2\right)\\-1\left(x< 2\right)\end{cases}}}\)

a) Ta có: \(5\sqrt{a}-3\sqrt{25a^3}+2\sqrt{36ab^2}-2\sqrt{9a}\)

\(=5\sqrt{a}-15a\sqrt{a}+12b\sqrt{a}-6\sqrt{a}\)

\(=-\sqrt{a}-15a\sqrt{a}+12\sqrt{a}b\)

b) Ta có: \(\sqrt{64ab^3}-3\sqrt{12a^3b^3}+2ab\sqrt{9ab}-5b\sqrt{81a^3b}\)

\(=8b\sqrt{a}-6ab\sqrt{3ab}+6ab\sqrt{ab}-45a^2b\sqrt{ab}\)

a)\(5\sqrt{a}-3\sqrt{25a^3}+2\sqrt{36ab^2}-2\sqrt{9a}=5\sqrt{a}-15\left|a\right|\sqrt{a}+12\left|b\right|\sqrt{a}-6\sqrt{a}=-\sqrt{a}-15a\sqrt{a}+12b\sqrt{a}\)

b)\(\sqrt{64ab^3}-3\sqrt{12a^3b^3}+2ab\sqrt{9ab}-5b\sqrt{81a^3b}\)

\(=8\left|b\right|\sqrt{ab}-6\left|ab\right|\sqrt{3ab}+6ab\sqrt{ab}-45b\left|a\right|\sqrt{ab}\)

\(=8b\sqrt{ab}-6ab\sqrt{3ab}+6ab\sqrt{ab}-45ab\sqrt{ab}\)

\(=8b\sqrt{ab}-6ab\sqrt{3ab}-39ab\sqrt{ab}\)

a) \(\sqrt{0,49\cdot a^2}=\sqrt{0,7^2\cdot a^2}=\sqrt{\left(0,7\cdot\left|a\right|\right)^2}=0,7\left|a\right|\) (với a < 0)

b) \(\sqrt{25\left(7-a\right)^2}=\sqrt{\left[5\left(7-a\right)\right]^2}=5\left|7-a\right|\) (với a >/ 7)

c) \(\sqrt{a^4\left(a-2\right)^2}=a^2\left(a-2\right)=a^3-2a\) (với a >0 )

Tớ mới học nên cx ko chắc chắn lắm nhé.