\(B=\left(120+20\right):4-2^2.3\)

\(B=140:4-4.3\)

\(B=35-12\)

\(B=23\)

Tick mik nha

\(A=120+20:4-2^2.3\)

\(A=120+5-4.3\)

\(A=120+5-12\)

\(A=125-12\)

\(A=113\)

Câu 1:

Thay \(x=-12\) vào \(\left|x-2\right|\)

\(\Rightarrow\left|-12-2\right|=\left|-14\right|=14\)

Câu 2: Chọn phương án A.

Câu 3:

\(\left|-120\right|+\left|20\right|=120+20=140\)

Câu `1`

` |x + 2|`

mà `x=-12`

`->  |-12 + 2|= |-10|=10`

`->B`

Câu `2`

`->A`

Câu `3`

`A = |-120| + |20|`

`= 120 +20`

`=140`

C=120+(20:4-23). 3

C=120+(5-23).3

C=120+(-18).3

C=120+(-54)

C=66

D= (120+20:4-2²) .3

D=(120+5-4).3

D=(125-4).3

D=121.3

D=363

Với \(k\in N;k\ne0\) ta có :

\(\frac{1}{\left(k+1\right)\sqrt{k}+k\sqrt{\left(k+1\right)}}=\frac{1}{\sqrt{k\left(k+1\right)}\left(\sqrt{k}+\sqrt{k+1}\right)}\)

\(=\frac{\sqrt{k+1}+\sqrt{k}}{\sqrt{k\left(k+1\right)}\left(\sqrt{k+1}-\sqrt{k}\right)\left(\sqrt{k+1}+\sqrt{k}\right)}=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}}\)

\(=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\)

Áp dụng ta có :

\(M=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{120}}-\frac{1}{\sqrt{121}}=1-\frac{1}{11}=\frac{10}{11}\)

ta có:1/n(1+2+...+n)=1/n.n((n+1))/2=(n+1)/2

=>S=1+3/2+2+5/2+...+10=43