Căn bậc hai. Căn bậc ba - Hỏi đáp

Đk:\(x\ge\sqrt{15}\)

Đặt \(\sqrt{x^2-15}=a;\sqrt{x-3}=b\left(a,b>0\right)\)

Thì \(a^2+b^2=x^2+x-18\) khi đó

\(pt\Leftrightarrow a^2+b^2+1=ab+a+b\)

Áp dụng BĐT AM-GM ta có:

\(\left\{{}\begin{matrix}a^2+b^2\ge2\sqrt{a^2b^2}=2ab\\b^2+1\ge2\sqrt{b^2}=2b\\a^2+1\ge2\sqrt{a^2}=2a\end{matrix}\right.\)

Cộng theo vế rồi thu gọn 3 BĐT trên ta có:

\(VT=a^2+b^2+1\ge ab+a+b=VP\)

Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}a^2+b^2=2ab\\b^2+1=2b\\a^2+1=2a\end{matrix}\right.\)\(\Rightarrow a=b=1\)

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x^2-15}=1\\\sqrt{x-3}=1\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x^2-15=1\\x-3=1\end{matrix}\right.\Rightarrow x=4\left(x\ge\sqrt{15}\right)\)

\(A=\frac{\sqrt{5}-1}{4}+\frac{\sqrt{9}-\sqrt{5}}{4}+...+\frac{\sqrt{2017}-\sqrt{2013}}{4}\)

\(A=\frac{\sqrt{2017}-1}{4}\)

Áp dụng bđt Svacxo ta có :

\(\dfrac{2017}{\sqrt{2018}}+\dfrac{2018}{\sqrt{2017}}\ge\dfrac{\left(\sqrt{2017}+\sqrt{2018}\right)^2}{\sqrt{2018}+\sqrt{2017}}=\sqrt{2017}+\sqrt{2018}\)

Dấu bằng xảy ra khi:

\(\dfrac{2017}{\sqrt{2018}}=\dfrac{2018}{\sqrt{2017}}\Leftrightarrow2017=2018\left(vl\right)\)

Suy ra không xảy ra dấu bằng

Vậy \(\dfrac{2017}{\sqrt{2018}}+\dfrac{2018}{\sqrt{2017}}>\sqrt{2017}+\sqrt{2018}\)

không thể cm

\(A=\left(\frac{\sqrt{a}}{\sqrt{ab}-b}+\frac{\sqrt{b}}{\sqrt{ab}-a}\right):\frac{\sqrt{a}+\sqrt{b}}{a\sqrt{b}-b\sqrt{a}}=\left[\frac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}+\frac{\sqrt{b}}{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}\right].\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{a}+\sqrt{b}}=\left[\frac{a}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}-\frac{b}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\right].\frac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}=\frac{\left(a-b\right)\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\frac{a-b}{\sqrt{a}+\sqrt{b}}=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}=\sqrt{a}-\sqrt{b}\)

Ngu vã đái 1s là ra ovs vật🤦‍♂️🤦‍♂️🤦‍♂️🤦‍♂️🤦‍♂️🤦‍♂️🤦‍♂️🦵👎👎👎👎👎👎👎👎👊👎👎😡😡😡👋👋👋👋👋👋✊✊✊✊✊

help me

\(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{ab+1}\)

\(\Leftrightarrow\frac{1}{a^2+1}-\frac{1}{ab+1}+\frac{1}{b^2+1}-\frac{1}{ab+1}\ge0\)

\(\Leftrightarrow\frac{ab+1-a^2-1}{\left(a^2+1\right)\left(ab+1\right)}+\frac{ab+1-b^2-1}{\left(b^2+1\right)\left(ab+1\right)}\ge0\)

\(\Leftrightarrow\frac{\left(ab-a^2\right)\left(b^2+1\right)+\left(ab-b^2\right)\left(a^2+1\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\ge0\)

