1+2+3+......+x=500500
giúp mình với TvT
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Những câu hỏi liên quan
HP
1
18 tháng 9 2021
1) \(\left(a+b\right)^2-\left(a^3+b^3\right)\)
\(=\left(a+b\right)^3-\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(=\left(a+b\right)\left(a^2+2ab+b^2-a^2+ab-b^2\right)\)
\(=3ab\left(a+b\right)\)
2) \(\left(x+1\right)^2-4\left(x+1\right)y^2+4y^4\)
\(=\left(x+1+2y^2\right)^2\)
KA
1
27 tháng 3 2020
Ta có f(x)=1-5x
=> f(1)=1-5.1=1-5=-4
f(2)=1-5.2=1-10=-9
f(\(\frac{1}{5}\))=1-\(5\cdot\frac{1}{5}=1-1=0\)
\(f\left(\frac{-3}{5}\right)=1-5\cdot\left(\frac{-3}{5}\right)=1+3=4\)
6 tháng 10 2019
8x3 - 27y3 = 23 . x3 - 33 . y3 = ( 2x )3 - ( 3y )3 = ( 2x - 3y ) [(2x)2 + 12xy + (3y)2 ].
10 tháng 10 2021
a, \(2x\left(x-3\right)-15+5x=0\\ \Rightarrow2x\left(x-3\right)-\left(15-5x\right)=0\\ \Rightarrow2x\left(x-3\right)-5\left(3-x\right)=0\\ \Rightarrow\left(2x+5\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x=3\end{matrix}\right.\)
b, \(x^3-7x=0\\ \Rightarrow x\left(x^2-7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\pm7\end{matrix}\right.\)
c, \(\left(2x-3\right)^2-\left(x+5\right)^2=0\\ \Rightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\\ \Rightarrow\left(x-8\right)\left(3x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Xem lại đề câu d
HT
27 tháng 10 2020
sửa đề câu a và câu b nhá , mik nghĩ đề như này :
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)
= \(\frac{1}{1}-\frac{1}{215}\)
\(=\frac{214}{215}\)
b, đặt \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{213\cdot215}\)
\(A\cdot2=\frac{2}{1\cdot3}+\frac{2}{3.5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)
\(A\cdot2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)
\(A\cdot2=\frac{1}{1}-\frac{1}{215}\)
\(A\cdot2=\frac{214}{215}\)
\(A=\frac{214}{215}:2\)
\(A=\frac{107}{215}\)
HD
Giúp mình với TvT
1
21 tháng 10 2021
\(a,\) Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
\(\Rightarrow\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2}{d^2}\\ \dfrac{a^2-b^2}{c^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\dfrac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\dfrac{b^2}{d^2}\\ \Rightarrow\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\)
b, Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
\(\Rightarrow\left(\dfrac{a-b}{c-d}\right)^4=\left(\dfrac{bk-b}{dk-d}\right)^4=\left(\dfrac{b\left(k-1\right)}{d\left(k-1\right)}\right)^4=\dfrac{b^4}{d^4}\\ \dfrac{a^4+b^4}{c^4+d^4}=\dfrac{b^4k^4+b^4}{d^4k^4+d^4}=\dfrac{b^4\left(k^4+1\right)}{d^4\left(k^4+1\right)}=\dfrac{b^4}{d^4}\\ \RightarrowĐpcm\)
c, Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
\(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2}{d^2}\\ \dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\dfrac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\dfrac{b^2\left(k+1\right)^2}{d^2\left(k+1\right)^2}=\dfrac{b^2}{d^2}\\ \RightarrowĐpcm\)
C
2
8 tháng 6 2019
\(x^2+y^2=xy\)
\(\Leftrightarrow x^2-xy+y^2=0\)
\(\Leftrightarrow x^2-xy+\frac{y^2}{4}+\frac{3}{4}y^2=0\)
\(\Leftrightarrow\left(x-\frac{y}{2}\right)^2+\frac{3}{4}y^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x-\frac{y}{2}=0\\\frac{3}{4}y^2=0\end{cases}}\Leftrightarrow x=y=0\)
Số số hạng
\(\left(x-1\right):1+1=x\)
Tổng
\(\left(x+1\right)\cdot x:2=500500\) \(\left(x\ge0\right)\)
\(\left(x+1\right)x=500500\cdot2\)
\(x^2+x=1001000\)
\(x^2+x-1001000=0\)
\(\orbr{\begin{cases}x=1000\left(n\right)\\x=-1001\left(l\right)\end{cases}}\)
\(x=1000\)
Ta có: \(1+2+3+...+x=500500\)
\(\Leftrightarrow\frac{x\left(x+1\right)}{2}=500500\)
\(\Leftrightarrow x^2+x-1001000=0\)
\(\Leftrightarrow\left(x-1000\right)\left(x+1001\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1000\left(tm\right)\\x=-1001\left(ktm\right)\end{cases}}\)