[x+1]+[x+2]+[x+3]+...[x+100]=7050
ai giúp em với ạ
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\(\left(1-\frac{3}{4}\right)\left(1-\frac{3}{7}\right)\cdot\cdot\cdot\left(1-\frac{3}{100}\right)\)
\(=\frac{1}{4}\cdot\frac{4}{7}\cdot\frac{7}{10}\cdot\cdot\cdot\frac{97}{100}\)
\(=\frac{1.4.7.10...97}{4.7.10.13...100}\)
\(=\frac{1}{100}\)
a, ĐK: \(x\ge2\)
\(\sqrt{2x+1}-\sqrt{x-2}=x+3\)
\(\Leftrightarrow\dfrac{x+3}{\sqrt{2x+1}+\sqrt{x-2}}=x+3\)
\(\Leftrightarrow\left(x+3\right)\left(\dfrac{1}{\sqrt{2x+1}+\sqrt{x-2}}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\left(l\right)\\\sqrt{2x+1}+\sqrt{x-2}=1\left(vn\right)\end{matrix}\right.\)
Phương trình vô nghiệm.
b, ĐK: \(x\ge-1\)
\(\sqrt{x+3}+2x\sqrt{x+1}=2x+\sqrt{x^2+4x+3}\)
\(\Leftrightarrow\sqrt{x+3}+2x\sqrt{x+1}=2x+\sqrt{\left(x+3\right)\left(x+1\right)}\)
\(\Leftrightarrow-\sqrt{x+3}\left(\sqrt{x+1}-1\right)+2x\left(\sqrt{x+1}-1\right)=0\)
\(\Leftrightarrow\left(2x-\sqrt{x+3}\right)\left(\sqrt{x+1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+3}=2x\\\sqrt{x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x+3=4x^2\end{matrix}\right.\\x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)
i don't now
mong thông cảm !
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\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
ta có :
\(\frac{1}{2^2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3^2}< \frac{1}{2\cdot3}\)
\(\frac{1}{4^2}< \frac{1}{3\cdot4}\)
...
\(\frac{1}{100^2}< \frac{1}{99\cdot100}\)
nên \(A< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< 1-\frac{1}{100}\)
\(\Rightarrow A< \frac{99}{100}< 1\)
\(\Rightarrow A< 1\left(đpcm\right)\)
nhiều qá lm sao nổi
c) x ⋮ 2; x ⋮ 7; x ⋮ 35
⇒ x ∈ BC(2; 7; 35)
Ta có:
2 = 2
7 = 7
35 = 5.7
⇒ BCNN(2; 7; 35) = 2.5.7 = 70
⇒ x ∈ BC(2; 7; 35) = B(70) = {0; 70; 140; 210; ...}
Mà 100 ≤ x ≤ 200
x = 140
b) Do x ∈ BC(21; 35; 99) và x nhỏ nhất, x ≠ 0 nên x = BCNN(21; 35; 99)
Ta có:
21 = 3.7
35 = 5.7
99 = 3².11
⇒ x = BCNN(21; 35; 99) = 3².5.7.11 = 3465
e) Do x nhỏ nhất, x ≠ 0; x ⋮ 12; x ⋮ 15; x ⋮ 20
⇒ x = BCNN(12; 15; 20)
Ta có:
12 = 2².3
15 = 3.5
20 = 2².5
⇒ x = BCNN(12; 15; 20) = 2².3.5 = 60
\(x\cdot\dfrac{3}{7}-x\cdot\dfrac{1}{2}=\dfrac{3}{5}\)
\(x\left(\dfrac{3}{7}-\dfrac{1}{2}\right)=\dfrac{3}{5}\)
\(x\cdot\dfrac{-1}{14}=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}:\dfrac{-1}{14}\)
\(x=\dfrac{-42}{5}\)
\(a,=\left(x-5\right)\left(x+5\right)\\ b,=\left(x-3\right)^2\\ c,=\left(3x-2\right)\left(3x+2\right)\\ d,=\left(x+1\right)^2\\ e,=\left(x-10\right)\left(x+10\right)\)
\(A=x^3-xy-x^3-x^2y+x^2y-xy=-2xy\\ A=-2\cdot\dfrac{1}{2}\left(-100\right)=100\)
\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=7050\)
\(\Rightarrow x+x+...x+1+2+3+...+100=7050\)
\(\Rightarrow50x+\dfrac{50.51}{2}=7050\)
\(\Rightarrow50x+1275=7050\)
\(\Rightarrow50x=5775\Rightarrow x=\dfrac{5775}{50}=\dfrac{1155}{2}\)