Trả lời:

1) \(\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=\left(\sqrt{x}\right)^2-2\sqrt{x}+\sqrt{x}-2=x-\sqrt{x}-2\)

2) \(\left(x+4\right)\left(x-2\right)-\left(x-3\right)^2=x^2-2x+4x-8-\left(x^2-6x+9\right)\)\(=x^2+2x-8-x^2+6x-9=8x-17\)

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

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

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

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

Hằng đẳng thức: \(\left(a+b\right)^2=a^2+2ab+b^2\).

b) \(\left(x+5\right)^2-\left(2x+10\right)\left(x-6\right)+\left(x-6\right)^2\)

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

\(=\left[\left(x+5\right)-\left(x-6\right)\right]^2\)

\(=11^2=121\)

Hằng đẳng thức: \(\left(a-b\right)^2=a^2-2ab+b^2\).

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

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

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

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

b.\(\left(x+5\right)^2-\left(2x+10\right)\left(x-6\right)+\left(x-6\right)^2\)

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

\(=\left[\left(x+5\right)-\left(x-6\right)\right]^2\)

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

\(C=\frac{x^2}{x-1}=\frac{x^2-1+1}{x-1}=x+1+\frac{1}{x-1}=2+x-1+\frac{1}{x-1}\)

\(\ge2+2\sqrt{\left(x-1\right).\frac{1}{x-1}}=2+2=4\)

Dấu \(=\)khi \(x-1=\frac{1}{x-1}\Leftrightarrow x=2\)(vì \(x>1\)).

Vậy \(minC=4\)xảy khi khi \(x=2\).

Ta có: \(C=\frac{x^2}{x-1}\)

\(=\frac{x^2-2x+1}{x-1}+\frac{2x-2}{x-1}+\frac{1}{x-1}\)

\(=\frac{\left(x-1\right)^2}{x-1}+\frac{2\left(x-1\right)}{x-1}+\frac{1}{x-1}\)

\(=x-1+2+\frac{1}{x-1}\)

\(=x-1+\frac{1}{x-1}+2\)

Nhận thấy \(x-1+\frac{1}{x-1}\ge2\sqrt{\left(x-1\right)\frac{1}{x-1}}=2\)

\(\Rightarrow A_{min}=4\)

Dấu "=" xảy ra khi :

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

\(\Leftrightarrow\left(x-1\right)^2=1\Leftrightarrow\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}}\)

Cre: mạng

\(1.\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)\)

\(=\sqrt{x}\left(\sqrt{x}-2\right)+1\left(\sqrt{x}-2\right)\)

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

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

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

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

\(=x^2-2x+4x-8-x^2+6x-9\)

\(=8x-17\)

đk: x > = 0

\(\left(\sqrt{x}-1\right)^2+\sqrt{x}\left(4-\sqrt{x}\right)=11\)

<=> \(x-2\sqrt{x}+1-x+4\sqrt{x}=11\)

<=> \(2\sqrt{x}=11\)

<=> \(\sqrt{x}=\frac{11}{2}\)

<=> x = 121/4

b) 4x²  - 4 = 0

<=> 4(x - 1)(x + 1) = 0

<=> x = 1 hoặc x = -1

Trả lời:

a, \(\left(\sqrt{x}-1\right)^2+\sqrt{x}\left(4-\sqrt{x}\right)=11\)

\(\Leftrightarrow\left(\sqrt{x}\right)^2-2\sqrt{x}+1+4\sqrt{x}-\left(\sqrt{x}\right)^2=11\)

\(\Leftrightarrow2\sqrt{x}+1=11\)

\(\Leftrightarrow2\sqrt{x}=10\)

\(\Leftrightarrow\sqrt{x}=5\)

\(\Leftrightarrow\sqrt{x}=\sqrt{25}\)

\(\Rightarrow x=25\)

Vậy x = 25

b, \(4x^2-4=0\)

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

\(\Leftrightarrow4\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

Vậy x = 1; x = -1

\(a)\)

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

\(b)\)

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

\(c)\)

\(\left(x-3\right)\left(x^2+3x+9\right)\)\(=x^3-27\)

\(a,4x^2-4x+1=\left(2x\right)^2-2.2x.1+1=\left(2x-1\right)^2\)

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

\(c,\left(x-3\right)\left(x^2+3x+9\right)=\left(x-3\right)\left(x^2+3x.1+3^2\right)=x^3-3^3\)

a) Ta có : n³ + 3n² + 2n

= n(n² + 3n + 2) 

= n(n + 1)(n + 2) \(⋮\)6 (tích 3 số nguyên liên tiếp) (đpcm)

b) A = 2⁰ + 2¹ + 2² + 2³ + 2⁴ + 2⁵ + 2⁶ + 2⁷ + 2⁸ + 2⁹ + .... + 2⁹⁵ + 2⁹⁶ + 2⁹⁷ + 2⁹⁸ + 2⁹⁹

= (1 + 2 + 2² + 2³ + 2⁴) + 2⁵(1 + 2 + 2² + 2³ + 2⁴) + ... + 2⁹⁵(1 + 2 + 2² + 2³ + 2⁴)

= 31 + 2⁵.31 + .. + 2⁹⁵.31

= 31(1 + 2⁵ + ... + 2⁹⁵) \(⋮31\)(đpcm) 

c) Ta có 49ⁿ + 77ⁿ - 29ⁿ - 1

= (49ⁿ - 1) + (77ⁿ - 29ⁿ) 

= (49 - 1)(49^{n - 1} - 49^{n - 2} + .... - 1) + (77 - 29)(77^{n - 1} - 77^{n - 2}.29 + 77^{n- 3}.29² - .... - 1) 

= 48(49^{n - 1} - 49^{n - 2} + .... - 1) + 48(77^{n - 1} - 77^{n - 2}.29 + 77^{n- 3}.29² - .... - 1) 

= 48(49^{n - 1} - 49^{n - 2} + .... - 1 + 77^{n - 1} - 77^{n - 2}.29 + 77^{n- 3}.29² - .... - 1) \(⋮\)48 (đpcm)