`( 5/6 -1/2 ) : 2/3`

`= ( 5/6 - 3/6) xx 3/2`

`= 1/3 xx 3/2`

`= 3/6`

`=1/2`

`@ `

`(5/6-1/2) : 2/3`

`= 5/6 : 2/3 - 1/2 : 2/3`

`= 5/6 xx 3/2 - 1/2 xx 3/2`

`= 15/12 - 3/4`

`= 5/4 -3/4`

`= 2/4`

`=1/2`

`------`

`(3/4 +1/2) : 3/5`

`= ( 3/4 + 2/4) xx 5/3`

`= 5/4 xx 5/3`

`= 25/12`

`@ `

`(3/4 + 1/2) : 3/5`

`= 3/4 : 3/5 + 1/2 : 3/5`

`= 3/4 xx 5/3 + 1/2 xx 5/3`

`= 5/4 + 5/6`

`=25/12` 

`(3/4 +1/2): 3/4`

`=(3/4 +2/4) xx 4/3`

`= 5/4 xx 4/3`

`=20/12`

`= 5/3`

`-----`

`(3/4 +1/2) : 3/4`

`= 3/4 : 3/4 + 1/2 : 3/4`

`= 1 + 1/2 xx 4/3`

`=1+ 2/3`

`= 3/3+2/3`

`=5/3`

20/24

C1= 8/15x1/4= 2/15

C2= 1/3 x1/4+ 1/5x1/4

    = 1/12+1/20

    =2/15

cách 1 : [ 1/3 + 1/5 ] x 1/4 = 8/15 x 1/4

                                       = 8/60

cách 2 : [ 1/3 + 1/5 ] x 1/4 = 1/3 x 1/4 + 1/5 x 1/4

                                       =1/12 + 1/20

                                       = 8/60

nếu ai thấy đúng cho minh k nhé

\(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15=0\)

\(\Leftrightarrow\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15=0\)

Đặt \(x^2+8x+11=y\Rightarrow x^2+8x+7=y-4;x^2+8x+15=y+4\)

Khi đó:

\(pt\Leftrightarrow\left(y-4\right)\left(y+4\right)+15=0\)

\(\Leftrightarrow y^2-1=0\)

\(\Leftrightarrow y=1;y=-1\)

Nếu \(y=1\Rightarrow x^2+8x+11=1\)

\(\Rightarrow x^2+8x+10=0\)

\(\Rightarrow-\left(6-x^2-8x-16\right)=0\)

\(\Rightarrow-\left[6-\left(x+4\right)^2\right]=0\)

\(\Rightarrow-\left(\sqrt{6}-x-4\right)\left(\sqrt{6}+x+4\right)=0\)

\(\Rightarrow x=-4-\sqrt{6};x=\sqrt{6}-4\)

Nếu \(y=-1\),ta có:

\(x^2+8x+11=-1\)

\(\Rightarrow x^2+8x+12=0\)

\(\Rightarrow x^2+2x+6x+12=0\)

\(\Rightarrow x\left(x+2\right)+6\left(x+2\right)=0\)

\(\Rightarrow\left(x+2\right)\left(x+6\right)=0\)

\(\Rightarrow x=-2;x=-6\)

Vậy \(x=-2;x=-6;x=-4-\sqrt{6};x=\sqrt{6}-4\)

\(\frac{3}{22}\times\frac{3}{11}\times22\)

1.\(=\frac{3\times3\times22}{22\times11\times1}\)

\(=\frac{3\times3}{11\times1}\)

\(=\frac{9}{11}\)

2.  \(=\frac{9}{242}\times22\)

\(=\frac{198}{242}\)

\(=\frac{9\times2\times11}{2\times11\times11}\)

\(=\frac{9}{11}\)

\(\left(\frac{1}{2}+\frac{1}{3}\right)\times\frac{2}{5}\)

1.\(=\frac{5}{6}\times\frac{2}{5}\)

\(=\frac{5\times2}{3\times2\times5}\)

\(=\frac{1}{3}\)

2.\(\)\(=\frac{1}{2}\times\frac{2}{5}+\frac{1}{3}\times\frac{2}{5}\)

\(=\frac{1}{5}+\frac{2}{15}\)

\(=\frac{3}{15}+\frac{2}{15}\)

\(=\frac{5}{15}=\frac{1}{3}\)

Have a good day

a: ĐKXĐ: \(\begin{cases}x-2\ge0\\ x+1\ge0\\ x^2-x-2\ge0\end{cases}\Rightarrow\begin{cases}x\ge2\\ x\ge-1\\ \left(x-2\right)\left(x+1\right)\ge0\end{cases}\)

=>x>=2

Ta có: \(3x + 14 + 5\sqrt{x-2} = 7\left(\sqrt{x+1} + \sqrt{x^2 - x - 2}\right)\) (1)

Đặt \(a=\sqrt{x-2};b=\sqrt{x+1}\) (ĐIều kiện: a>=0; b>0)

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

\(b^2-a^2=x+1-\left(x-2\right)=3\)

\(b^2-1=x+1-1=x\)

=>\(3x+14=3\left(b^2-1\right)+14=3b^2+11\)

(1) sẽ tương đương: \(3b^2+11+5a=7b+7ab\)

=>\(3b^2-7b+11+a(5-7b)=0\)

=>\((b - 2)(40b^3 + 52b^2 - 133b + 98) = 0\)

=>b-2=0

=>b=2

=>x+1=4

=>x=3

b: ĐKXĐ: 7/3<=x<=7

\(7\sqrt{3x-7}+(4x-7)\sqrt{7-x}=32\left(2\right)\)

Đặt \(a=\sqrt{3x-7};b=\sqrt{7-x}\)

=>\(a^2+b^2=3x-7+7-x=2x\)

\(3a^2+b^2=3\left(3x-7\right)+7-x=9x-21+7-x=8x-14\)

\(3x-7=a^2\)

=>\(3x=a^2+7\)

=>\(x=\frac{a^2+7}{3}\)

=>\(4x-7=4\cdot\frac{a^2+7}{3}-7=\frac{4\left(a^2+7\right)-21}{3}=\frac{4a^2+7}{3}\)

(2) sẽ trở thành: \(7a+\frac{4a^2+7}{3}b=32\iff21a+(4a^2+7)b=96\)

mà \(a^2 + 3b^2 = 14\)

nên \(a=\sqrt5;b=\sqrt3\)

=>3x-7=5

=>3x=12

=>x=4(nhận)

Áp dụng công thức dãy số cách đều, ta có tổng của dãy trên là: 

(1 + 99) x 99 : 2 = 4950

Đáp số: 4950

*Dãy trên có 99 số hạng.