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Ta có: x(x+3)=6

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

=>\(x^2+3x+\frac94-\frac{33}{4}=0\)

=>\(\left(x+\frac32\right)^2-\frac{33}{4}=0\)

=>\(\left(x+\frac32\right)^2=\frac{33}{4}\)

=>\(x+\frac32=\pm\frac{\sqrt{33}}{2}\)

=>\(x=\pm\frac{\sqrt{33}}{2}-\frac32\)

\(x^2-2x-120=0\)

=>\(x^2-12x+10x-120=0\)

=>x(x-12)+10(x-12)=0

=>(x-12)(x+10)=0

=>\(\left[\begin{array}{l}x-12=0\\ x+10=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=12\\ x=-10\end{array}\right.\)

\(x^2-2x-120=0\)

\(x^2-2x+1-121=0\)

\(\left(x-1\right)^2-11^2\) =0

\(\left(x-1+11\right)\left(x-1-11\right)=0\)

\(\left(x+10\right)\left(x-12\right)=0\)

\(\left[\begin{array}{l}x+10=0\Rightarrow x=-10\\ x-12=0\Rightarrow x=12\end{array}\right.\)

Vậy x = -10; x = 12

\(-\frac{313}{370}=\frac{-370+57}{370}=-1+\frac{57}{370}\)

\(-\frac{314}{371}=\frac{-371+57}{371}=-1+\frac{57}{371}\)

mà \(\frac{57}{370}>\frac{57}{371}\left(370<371\right)\)

nên \(-\frac{313}{370}>-\frac{314}{371}\)

−370313​=370−370+57​=−1+37057​

\(- \frac{314}{371} = \frac{- 371 + 57}{371} = - 1 + \frac{57}{371}\)

mà \(\frac{57}{370} > \frac{57}{371} \left(\right. 370 < 371 \left.\right)\)

nên \(- \frac{313}{370} > - \frac{314}{371}\)

a: \(\left(2x-1\right)^2-x\left(4x-3\right)\)

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

=-x+1

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

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

\(=x^2-3x+2-x^2+6x-9=3x-7\)

c: \(\left(x-2\right)^3-x\left(x-1\right)\left(x-2\right)\)

\(=x^3-6x^2+12x-8-x\left(x^2-3x+2\right)\)

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

\(=-3x^2+10x-8\)

d: \(\frac12x\left(x-2\right)-\left(2x-3\right)^2\)

\(=\frac12x^2-x-\left(4x^2-12x+9\right)\)

\(=\frac12x^2-x-4x^2+12x-9=-\frac72x^2+11x-9\)

(a + a) x 4 : 8 = a

[ a x 1 + a x 1] x 4 : 8 = a

a x (1+ 1) x 4 : 8 = a

a x 2 x 4 : 8 = a

a x (2 x 4 : 8) = a

a x (8 : 8) = a

a x 1 = a

Vậy a là mọi số khác 0

\(\frac43+56=\frac43+\frac{168}{3}=\frac{172}{3}\)

\(=\frac43+\frac{56}{1}\)

\(=\frac43+\frac{168}{3}\)

\(=\frac{4+168}{3}\)

\(=\frac{172}{3}\)

Ta có: \(\frac{x}{20}+\frac{x}{30}+\frac{x}{42}=3\)

=>\(x\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)=3\)

=>\(x\left(\frac14-\frac15+\frac15-\frac16+\frac16-\frac17\right)=3\)

=>\(x\left(\frac14-\frac17\right)=3\)

=>\(x\cdot\frac{3}{21}=3\)

=>\(x=3:\frac{3}{21}=21\)

\(\frac{x}{20}+\frac{x}{30}+\frac{x}{42}=\) \(3\)

\(\frac{21x}{420}+\frac{14x}{420}+\frac{10x}{420}=3\)

\(\frac{45x}{420}=3\)

\(45x=3\times420\)

\(45x=1260\)

\(x=1260:45\)

\(x=28\)

A = 1.2 + 2.3 + ...+ n(n + 1)

1.2.3 = 1.2.3

2.3.3 = 2.3(4-1) = 2.3.4 - 1.2.3

.............................................................

n(n + 1).3 = n(n + 1).{(n + 2) - (n-1)} = n(n+1)(n+2)-(n-1)n(n+2)

Cộng vế với vế ta có:

3A = n(n+1)(n+2)

A = \(\frac{n\left(n+1\right)\left(n+2\right)}{3}\)

A = 1\(^2\) + \(2^2\) + ...+ n\(^2\)

A = 1 + 2.(1+ 1) + ...+ n[(n - 1) + 1]

A = 1 + 2.1 + 2 + ...+ n(n-1) + n

A = (1 + 2 + ..+n) + [1.2 + 2.3 + 3.4 +...+(n-1)n]

Đặt B = 1 + 2+ .. +n

C = 1.2 + 2.3 +..+ (n -1)n

B = 1 + 2+ ...+ n

B =(n + 1).n : 2

1.2.3 = 1.2.3

2.3.3 = 2.3.(4-1) = 2.3.4 - 1.2.3

3.4.3 = 3.4.(5- 2) = 3.4.5 - 2.3.4

................................................................

(n -1).n.3 = (n - 1).n.[(n +1) - (n - 2)] = (n-1)n(n+1) -(n-2)(n-1)n

Cộng vế với vế ta có:

3B = (n-1)n(n+1)

B = \(\frac{\left(n-1\right)n\left(n+1\right)}{3}\)

A = B + C

A = \(\frac{n\left(n+1\right)}{2}\) + \(\frac{\left(n-1\right)n\left(n+1\right)}{3}\)

A = n(n+1).(\(\frac12\) + \(\frac{n-1}{3}\))

A = n(n+1).(\(\frac{3+2n-2}{6}\))

A = n(n+1).\(\frac{2n+1}{6}\)

A =\(\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)