# Violympic toán 9

Hôm kia lúc 14:16

a) $\sqrt{\dfrac{7}{27}}$ $=\sqrt{\dfrac{21}{81}}$ $=\dfrac{\sqrt{21}}{\sqrt{81}}$  $=\dfrac{\sqrt{21}}{9}$

b) $\sqrt{\dfrac{5}{11}}=\sqrt{\dfrac{55}{121}}$ $=\dfrac{\sqrt{55}}{\sqrt{121}}$ $=\dfrac{\sqrt{55}}{11}$

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Hôm kia lúc 15:02

Có $x^2+x-1=\left(\dfrac{-1-\sqrt{5}}{2}\right)^2+\dfrac{-1-\sqrt{5}}{2}-1$

$=\dfrac{6+2\sqrt{5}}{4}+\dfrac{-3-\sqrt{5}}{2}$$=\dfrac{3+\sqrt{5}}{2}+\dfrac{-3-\sqrt{5}}{2}$

$=0$

$\Rightarrow x^2=1-x$ và $-x^2-x=-1$

Có $M=\left(4x^5+4x^4-5x^3+2x-2\right)^{2018}+2019$

$=\left[4x^3\left(x^2+x-1\right)-x^3-2\left(1-x\right)\right]^{2018}+2019$

$=\left[-x.x^2-2x^2\right]^{2018}+2019$

$=\left[-x\left(1-x\right)-2x^2\right]^{2018}+2019$

$=\left(-x^2-x\right)^{2018}+2019$

$=\left(-1\right)^{2018}+2019$

$=2020$

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Hôm kia lúc 15:16

Đặt $4x^5+4x^4-5x^3+2x-2=A$

Xét pt $t^2+t-1=0$

$\Delta=b^2-4ac=1+4=5\Rightarrow\left[{}\begin{matrix}t=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-1-\sqrt{5}}{2}\\t=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-1+\sqrt{5}}{2}\end{matrix}\right.$

$\Rightarrow x$ là nghiệm của pt $t^2+t-1=0\Rightarrow x^2+x-1=0$

$A=4x^5+4x^4-4x^3-x^3-x^2+x+x^2+x-1-1$

$=4x^3\left(x^2+x-1\right)-x\left(x^2+x-1\right)+x^2+x-1-1=-1$

$\Rightarrow M=\left(-1\right)^{2018}+2019=2020$

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Hôm kia lúc 14:36

Đk: $x\ge3$

PT $\Leftrightarrow25.\dfrac{\sqrt{x-3}}{5}-7.\sqrt{\dfrac{4\left(x-3\right)}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9\left(x^2-9\right)}{81}}=0$

$\Leftrightarrow5\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0$

$\Leftrightarrow\dfrac{1}{3}\sqrt{x-3}-\sqrt{x^2-9}=0$

$\Leftrightarrow\sqrt{x-3}\left(\dfrac{1}{3}-\sqrt{x+3}\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\\dfrac{1}{3}=\sqrt{x+3}\end{matrix}\right.$$\Leftrightarrow\left[{}\begin{matrix}x=3\left(TM\right)\\x=\dfrac{-26}{9}\left(L\right)\end{matrix}\right.$

Vậy..

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Hôm kia lúc 14:37

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Hôm kia lúc 14:27

$\sqrt{18x+9}-\sqrt{8x+4}+\dfrac{1}{3}\sqrt{2x+1}=4$ (đk:$x\ge-\dfrac{1}{2}$)

$\Leftrightarrow\sqrt{9\left(2x+1\right)}-\sqrt{4\left(2x+1\right)}+\dfrac{1}{3}\sqrt{2x+1}=4$

$\Leftrightarrow3\sqrt{2x+1}-2\sqrt{2x+1}+\dfrac{1}{3}\sqrt{2x+1}=4$

$\Leftrightarrow\sqrt{2x+1}=3$

$\Leftrightarrow x=4$(TM)

Vạy...

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Hôm kia lúc 14:25

b) Có $\dfrac{5}{2}\sqrt{\dfrac{1}{6}}=\sqrt{\left(\dfrac{5}{2}\right)^2.\dfrac{1}{6}}=\sqrt{\dfrac{25}{24}}$

$6\sqrt{\dfrac{1}{37}}=\sqrt{\dfrac{6^2}{37}}=\sqrt{\dfrac{36}{37}}$

Có $\dfrac{25}{24}>1>\dfrac{36}{37}$

$\Rightarrow\dfrac{5}{2}\sqrt{\dfrac{1}{6}}>6\sqrt{\dfrac{1}{37}}$

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Hôm kia lúc 14:22

a)$A=\sqrt{12}+3\sqrt{27}-5\sqrt{48}$

$=\sqrt{2^2.3}+3\sqrt{3^2.3}-5\sqrt{4^2.3}$

$=2\sqrt{3}+9\sqrt{3}-20\sqrt{3}$$=-9\sqrt{3}$

b)$B=3\sqrt{a^2+3}-3\sqrt{16a^2+48}+4\sqrt{25a^2+75}$

$=3\sqrt{a^2+3}-3\sqrt{16\left(a^2+3\right)}+4\sqrt{25\left(a^2+3\right)}$

$=3\sqrt{a^2+3}-12\sqrt{a^2+3}+20\sqrt{a^2+3}$$=11\sqrt{a^2+3}$

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11 tháng 6 lúc 23:57

Đề bài sai, bạn kiểm tra lại điều kiện $a^2+b^2+c^2=1$

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11 tháng 6 lúc 23:57

Đề bài sai, bạn có thể thử kiểm tra với $a=1.0001$ và $b=0.9999$

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