Đặt \(A=1\cdot2+2\cdot3+\cdots+99\cdot100\)

\(=1\left(1+1\right)+2\left(2+1\right)+\cdots+99\left(99+1\right)\)

\(=\left(1^2+2^2+\cdots+99^2\right)+\left(1+2+\cdots+99\right)\)

\(=\frac{99\left(99+1\right)\left(2\cdot99+1\right)}{6}+\frac{99\cdot\left(99+1\right)}{2}\)

\(=\frac{99\cdot100\cdot199}{6}+\frac{99\cdot100}{2}=\frac{99\cdot100\cdot199+3\cdot99\cdot100}{6}\)

\(=\frac{99\cdot100\cdot202}{6}=33\cdot100\cdot101=333300\)

Ta có: \(x^2+\left(x^2+1\right)+\left(x^2+2\right)+\cdots+\left(x^2+99\right)\)

\(=\left(x^2+0\right)+\left(x^2+1\right)+\cdots+\left(x^2+99\right)\)

\(=\left(x^2+x^2+\cdots+x^2\right)+\left(0+1+\cdots+99\right)=100x^2+\frac{99\cdot100}{2}=100x^2+99\cdot50=100x^2+4950\)

Ta có: \(\frac{1\cdot2+2\cdot3+\cdots+99\cdot100}{x^2+\left(x^2+1\right)+\left(x^2+2\right)+\cdots+\left(x^2+99\right)}=50\frac{116}{131}\)

=>\(\frac{333300}{100x^2+4950}=50+\frac{116}{131}\)

=>\(\frac{6666}{2x^2+99}=\frac{6666}{131}\)

=>\(2x^2+99=131\)

=>\(2x^2=32\)

=>\(x^2=16\)

=>\(\left[\begin{array}{l}x=4\\ x=-4\end{array}\right.\)

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}\)

Áp dụng t/c dtsbn:

\(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}=\dfrac{x^2+y^2+z^2}{4+9+16}=\dfrac{116}{29}=4\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=4.4=16\\y^2=4.9=36\\z^2=16.16=16^2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=4\\y=6\\z=16\end{matrix}\right.\\\left\{{}\begin{matrix}x=-4\\y=-6\\z=-16\end{matrix}\right.\end{matrix}\right.\)

thank bn nha!

\(\frac{30}{13}\)× \(\frac{52}{29}\)×\(\frac{116}{240}\)= \(\frac{3×10×13×4×29×4}{13×29×2×3×4×10}\)( gạch số giống nhau đi )=\(\frac{4}{2}\)= 2

k giúp mik nha

bn viết lại đầy đủ đề bài và pt của các câu

a) A= 100 - 116 : 4

=100-29

=71

b) B= 100 - [ 116 - (13 - 5)²]

=100-[116-8²]

=100-(116-64)

=100-52

=48

c) C= |- 65| + (- 42) + |65|

= 65+(-42)+65

= 88

\(100-\left(116-\left(13-5^2\right)\right)=100-\left(116-\left(-12\right)\right)\)

\(=100-\left(116+12\right)=100-128=-28\)

100-[116-(13-5²)]

=100-[116-(13-25)]

=100-(116-13+25)

=100-(103+25)

=100-128

=-28

\(100-\left[116-\left(13-5\right)^2\right]\)

\(=100-\left(116-8^2\right)\)

\(=100-\left(116-64\right)\)

\(=100-52\)

\(=48\)

=100-(116-(9.9)

=100-(116-81)

=100-35=65

5) Ta có: \(\dfrac{\left(5\sqrt{3}+\sqrt{50}\right)\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}\)

\(=\dfrac{5\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)^2}{5\left(\sqrt{3}-\sqrt{2}\right)}\)

=1

cảm ơn nha