HOC24
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Môn học
Chủ đề / Chương
Bài học
a)Ta có \(y=\sqrt{x^2-2x+5}=\sqrt{x^2-2x+1+4}=\sqrt{\left(x-1\right)^2+4}\ge\sqrt{4}=2\)
Dấu = xảy ra <=> x = 1.
b)Ta có \(y=\sqrt{\dfrac{x^2}{4}-\dfrac{x}{6}+1}=\sqrt{\dfrac{x^2}{4}-\dfrac{x}{6}+\dfrac{1}{36}+\dfrac{35}{36}}=\sqrt{\left(\dfrac{x}{2}-\dfrac{1}{6}\right)^2+\dfrac{35}{36}}\ge\sqrt{\dfrac{35}{36}}\)
Dấu = xảy ra <=> \(x=\dfrac{1}{3}\)
Vậy bạn thêm TH -x nữa vào là được.
Ta có \(x+\dfrac{2x\sqrt{6}}{\sqrt{x^2}+1}=1\Rightarrow x+\dfrac{2x\sqrt{6}}{x+1}=1\Rightarrow\dfrac{x^2+x+2x\sqrt{6}}{x+1}=1\Rightarrow x^2+2x\sqrt{6}=1\Rightarrow x^2+2x\sqrt{6}+6=7\Rightarrow\left(x+\sqrt{6}\right)^2=7\Rightarrow x+\sqrt{6}=\pm\sqrt{7}\Rightarrow x\in\left\{\sqrt{7}-\sqrt{6},-\sqrt{7}-\sqrt{6}\right\}\)Vậy \(x\in\left\{\sqrt{7}-\sqrt{6},-\sqrt{7}-\sqrt{6}\right\}\)
Vì \(\dfrac{\overline{ab}}{b}=\dfrac{\overline{bc}}{c}=\dfrac{\overline{ca}}{a}\Rightarrow\dfrac{10a+b}{b}=\dfrac{10b+c}{c}=\dfrac{10c+a}{a}\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=\dfrac{a+b+c}{a+b+c}=1\)(theo tính chất của DTSBN)
Khi đó, ta có \(a=b=c\)
Thay vào, ta có \(\left(\overline{abc}\right)^{123}=\left(\overline{aaa}\right)^{123}=\left(111.a\right)^{123}=111^{123}.a^{123}=111^{123}.a^{40}.a^{41}.4^{42}\)(đpcm)
Gọi số cây tổ 1 là x, số cây tổ 2 là y, số cây tổ 3 là z (x,y,z ∈ N*)
Từ GT, ta có \(\left\{{}\begin{matrix}\dfrac{y}{8}=\dfrac{z}{9}\Rightarrow\dfrac{y}{24}=\dfrac{z}{27}\\\dfrac{x}{5}=\dfrac{y}{6}\Rightarrow\dfrac{x}{20}=\dfrac{y}{24}\end{matrix}\right.\Rightarrow\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{27}\)
Lại có \(2x-z=26\).Lại có \(\dfrac{2x}{40}=\dfrac{z}{27}=\dfrac{2x-z}{40-27}=\dfrac{26}{13}=2\) (theo tính chất của DTSBN)
Thay vào, ta được \(x=40,y=48,z=54\).
Ta có \(A=3\left(\dfrac{1}{5}+\dfrac{1}{5^4}+\dfrac{1}{5^7}+...+\dfrac{1}{5^{100}}\right)\Rightarrow5^3A=3\left(5^2+\dfrac{1}{5}+\dfrac{1}{5^4}+...+\dfrac{1}{5^{97}}\right)\Rightarrow5^3A-A=3\left[5^2+\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\left(\dfrac{1}{5^4}-\dfrac{1}{5^4}\right)+\left(\dfrac{1}{5^7}-\dfrac{1}{5^7}\right)+...+\left(\dfrac{1}{5^{97}}-\dfrac{1}{5^{97}}\right)-\dfrac{1}{5^{100}}\right]\Rightarrow624A=3\left(25-\dfrac{1}{5^{100}}\right)=3.\dfrac{5^{102}-1}{5^{100}}=\dfrac{3.5^{102}-3}{5^{100}}\Rightarrow A=\dfrac{3.5^{102}-3}{5^{100}.624}\)
Ta có \(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\Rightarrow2A=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\Rightarrow2A-A=2+\left(1-1\right)+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{2^2}-\dfrac{1}{2^2}\right)+...+\left(\dfrac{1}{2^{99}}-\dfrac{1}{2^{99}}\right)-\dfrac{1}{2^{100}}\Rightarrow A=2-\dfrac{1}{2^{100}}=\dfrac{2^{101}-1}{2^{100}}\)Vậy \(A=\dfrac{2^{101}-1}{2^{100}}\)
Ta có \(16^2+17^2+18^2+19^2+20^2-15^2-14^2-13^2-12^2-11^2=\left(20^2-15^2\right)+\left(19^2-14^2\right)+\left(18^2-13^2\right)+\left(17^2-12^2\right)+\left(16^2-11^2\right)=\left(20-15\right)\left(20+15\right)+\left(19-14\right)\left(19+14\right)=\left(18-13\right)\left(18+13\right)+\left(17-12\right)\left(17+12\right)+\left(16-11\right)\left(16+11\right)=5.\left(35+33+31+29+27\right)=775\)Vậy VT = 775.