\(\Leftrightarrow\frac{-a\left(b^2+1\right)\left(a-b\right)+b\left(a-b\right)\left(a^2+1\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\ge0\)

\(\Leftrightarrow\frac{\left(a-b\right)\left(-ab^2-a+a^2b+b\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\ge0\)

\(\Leftrightarrow\frac{\left(a-b\right)\left[ab\left(a-b\right)-\left(a-b\right)\right]}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\ge0\)

\(\Leftrightarrow\frac{\left(a-b\right)^2\left(ab-1\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\ge0\)( luôn đúng )

Dấu "=" xảy ra \(\Leftrightarrow\left[{}\begin{matrix}a-b=0\\ab-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=b\\ab=1\end{matrix}\right.\)

Xem lại đề đi!

ĐKXĐ: \(x\ge2\)

\(\Leftrightarrow\sqrt{5x^2+27x+25}=5\sqrt{x+1}+\sqrt{x^2-4}\)

\(\Leftrightarrow5x^2+27x+25=25x+25+x^2-4+10\sqrt{\left(x+1\right)\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow2x^2+x+2=5\sqrt{\left(x^2-x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow2\left(x^2-x-2\right)+3\left(x+2\right)=5\sqrt{\left(x^2-x-2\right)\left(x+2\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x-2}=a\\\sqrt{x+2}=b\end{matrix}\right.\)

\(\Rightarrow2a^2+3b^2=5ab\Leftrightarrow2a^2-5ab+3b^2=0\)

\(\Leftrightarrow\left(a-b\right)\left(2a-3b\right)=0\Leftrightarrow\left[{}\begin{matrix}a=b\\2a=3b\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-x-2}=\sqrt{x+2}\\2\sqrt{x^2-x-2}=3\sqrt{x+2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-x-2=x+2\\4\left(x^2-x-2\right)=9\left(x+2\right)\end{matrix}\right.\) \(\Leftrightarrow...\)

\(\left(4x-1\right)\sqrt{x^2+1}=2x^2+2x+1\) (1)

Đặt: \(a=\sqrt{x^2+1}\) ĐK: \(a\ge1\)

\(\left(1\right)\Rightarrow\left(4x-1\right)a=2a^2+2x-1\)

\(\Leftrightarrow2a^2-\left(4x-1\right)a+2x-1=0\) (*)

Phương trình (*) là phương trình bậc 2 ẩn là a.

\(\Delta=\left(4x-1\right)^2-8\left(2x-1\right)\)

\(=16x^2-8x+1-16x+8\)

\(=16x^2-24x+9\)

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

(*)\(\Leftrightarrow\left[{}\begin{matrix}a_1=\dfrac{4x-1-4x+3}{4}=\dfrac{1}{2}\left(L\right)\\a_2=\dfrac{4x-1+4x-3}{4}=2x-1\end{matrix}\right.\)

\(a=2x-1\) (\(x\ge1\))

\(\Rightarrow\sqrt{x^2+1}=2x-1\)

\(\Rightarrow x^2+1=4x^2-4x+1\)

\(\Leftrightarrow-3x^2+4x=0\)

\(\Leftrightarrow x\left(-3x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(L\right)\\x=\dfrac{4}{3}\end{matrix}\right.\)

Vậy phương trình (1) có nghiệm \(x=\dfrac{4}{3}\)

\(\dfrac{a}{4b^2+1}+\dfrac{b}{4a^2+1}=\dfrac{a\left(4b^2+1\right)}{4b^2+1}-\dfrac{4ab^2}{4b^2+1}+\dfrac{b\left(4a^2+1\right)}{4a^2+1}-\dfrac{4a^2b}{4b^2+1}\)

\(\ge a-\dfrac{4ab^2}{4b}+b-\dfrac{4a^2b}{4a}\) (bđt Cô-si)

=a-ab+b-ab=a+b-2ab=4ab-2ab=2ab

Lại có a+b=4ab \(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=4\ge\dfrac{2}{2\sqrt{ab}}\Rightarrow4\sqrt{ab}\ge2\Rightarrow ab\ge\dfrac{1}{4}\)

\(\Rightarrow2ab\ge\dfrac{1}{2}\Rightarrow\dfrac{a}{4b^2+1}+\dfrac{b}{4a^2+1}\ge\dfrac{1}{2}\)

Dấu ''='' xảy ra khi \(a=b=\dfrac{1}{2}\)

\(\dfrac{a}{4b^2+1}+\dfrac{b}{4a^2+1}\ge\dfrac{1}{2}\)

\(\Leftrightarrow a-\dfrac{a}{4b^2+1}+b-\dfrac{b}{4a^2+1}\le a+b-\dfrac{1}{2}\)

\(\Rightarrow\dfrac{4ab^2}{4b^2+1}+\dfrac{4ba^2}{4a^2+1}\le4ab-\dfrac{1}{2}\)

\(\sum\dfrac{4ab^2}{4b^2+1}\le^{CS}2ab\)

\(\Rightarrow CM:2ab\le4ab-\dfrac{1}{2}\Leftrightarrow ab\ge\dfrac{1}{4}\)

Từ GT \(\Rightarrow4ab=a+b\ge2\sqrt{ab}\Leftrightarrow ab\ge\dfrac{1}{4}\)

\(\Rightarrow dpcm\)

\(x^2+3x+1=\left(x+3\right)\sqrt{x^2+1}\)

\(\Rightarrow x^2+3x+1-x\sqrt{x^2+1}-3\sqrt{x^2+1}=0\)

\(\Rightarrow\sqrt{x^2+1}\left(\sqrt{x^2+1}-3\right)-x\left(\sqrt{x^2+1}-3\right)=0\)

\(\Rightarrow\left(\sqrt{x^2+1}-3\right)\left(\sqrt{x^2+1}-x\right)=0\)

Xét \(\sqrt{x^2+1}-3=0\)

\(\Rightarrow x^2+1=9\)

\(\Rightarrow x=\pm2\sqrt{2}\)

Xét \(\sqrt{x^2+1}-x=0\)

\(\Rightarrow x^2+1=x^2\)

\(\Rightarrow1=0\) ( vô lí )

Vậy nghiệm của pt là \(x=\pm2\sqrt{2}\)

Xét

Xét

( vô lí )

Vậy nghiệm của pt là

√x+1+3√x−1=3√5x⇔x+1+33√(x+1)(x−1)(3√x+1+3√x−1)+x−1=5x⇔33√5x(x2−1)=3x⇔3√5x3−5x=x⇔5x3−5x=x3⇔4x3−5x=0⇔x(4x2−5)=0⇔x(2x+√5)(2x−√5)=0⇔⎡⎢⎣x=02x+√5=02x−√5=0⇔⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣x=0x=−√52x=√52

\(\sqrt[3]{x+1}+\sqrt[3]{x-1}=\sqrt[3]{5x}\\ \Leftrightarrow x+1+3\sqrt[3]{\left(x+1\right)\left(x-1\right)}\left(\sqrt[3]{x+1}+\sqrt[3]{x-1}\right)+x-1=5x\\ \\ \Leftrightarrow3\sqrt[3]{5x\left(x^2-1\right)}=3x\\ \Leftrightarrow\sqrt[3]{5x^3-5x}=x\\ \Leftrightarrow5x^3-5x=x^3\\ \Leftrightarrow4x^3-5x=0\\ \Leftrightarrow x\left(4x^2-5\right)=0\\ \Leftrightarrow x\left(2x+\sqrt{5}\right)\left(2x-\sqrt{5}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\2x+\sqrt{5}=0\\2x-\sqrt{5}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{\sqrt{5}}{2}\\x=\dfrac{\sqrt{5}}{2}\end{matrix}\right.\)